¢ he Moti 1g eh ‘ ' ~thel a . s sskae . . stingrays Sashes = pie ’ i last fhets Ap * : pe Sea : ‘ " re 7 piapaine gil te ; : rd ied <9 . ; = Boe As 2 PI me gD) etl es ® aren: . 1 ar SPE Saco a By ares ; siskees OS Rr ke cree vote be ies * F: anit a3) 2S ital eae yin Sete = =o 2 , 3 chat poi Seta ae i> pra a aol oA Sas 9 Salida ae tO tate ee Tate APOE Say beh sie eR * ary nome ithe? 3 Lad et lias r Rasy EN NEP ate band pe ott nh re hala wasp ati MB SE rab bee ss , = . - 2 : ; 4 ‘ a j re : ~ ss a wy x . *, ‘ - . > > = . : : ; - ‘ > iy, > a - ae = - * os . : 2 “ ~ * . z * > > .. A - = . . * , . 5 me x « > - x - © 2 te ied a 3 f, ae ‘ J oo — a al ba te “. > i : es Seamer ne ~ M eee : ; A v et y —e . . i ibe 2 ; 7 Fa avra yt ~ , ha Vac ter ab - va ~~. t ny Att omna = teens Paras 4 bs 3 ge PR, wee vane OR Sat Fa 6 aS a ‘ einai MATHEMATICS LIBRARY i i) eta iy) tee Th vey 7 SCHOOL CALCULUS ete Ee ~—. 7 % ae ‘ : : N.B.—For the convenience of teachers, the Answers to the Miscellaneous Examples — on pages 323 to 351 are not included in this volume, but must be obtained — . separately (Price, 6d.; post free, jan 7 = i ; ; ' L ‘ i A al a | > * CoP . 4 f z : YD Aye ah S - rs a 8@ ei 2 es re PRINTED BY = HAZELL, WATSON AND VINEY, LD., LONDON AND AYLESBURY, | a Ost Vo Sis MATHEMATICS Ligeaey M26» Bale bier Alor: THE object of this book is to present the fundamental principles Peis of the Calculus in as simple and intelligible a form as possible with a view to their immediate use in elementary applications. The treatment adopted has been to employ definite numbers instead of algebraic quantities for preliminary explanations, more especially in the earlier stages, so that the real meaning of new and unfamiliar ideas, limits for example, may be the more readily grasped by the beginner: and as little previous mathematical knowledge as possible has been assumed. While the book is primarily intended for Candidates for the Army and similar examinations which require a somewhat practical knowledge of the subject, it is hoped that the definitely numerical treatment we have employed may enable boys to begin reading the Calculus more generally and earlier than is customary at present ; and that even those who contemplate a fuller and more theoretical course may reach a clearer under- standing of the formulae and symbols they employ, and of the units and dimensions of the results they obtain from the emphasis laid on these points throughout. Two short introductory Chapters on Graphs were considered necessary : the next two contain the principles of Differentia- tion and Integration, and the rest of the book is devoted to applications, Chapter IX. in particular dealing with Mechanics at some length. Numerous examples have been given throughout, and in . addition to these there are at the end of the book 200 eed 6 vi PREFACE Miscellaneous Examples on the whole course, divided into sets of five, whose answers may be obtained separately. We have confined ourselves to the use of four figure loga- ‘rithms, and have alsoincluded a table of Napierian logarithms, since they are so frequently required for the evaluation of Definite Integrals. The use of a Slide-rule is recommended for shortening calculations where convenient. As all the work has had to be done in spare moments, we cannot expect to have avoided mistakes ; but we hope never- theless that the work may in some ways meet what is, still, an obvious need. A. M. MoNzize. J. D. McNBEILE. May 1911. 6 4 8 2. CONTENTS CHAPTER I GRAPHS FROM GIVEN CONDITIONS Points needing Attention Irregular Rectilinear Graphs . Irregular Curved Graphs Experimental Graphs Graphs of Mathematical Aaa Definitions of Variables and Functions Explicit and Implicit naeeions 3 2 Graph of ees. Ree Conditions for Infinite Values of Algebraic Functions . Constants. Absolute Term . 10, 11. Quadratic Functions 12. 13. 14. 15. 1 2. 3. 4 5 Graphical Solution of Semattaneoae mynation® Graphical Solution of Quadratic Equations . Graphical Solution of Cubic Equations . Compounded Graphs : : CHAPTER II EQUATIONS OF GIVEN GRAPHS Distinction between Different Cases Two Kinds of Graphs having Equations (i) Mathematical Graphs . ; (ii) Experimental Graphs -8. Graphical Tests for certain Taree of Meustions vil 24 24 24 29 36 Viil SECT. Sear cme Ye fate Ne 26. CONTENTS CHAPTER III DIFFERENTIATION Introduction . Notations for Runcconn Three Important Series and Tania Three Explanatory Examples on Limits Definition of Limits Evaluation of certain Simple lamnte Notation for Small Increments of the Variables Rate of Change of a Function Definitions of Differential Coefficients and Differential Cie dy dx Tangent and Normal at any Point of y = f(z) Differential Coefficients of certain Standard Functions . List of Standard Differential Coefficients Differentiation of Combinations of Functions Logarithmic Differentiation . : Inst of Differentia! Coefficients of Combingiing of Funehiote eS when 2 = f(¢) and y = ¢(¢). - Parameter Differentials . #4 for j(z.7) Use Tangent and Normal to any Curve Lengths of Tangents, Normals, etc. Differential Coefficients of Inverse Trigonoiiettical Funsueee Hyperbolic Functions : Differential Coefficients of Hee alic Ane OnE : Differential Coefficients of Inverse Hyperbolic Functions Asymptotes Geometrical meaning of and 27-29. Undetermined Horne and nice be Ditsrennatan 30. apt Seidel Evaluation of some Limits without Differentiation CHAPTER IV INTEGRATION A Process for reversing Differentiation Constant of Integration Sign of Integration and dx Integration of Standard Forms 100 104 104 109 109 10. CONTENTS List of Standard Integrals ; Integration of Sum and Difference of uncrions : Integration of Product of Two Functions Integration by Change of Independent Variable Certain Standard Substitutions for Integration Integration of f(x): /’(x) 11-20. Integrations of Standard Miocene INGneriOna 21-25. Integrations of Standard Trigonometric Functions . 26-28. Integration by Parts 29. Additional Useful Integrals . 30-40. Areas by Integration 3l. Meaning of dx in /. p(x) dx 32, 33. Definite Integrals 34. Change of Independent Variable in psnnite Tateprals ‘ 35-38. Area and Definite Integral as the Limit of a Sum . 39. 40, 41. 42, 43. Pe Se) St Sebo Integral Curve, and Signs of Definite Integrals Infinite Limits of Integration, and Infinite Ordinates . Differential Equations: Variables Separable Tmportant Case : eb ea ky dx dy Solution of © st ke +ly=0 CHAPTER V COMMON CURVES Introduction, and Notes on the Differential and Integral Curves Straight Line. Circle Parabola Ellipse . Hyperbola Rectangular I oerbotan Cubic Logarithmic ees Exponential Curve Cycloid . Catenary : Catenary and Ee ole Seripared 158 160 160 162 163 164 166 167 169 170 171 174 177 x CONTENTS CHAPTER VI PLANE AREAS AND LENGTHS I. Puanr AREAS SECT. 1. Introduction . 2, 3. Area of Circle Area of Parabola . 5. Area of Ellipse 6. Area of Hyperbola 7. Area of Rectangular Hyperbola 8 9 i Area under the Curve yx” =c : : OF ARCS Area between wy = a, yx’ =c, xy = b, and yx” =d 10. An Area which apparently = 0 11. Area for Curve x = f(é), y = p(t) 12. Area of Cycloid : 13. Area of Parabola in nea of one PaRaneiee Trapezoidal and Simpson's Rules : 14. ‘Trapezoidal Rule . 15. Simpson’s Rule 16. Evaluation of Definite Integrals oreehigiv’ II. Lenatus or Arcs 17. Only found by Integration 18. Formulae for .s 19. Are of Parabola 20. Arc of Parabola near Ve a 21. Arc of Catenary F : 22. Are of Curve x = f(t), y = p(t) 23. Arc of Ellipse in Terms of Parameter 24. Arc of Cycloid . : : : CHAPTER VII VOLUMES AND SURFACES I. VoLuMES General Volume, and Volumes of Revolution: Volume of Sphere, and Cap . ; Ellipsoids of Revolution or Spheroids 5 7. Hyperboloids of Revolution 8. Guldin’s Theorem on Volumes 9. Application of Simpson’s Rule to Volumes Ale Volume of Right Circular Cone by Integration Formulae PAGE 181 181 183 184 185 186 188 189 191 193 193 195 197 198 201 206 206 207 208 209 210 210 210 212 213 214 215 216 220 222 CONTENTS II. ARgas or SURFACES sner. 10. Surface of Cone by Integration 11. Area of any Surface of Revolution 12. Surface of Sphere . 13. Surface of Paraboloid 14. Surface of Spheroid 15. -Guldin’s Theorem on Surfaces CHAPTER VIII MAXIMA AND MINIMA 1, 2. Introduction 3. Definition 4, An Example . 5. General Discussion of Maca and Minima Gey chicells 6. Maxima and Minima by Differentiation : : 7. List of Rules for Discriminating Maxima and Minima . 8. Practical Examples 9. Points of Inflexion CHAPTER Ix APPLICATIONS TO MECHANICS I. CENTROIDS AND CENTRES OF GRAVITY 1. « for Isolated Masses 2. x for Continuous Bodies 3, 4. Centres of Gravity for Bodies 5 Variable Density II. Moments or INERTIA Kinetic Energy ; Moments of Inertia; Radius of Gyration Theorem of Parallel Axes § , Theorem [z= I,+ I, Theorem I = I, ia Ty sin? pe ee te III. FormutArE or MorTion By INTEGRATION 9. When the Acceleration is Constant 10. When the Acceleration is Variable and ll. (a) P= g(t) 2. (8) P= ¢(s) 13. (y) P= d(v) 14-21. Harmonic Motion 22. Damped Harmonic Motion xi PAGE 223 224 225 226 227 228 230 231 231 233 235 237 241 249 252 252 257 260 262 263 266 271 272 273 275 277 279 285 Xli CONTENTS IV. ACCELERATIONS AND VELOCITIES GRAPHICALLY SECT. PAGE 23. (i) Space, (ii) Velocity, (iii) Acceleration Curves . ‘ ; . 292 V. Work anp Power 24. When the Force is Constant . : ‘ 7 , : : <1 teed 25. When the Force is Variable . ; : : : : : <1 298 26. Rate of Working . : : : : : - 299 27. Work done by any Force, oraphic tea ‘ , ; : ; ye a01 28. Work done by an Expanding Gas : : ; , : . 304 29. Indicator Diagram : : : ; : : ; , . 306 VI. VirtuaL WorK 30. Explanation of Theory . ; : : : | opel], 31. Negligible Forces, and Forces counivon ini fren eane : . 309 VII. CrntTres or Pressure ~ 32. Formulae for x for Liquids of Constant and Variable Densities 313 MISCELLANEOUS EXAMPLES ; : ? ; : : - 323 ANSWERS . : , ; : : ; : . : : . 353 TABLE OF NAPIERIAN LOGARITHMS , : : ; . v72 Page 22, line 14.—For “especiallyt rigonometrical ’’ read ‘“‘especially tri- o> ERRATA gonometrical.”’ 43, Ex. 9.—For ‘‘ Tke’”’ read ‘ The.”’ 66, line 3 from end.—For “‘equations and normals of tangents and normals.”’ 29 69, line 3 from end.—In denominator for “ cosx’x”’ read ‘ cosa dz.”’ ce 71, line 1.—In numerator for “ y’’ read “‘ dy.”’ 73, last line.—Insert —sign between fractions. 83, line 7 from end.—For “‘ £9 >> read « 8Y dx ba 1 118, line 6.—For “ = fa” read ie jde.”’ : x ; es: 118, line 8.—For “ tan oy > read “ tan-'.’ 99 127, line 1.—For ‘“‘ sin?6@’”’ read “‘ sin7@.”’ 132, lines 3, 4.—For “ a2(et—1)’”’ read ‘‘ e%(~—1).” 142, line 18.—For “ Li, =,”’ read “ Lt; <9 be 144, line 7 from end.—For “‘ always 145, line 1.—-For ‘‘ 2nd and 38rd ”’ read ‘“‘ 8rd and 4th.”’ 145, after line 23.—Add “ wg being always taken algebraically > 2.” : AG pA ly 1x 160, line 9 from end.—For “ sin 5 ” read “‘ sin 3 ey 185, line 5 from end.—For “‘ TE ” read “‘ me a 185, last line.—For “ se "read “1 x % 186, line 3 from end.—For ah 59 read ‘ 2» v we Vy 208, lines 11, 12.—Read “‘/ ”.” eo} 209, line 2.—For “3 feet’ read ‘‘ 4 ft. 6 ins.” read ‘‘equations read “in the first quadrant.” ERRATA d Page 236, lines 3 to 5.—Read “‘ while at such points as D and # “Y becomes > dx either infinite or 0, without passing through those values and changing sign.” 236, line 18.—For “‘ ¢’(x)”’ read “ $’(x).”’ 243, lines 3, 5, 6.—For “‘ 39-#2”’ read “ 39 —5#2.” 262, line 7 from end to p. 263, line 1.—Replace by: “‘ Take the centre of gravity as origin, the given axis through it as axis of 2z, and the plane through it and the parallel axis as plane of xz. Let the square of the distance of any element dM of the body from the first axis be x?-+ y?; then the w- co-ordinate referred to the new axis is x-kd, the other co-ordinates being unchanged, so that the moment of inertia of the element about that axis is dM {(a-+d)?+y?}. . T=f((wkd? +y}dM. = {(x?+-y?) dM + 2d/xdM +d? /dM. Now /(x?+y?)dM =Io;”’ etc. 269, lines 7 to 9.—Delete sentence in brackets. 288, line 1.—For ‘‘ 12”’ read ‘‘ 5.”’ 288, lines 10, 11.—For 90 Ay? and «fd oP) read (73 20—y oe) and “é —@ 288, lines 17 to 19.—Read: ‘‘amount to : poundals acting on 5 lbs. mass, so that when the velocity is v, or dy a inches per sec., they produce Pie OY ds an acceleration = winches per sec. per sec. Therefore,” etc. 301, line 1.—For ‘“‘.2a’’ read “‘a.”’ 303, line 7.—For ‘‘ forces... change of shape ”’ read “‘ forces per unit area . . . change of shape per unit length.” 303, line 27.—For “force P” read “stress P acting on unit length.” 319, line 13.—For “ab?” read “abi,” 331, Ex. 61 (i).—For ‘ E 2 read “2 thirds x3? 2 “2 2 340, Ex. 119, line 3.—After “ P”’ insert “in a plane.”’ A SCHOOL CALCULUS CHAPTER I GRAPHS FROM GIVEN CONDITIONS 1. In drawing graphs, special attention should be paid to the following points : (i) Numbers must be clearly marked along each axis to show what length of line is being used to represent one unit, that is to say, the scale of the drawing must be clearly indi- cated ; and the units employed should also be stated on their respective axes. (ii) The graph must be large enough for results to be read off to the degree of accuracy required. (iii) Clear verbal explanations of methods employed, and results obtained, must always be given. (iv) The nature of the graph should be determined (as explained in this chapter), before joining up any points, as in some cases they may require to be connected by straight lines, and in others by a smooth curve. (v) Any part of a graph which is required to be more accurate should be separately plotted on a larger scale, and more points on it first obtained by calculation, if possible. 2. Irregular Graphs consisting of Straight Lines. In some cases where we are given a set of values for tabulation, if they are merely isolated values connected by no mathematical law, and if further from the nature of the case no intermediate values can be obtained, the points corresponding to the given values are then joined by straight lines and not by a curve. For the points alone are of any importance, and the lines which join them merely serve to give them prominence, and to exhibit graphically a general upward or downward tendency. ExampLE 1. Draw a graph exhibiting the following total 1 [CHAP, I 1899 | 1900 | 1901 | 1902 65 | 71 | 67 | 66 1898 5°8 5°8 6°1 6th 5th ~ | ~ PPE LE SA Shs Volt SURE ols ee MiWwADMSBEUBRE |_| | eee SSS [2 Serene teleport Af | | | ci i | 4th 3rd JSR SCR T Aas es Sil oie balan ere bee Nel ips Cot et Ee aoa Pee [eve a BERS RESRReE See PP rH epee eae 2nd TELELETLLDL Tl sl st 1,510 | 1,625 1,540— 1,675 | 1,780 | 1,780 Fia. I. 1 1895 | 1896 | 1897 5'8 SCHOOL CALCULUS British exports for 8 consecutive years, given in £ per head of population. Exports per head in £ Year Show graphically the following table of expenses of a school : el ete ine aie tel [oll] tye a | Tis Trrerrscirrs AoC Bereeee | Ei Ein Bee ein eer Gis COGS O NPSL 4 He Bid : 4a | j EXAMPLE 2. Number of Term Expenses in £ .. Fia. 2. CHAP. I] YRAPHS FROM GIVEN CONDITIONS 3 In these two cases we have joined the points by straight lines because in the first the statistics are published yearly, and in the second the accounts are made up at _ the end of each term so that intermediate values cannot be obtained. It should be noticed that intermediate ordinates (such as PM) in each figure are practically meaningless, and a curved graph through the points would therefore be useless. 3. Irregular Curved Graphs. Sometimes the values to be plotted form part of a continuous set, although they follow no known mathematical law. In this case intermediate values could be found, and intermediate ordinates have a definite meaning. We therefore join the points by such a continuous curve as seems best to represent the conditions given. EXAMPLE 1. Time ../9a.m.j/3 p.m. 9 p.m.| 3 a.m.) 9a.m.| 3 p.m. 9 pm. 3 am. 9 a.m. | 30°51 | 80°23 | 30°13 | 30-00 | 29°60 | 29°75 Height in inches .. | 29°95 | 30°40 29°68 | | | | H aches tt BUSeRee VU Oea eae as E na BEE pereseie rece eee BH pare poe teh et Fr 4 ++} Sauk t+ Bae -HAAEHA-HE-H +H RA aORGeneee H - aaae CE RRARRERe be EEE revere ehi ten ALF P: Potep el ofc e ie iF 9 pepe tha mat fot Bhi CGS RaRSE eee kane ae HHS HEEB EE EEE EEE ide Hb aT Lh pS US eee eS a08 | AR Ro ARAB Pp EE BEB S Cee OUI CER SB eRe ee ES pt ie pp} Lb J So aoe oes aide ae a eee RAR eRe ‘Seem wen H+-H ELSE peeeoe usceeooeeoe by pL aBSSea I++ BRR et ee taal ‘) ja RRR SBI } | ao c BE HN PH Perepetepevel oT TT ee Rae mw eee aeeeaeen Pm DAVES eae oReaAwe sus ovae aude usees Javae LESETVGUEECEVUE agar uat i UGS ARERR Sahee ma fag Ba bees | This graph represents the readings of a barometer for two days, as given in the table, as nearly as we can judge, though possibly omitting intermediate fluctuations. The probable reading at 6 p.m. on the first day was approximately 30-54 ins., and at midnight after the second day, it was approximately 29-62 ins., as we see from the two ordinates drawn. EXAMPLE 2. Draw a graph of the population of a certain 4 SCHOOL CALCULUS [CHAP. I town for the years 1830-1900 as given in the following table. Vea _. | 1830| 1840/ 1850! 1860 | 1870| 1880! 1890| 1900 Population in thousands. . | 31°9 | 33°6 | 36:0 | 39°5 | 40°7 | 43°3 | 47-2 | 50:2 el SRR RSME ENENERERRRERARAAe BRESo 150] TAhousands| | | Tt] ttt TTL TL a a id BRERERER BERBER RRERERRAR UU PET TEL ETT CET Tree ET es aie BRRERRAARRRR ARTA S see EDPABERRS. gst oo} fete pe iA | eee (laser Ti ET TTT TET TTT TT TL el aA eae TTI TI tele CLC LCL Le a see PE TTT CPT TT TE ee eee ei SREREERRRRRRESEREREERRERZARERA 4 PELE TTT TET Tis eee ee Se eee ese ARERR eoo PETE TTP TEE LT Pree a a ee y TTT ETP TTT YC Dee ea ae eee BRR RRR EP CIARA UU é tt ae tt EEE EE rot TT ae Le ee ae ee PEL TT Pee TT Ee Ce St + 4] +- HH bas] SEG vat tte} et EE ax soft EH HE} ISO TTT TTT TTT TT Oe a ae | | tye30Lll4ol]llso0lt Tieo live LTT l60lLl_l lsat TI irgec(_ mae Pt tet eel, ELLE, Lei ee a el Pt BRREREERERRRERRRREREDS UU Higa In this case we see the population increases on the whole, as we should expect, though irregularly ; and intermediate ordinates represent the populations for intermediate years. Thus we see in 1857 the population was 38-6 thousands and in 1892 it was 48 thousands: and these were the populations in those years as nearly as we can determine them by a graph. 4, Experimental Graphs. In these cases some natural law is under investigation, but any particular set of experiments is liable to inaccuracies due to errors of observation and imperfections in the apparatus employed. Our object is to draw that curve which shall best represent the law under investigation ; this we know is a continuous smooth curve without abrupt changes of direction. It should therefore be drawn evenly among the points, and not necessarily passing through all of them. In this way we shall best eliminate the errors, and hope to reproduce the true curve. A single set of experiments cannot be considered trustworthy, but by in- creasing the number of experiments and by varying the CHAP. I] GRAPHS FROM GIVEN CONDITIONS 5 conditions under which they are made. greater accuracy can be obtained. EXAMPLE l. Weight in kilos Se Ie 2h j1°5 1-75 | 2 12°25) 2°5 12°75) 3 13°25) 3°5 Extension inems. .. |'28/°51| ‘56 |°'70| ‘82 “00 1°09 |1°21 |1°30 |1°5 |1°56|1°70 This table gives the extension in cms. of a spring, 30 cms. long, loaded with various weights. Draw a graph to represent the law of extension. Also find the probable true extension with a load of 2 kilos, and the weight required to extend the spring to a total length of 31-5 cms. In this case the curve lying most evenly among the points given is a straight line, which therefore represents graphically the law under investigation. Loo nee SPAS SRR nan eee eee eee Ee er ee Lad meee ristei | tt Eee TEE err tT mroig ere ere ee eae ere tee eer eee ream sesame Yoleislei mate) tel PLT EP ET oJ. TS SR RSE R SERA Sees RRR Ae eee re ei sleet ise eT tebe ET eee So agh hope ee tte pep beet te be mens jer reisiei ote tet | 1 tpt dT Pt TE Ty Tr ee ee ee eee JSD RGS SRR eS Ack eee eee eRe Z oS BRB RB RS Sep ate ee ERE REA a rere alee ee) eer iota t tel | Tet PTT er J JI Sas S 2a Ee eee AeA R eee . . DRESS SERRE RR AeA itieieie il | tl teie obi LP lteh Ey LEE EL ee 1.) Ee 2S RB RERRRRRERAe aa ett tt pb be teh ttt atts gee Sia ett Ee ET BSCE SRA os] | || mala re fir ft ERE EEE eerie ereielatat al} fof tials ee Ve he Sted [ate ies a Free 5; We see the probable true extension for 2 kilos is -98 cms. : and the weight required to extend the spring to 31-5 cms. Is 3°08 kilos. Exameie 2. The accompanying table gives the volume in c.c. of a certain quantity of gas at constant temperature and at various pressures, measured in kilos per sq. cm. Pressure in kilos per sq. cm. .. | 2°5 | 3°5 | 5-0 | 7-5 | 10-0 | 12°5 | 15:0 | 175 88 35 171 | 124 62 | 46 Volume in c.c. .. oo Ae ‘249 43 6 SCHOOL CALCULUS [CHAP. 1 We see from the graph that when the volume is 100 ¢.c. the pressure is apparently about 6-2 kilos per sq. cm.: other results can similarly be read off. TE SI SS Pe is] tel ele ERE) ie Conau BeBe eSURUUUUUEESSS = PERCH EEE Maer: See ooeeeeee ERBRM eS! tbat tH Hitt ttt ttt BBhi RRM ARERR Eee CRT Te Wels fe Pee ST ey shot ees cs pt tate et EE Pe | cede pt pA LT ii ot oat tte et tt +E EEE | H-E- OR RERR SREB tpt AEE pap Pt See cee eee eee Seeee Bue: SSR EARRAwSRaSE Sen Eee dat tH NN iW UU PEERBREERBEMaSRASse Ie BRRRBEARENERAREOERAeesA ERREMREREENREBREOeSe CLE mbt td pp Nf tet HRSA Roe a tt Se ees aa ee CeCe eC eee ee BERBER ERRERBRSREA ees PE ee ae ee — of 4-44 aes LE Cr ee 0 a a ES ol Be a BEHRBELL. CEE EET kites per 6q-¢m-[ tt tt ee abe SRRReReenne SESGGRREEO Fia. 6. In each of these examples we know that some natural law exists, and we have drawn a curve lying evenly among the points plotted (which in Ex. 1 is a straight line): Such a curve is often called a ‘‘ smoothed ”’ curve, and of those points through which it does not pass about as many should lie on one side of the graph as on the other, since experimental readings are about equally likely to be too big or too small. EXAMPLES 1. Plot graphs from the following data : : Times -./6am. 7 | 8 110) 12 |2 p.m eee Temps. in deg. Fahr | 30°9 |32°5 |32°9| 39|42°8| 43:4 |36°1| 32 31:5 CHAP. 1] GRAPHS FROM GIVEN CONDITIONS 7 2. The following table gives the number of men in the British Army in thousands for the years stated : [xet, .. —.. | 1900 | 1901 1902 | 1903 | 1904 | 1905 | 1906 | No. of men : .. | 480 | 450 | 420 | 236 | 227 | 221 | 204 3. Average annual premiums per £100 of an English Life Assurance Co. for whole life policies, with and without profits. Age next birthday] 25 | 30 | 35 40 45 50 60 | | | 7o tur) — Qu wt So de Premium 6 74 Do. (with profits)2 3 12 Seo 15 83 4 6315 9411 67 0 8 4, Average prices of Wheat, Barley and Oats per Imperial Quarter in the years given. | ‘Year .. .. | 1870 | 1875 | 1880 | 1885 | 1890 | 1895 | 1900 | 1905 | Wheat... 46/11 45/1 44/4 | 32/10 31/11 23/1 | 26/11 29/8 “Barley... .. 34/7 38/5 | 33/1 | 30/1 | 28/8 21/11 24/11 24/4 Oats .. .. | 22/10 28/8 | 23/1 | 20/7 | 18/7 | 14/6 | 17/7 17/4 5. The prices of kettles of various capacities are listed by a certain maker as follows : Capacity in pints Ae 4 6 8 12 14 18 20 Prices .. Ss 7a) 4/1) 4/5 | b/= | 6/2 | 6/9 | Z/ll | 8/9 Find the probable prices of 10 and 16 pint kettles. 6. The amount of £1 invested at Compound Interest for different times at different rates is as follows: | 5 | 10 15 20 3040 at24%.. .. 11314 1-2801 | 1-4483 | 1-6386 | 2-0976 2-6851 at4% .. .. | 12167 | 1-4802 | 1-8009 | 2-1911 | 3-2434 | 4-8010 hep, .. =. +| 1 -2768 | 1-6289 2-0789 | 2-6533 | 4-3219 7-0400 Hence find the Amounts of £1— (i) at 24% in 3 years and 25 years. (ii) at 4% in 12 years and 27 years. (ili) at 5% in 7 years, 14 years and 21 years, 8 SCHOOL CALCULUS [CHAP. 1 5. Graphs of Mathematical Functions. A Function of x is any quantity which contains 2, and which assumes a definite value when x does so. 7x3 — 9 - aT f 3a qa r/ax + b, log sin2% are func- tions of x. If we give to 2 various numerical values, it is clear that any such function will also assume different, but definite, numerical values: and it will be found as a rule that if the different values assigned to 2 increase by small amounts, the corresponding values of the function under consideration will also increase or decrease steadily by small amounts. A function of x is usually denoted by f(x), F(x), o(x), or some other similar symbol; e.g. if we are considering the For instance x -+ 2, 3 au function = ie = we should write eo cn a = f(x): and then (Le orte 1x3 5 >i f(a) means Sa A /(2) means eee 51; ier se z f(—:5) means Citar es 5g 412 ee = Soe | 3(—.5) + 4 «Od 8 ee and f(0) means ues — = —]-25. O+ 4 If now we write y = f(x), since f(z) changes in value when x changes, therefore y also changes in value when x does; that is to say x, f(x) and y all change in value together, and they are al] therefore called Variables. But we may assign absolutely any value we please to x, and when this is done the corresponding value of y, i.e. f(z) which is dependent on 2, can be calculated. On this account x is called the Independent Variable and y the Dependent Variable. xis also sometimes called the Argument of the function y or f(x). 3 It is of course possible to have more than one independent variable : e.g. in such an equation as z= 2x2 — ay 4+ 3y?+ 5 (which may be written z= f(x, y) to conform to our previous notation) x and y are independent variables, that is to say each of them may quite independently have any values we please, and the value of z is then determined by calculation, so that z is the dependent variable. We shall however have CHAP. I] GRAPHS FROM GIVEN CONDITIONS 9 no occasion to consider functions of more than one independent variable. It should also be noticed that in the equation y = f(x) we may if we please assign values to y and then calculate the corresponding values of x; y then becomes the independent variable, and a the dependent variable, and it is therefore a mere matter of convenience which is chosen as the inde- pendent and which as the dependent variable. 6. Consider the equation 2ey + 3% — Ty+4=—0.. (1) Such an equation is said to be Implicit in x and y. We may however solve it for y, and so write it in the form _ d@+ 4 7 — 2x and in this form it is said to be Explicit in y, or y is said to be given as an explicit function of x. x would of course now be taken for the independent variable, and y for the dependent. Again we might solve equation (i) for x, and so write it in the form y AG tees 2244-3 and in this form it is said to be explicit in a, or x is said to be given as an explicit function of y. y would now be taken for the independent, and a for the dependent variable. Theoretically any equation in x and y can be solved so as to be explicit in either x or y; and as a particular case, any equation explicit in y, say y = f(x), can be solved so as to be explicit in x, becoming say «= ¢(y), though in practice this process is not often necessary and is frequently troublesome. 7. Suppose now we require the graph of any function /(2), xe +. 242 + 8 oe a This function assumes different values, when different values are assigned to a. If therefore we write y = f(x), gy? -+ Qo2 + 8 e2— 4 and represents the different values of this function. To obtain these, we plot a table of values of x and y, assigning any convenient numerical values to x, and thence calculating _ (iii) such as Le. here y = , y becomes the dependent variable 10 SCHOOL CALCULUS [CHAP. I those of y. The equation should be first simplified if possible. Thus we here divide by #2— 4, and write ~ 7 a ai? sh 2... w«t—7|—6| —5| —4| —3 || <2) Sipe c+2 ..| 8} -4|3|—2| 13) 01 e+ 4) -| 12/8] 4] | 4 | 6 | | x2 —4 45 | 32 |~ 21 5 | 225 1°75 3°75 y .. .. \-8\-43|-30| 2 | 22 | 20) eee : . .|0)| 8 | 1 | ie) 2 /a) anon ieee +2 ..(2|25 | 3 | 83 141616) 7 eae 4e+4) |_,| 18) 2) 22) | 28| 8 | 36| 5 | 44) 4 x2 —4 375| —°3 |~ 1-78 6 | 3 |21| 4 | 45) & ys .. '-2}-23| -37|—91 | oo 10°6| 8-7 8-71| 92 | 10 [108 Note that values of x have been chosen differing by unity, except between —3 and 2, where more points were required to ensure accuracy in drawing the graph, and two more points were also subsequently added, namely = |—2°25 | 2°5 y=d 64516 | 3 2 This curve is called the graph of a ‘ . and 34.292 1 8 , a _ =a Te is said to be the equation to the curve. In general it is immaterial whether we say we require to draw the graph of f(z) or to trace the curve y = f(a). 8. When speaking of a function in § 5, it was stated that as a rule if x increased steadily by small amounts, f(x) also increased or decreased steadily by small amounts. ‘This is generally true, but it sometimes happens that near a par- ticular value of v, a small change in x produces a large change in f(x), Le. in y. 11 that is, a change in w of amount GIVEN CONDITIONS FROM GRAPHS There are two such values of x in Fig. 7, namely x =-+2. For when « is a little less than —2, for instance — 2-001, y is large and positive, being 6,000 roughly ; while when a is a little greater than — 2, say — 1-999, y is large and nega- tive, in fact about — 6,000 : -002 produces a change in y of about 12,000. CHAP. 1| 28 Seesaw Boba | | rsfabeiaiels ee [Io eats [iele tale late bale la laos [wiaialal cle Shoo R COIS aoe Ree ERRee ons RNa nual ereleniciaieieiee \ [sh eles eee [eae se ia Sie ere ese eal DARN SIZ RA SERS PRR eae nee ie ae) Ae PPS eh e hehe o/s set ope be 18] ei Te Peres a TS ae a es eieieie isis | \intclelossleb bel alee setae Shee elae ee REQR CCR GOR SR ARR ARE eae Ras ees oe JOO DOD SD COUEE BBO RE ROBBEN ee sobs eae wine iO te ele et es S1e els lenis a] ieee eel ee) stan elas ee) he ee pint aladad efule lobelatels (iy io] fale le[ feel ebaditelmiale arate ta bal aS SOO OS BARS Ree eons eas DA SRM GO lS 393 2NG ESS Pee Be sea eRe esses Fe meas i Ta bere i fhe | a od Fee Pins Pal ee Le RE Bort oR eee ee eee ahs Pease a aes BES REEO Eee asESRR eee aee Baa ee BeenueSes jae BSCE RRO Seog esol Ae SSNS eae Roe anee Sooo lsfe slabs ele iat ei etalebe isle palwiay ele slop tate! SARE BO Se OO eo Gee Re Con Soe ee 48 Lit] OL Lt bg tS ee BERS Sa Slee eee Saeed eee SS SESS S BSR SS SSeS s eee sees Seen ee eee BEEBE RAS Roe wea eae see = See CCOOCCCCCC CEE EEE CCCCCE Eee Ss) ei al rela a serene che meade lei: seg [es) oele |S [oleh eer REBSOSRU SES Seb OR ae OUD BRA ees aR BIE SRaBsC aces aoe See Behe Reece DSHESe SSeS SS e one P ES ae ea ee Fae CIC Cc 7 ASSN eRe oe VSG RP SR SMeeh Sw ennee SASS OR e ae HH | EE ras Pobat att eal ata} Fia. 7. These finite values of « which make y infinite (and near A similar large change occurs in y near x= 2. When 2 which a small change in x produces a large change in y) can is actually equal to + 2, y is infinitely large. “*— 4=— 0; and it is further obvious that only such functions be obtained by making x? — 4= 0, ie. by solving the equation 12 SCHOOL CALCULUS’ [CHAP. I as are fractions can become infinite for finite values of the variable, and then only by the vanishing of the denomin- ator; and these values are to be found by solving the equation denominator = 0. It is obvious that algebraic functions of #, which are not fractions, can only become infinite, if at all, by making x itself infinitely large. 9. Constants. Consider the function az+ b%-+c. This is a particular function of « which changes in value when x changes, but a, 6 and c are fixed quantities representing numerical coefficients : if however a, 6 or ¢ are given different values we get different functions of # having different graphs. Suppose then we put y= aa+ br+c, by giving different values to x and therefore also to y (keeping a, b and c unaltered) we obtain a series of points on the graph of y = ax? + ba+c: but if now we alter the values of a, b, c, we obtain the equations to a series of different curves. In other words, giving different values to w and y gives different points on the same curve ; whereas giving different values to a, b, c, gives us the equa- tions to different curves, so that for any one curve a, b and c are invariable. So then, just as ~ and y are called the Variables of any equation such as y= ax? + bx-+c, a, b and ¢ are called the Constants of the equation, and c in particular is called the Constant or Absolute Term becaus: that term is unaltered by any change in x or y. 10. Quadratic Functions, and Solution of Quadratic Equations. Any expression of the form az? -+ bx-+c is called a quadratic function of x: which means— (i) Its highest term ax? is quadratic, or of the second degree, its other terms being a linear term bx, and a constant term c. In particular cases 6 and c may either or both of them be zero, but not a, for then the function ceases to be quadratic. | (ii) It changes value when x changes, as already explained for any function of x, but here, as always, it should be remembered that it does not necessarily increase always when « increases, or decrease when 2 decreases. Consider the function 322— 7~a— 8. To draw this, let y = 30% — Tx — 8, and we get the following table of values [using the form y = 3x2 — (7a-+ 8)]. 13 29 | 36 | 43 12 | 27 | 48 | 75 22 6°75 18°5 FROM GIVEN CONDITIONS 40 | 18 | 2 | —8|—10-75| —12|—-11-75 | —10) ~2/ 12 | 32 GRAPHS A rough plotting of the points showed that two [ Note. extra ones corresponding to «= °5, and «= 1°5 would be useful for drawing the curve with greater accuracy. | CHAP. I] 3x2 ars an eile -- picieecielsls C FCO 0ST SE LE ES OE aera cteiciaiceincte alate! eis h els ep ees tebe acs ier: iste EEE EEE EE EES EE Sa SS PAss eS seeeesaeReaSoNos Da cvRR SACO Rese Reee oo Rib BERR SE ieee eee tye Ht EE pete Bitawh es Mie Ree et Se ede ee aie ete bode CL eer feels] eielemdae setae isieintwl |: lalaiel | tal load pat a BAS hASRMa Road aaa BASS US PSO R AARP AeA BAB PS FY ON a a a OO estate Clair, ieisele rt tat iets (a tt ee its ele tS e eee rereletel (ol telat is ie pcph le ep peed She Witla Pin LA ate el Teh ie eRe tes ta ERASER ASR SORA Aes eat yt ter TT TT SBP SSRReEReInmsae tn ad Rg RRR SE eres Sana es graph may Lal Fia. 8. (i) It is a curve joining the tops of ordinates which represent (See also § 11.) by their length the various values of the function 3x2 — 7x — 8, (ii) It is a curve passing through all points whose co- There are two points of view from which this with advantage be considered. whose changes of value are thus graphically tabulated in a manner more readily appreciable than from a merely numerical values of x between — 3 and 5, not only those previously table, giving as they do the values of the function for all calculated. 14 SCHOOL CALCULUS [cHaAP. I ordinates satisfy by their lengths the equation y = 3x2 — Tx— 8, due attention being paid to the scales on which the co-ordinates are measured. (See also §12.) 11. Consider further the first point of view. We see from the graph that the function 3%2— 7x — 8 decreases steadily as x increases from negative values up to about 1:2, but that with further increases in « the function steadily increases. Again as the curve is a graphical table of values of 32 — Ta — 8 for different values of x, so conversely it enables us to find what values of x make this function assume any particular value. For instance, what values of x make 3a2 — Tx — 8 equal to 20% ‘To find these, we have to look for ordinates whose lengths are 20 units, and whose tops lie on the curve: we therefore draw the line through the tops of all ordinates of length 20 units, and this line cuts the curve at the tops of the ordinates required. The abscissae of these ordinates, namely #«=-—210 and w= 4°43, are the values of « which make 3x2 — 7a — 8 = 20, as nearly as can be read from the figure. Similarly values of « can be found which make 32? — 7a — 8 equal to any assigned quantity, and in particular those which make 322— Tx—8=0. These are x = —‘83 and = 3°17. This gives a graphical solution of the quadratic 3x2 — Tr — 8= 0. It should be noted that the values of x which make 3x2 — Tx — 8 = 20 are the solutions of the quadratic equation 322 — Te — 28=0; and in the same way any quadratic which differs from this in its constant term only (i.e. is of the form 3a7 — 7a-+-a= 0), can be at once solved from this same graph by writing the equation in the form 342 — Tx— 8 = Fa— 8 and then finding those values of # for which the ordinates in the figure are --a — 8 units long. This method should however only be employed if for some reason a graph similar to that above has already been drawn. A better method is that explained in § 13. 12. Considering now the second point of view at the end of § 10, we have a method of solving graphically any two simultaneous equations. For if we plot their graphs, each graph contains all points whose co-ordinates satisfy its own equation, and where these graphs cut one another we have 15 Se. UG & Rey eR FROM GIVEN CONDITIONS Solve the simultaneous equations 3x" + 12¢— 3 4uy — 2x — 15=0 é 3x2 -+ 1l2xa— 5y—3=0.. GRAPHS These can be expressed explicitly in y, and become points whose co-ordinates satisty both equations simultaneously. KXAMPLE. The co-ordinates of the common points give us therefore the solutions required. CHAP. I} | {Prt oe fee =| i } } eat cle ele to pas Oy lobe | ses el Sta lehs melebelele Lata eats pera BeBe saarenls / ao Rea eZee, BRA RBESRRBeeoORs BSERRESSaSaRa OAT Te eteeaiote lel sheet] |Srereigie ys @s Bh Bae Rea REM eR Rew aeees ASRREERORS CEE SRR ROS, AGAR ee BORER Se oUarceena BREEREEo ns Veta ete be elt bee Pe a td et ate ot he ele op ed adi RARER yee eR eee eee ee Oe ee ee SR OG pe™ «SSSR n Se eens eee GOs SA tins 20 Peete ie el ee at a eet De Fie] ae eee Sas Bakar PS Pig etter tel Sheba ee ele ere See Ole ae Sia sis pel ABBORSS SeBVEARSS ADEE RESRER Roar, BGR SSRSAshnGes DRE DORA EROR Sean onoe.= BD VRRGEMAny BGS. BRR RMNRE hee oaeeboee mM >. bathe e te ble tent e Sae ated pee aaa | al elo ae Hmgageeoo ep et NL EL reste sae oie ee bor lel eter: (ele elsleleiel (seh iste leie obs ier AY Sete he eel eles aa eiere betete Tel {eel ebb abel N [se erie bere Ea RAMS aE MRIS eRe SE RARE RE RRARAAPNB ARABS Peed epee ee ere eye per ole. lets eet olatat io pele | [aS ean cre cate perme Te ae eee eis rears eye rele pts bel 1 cls Si es iene eect ores Perna sperey= als tele) oes te ial etd de | bets pel ee Nee pete ae a ate eee Peete ed oS bal ST Lotte bebe trite eis BBMESS HOURS CSRS eRe Skee, Reena BRAGS elpore pele le aleie Sie Mi eiels |e dett abet bee Ut ie edaaly, BE MiD est WIG Rn CORSA eee eee aoRe Ny, DRE Sarasa s ete pesten eDiets abe tr cba alee epee tar ae Te ate bel ela ie) ARAB ADRS esiaepa teed er tae ag 1 ete Pel oth letiet Fol lal hele eet REBEZHEBEY RALER SRR SSRs eeRE eRe, HAD“ZREARER RRAESTA MRR Sea E SE, MARR OesUee oP e ie A Ps ec eel eet eeiel eal otel [oisisy RPSL AP ange ed See leer aled s]eted eboled ol 1s Loh able Weds 1) Fae ae ie ae aa BRSHOLSTAR MAUR Ree EL et ppb ta aa OC SeOn nA Raa ReDSa RiP anh Reariissesnshi BESADHEO VEC OSA RAPER SE AcReNRRE Rosas et ttt tt tt tt eer tt. ttt tt tt eer Fie BRO BRES =e { re Fia. 9. 16 SCHOOL CALCULUS [CHAP. I We have the following tables. For (iii) : x .. ...|—6|-4|-3| —24 -1 | 0.) 7a Or 1 Go. SAM eee tate eee 9 : 1 | 13 |15| 17 |) 18am 4x .. -.. 1290/16/12) <3 | U4 | 0) 2 PY se ae |=25|—-44 | —-75]—1:38|—-3-25| 00 | 425 | 3 2°38 1-75. For (iv) le .. ..|-3}—4123| 2) 21 de) | 5. |< ee ee 3224 12% —3| 12 | —3 —12, —15 | —12 | —3/) 3°75 | W228 ores Wwe oe .. | 24.) —6 |—2°4| —3 | —24|—-6) 7b") 24a These curves contain respectively all points whose co- ordinates satisfy the two equations. The points (—4°08, —°41), (—1:20, —2°6) and (1°28, 3°43) where the curves intersect satisfy both equations at once. The required solutions are therefore x= —4°08) % = —1°20) x= 1°28) y= —'4lJ y = —2°6 J y = 3°43) This method of solving any pair of simultaneous equations can theoretically always be used, and in practice will often be found very convenient. EXAMPLES Solve graphically the following equations : 1. 8:22= 3-3y-+ 5. 3. 22+ y? = 25. 4:3¢-+ 7-5y+ 3-7= 0. xy — 3y= 4. 2. 4-5¢4 — 3-Ty= 10. 4, SY 3xy = 14. 3x? + 8y= 13. General Method of Solving a Quadratic Equation graphically. The method will be best seen from the following example. Solve the equation 2%? + 7a— 3= 0. We first write the equation in the form 22 = _# -|- : and draw separate graphs of each of the two functions x2 and GRAPHS FROM GIVEN CONDITIONS 17 CHAP. 1] 4 2 with the same axes and the same units; and then “fb find those values of x which make the two functions equal to one another. 7x 2 graph of the function #2 we put y= v?, and draw its graph immediately without making To draw the we can ‘then ! BECP EEE seaeaeae pistes PEERS Fa Sl aes a | SS Slats|efela telat CERROSEEE = iB eS slats elas ietatet fetlate lt dalctata at etal] ts iaaiolebeb ak eee Bde] s[atere| alee tel dadate pap sel etete is) feletefeteieletebiniaieisietat | lelelatal Lis a ot HAG Rin led Sipe) 4 104 es i IN an Gis mie elf etl ea ieee actos ett a niles He tt ttt ee pet HOSE BERIT SAE Eroeg te arb el toned ete etet cet otot pe elsiols: leet eee aisle eae tt ABECE RS mis ttt ‘ Bera 2a BERE SSeS Sees Fie. 10. any table of values, as the values of y corresponding to any value of x can be mentally calculated and at once plotted. , we put 2 2 To draw the graph of the function Ee and since we know this is the equation to a but to check our working and insure accuracy three be taken, and we therefore get the following table. straight line it is only necessary to plot and tabulate two points, should x ios | 18 SCHOOL CALCULUS [CHAP. I The straight line joins the tops of all ordinates whose lengths represent the different values of 2 -+- : for different values of 2. The other curve similarly joins the tops of all ordinates whose lengths represent the different values of a? for different values of x. The points where the two curves cut one another are therefore the tops of ordinates which represent simul- taneously x? and e+ 5 i.e. which make x2 = ee, - and for these ordinates « = —3°89 or x= °38: i.e. these are the values of # for which 2? = — -+- : ; that is to say these are the roots of the quadratic 2a? -+ 7a — 3= 0. It is clear that any quadratic equation can be solved in this way, and its advantages are that— (i) We need draw no other curve than y = x? in the case of any quadratic equation, a curve which requires no tabula- tion but can be immediately drawn. (ii) The other graph is only a straight line, whose tabulation is very simple. (iii) If we wish to solve several quadratic equations, the same curve y= 2x? does for them all, and we have only to draw several straight lines, one for each equation, to complete the solutions. EXAMPLES Make a short table of values for the following functions, draw their graphs and hence determine for what values of x between 10 and —10 the functions are increasing and decreasing, and find approximately the values of x at which the change takes place. l. v?+ 3a+ 5. 3. 5+ 3a — x. 2. a7 — 3x — 5. 4. 5 — 3x — 2x. Solve graphically the following equations : §. #7+7a+5=0. | 7. 3:82? — 5-7 = 2-32 6. 192+ 11= 32?, 8. 2-1xz?-+ 11-22= 7-5 14. General Solution of a Cubic Equation. Any equation of the form ax? + bx? -+ ca+ d= 0 is called a cubic equation, because its highest term contains #3. As we shall occasionally require to solve such equations, and the algebraical solution is a troublesome process, we give two methods of solving 19 of xz, that is, = #3 + 3x42 — 8% — 11, and therefore the function, 0. s are exactly similar to those osses the axis Consider the equation v3 + 342 — 8a4— 11 GRAPHS FROM GIVEN CONDITIONS We plot the curve y First Method. and find the points where it cr the values of « which make y, them graphically. The method used for quadratic equations. CHAP. I] FEEEEEEEEEE He EEE EEE HEBSRUME SESS, PRURAR SSeS PHA ARS eRe Basse BV OSRwS Bs ORE Re ae See Ss a ee PREEL ES SCE PEEL ELEC EEL EEE LE EEE CEE et SSCL RAV ABAGRE RAR PSER AAEM RReSEe ees BA BABNG BS). BAe SESE EERE RAEP ARS aS ASO e HS SEGNe DANE DAREN DRE REDDER PRE eEsees mec eats beta E te cegtet helo bal oie be betebopet pebs oie ote ee] sheleie bas Lae) aia PR EUPRACARL ACTRESS aN ARERR eee Rees FEEEEEEEEEEREEEREEE EEE EEE EEE EEE RCE EDARULS UTR RRS eR AVA sae REPEAL ao ee Nae ee eee eee eee BUM ROO SL eR ORE SASSER AOR aee AS Pht eT NT ott it sit tT eet Tt ott tT ott tt —— | Ss AS CONTE EER ERa= feo aR SERRA RUNS CLR SR ERO VEROO Nee eRe SE RE BEG ERS 8 FABRE RSA RSE CRRA LBA Sase BORER RARE USREREP RAMUS OCS SRS ees BOARDED EAY,, (SRNR ENR eNOS RRs ERREBERRANE SU Io EOS SRA c See RE BRBRAUERAC SR OL Cee ROM Me Rea ke Sea BRHESERRREUR HED SMUT eUeRa Ras RES SSSREBERST ARN BoE LARP oR aaa eee eee eho ebpiels) Oe & bo 202 b i logio aly CHAPTER II EQUATIONS OF GIVEN GRAPHS 1. The converse problem of finding the equation of a given curve, or of a curve passing through a number of given points, occurs very commonly in applied mathematics. This is the same thing as finding the relation connecting two varying quantities when several pairs of simultaneous values of those quantities are given. For instance in Chap. I., § 4, Example 1, we are given twelve pairs of values connecting the extension of a spring with the weights which produced those extensions ; and the problem now before us would be to determine the relation connecting weight and extension, that is, to find the equation of the graph which we drew in that example, and which we found to be a straight line. It is obvious that in cases such as those in Chap. I., §§ 2, 3, there is no mathematical relation connecting the quantities, and therefore no equation can be found. 2. There are two cases to be distinguished when finding equations to suit given data. CaseI. When all the values given are exact, or known to be connected as they stand by a mathematical equation. Case II. When the values given are the results of observa- tion or experiment, and so liable to errors ; for it often happens that the same experiment or observation, repeated with the greatest care, will give slightly different results. 3. Case I. When the values given are exact, after plotting them carefully and drawing the curve passing through them, if the form of the required equation is not given, we endeavour to decide from the graph what the curve probably is, and assume its equation with undetermined constants. We then substitute the given values in this equation, and see whether the constants so determined are always the same. The method will be best seen from the following examples. 24 CHAP. IT| EQUATIONS OF GIVEN GRAPHS 25 EXAMPLE 1. Find the relation, which is an exact one, connecting the velocity of a train with the time, from the following table. The train was moving with a velocity of 50 m.p.h. when the brakes were applied. Find also when the train comes to rest. | Velocity in m.p.h. | | Time in secs. after application of brakes ere Nhe LaCveiea seus aeosares NinieiG ie | Sheeting Pee sie he Te Eales Taped | M0l mph. SCE CEE EE aw eee rape eel Seles spo sels ep ta iab pels boley 4 tt tt ft tt mre ere rmiepeie rel eeieteleie | pole) Pt ETE Ble ag) toe = AK Oia eat pete be Le] ep Neveereteer a) tet styhet TP rei mere eo ercorenae reli l | edetet deeb ed 4 _ JJ Oca Se ees SRR RRR eee eto serie INSet Tt le Pit Ee J. (50 (Gee e seat hee RR eee “oo JOS RSs se Sova eRe RRR eee of tee EE I ff merece emia |e) INT Ete Ebel . TARO eS Sse eae eee VERE _) SS SEO RSE RGGG RE EER eee eee paereorminprie eiete rome eyed | Te ENE TTT Ey ee ew ei slewieieisepeie is |b betel bp Ne LEE EL er ee ie ie isin recite ie LT TINT TTT ee ite eee er EP EL ET NI Ld FoR tH EE NEBR Pee miele eh ips elstaet [| | lop |} SeCS.. NEE +t et tt iol {| isjol | | i4iot | | jolot | | eto ee ee Ceeeee APA ae ey lod eal o Loe Psi) Fie. 14. On plotting these values of the velocities and times, say V and 7, we see that the graph is a straight line. The equation is therefore V = aT + 6b, where a and 0 have to be determined. Substituting the above values we get 42=10a+b .. (i) 34=20a+ 6 .. (ii) Poe 0a bl {iit} VS) 4a-60 aio (iv) On solving any pair of these equations we shall find 4 cocaine i" b= 50, 26 SCHOOL CALCULUS [CHAP. II Therefore the equation V + ih = 50 or 5V + 47 = 250 satisfies all the conditions, and is the relation required. To find when the train comes to rest we put V = 0 in the formula, which becomes 47’ = 250. ee O22) Hence the train is at rest 624 secs. after the brakes were applied. This also agrees with the figure, as the graph when pro- duced cuts the axis of 7 at 62:5. Note. When we are certain of the shape of the curve whose equation has to be found, and can therefore assume immediately the right form, we need only substitute as many of the given values as are necessary to find the undetermined constants, for instance two in the last example: and when the given or tabulated values are known to be exact, that is to satisfy some mathematical equation exactly as they stand, it is only when there is any uncertainty as to the shape of the resulting curve (as some curves, especially when we have only a small portion of them, are not always easily recognisable from a figure) that we need make sure that all the values given satisfy the assumed relation, that is that we need use all the equations for determining the constants, and verify that the results are always the same. EXAMPLE 2. Find the equation to the curve passing through the following points : a x 6 —3 —2 | —5 |¥ On plotting these points we see that they lie on a circle. We therefore assume the equation to be v2 + y? + ax+ by+c= 0, which is the general equation to a circle. Substituting the first three given values we have 1+ 9+ a+3b+c=0 ..° (i) 1+ 49+ a—7b+c=0 ... (i) 16+ 4+ 44+ 2b+c=0 .. (iii) For these are enough to find a, 6 and ¢: and on solving we get a= —2, b= 4, c= — 20, EQUATIONS OF GIVEN GRAPHS 27 CHAP. It] 2_ 94 + dy — 20 = 0, and The equation is therefore x + y it will be found, on trial, that this is satisfied by any ot her of the given points, though the curve in this case 1s so apc eoinis at sisi eiaial etate t= |p Nele[set at HGS o4oe oa Bao eno BORNE: ACMA ID Seno Leena BAAN eee ehiflalelelain|atalataleleisieleimieta| fala iabe leaded ZS aa eee eee ioe Skee nea pS a TBs Pee ee Doe Te 8 be eee rote fe lobeps echelaeie ls belie bay pate ed cle es Tae La a nd eee op ha BREE ASS PE Ber SDR Ce CBOs anes 0 Bd a AS GN Sas SP dB [ie AUN PARLE Reena Tee Ghia a HERE H+ Fie. 15. certain that it is almost unnecessary even to test one more point. Find the equation to the curve passing EXAMPLE 3. through the points : 5 PA | , and assuming the curve to be a incl | a. On plotting these points parabola through the origin, be y= ax + ba? somewhat as the continuous line in the figure, we take its equation to where a and b have to be determined. Substituting any two convenient points, say the first and =) 9 S ; Peo a oe ll i~ =, 8 | — Oo ae oD) ee Pee ts f g Bs ai be 2) Cre 28 SCHOOL CALCULUS [ CHAP. IT Substituting one other point, say (1°5, 2°7) to test this result, we have 27=°2x 15+ ‘9x (1°5) = ‘3+ 2°025 = 2°325 which is not true. The curve is therefore not a parabola as assumed, and eb tee eae CHEERE EEE Cee See ee || is i ne BR Re He ai | | | | rf ii | | am wits era ae | | me a a EY te be | NJ | | Aoi Al el Siete opel ele lea e) ST 4 -~ SAREE DSS CS eee Wilts Tsiats HABSSaSanes Fig. 16. we will therefore try if it is the cubic y= ax -+ ba? + ca (which also passes through the origin). Substituting now any three points to find a, b and c, say the same two as before, and (2, 5°2), we have ‘7T=—-a+b—ce setna lil) ll=a+b+e Ra Cy 52= 2a+ 46+ 8c .. (v) from which we get a= 0, b= -9, c= -2, Substituting now one other point, say (15, 2°7) to verify, we get 2°7=°9x (1°5)2+ °2 x (1°5)8 = 2025+ °675 = 2°7 | il CHAP. IT] EQUATIONS OF GIVEN GRAPHS 29 which is correct ; and on trial the remaining pojnt (-5, -25) will also be found to satisfy the equation. The required equation is therefore . y = “9x 4- 2x3 or 10y = 9x2 + 223, and the curve is therefore somewhat as the dotted line in the figure. | 4. Case II. When the values given are not exact, but are the results of experiment, after plotting the points, we draw that curve which seems to lie most evenly among them fas in Chap. I., §.4], and endeavour to find its equation as accurately as possible, as in the examples that follow. We are of course only dealing with experiments in such Natural Phenomena as are subject to some law, and can therefore be expressed in terms of some mathematical equation. If we are not given the form of this equation we can only speculate on its probable form from the curve we have drawn; and we should not be satisfied until we have found an equation which represents the curve with considerable accuracy. Exame.eE |. Find the relation connecting the pressure and volume in Ex. 2., Chap. [., § 4. We have already in that example plotted the points and drawn the curve which lies fairly among them. Also from Boyle’s Law, we know that the equation is of the form pv = constant, = ¢ say. | We now see that the point given by the values p = 10, v = 62 is apparently on our curve, as also that given by p=17°5, v=35. Calculating pv in each of these cases, we get the values 610 and 620 respectively for ¢. Hence we conclude that the required equation is very approximately pu= 615, with a possible error of 5 in more than 600, or less than 1%. As a test, read off one more point from the curve, say p= 3, v= 204. This gives pu or c= 612, which agrees very closely with the above result. Example 2. The Volume V of Carbon Dioxide dissolved in unit volume of water at temperature 7° Centigrade, is determined by experiment as follows : | Ses 5° 10° 15° V 1-76 | 1-49 | 1-22 96 Find the relation between V and T. 30 SCHOOL CALCULUS [ CHAP. II On plotting these values, we see that the graph lying most evenly amongst them is a straight line. We therefore assume its equation to be V=a- b7. Substituting all the given values in turn, we obtain the following equations to determine a and 0. 1] Oe ee i (1) 174926 shh Sen 1-23 ¢-+ 100 22. Fin) -96=a+15b .. (iv) Litt -CESSEEEEEEEEEEEEEE Hes bt be CHEF ESSEFEEE EH Fat p be TEFL NEE SaPEeseOPaucdcaanee FEEECEE-EE-EE-EEERS =H EEE ao) Sa Fia. 17 From (i) and (ii) hes ills b = —'054 », (ii) and (iit) Gaia LO b = —'054 ,», _ (ii) and (iv) a= 1-74 b = —'052 Hence we take as the most probable values a= 1-75 and b= —‘053 so that the relation required is approximately V = 1-765 —-0587. Note. This result cannot of course be relied upon beyond the limits of the experiment, as for higher temperatures the law may be very different. Similarly nearly all such relations founded on experiments 31 1,000 600 400 200 100 15 8 The distances approximately in nautical miles at which objects are visible from a height H feet above EQUATIONS OF GIVEN GRAPHS sea level being as follows : D miles oF foot On plotting these points we see that the curve lying evenly among them appears to be a parabola touching the axis of D at the origin. must not be considered trustworthy beyond the limits of EXAMPLE 3. those experiments, though they might prove so on further trial. Find the relation between H and D which is best satisfied by these values. CHAP. It] q BHO Bons Eat ie ti fals1e lee alah eis es atest ates hale [aye Stelle [a otf x baci saab ie (pat Ki 8 ol eho) sled shel Nie Sie lol alae Oat SIS ioe SS ea PLE Wel SS NGS rea se 0 | ca sd eps at | a 2 RP eee Ses OSLO OS SPSS ER SO SCM Ea Meeenenoea PEEEEEEE EERE EEE po OS ade ee ee ie (e see eo eae 1 1d Sahm ad ol eae Ne NC Ta Le ee AS RR BSHGR BURR eR PANS RES EMail. REBBS GRRE Nee eeasAnh Ronan ASS SSH SRERNRRSBESeSoe ee eed Ese ss) eae SR SSE Seqqegcac— SARA SRS SI5 SERRA KOSRMRseS Ameer mevemin? (at lol alal aN [ateltobal or 4 208 ae Sa ee eee QM SVOD CSR eRe BARA UREREES ere eer een eye el eee) spate [et teh TAIN EL Pop eEe ea Th b oS oe SRR AUEm Cc iia e esi sb ee 2 [sabe Elelalgpet\ ope iahs TENN ToSee aR ARReeEaeAInAns DAR WEIR eR aaa Ree BAER So OnE aR sea ease eee Cee Shes eh) ES EDP ESTES tO BS BY ES CRE keane Bessa sas me ze gee p age giving us a mean ities. Although this value of a differs by just over Fig. 18. WEAN Hoo, W181, “157, We therefore assume the relation to be H = aD?, and on "15, substitution we get the following values of @ in order °757, value of -766. 782, 32 SCHOOL CALCULUS [cHapP. II 2 per cent. from the greatest and least individual values obtained, yet the distances given are clearly not very accurate, and could not from the nature of the case be exactly measured, so that the equation obtained, viz. H = -766D? may be con- sidered to represent the relation with sufficient accuracy, and more nearly than any other equation that we could find. Exampue 4. It was found experimentally that the volume V c.c. of a certain liquid, forced through a capillary tube of given length by a given pressure in one second for different radii r of the tube is as follows : “2 -00314 a 3 “4 5 -. | -000196 .0159 | -05024 |-1227 Determine whether an equation of the form V = ar” satis- factorily represents the relation between V and r, and find the volume forced through a tube of radius -35 mm. in one minute. - Since there is an unknown index to determine, we take logarithms, and so get log V = log a+ n log r which is the same relation in another form. Substituting the given values, log -000196 = log a + n log (-1) or 4:2923 = loga— n er: (i) and similarly 3:4969 = loga— -6990n .. (ii) 2-2014=loga— -5229n .. (iii) 2-7010'= log a—, -3979 " (-. 2 eee 1-0888 = log a— -3010%" >. From (i) and (ii),n = 4-001 log'a = -2923)0— ian From (ii) and (ili), n = 3-997 log a = +2910, =a From (iii) and (iv), n = 3-998 log a = -2934, a= 1-965 From (iv) and (v), 7 = 3-999 log a= -2918;a = 1-958 Taking the averages of these values for nm and a, we have V = 1-9595r3°999 or V = 1-96r4, very nearly which is a satis 33 U In Fig. - 1252, EQUATIONS OF GIVEN GRAPHS CHAP. IT] 209%. To find the volume that is forced through a tube of radius factory formula since no error is greater than ‘005 in 1-96 -35 mm. per minute, we have or approx. 1-96 x (-35)4 = -02941 ¢.c. per second. V = This agrees very well with the figure, and multiplying by 60, the required volume per minute -02941 x 60 c.c. = 1-7646 c.c. PaRS Uae PETES Co ECO a Sheer keh ah ea eae ee eee ele hele ebelag [ele ERR Coss ana Eees ab | ra PEPER CERES Fie. 19. -312522, y -063z?, have been drawn with dotted lines (all cutting the former curve y = 1-9624* at the same -781a3, y We notice that the greater the value of n, Note on the shape of curves of the type y = ax”, as their general shapes are not affected by the particular values point), as an aid to recognising curves of the type y= a2”, 19, the 5 graphs of y y= :079a and y of @ in any case. 34 SCHOOL CALCULUS [ CHAP. Ii the closer the curves lie to the axis of x near the origin, and the more sharply they rise later: and when 7 is fractional, the smaller n is the nearer do the curves lie to the axis of y near the origin, and the flatter they become later. ExampLE 5. When a belt passing round a rough pulley ' turns and transmits power, the tensions 7 and 7'g of the parts of the belt are not equal, and their ratio varies with the angle subtended at the centre of the pulley by the part of the belt in contact with it. In some experiments, with a coefficient of friction -4, the following results were obtained. 2 1-150 | 1-322 | 1-520 1-748 | 2-010 | 2-311 | 2-658 2 Angle of contact in degrees =0| 20° 40° 60° 80° 100° | 120° | 140° Show that these can be very closely represented by a relation of the form a = A-B*, Find A and B, and calculate from 2 the formula the value of 7 when @ = 300°, which was ex- 2 perimentally found to be 8-121. Assuming the given formula, and again taking logarithms, as the formula contains an index, we get T logio T, = logi9 A + @ logio B. Substituting the given values of ae and @ in turn, T's logi9 1-150 = logig A + 20 login B 1.6. -0607 = log A+ 20 log B eee and similarly -1212=log 4+ 40logB .. (ii) -1818 = log A+ 60 log B .. (iti) -2425= log d+ 80 log B .. (iv) and three more similar equations. Solving these we find from (i) and (ii) 20 log B= -0605, log B = -00302 and log A = == +0002; .. A= bh: OOM aie from (ii) and (iii) 20 log B = -0606, ‘log B = -00303 and log A = 0; : A= eee B= 1-007 from ey and (iv) 20 log B = -0607, log B= -00303 and log 4 = 0; -, A= 1-000, 61007 Taking also say the last given pair of values as a test, they 3D Si = 4 0 7, = (1-007) -00303 as before. EQUATIONS OF GIVEN GRAPHS give -4245 = log A+ 140 log B, and if we put log A = 0 we 140 The required formula is therefore -4245 where @ is expressed in degrees. CHaP. 11] get log B i it HOMCEERE RSS SSE Beats MSBAMH RE x | SVE oN MSR sekess SRR Ree Annee s LUC SONU SRST RARE ERM SORARBRRe SRS RRA Roe et aaa IS et ie ee ea hse she ie BEC AGS hh Aa SOR Ree ee rl stare le NE See te ate | or eal here ve el eles Clee ah ees peer eine Sheer be bet ahs B ee es RN Re NS pd an pes Pe Pe oes eee | tN ia ote iia Sei cieeNe (pets Sl el elle eee RUESRSSaR Sea wea RRR Ake SoM ae SS SRRRRRSRAVERRR ee SR eRe Ieee leisy aleve astern tet Ni spel oid Shieh Sl aio sive pet eas et a SORE SARA el et ete bet Noise ao lobe atl apap ep ore ls) ape A a >| | SeRE Eas Ls beh ohe Palais lado Cie are ad beep Stet ot | oh op PUSUN A ee et erel hep led fe bet 1 eps hehe eee een arm etche eran here tT oe |b. SREB RON TORE oe GAS eae eel ES ROMMASO NAM ONE RUBBER ERR Lo ef ike bse TAR te ne CN omens Site tts] fot ge | lis RRB eaES HEE PL tt bn eps een hel ota Wied Poet ciated” [ote ct Ris Rie Hf ty sc eH EEE HA | HHH N BRORRESRNSMAVES A eee eee eae : SABER SSeS eMC eae RRR eae HH Ee pies | Ab 2 tS | Se Pete eels Pale tebe eter tal oie Sial SR RAPA ME ANAS OSE Aa ae Meee bh ciel sbebel abel S19 cAe eee eke! faba intel ied BRS eAsIe wes. aS OSS e eed Ror ete Le) Sobre PMs heel ey cate cise ioe lal |e PACCOCCEECY aCe EEE si ~ BURecdeneor ce (coceeManeacrad J Fia. 20. 300 x log 1-007 ' when 6 = 300° we have 44 log Ts 2 wane mn t (Suitable -909 = log 8-11 approx. 300 x «00303 EXAMPLES Plot the curves lying evenly among the points given by the following tables, and find their eq uations in each case as accurately as you can. which agrees fairly well with the experimental result. 36 SCHOOL CALCULUS [cHAP. Ii units should be carefully chosen for each, so that the curve may always cover as much as possible of the space at your disposal.) 1 . ude 1 3 5 "7 ye 2.75 | —1-98 | —5-04 | —Ss1eueninEee 2, t i 3 5 7 9 5) =) \7s86 | 11-7 era mons 25-25 | Py My 2 —3 | —2 | —1 | —-5 | 0 | 2-5 ly. 99 | 14 | 5 | 9:78 | 2 |) O-75 yea 4, E Wen ed or te Ss eo: | F 0. | so: Wiseae ton eas | 5. | . | 3-> oe ee 5 | 76 coon y . | 1-35 | —-4 | —-006 | 0 | -006 | -021 | 4 | 1-35 eae ue 1/21/3441 5 | 6 | =n | Sem: 2 De 43 | 78 | 108 | 136 | 160 | 174 | 184 | 192 | 190 | 5. Graphical Tests for certain forms of Equations. It will be seen in the last three examples that, after plotting the given points, there is considerable labour involved in solving the set of simultaneous equations to determine the constants, and in many other cases this would be still greater. Further, if the proper form to assume is not given, we may select an incorrect one (e.g. in Ex. 3 of § 3), and find when the equations have been solved that the results do not agree. Now there are certain forms of common occurrence, which can all be tested by means of a straight line: and further, if the plotted points lie almost exactly on that straight line, so that its position is certain, the necessary constants of the required equation can be determined merely by finding the equation of that line. CHAP. IT] EQUATIONS OF GIVEN GRAPHS 37 These forms are y = Ax" + B, where n is a positive integer. y = Ax”, for any value of n. iA 1s”, 6. Graphical Method of testing tor the form y= Axr+B This method should be used if we have any reason to know that » is a positive integer. Thus the graph of Ex. 3, § 4 Hoes Pie titelats [op lors bree oO ee eee eee anes fet bt et | fp EE EP at tt LON SS GRR ee Ree eee a STS ES lS aM 9 a eee eet ieee shee .. JI SURES RO ee Ree eee SES ee ee ee eS 2a eee erence) hea | pr ere eee a tel PTT Tr emerald tt I DPA aR eR SP ae eee eee Sei isi eT PEE mete eel sisiget ie et ee et ee J Ree ARS eee reese! eisai tt eee EEE SSD 4h ee Se eee HH SOO OS Ae ee eee _.. ~L ODA BSAC BRERA EERE pay mb | his Foe | mares jot | el eabelede aay |< (a! HEGSeaasane i eee hs) SS pee) Slew isieiete isl le isie tate JR a eee bah eye eae eee fat hear | abe ashe alee _ JSR Bee Re ae eee eee eee Oly | | [loo | | [4loo | | \6joo | | [800 | | [tooo | jz\00 | | 14/00 LYE get Fia. 21. looks very like a parabola. We therefore assume the equation to be H= A-D2+ B. (Of course if we are certain that the curve touches the axis of D at the origin, we may put B=0 at once; but this is not necessary as the new graph will settle the point automatically.) Now tabulate the values of H and D?, and graph them, looking on D? (instead of D), as a new variable, say z. JI j the assumption is correct the graph will be a straight line representing the equation H = Az-+ B. From this graph A and B can be immediately determined since A is its slope, and B the intercept on the axis of H; 38 : SCHOOL CALCULUS [ CHAP. II or we may find the equation to the line by substituting the coordinates of two points on it, in the usual manner. If however the graph is still curved, we have taken a wrong value of n, and must then repeat the process with another integral value. Thus in the above mentioned Ex. 3, § 4, we have the follow- ing table : Rei Hs 50 | 75 | 100 | 200 256 400 529 600 784 25 1000 D2 : 33 | 64 | 100 | 132 1296 The values of D? can be rapidly found with a slide-rule, with quite sufficient accuracy. On plotting these points we obtain a straight line, whose slope is -766= A. Also it passes through the origin so that B= 0. Hence the required relation is H = »766D? as already found. 7. Graphical Method of testing y= Ax". If n is not known, or is known to be other than a positive integer (in which case the tabulation of the last method would be tedious), taking logarithms of this equation we get | log y= log A+ n log x. Therefore tabulate logy and logx and plot their values, looking on them as new variables, say Y and X. If the assump- tion, that y= Ax" is the equation required, be correct, we shall again obtain a straight line, namely the graph of the equation Y =n X + log A. From this, as before, the values of n and log A can beimme- diately obtained, and thence from the log. tables the value of A. EXxameLE. In certain experiments on lamps for lght- houses it was found that the consumption c of oil in grammes per hour with a wick of diameter d cms. was as follows : “egrms. 56-13 | 106-3 | 174-7 | 262-0 495-7 | 803-2 22°83 dems. 2 3 4 5 6 8 | 10 Assuming the connection between c and d to be c= Ad", find A and n. We have the following second table of logarithms: loge .. 1:7492 | 2-0267 | 2.2492 | 2-4182 | 2-6952 | 2-9102 logd.. 1-3585 -3010 4771 -6020 | -6990 | +7782 -9031 1 CHAP. IT] EQUATIONS OF GIVEN GRAPHS 39 The graph of these is shown in Fig. 22, and we see that it is a straight line as we expected. The intercept on the axis of log c is -69 as penny as we can read it. °. log A= -69 and A = 4-9 approx. : a= OU ; Also its slope is (taking the ends of the line to deter- 2.3] 1-04 a ee ee cero ace Peete ys [ep] MEATUS Baa SARs A tog EEE ae Elid Bs || |_| oe fe dae mw A A, LN C ental eee Re eta i ial |_| gah a5 BES ROR AE BEE saan te Ee Te ® be Pel Spe Te BEES E ae Beass EEEECEECEEEERSEEE ster |_| AG SEEM an gaa Ee Da D4ee Raa leans Patt be ny ap WO SAR EAaS BRO SoRae SAS ah e> coor Geogr Osos Att bE eye ere la er Popeater eels Roem ese isipiacas sas pero t a1 | pe ole be Ps |e || NS erase eaeaaes noe Sed ae Dea e tee T | dial as] HVA eee eee HERERO S ae BN LTRS fy eae at fn) 7 eves] BN tla eo OTE Sala se aay aC er Rt tt fe PS] SETS ECLA OTE ee Pee pale Pelee ENE yt Sia ee 1S Peas BeRanos Ny BEEEEEEEEIUOUTEEERES eC a et ee - aH lo BE Fig; 22. Hence the required relation is c= 49d 222 and this is, as a matter of fact, as nearly as possible correct, as we should find by solving the equations. 8. Method of testing y= A-B* graphically. Taking loga- rithms of this equation, we have log y = log A + wx log B, 40 SCHOOL CALCULUS [CHAP. II Therefore tabulate the values of log y and x and plot them, using log y as a new variable, say z, instead of y. If the assumption be correct, that y= A-B* is the proper form, the graph will again be a straight line, namely the graph of the equation, z= Cz+ D, where C stands for log B, and D for log A. We can then as before determine C' and D from the graph, and hence A and B by reference to the logarithm tables. Exampie. Test in this way the data given in Ex. 5, § 4. Since - and @ take the place of y and « in this case we 2 have to tabulate the values of log 7 and @. iy log -0607 | -1212 | -1818 | -2425 | -3032 | -3638 | -4245 2 0 ste a 20° 40° 60° 80° 100° | 120° | 140° i BS BEREREESES SOD. Pifog= TT Tet DT TEP Es ee | Afb ey Pp ep BL Zee REREAD UCU PET PPE ee SRSA! te | EE BHRERERRERRRERRRRERRERAARERAREAL.. LbeSl PT TT ET eee PETE TTT eee PPP Pe aA Ee Pt PCT TR Oe Se BRRERREERRRRRREREEREZ ARADO SCMERERREREREEEERS SERRE... UCC BRRERRERRRRRRERPZGRERRREERAE. | UU BREEDER ARERERERRERE BERR REREEEPZRRRRRAUREAEReeS UU BERESREREREAGRREBREEERESER EUMRERRERP ARREARS ee ees [tee an el as eee REaRwoe PET ep eee BRERP ARERR RERRRARASPARAA UC So EE ~- a | | lelot | Tt slot | tT tejol Tt faiol tt fojoy | | 7zjo| | Tt fio) BASS Baw eeerue Dee EEERD TOL Fra. 23. On plotting these points we again get a straight line through the origin. Its slope is -00303, which represents C, or log B. Hence B = 1-007 Also D= log A= 0. pwAs ls CHAP. IT] EQUATIONS OF GIVEN GRAPHS 41 Hence the required relation is a = (1-007), as already found. 2 Note. The methods of §§ 7 and 8 cannot be applied to equations of forms y=Aar+B y=A-Be+C as we cannot take logarithms of these equations. EXAMPLES Find the equations to the following curves by the methods of §§ 5-8. 1. P |i | v6 [2 | 26 | 3 | 3-5 ae 2 | 6-75 | 16 | 31-25 | 54 | 85-75 Use § 6. 2. x Orie2 5 10 20 50 ves 0 | 16 | 21-71 | 27-36 | 34-47 | 46-79 Use § 7. 3. b. °05 2 5 *75 -9 A :004 | -07 °43 *92 | 1-41 Use § 6. 4, oe. tl 4 7 1-5 | 1:8 2 Jie 5:53 | 7°46 | 10-1 | 22-4 | 30-2 | 36-9 Use § 8. 5. |v 4 5 | 5-8 | 6-5 |.7-25 | 8 | 10-5 p 95°5 | 70-65 | 57°85 | 49-5 | 42-7 | 37-7 | 25-95 Use § 7. 6. t 1 L251 eli Sail: 75. to.2:0 v +825 | 1-41 | 2-42 | 4-13 | 7-08 | 42 SCHOOL CALCULUS [CHAP. II MISCELLANEOUS EXAMPLES Plot the curves lying evenly among the points given in the following examples, and find their equations. 1. D is the twist of a wire in degrees, under the action of a couple of P foot pounds. P.. | .. 27 | 9:94 4:05u 629) agree 9-6 | 13-9 | 17-7 | 22-1 | 26-2 2. In a certain system of pulleys the following forces F cwt. were necessary to lift weights W cwt. . Wie el 868 105 | 126 | 186 1 185 F.. .. | 16-2 | 18-05 | 19-1 | 20-6 | 22-05 3 » OF |e 7 5 ata D) 3 4, The horse power required to drive a certain vessel at V knots per hour is as follows View) oe bBo) 1015 0 eames HP. ..| 33 | 43'| 73 | 131 | 226 5. A body moves s feet in ¢ secs. as follows eae a -68 | 1-32 | 1:89 | 2-05 | 2-55 Sts ae | 7°44 28 | 57°4 | 67-6 | 104-6 — — : Find also the acceleration. 6. The prices of certain tables of length / feet are given as follows: l | 5 6-5 7:5 8 10 P £718. 0d. | £10 7s. Od. | £13 1s. Od. £14 10s. 6d. | £21 9s. Od. Wetee | ae 1 |a33°) 1-6] S1seueneee | a Try if this satisfies the form y= Aan + Bb. | 1-5 5:69 | 12-34 | 22-01 | 35-27 CHAP. It] EQUATIONS OF GIVEN GRAPHS 43 8. ae eat Lata hee sap 3°2 4 4°86 Parte Aner Ob 4°8 4-2 3°75 3°4 Assume the form y= Az”, 9. Tke pressure of air in grammes per sq.cm. is connected with its volume in litres in a certain experiment as given: Pare | 8 4) SG) A-teb ol 6 17-8 | p.. «| 107-5 | 86-5 | 62-6 | 50 | 40-5 | 27-5 10. The specific gravities of a number of balls of the same weight but different substances and their weights x grammes in water are as follows: ek, re 26-52 25:9 | 25-44 | 24 | 19-52 | 18 19 “13-4 11 fi 3°3 |2°8 The curve is a rectangular hyperbola. ll. Pee ROR ee | ee O81 .3-4< |. 5 Y + xe 5 2-92 2-036 1-808 1-715 | 1-6 Plot u and I x y 12. The following table gives the temperature in degrees Centigrade of some boiling water allowed to cool, readings being taken every 2 minutes. Find the relation between @ and ¢. 6° ie -- | 87-1 | 75-8 | 66 | 57-4) 50 Time in mins. .. 2 4 6 8 10 Use §8. 13. If a uniform rod be laid symmetrically over two supports in the same horizontal line, and loaded with equal weights at the ends, it can be shown that between the supports the rod will bend into an are of a circle. In some experiments with a rod one metre long, and loaded with one kilogram at each end, the radius (r cms.) of that arc and the distance (% cms.) between the supports were found to be as follows : ee os. 30 | 40 50 60 | 70 | 80 ree. | 2907 84-7 | 41-6 | 52 | 69-4 104 Find the relation between x and r. (Plot x and ) 44 SCHOOL CALCULUS [CHAP. II 14. The following table gives the observed pressure (p lbs. per sq. in.) of the atmosphere at heights (h feet) above sea level. hice, .. | 1,000 | 2,000 | 3,000 | 5,000 | 7,000 | 8,000 Des »- | 14-01 | 13°35 | 12-72) 11-56; 10-50) 10. Find the relation, which is of the form p= a(10°"), and assuming the same law to hold, find the pressure at 20,000 ft. above sea level. CHAPTER III DIFFERENTIATION 1. The definitions of a Variable and of a Function of a Variable have already been given in Chap. L., §§ 5 ff. 2. It was also there explained that f(x), F(x), d(x) and similar notations are used to denote any function of x, and that f(z) is not to be considered an ordinary algebraical expression indicating that “‘ f’’ is a multiplier, but the complete symbol f/(~) must be taken as a whole, meaning “* some function of «,’’ whose actual nature must be independently stated in each particular case. Also, as before explained, if f(x) denote some particular function of x, in that case f(a) will mean the value of the same function f(x) when x has the value a, or when a is written instead of x. But it must be noticed that f(x) and /f’(x) denote two different functions of the same variable which may have no connection with one another. As a matter of fact two quite independent functions would be represented by two more distinct symbols as f(#) and F(x), or f(x) and $(x), while f(z) and f’(#) or F(x) and F(z) would be used to represent two functions which had some connection with one another though quite different in form. This point will become clearer as we proceed, and it will be found easy to distinguish the uses of the functional symbols. 3. The following important Series and Limits are here collected for reference as they will frequently be required : but their proofs may be found in text-books on Algebra and Trigonometry, and will be assumed. Goizass (1 +1)" =e (ii) Ltz—o -— : —log.a an i t (iii) Ltz—o = oe ls ih ae eat =) a being measured in radians. 45 46 SCHOOL CALCULUS [cHapP. Iii Although it is assumed that these statements are known and understood (and we Gannot proceed much further without using them), yet so much now depends upon a clear under- standing of what is meant by a “Limit” or “ Limiting value’ that we proceed to consider the matter more fully. And first it should be noticed what results we obtain by direct substitution in the above functions. Thus in (i) when #=oo direct substitution would give us ¢ +5) =( +0)%=1°% Similarly in (ii) we should get are apsk = = iain SLL eee and in (1i1) ge Hence it will be seen that in each case we get a result whose value is ‘‘ Indeterminate ’’ at first sight ; but it will be found that as the variable is made to approach indefinitely to the value required, the function also approaches indefinitely to another particular value called its “limit” or “ limiting value.” 4, Examples to explain Limiting Values or Limits. We will take three examples to explain more clearly what is meant by these limiting values, and what is their nature. sin x (i) Consider the Ltz-9 ———, which means the limit or limiting value of the fraction See when x approaches in- definitely to the value 0. The following table gives to 7 places of decimals the circular measure of the six angles 1°, 50’, 40’, 30’, 20’, 10’, in that order with the corresponding sines. Gas) 0174533 | .0116355 -0029089 0174524 0145444 - 0087266 | «0058178 sini ee! 2 -0145439 | -0116353 | «0087265 | «0058177 -0029089 from which we see that as x decreases towards the value 0, sin x and x approach one another in value, until when « is an angle of 10’, its circular measure is the same as the sine to seven places of decimals : and we can see that as # approaches the value 0, the values of # and sin x become still more nearly cap. 111] DIFFERENTIATION 47 sin 2 equal, so that the less a differs from 0 the less does ——~ differ from 1. This is the meaning of the above statement that Ltp—o = aay (ii) Again, consider a triangle ABC, whose sides AB and AC are cut at D and EF by a line parallel to BC. Then the oe AD Ab ae ratio AB equal to the ratio AC for all positions of the cutting line: and this is true however near to A that line be drawn, so that the limiting value of os when AD or AE is indefinitely diminished, is still equal to the finite quantity oe AC” When the cutting line actually passes through A there are of course no parts AD and AEF at all, and there cannot be a ratio of two things which do not exist: but no matter how small we make AD and AZ, even if both be absolutely negligible compared with the shortest imaginable finite line, Va nS eae _ which measures the ratio of these two infinitesimally AE ale ; AD AB small lines is itself still finite, and Lt, ,~0 AB AC" (iii) Again, take the case of a body moving with a con- tinually changing velocity, whose velocity we wish to find at some particular moment t, seconds after it started moving say; let ¢ seconds be an interval of time always containing this moment, and let the distance described in that interval be s teet. Then the average velocity during that interval ae is | ft. per sec. If ¢ be small the average velocity : ft. per sec. is nearly equal to the velocity after t, seconds, and the shorter this interval be taken, i.e. the more nearly ¢ approaches the value 0, the more nearly is ; ft. per sec. the velocity after t, seconds, until in the limit when t=0 we have Lti=o : =velocity after t, secs. in ft. per sec, 48 SCHOOL CALCULUS | [CHAP. It! giving us, that is, the velocity at the particular moment required. 5. We will now state a Definition of Limiting Values or Limits. If f(x) can be made to differ from a definite number “A” by less than any assignable quantity however small by making x approach to a definite number “a’’ as nearly as we like, then “A” is said to be the limit or limiting value of f(x) as x ap- proaches ““a’’ as its limiting value or limit. This is usually expressed more shortly by saying that in this case ‘‘ The limit of {(~) when x=a, is equal to A’”’: and this also is what is meant by Ltzoa f(x) =A. If we consider this definition in relation to the examples taken in the last article, we see that in the first case we can sin x make differ from 1 by less than any assignable quantity however small, by making x approach sufficiently nearly to sin x the value 0, so that Lt, 6 as In the second case the ratio a is always equal to eh and remains the same however nearly we make AH or AD approach the value 0: and in the third case we can make : differ from the velocity of the body at any particular moment by as little as we please by taking the period ¢ secs., which contains that moment, small enough : and in general in every case (except those like case 2 above, where the ratio Se is constant, and therefore always exact) the more nearly we make the variable, x say, approach its limit “a,” the more nearly does f(x) become equal to A; and this is expressed by saying Leen] (eae 6. The Evaluation of Limits. We have already seen in § 3 two of the forms which may be assumed by functions when some particular value of the variable is directly sub- F : ; 0 stituted in them, viz. 1° and 0° Such forms are called Undetermined or Indeterminate forms, and many others such CHAP. III] DIFFERENTIATION 49 as =, 0”, 0x co, etc., also occur. But of all these 5 is the most common and important, and the other forms may be made to depend upon this one: e.g. 0x co=0x = : Cone LeU oo and we shall therefore only consider at present the one case of evaluating f(x) when this assumes the form ‘ upon directly substituting =a. The method for evaluating such a limit is to substitute a ++ h, instead of a, for x, where hf is a small quantity: it will then be found in general that h divides out from the numerator and denominator, leaving us a result whose limit can be immediately seen. For instance to find Lity=3 a: if we write 3 for a this assumes the form vi e Substitute therefore 3 -+ h ae a, where hf is small : Po ae pk then Lt,-3 ——_, pi 2 aL 0 aa 3 6h +h? h =Ltya9 (6+h) Now 6+ approaches indefinitely to the limit 6 as h ap- proaches the value 0: therefore 6 is the limit required so x2 —9 that Lt.—s3 =| tf x—3 In this it might seem that we have divided out from the numerator and denominator a factor # which is 0, and which cannot therefore be so treated: but this is not the case, for h is never actually 0, as we are endeavouring to find to what =Lth=0 ie number 3 we have found that the more nearly x approaches 3 the more - approaches in value as x approaches 3, and aes nearly does — : approach 6, which is therefore the required limit. We sometimes require to find the limit of f(z) when x=oo, + 50 SCHOOL CALCULUS (CHAP. iii that is when x becomes indefinitely large, and then the defini- tion in § 5 would no longer be applicable, as we cannot now 6¢ 99 make x approach a definite limiting value “a”: in these ute ieee : cases, however, we write — instead of x, f(x) thus becoming i(,) and we then have to find Lt,—o i(,) which may be done as before. (2% -+1)2—4a2 x ye ey [ieee 4x 2 2 64 ele G y" : ] aye 2A Ee a (of the form 5) ioe ET oe 4 EXAMPLE 1. To find te ates Put eee Then y = =Lty=0 (4+y) Note. If in this example we substitute 0+A for y, we (2+h)—4 (2+y)2—4 gy only get instead of : the substitution h is therefore unnecessary, and we merely consider y itself to be a small quantity whose value ultimately approaches indefinitely to 0. Another important limit is here added as an example. Gh EXAMPLE 2. To find Lt,—1 a 0 east This again becomes 9 on direct substitution of 1 for 2, we therefore put «=1-+h, where h is small. Then Fel oe ey ee eal ee a—l h 1m pO) p24 a =Lin=o 7 CHAP. I1I] DIFFERENTIATION 51 al AheG " ia h--higher powers of h| = This important limit is true for all values of n: for h being as small as we please can always be less than 1, and there- fore the expansion of (1-+h)" by the Binomial Theorem is always convergent and valid. A further discussion of the limiting values of undetermined forms will be found in §§ 27-30; but the method of substitu- tion explained above, together with the use of the three limits at the beginning of § 3, will be sufficient for the present. EXAMPLES Find the following limiting values. 322 — 22 — 8 4, [tp (WV 24+ 1 — V2}. hy, Ltr—9 —— ee t= ~- 5a — 10 ‘ ore 6° ips apne cee : ame 42 o. Lite=o a 21? —¢ (n —1)2 — n2 eh Ge lta 7. Notation for Small Increments of the Variables. Let f(x) represent the function 3x2—2x—7 of the variable x: Then f{(9)=3 x 92—2x9—7 and {(9-001) =3(9-001)2--2(9-001) —7 and /(9:001)—/(9) is a definite small numerical quantity corresponding to the increase -001 of x from 9 to 9-001. Similarly /(12-001)—/(12) is another small quantity corres- ponding to the same increase -001 of x from 12 to 12-001. But [f(9-001)—f(9)] and [f(12-001)—f(12)] have different values, although the amount of increase of x is the same in each case: we therefore require a notation which shall state not only the small increase of x but also the definite numerical value of x from which that increase is reckoned. When therefore x increases from the value 9 by a small amount (such as -001 or any other small quantity), we will call this small increase 59, so that x increases from 9 to 9+69: similarly $12 will denote a small increase of « above the value 12. Also 69 and 612 may be any small quantities not neces- UNIVERSITY OF ILLINOIS LIBRARY 52 SCHOOL CALCULUS fcHaP. It sarily equal: and even if they are equal [f(9+69) —f(9)] will most probably not be equal to [f(12+612)—/(12)] whatever function of « may be denoted by f(z). Similarly if x take any particular values 2 and z2 not numerically specified, then small increments of x will be denoted by dx; and Sze. | } Suppose we put f(x) equal to y, so that y=f(x): then when 2 =2X1 f(a) becomes f(x%1) which is a particular value of y, say y1: — similarly y2=/f(x2), etc. Now f(%1+6a1) is a particular value of y differing by a small quantity from /(#), i.e. from 4: this small difference from y, we denote as before by dy; so that we get y; 4-6y1 =f(a1 +62) : similarly yo+6y2=f(x2+629), where it is obvious that dy; is not equal to 641 nor dy2 to dx2 ; but each only denotes some small increment of x or y. To take the particular case at the beginning of this section, if y==3a?—2a—7, and x1 is 9 and 22 is 12, and for convenience in this case 6a, and da are each taken equal to -001, we find yi i.e. {(9)=243—18—7=218 yi +Sy4 i.e. f(9-001) =243 -054—18-002—7=218 - 052 yo ie. f(12) =432—24—7=401 ya +- dye i.e. f(12-001) =432-072 —24 -002—7 =401-070 Hence dy, =-052 and dy2=-070, which are neither equal ‘to one another nor to 69 and 612 respectively. EXAMPLES 1. If y= 3a? — 7a+ 4, find dy when x increases from (i) 2 to 2-01 and (ii) 0 to -O01. 2. The distance S feet described by a body in time T secs. is given by S=57'+ 1672; find the small distance 5S feet described during ~, sec. (67') immediately after 2 secs. of motion, and the value of Be 63 3. Ifs= 0 — 3° find the values of 6s when 6 increases from (i) -5 to -51, and (ii) 3-5 to 3°51. 4. Show that if y = az-+ b, 4 has always the same value for any increase x ¢ da, of x from any particular value a, of 2. 8. Rate of Change of a Function. Let us now consider the rate at which the function 37%2—2x2—7 changes in value as # increases steadily by small amounts. When x increases from 9 to 9-001, 3%2—2x—7 [or f(x)] increases by -052 approxi- CHAP. IIT] DIFFERENTIATION 53 mately, i.e. it increases by about 52 times as much as x: when x increases from 12 to 12-001 (i.e. by the same amount as before), {(a) increases by -070 approximately, i.e. it increases Oy by about 70 times as much as 2; so that Se Oe nearly, 7 1 - and 0 nearly. Now oy; and 6a, are the actual 2 increments of y and # respectively while x changes from ) 9 to 9-001, so that = measures the average rate of increase 1 of f(x) (which is y) compared with x, during that small interval. For consider the case of a much larger interval, say while « changes from 9 to 12: f(x) during that interval increases from /(9) to f(12), i.e. from 218 to 401, so that its total increase is 183 while w increases by 3, and therefore the average : . 183 3 ahay increase of f(a) is 3 «OF 61 during each unit increase of 2, {(12)—f(9) € that is to say measures the average rate of increase of {(x) compared with x over the interval from 9 to 12. Exactly similarly Ee gee (9) or su measures the average rate of increase of f(x) compared with x over the interval from 9 to 9-001. This is not the actual rate of increase of f(a) either when x=9 or when x=9-001, but if we diminish the interval -001 to an indefinitely small quantity, the limiting value of Be when 6x, (and therefore also d6y,) becomes indefinitely 1 small measures the rate of increase of f(%) compared with x when x is 9, and this limiting value may be written Ltsx,=0 su in conformity with our previous notation for limits. Similarly Lt5z,—0 measures the rate of increase of f(a) compared with #« when 7=12. ExamMpPLeE 1. To explain more clearly the nature of Lts2,=0 sh the following table gives the values of dy; and a 54 SCHOOL CALCULUS [CHAP. III corresponding to different small increments 62 oF x above the value 9 when y=3x?—2x—7: ee S| 05 Ol -005 -001 -0001 by1 5-23 | 2.6075 | -5203 | -260075 | -052003 | 00520003 on . ,. | 52-3 | 52-15 | 52-03 | 52-015 | 52-003 | 52-0003 ey from which we see that as 62; approaches the value 0, so that v approaches the value 9, ee approaches continually 1 to the value 52, and as a matter of fact if smaller values ) still were given to 6a, then ee would approach still more 1 closely to the value 52, which is actually its limit when a =9. This limiting value of out is written dys that is to say Ox4 ax Lt8x,=0 oy is written more briefly dy : similarly : Ov} adx4 oy2 __ dye Lt eee Sa date’ and it is with the evaluation of such limits for different functions of x that we shall now be concerned, the method to be employed being that explained in § 6; for the quantity dy in general, corresponding to any particular value of zg, dx is always the limiting value of a fraction - whose numerator and denominator both vanish on direct substitution; but to find this limit on each particular occasion by making a table such as that just given would be most laborious, nor would the resulting limit be always certain. Omitting now the suffixes by which we denoted a par- ticular value of a, and the corresponding value of y, and taking x and y in general to be two corresponding values of the variables x and F(x), we see that since ct ere) —/(2) CHAP. II] DIFFERENTIATION 55 the finding of Ltsr~0 e is similar to the finding of the limits in the Examples of § 6, while the meaning of this limit, i.e. ae is the rate of increase of y, or F(x), compared with the corre- sponding increase of x at that moment. EXAMPLE 2. Find the rate of increase of the function 2u*—9x+4: (i) when «=4, (ii) when x=1-5. (i) The rate of increase when x=4 is Lts4~0 Me malt, 4 aed eos a —Lpgao 1244-184) ese) el (2x 42—9x 444) eo Os — —4 Penta 32+ 16(64) +2(64) a 9(64) +4—324 36 7(84) +2(84)2 =Lt84=0 Sy yes =Itsa=0 {7-+2(54)} =7 Therefore 2¥2—97+4 is increasing 7 times as fast as 2, when +=4. f(1-5+61-5)—f(1-5) (ii) Lts1-5=0 Sine —L4dr5_9 P25 +515)? —9(1'5 +815) +4] —[2(1°5)? —9(1'5) +4] ce S15 — Ths g ~3OE BS) +2015? 61-5 = [181-5 =0 (—3+2(61-5)) = So that 27?—9x+4 is decreasing 3 times as fast as x is in- creasing when x=1-5. Now just as in this example we have found the two par- ticular limits separately, so it will be seen that we could first find the value of ey for any pair of corresponding values of x and y (where y=2x2—92+4), and substitute the particular value of x required in the result, and so avoid repeating the process, thus ; 56 SCHOOL CALCULUS [ CHAP. IIT dy _¥ | oy ae a Ltsx=0 Sa: ae DSSS. f(z+6x) —f(x) ox . [2(e+éda)2—9(¢-+6x)+4]—[227—92-+ 4] = Ltdx=0 Se > 7 Duptes 4x 8% +2(dx)2— 95a 2 = Itdz=o [4a—9+2(8z)] = 4%—9 Whenza= 4, 4r%—9=7 and when z= 1-5, 44—9=—3 as before. Similarly we can find the rate of increase of any function f(x) for any particular value of x by obtaining the limit of zl Ee i eat te f(a-+-6a) —f(x) 52 and then substituting that particular value of x in the result. 9. Differential Coefficient. Definition. If f(x) be any function of a, then Lazo (*t°I— I) is called the Differential Coefficient of f(x) with respect to wx, and the process of finding this limit is called Differentiating f(z). The limit is also sometimes called the First Derived Function of f(z) with respect to 2. | The notations used for the Differential Coefficient of f(x) 1; d2=0 ; af (x) df. ae ie are ['(x), a and sometimes ree and if y=/(x) it is also ; d written a or 1. Differential Curve. If the differential coefficient f’(x) be found of any function f(z) of a, it is in general (as in the last example) itself a function of x: the graph of this function is called the Differential Curve or the First Derived Curve of the function f(x) or of the curve y=/(x). (See also the end of § 10.) Second and Higher Differential Coefficients. Since f(x) is itself a function of 2, we can again differentiate it, and the result is called the Second Differential Coefficient of f(x), and CHAP. lt] DIFFERENTIATION 57 is written /’(x), or Ee) the latter being generally d2 f(x) : : expressed as : or if y=f(x) the second differential dx? 2 coefficient of y is written os or yz. Similarly higher differ- ential coefficients are written aay Sen, ae and this must da8 dzt be carefully distinguished from Or which means the nth dy 4 power of Jog OF Gay - Second and higher Derived Curve. Just as ie is a function of x whose graph is the first derived curve of f(x) or of y=f(z), 2 so f’’(%) or 2) is again a function of x, and its graph is called the Second Derived Curve of f(a”) or of y=f(x), being also, of course, the first derived curve of /’(x). Similarly the graphs of zt 2 58 SCHOOL CALCULUS [CHAP. 10 a or of a Differential Coefficient. Let us now consider the geometrical meaning of 10. Geometrical Meaning of dy dx’ for this purpose we will draw the curve y=/(x), for instance y =2u?—9e-+13. Draw the ordinates at x=2-5, and «=3, and complete the construction as in the figure. Here MN or PR is 62 (-5 units in this case), while NQ—MP, or RQ, is dy (1 unit, and hf le hes [il | 4) ERG e Shee z eth: apatites | ers h a ela le hat al Sasi cecneeeeee eT Eds | Rap Se ets Reema ee ot bal PST TAN \\ IN ad clei eerste bela eh ele ele Lert ale ei a i a BSR e es Awe aNe rR RAR AEA SNe RMB REBHDS SOR ERY SMiREeeBhEas Fie. 24. as we see from the figure). ‘Therefore “V9, and this being ae is the tangent of the angle QPR, i.e. tan 6, where @ is the inclination of the chord PQ to the axis of w If we now diminish $2 indefinitely, so that @Q approaches indefinitely close to P, oy becomes Bs and PQ becomes the tangent to Ox the curve at P, so that 2 is the tangent of the angle that the tangent to the curve at P makes with the axis of 2, CHAP. III] DIFFERENTIATION 59 i.e. tan ¢ in the figure, remembering that dy here means the da differential coefficient of 2%2—9xv7+13 when 7=2:5. To find its value {2(a-+8a)2— 9(~+6x)+13} — {2u02—9x7+13} Ltdx=0 Sa: 2eF, ae 4a; §%-+2(dx)2— 96x . Ox =Ltd2=0 (4% —9-++-2 62) =4y7—9 and 4%—9=1, when x=2-5 so that tan ¢=1, as we can also see from the figure, allowing for the different units used for «and y. Thus Zz at any point of a curve is the tangent of the angle made by the tangent to the curve at that point with the axis of w, i.e. it measures the slope of the curve at that point. It may happen that - is negative for a particular value of x. Thus in the above curve when x=1:-5, as we have _“ =4¢%—9, we get in this case ~“=—3 Gree s-8 pe eda ; whose geometrical meaning is obvious from the figure. For as X increases near this point, y decreases, and the tangent makes with the axis an obtuse angle ¢’, whose tangent is negative. already found that In any case we see what is meant geometrically by dy as dx the limiting value of 2 when 6v=0: for as 6x decreases continually towards the value 0, oy (i.e. ne in the figure) Sx PR becomes more and more nearly equal to tan ¢ ; a other words the less 6% or PR differs from 0 the less Shee = or a differ from tan ¢, so that ou (or Liss =0 a) is tan d. If therefore we want to find tan ¢ at a point P on a curve we must find au at that point, by finding first of all the general value of oe 60 SCHOOL CALCULUS [CHAP. III for the curve y=/f(x) and then substituting in the result the value of x at P. | It should also now be noticed that in finding the value of # at any point on a curve, we are in fact really considering some definite point P on the curve, then we calculate the tangent of the angle made by the chord joining P to another point @ near to it with the axis of x, and lastly gradually decrease the length PR until P and Q coincide: that is to say that while calculating the general differential coefficient a) x is kept constant but 6x is variable, just as when finding yh os Ltz=3 = oa AEE (in § 6), x was constant and equal to 3, while h (which was the same as what we now call 6%) was variable. We see then that ws being the tangent of the angle made by the chord PQ with the axis of a, measures the inclination of PQ to the axis of x, or measures the slope of the chord PQ, or measures the average change in value of y compared with x during the interval under consideration, as we saw inde- pendently in § 8: for if y changed uniformly from MP to NQ while x increased steadily from OM to ON, the tops of the ordinates whose lengths represented the values of y would lie on the chord PQ instead of on the arc PQ. The smaller we make MN or 6a, the more nearly does the chord PQ approximate to the arc PQ, and so the average rate of increase of y compared with x becomes more nearly equal to the actual rate of increase of y compared with x, until in the limit when MN or 6x is indefinitely small, PQ becomes the tangent to the curve at P and so momentarily coincides with the curve at P, and the slope of the line and the curve are the same, enabling us by means of su which is the tangent of the angle made by this line with the axis of x, to measure also the slope of the curve at P, i.e. to measure the actual rate of increase of y compared with x at P. 3 The ordinates therefore of the Differential Curve (see § 9) of y=d¢(x) measure the slopes of the tangents to the curve y=d(x) at the points whose abscissae are the same as those of the ordinates considered of the Differential or First Derived Curve, CHAP. IIT] DIFFERENTIATION 61 ExamPLE 1. Find the rate of change of the ordinate of the curve y=322—2x—7 compared with that of x when x=9. Also find when this rate is measured by the quantity L Here dy or f’ (x) da [3 (a +-6a)2 —2(%-+6x) —7] —[38a? —2a —7] =Lt32=0 : ox ig 1, 3x2 + 6ada + 3(d6x”)? —2a —2Q64 —7 —3a?242¢-+7 seit ic 0 ; on Je 2 ae 6x da —26% + 3(dm) bx =Ttd,z-0 (6% —24 36x) =6a —2. ..f/(9)=value of 6%—-2 when 2 is 9 =—5§4—2=52. The ordinate y is therefore increasing 52 times as fast as the abscissa x, when #«=9. : i eee dA —2=- when 6%=—, i.e. =, Again 6%—2 5 when 6r——], ie when «& 19 The ordinate or 3x2 —2% —7 is therefore increasing | times as fast 1] dy > : as # when #=F5) for at that moment da OF f(x), which we have seen is always 62 —2, will be equal to g ExameLe 2. In the curve 3a2—4y?2=7 find the slope when x=2, also what is the value of x when y is increasing at the same rate as «. We may write the equation to the curve in the form 3x2 —7 a erp, 4. Veto —V eet fan = 26x Pik [302 —7 + 6adx + 3(dx)?}? —[3a2 —7]? as 26% 352) Sa\ (sa2— 7) (14S) (8a meg 0 =F bx 62 SCHOOL CALCULUS [CHAP. TIT (62% + 362)da 302 —7 must be less than 1 since 5% may be as small as we like, we get this equal to using the Binomial Theorem to expand this since i (6%-+362)da . terms involving at (3a —7)8 [i+ ay 2(322 —7) T- Jeast (ox)2... —I Ltdx=0 ae - 262 (322 —7)}- hte de NF rdary a 2(3x”- —7) ey 262 See a ae A% 3x 4(3a2—7)? ~~ 94/392 —7 The value of this when x=2 is a = esl: which therefore measures the slope of the curve when x=2, being the tangents of the angles made by the tangents to the curve at these points with the axis of #; for two points on the ah curve have their abscissa equal to 2, since y then is --—— - must a Ly or the tangent to the curve at the point (or points) required must make an angle of 45° with the axis of a. Again if y increases at the same rate as 2, To find when shes we must solve the equation dx Oe tt zi 2/342 —7 *, 942 —4(3x2 —7) 27 21 wag P21 Hence y is increasing at the same rate as « when ne 2/21 ates or when «= — The meanings of these results will be clear if we plot the curve given: or otherwise, we know the equation is that of CHAP. 111] DIFFERENTIATION 63 a hyperbola, whose centre is at the origin, so that the two values of = when #=2 give us the slopes of the tangents at the ends of the double ordinate whose abscissa is 2; and in the second part of the question we see there are four points on the curve whose abscissae are neve) : but at 2 of these dy =1 and at the other two ey eaat for the solution of dx dx ee —l, i.e. of + gi = —1 would give the same results da 2V 3x2 —7 for x as did the solution of oY a1, Hence to really complete the solution of the second part of the question given we should require a rough graph of the curve, though it is truly answered by saying that y increases at the same rate as x 221 when 3 ae EXAMPLES 1. P is a fixed point on the curve y = 3-52? — 2x whose abscissa is 2-5. Q is a variable point whose abscissa is given in the following table: by com- pleting that table find the limiting value of the slope of PQ when it becomes the tangent at P. Abscissa of Q | Ordinate of Q | Slope of PQ 2°55 2°51 2-505 2°501 2. If y= 22+ se , complete the following table, and so find = when z= 2. x by 2 da by se 64 SCHOOL CALCULUS [ CHAP. Iti 3. If y = e*, find the value approximately of - when 2 = 2 by completing as the following table. by bx éy ne 5 *25 -] sd In Examples 4-6, prove approximately, by a similar method of tabulation, that : 4. If y = 10sing, she 10cos%, when x= -5 and also when x= 1:3, all E angles being measured in radians. Meret 5. If y= 2, = 3e8* when «= +75. o 6. The slope of the tangent at the point whose abscissa is 2-5 to the curve s = 16? continually approaches the value 80. 7. Find similarly from the following table the velocity of a body after 3 secs., if its distance in feet from a fixed point after ¢ secs. is given by 8 = 3t-+ 5t?. dt és Average velocity during the interval dé secs. Seas 8. Find the slope of the parabola y = 2a? -- 7a when (i) # = 1-25, (ii) 7 =3-5; write down the coordinates of the points at which these are the slopes. 9. The velocity of a body in feet per sec. after ¢ secs. is given by v = 5-5é. What does the slope of this curve represent ? Find the acceleration of the body graphically. 10. A body moves s metres in m secs., where s = 40n+ 16n?, What does the slope of this curve at any point represent ? Find graphically the velocity after (i) 7 secs., (ii) half a minute. 11. The number of tons 7’ unloaded from a barge in 2 mins. is given by T=1-5% — Find the rates of unloading (i) 2 mins. and (ii) 25 mins. wed 1000° from the start. Also find graphically the time taken to unload the barge if it holds 50 tons, 12. olf p= ee find the value of ne when v=5. If the pressure (p lbs. per sq. in.) and the volume (v cubic feet) of a certain quantity of gas are connected by the equation pv = 500, what is the meaning of your result ? CHAP. III] DIFFERENTIATION 65 11. Tangent and Normal at any point of the Curve y = f(x). We can now write down the equation to the tangent and normal at any point on a curve given by y=/(x) : for consider the curve y=2x?2—9x%+7: at the point where «=4, y=3, as we saw in § 9, the tangent to the curve makes with the axis of x an angle whose tangent is 7: the tangent required is therefore a line passing through the point (4, 3) and making with the axis of x an angle whose tangent is 7; its equation is therefore y —3=7(x —4) and the normal is a line through the point (4, 3) at right angles to this: its equation therefore is Peat UP aetar 7 = 0 or T(y—3)+a2—4= 0 Similarly we can write down the tangent at any point (%1, y1) on the curve y=/(x), and its equation clearly is y —y1 =f" (#1) (% —21), for this line passes through the point (a, y,) and makes with the axis of 2 an angle whose tangent is /’(2), which is the same as the slope of the curve at the point (7%, y1). Notice that (x1, y1), being a point on the curve, y,; might instead be written /(x%1), as y; is determined when wx takes the particular value 7. So too the normal at the same point is f(%1) (y— 1) +(% — #1) = 9. These equations are of course the same as d J ae (x — #1) dy d mh at otk. an We (y¥— 41) +(4#—%1) = 0 where dys means the value of dy when +=21. day da ExamputEe 1. Find the equations to the tangent and normal at the point (4, 54) on the curve y=2a +5 dy 9 i [2(a-+ dx)3+-5]—[2a3+ 5] dx bx meee 1002 Ox (O0)o 20x)? b 2a" —5 Ox =Ltdz-9 [6x2 + 6ada +2(da)?2] = 6a? 5 66 SCHOOL CALCULUS [CHAP. IIL d Therefore when x = 3, a == 3. x Hence the equation to the tangent is y—54=3 (v4) ie. 4y —21=64 —3 1.e, 4y=6x+18 or 2y=32-+9 and similarly the normal is } (y —5}) +a —4=0 or 12y+8x=67. [Notice that the point (4, 54) 7s on the curve, for when =, y=2(4)9+5=4+5=54] EXAMPLE 2. At what point on the curve y=2a2—9x+13 is the tangent parallel to the line 2y=5x%+-7 ? As in § 10 Ex. 1, Sas! —9, and the line given us makes 3 : . eo with the axis of x an angle whose tangent is —. Y We must therefore have 5 Agus Oy os 529 207 ya 2 y ne 64 3 +13 529 —828 ae ee Se yim Te 299 mr apy Therefore the point required is (22, 355). dy dx by elementary methods: but to write down the equations and normals at any points on any curve y=/(x), we shall require the differential coefficients of any function of 2 that Note. In these two Examples we have been able to find CHAP. lil] DIFFERENTIATION 67 may occur, and we therefore proceed to show how they may be found. Many curves have of course equations which are implicit both in x and y; but as previously explained (cf. Chap. I., § 6) these can all theoretically be so expressed as to be explicit in y, and we shall at present only consider curves which can be so expressed. EXAMPLES ‘ J. Find the slope of the parabola y = 22? — 3 at the point (3, 15); hence find the angle at which the line 57+ 2y = 45 cuts it at that point. 2. Find the angles between the curves y = 3a°+ a” and y= 3a — 2? at their points of intersection. 3. Show that the abscissae of the points on the curve y = ¢(a) at which the tangent is See to the lme y= ma-+c is given by the roots of the equation p’(x) = 4, If (1, y1) oe any point of intersection of the curves y = f(x) and y = (a), show that their angle of intersection at that point is tan—1 Mer) — (a), 1+ f'(x1) $(@1) 12. Differential Coefficients of certain Standard Functions. I. Aconstant. Let y=c. dy C—C aan = [it a ° Then ae d2=0 —<— = ==() This is also obvious from the definition of a constant quantity, for as it does not depend on wz, a change in x pro- duces no change in the constant; .°. dy is always zero, and therefore also ui and are also zero, no matter what dz may be. Or geometrically, y=c is the equation of a line parallel to the axis of x, and Z measures the slope at any point of the line; but this is everywhere zero. exe.) Let y—2". dy (w-+-da)" —x" Then a 4 Bt a Se an 4-nan—t - dar eee 2 Bee)? we gen =Ltdn=0 du =/1s..0 (nen + SE det... ) =n | 68 SCHOOL CALCULUS [CHAP. III [Note. It will be seen that when a function has to be expanded in powers of 6a as in the above example, the square and higher powers of 6x may be omitted from the start ; for after dividing out the 6z, all such terms will still contain dx, and so will disappear 1 in the limits when éx=0. We shall therefore neglect them in future. | III; .a%. Let y—a~ Cy qrtsx ax The SS te ” dx Lt 3n=0 bx ogee Ox =a" loge a [§ 6, 2]. Corr lireliay ae. ; Bis ae a ae loge €=-e% Cor.2. It y=10* Ly ae 100) @ dx Le Og eas logige +4343 IV. log, x. Let y=loga Pied dy =Ltdr—0 loga (v+6x) —loga x dx Ox 6x (+) er ale loga x Ox 6x 1 Now put Tare ae that z=oco when é6x=0, and we get Whee w= loge (1+; :) 1 Hira, = "Toga (1+2) 1 : =" -loga¢ [§ 3, (i)} Cor. 1. If ae oz dy 1 dx x , loge a Cor.2. If fares L, 7 dy _ -4343 ia, ! logs 10 rig a dx CHAP. IIt] DIFFERENTIATION 69 V. sinx. Let y=sin 2. Then dy Dah Ss sin (w+6x) — sin % dx bx 2 cos (x+3) sin og =Tt5, Mae SW Jo ae Ou sin os —=Tt§,—9 COS (+5) eae z eS) —=COS &. . 08 ain 4 ieee sc ii.), since bx _g Se 1 by § 3, (ii1.) 2 VI. cosx. Let y=cos z. Then ty rie cos (w+6x) —cos « dx ie 2 sin (x+5) sin (-3) 2 2 Ou sin — te aol (« +3) — =-—sin xv. [§ 3, (ili).] VII. tanx. Let y=tan x. Then dy ae re tan (#+62)—tan # da 6x sin (w+é6xz) sin x cos(x+éx) cos x Se a sin (v7+éa) cos x—cos (x-+6x) sin % cos (v« +éx) Cos °x i sin 62 cos (v+6x) cos con = —~—=sec2a. [§ 3, (iii).] tre Be) 70 SCHOOL CALCULUS [CHAP. III VIII. cosecx. Let y= cosec 2. Thon te Tds 4 Coe ene oe ee dx Ox by eee gs Gok sin (x fea sin @ ae ee sin «—sin (%+6z) sin (+62) sin #-da 2 cos (2+) sin (—*) =Ltdx=0 sin (w+6x) sin x ox cos ( +3) sin a aay anes = ee=0 sin (@--ox) sin a | ba 2 =a a =—cosec x cot x. [§ 3, (i1).] IX. and X. sec x and cot x. These two are differentiated in a similar manner to the foregoing, and the results are sec # tan x, and —cosec? x, respectively. [As an aid to the memory, it may be noticed in the above that the differential coefficients of the functions beginning with co-, i.e. of the cosine, cosecant, and cotangent, are negative. This is as we should expect, for these functions decrease as x increases. Also that the differential coefficients of sec x and tan 2 contain only sec w and tan 2, and similarly with cosec x and cot x. | XI. sina. Let y=sin —a. Then y+6y=sin —1(¢ +62) eR aa asen tL), x-+dxu=sin (y+6y) . de=sin (yt+6y) —sin y. by ~ilte=0 Sin (Y-Fby)—sin y CHAP. 11] DIFFERENTIATION 71 y 2 cos (y+) sin x ==[t5, =0 [Since y=0 when éx=0] oy 1 oF ==LtGp =n EEE SEES COS (y +) sin oy 2 1 1 1 ig kee = a iti: COSY V/1—sin2y ~W/1—2? ial: Note that a slightly different method of proof is here adopted. The other inverse functions might be similarly treated, but the method is cumbrous, and a better one is given in § 22, while the results are tabulated with the others in § 13 for convenience of reference. 18. List of Differential Coefficients of Standard Forms. 1 — _4 —() Are dx dy 2. y=axn ey red YY =X =i NX dy 3. y=ar gt oO ae a* loge a dy —per + as a ey tim dx dy 10” = 10* SS = Be SS y a 10” log. 10 Sete dy 1 4, —| fs see ee Y =1084q © ie 7 [08a e 4] ay 1 ns 4343 Y =10210 & a logio € Meo a! 1 Bey ee vem dx «2 5. y=si dy Y¥=SsIN wv =——==(C0S 77 dx 6. y=cos x ci Oe ve 16. SCHOOL CALCULUS _ [CHAP. 11 . y=tan x . y=cosec x . y=sec X& . y=cot x pani Oa wei COs mee, Oi) be ee if —COsCC gant Y= see Fa J COteae iene ny x dy —cosec % cot & da eben. x tan x dx a —cosec? x dy _ 1 dx /] — dy 1 da Pinhl —x? dy aah dx 1+4a2 dy 1 Ou. StA/ 42 1 dy _ 1 dx ur/y2—1 dy _ 1 dx 1t22 The following additional Differential Coefficients of Standard Forms are here added for convenience, and their proofs will be found in §§ 24 and 25. 17. 18. Lo) 20. 21. 22. 23. y =sinh x y=cosh x y=tanh x y=coth x y =cosech x y=sech x 7 sip dy dat =cosh x dy _ a =sinh x d ; ce sech2 x dy = —cosech? x dx dy —” — —cosech x coth x dx dy _ —sech x tanh x dx dy 1 CHAP. III] DIFFERENTIATION 73- 24. y=cosh ! x — Poo y—tanh, } 2 oy 5 (aes) Zou —coth tx, oo es (x > 1) oy g=-cosech. =a ae yee dy 1 2 Jeo =I het aus Poo -—sech— | 7 ah ERE 14. We have now seen how to differentiate the simple functions of x of common occurrence. It remains to consider how to differentiate combinations. of them. These will come under the following heads : A sum. Such as sin 7+" A difference. Such as sin 7 —2"” A product. Such as 2” sin x nh A quotient. Such as z sin x A function of a function. Such as sin (#") We proceed to consider these in order, for any general functions such as f(x), $(x), ete. I. Asum. Let y=/(x)+¢(2). Then aed f(x+6x)+¢4(x+6a) “. dy=/(x + 6x) —f(x) +4(u+dx) —$(x) . SY _f(u+édu) —f(x 2) 9 NS SAS, 0D Ox Ox and in the limit OY —f'(0) +9"(@) The proof can obviously be extended to more than two func- tions ; the rule is therefore, Differentiate the functions separately,. and add the results. II. A difference. Let y=/(x) —¢(z). Then y+éy=f(x +é6x) —d(# +62) ". dy =[(% +6x) —f(x) —( (x +a) —G(x)} “. Oy f(e+ox)—f(x) o(%+6x) —4(@) Ota, 62 Ox 74 SCHOOL CALCULUS (CHAP. III and in the limit he (x) —$’(x) and again this can clearly be extended to more than two functions: the rule being Differentiate the functions separately and subtract the results. Combining these two we see that if y =f(x) —F (x) —o(x) + then oY #2) F(x) 6a) III. Aproduct. Let y=f(x)-d(z). Then y+6y=f(%+ 6x) -6(x +62) ". dy =f(w+dux) - p(x +x) —f(x) - (pa) =f(x--6x) (p(x 6x) —G(x)} +(x) {f(x +dx) —f(x)} [Notice this step. The term /(v+6z)-4(x) is added and ea to produce the form we require. | Baja ee — 4-4(2) H+ 82) =H) and in the limit OY (x (2) >’ (x) +(x) -f’ (ax). If c is a constant it follows that if y=c¢(z) dy dx? Hence the rule is, Multiply each function by the differential coefficient of the other, and add the results. This can also be extended to a product of more than two functions, but the process and the result are both complicated. A better method of dealing with such. products is given in § 15, I IV. Aduotient. Let y= Tay Then y+édy= /(x-+62) (e+ 82) _ gy a feeb) _ flo p(x+ou) (par) _ at+dx)-b(x) —f(x)-$(@ +6x) p(x + 62x) p(x) _$(%) thet) —H(x)} —Ka) {b(%+6x) —G(2)} p(x + da) - h(x) CHAP. III] DIFFERENTIATION 75 [Notice this step again. The term f(x#)-¢(x) is added and subtracted. ] en fhe a hey ae and in the limit dy (x) f(x eee AZ) dx { p(x) }? Hence the rule is, Multiply the denominator by the differ- ential coefficient of the numerator, and the numerator by the differential coefficient of the denominator. Subtract the second from the first, and divide the result by the square of the denominator. V. A function of a function. Let y=d¢(u), and u=f(z), then y=¢{f(x)} so that y is a function of f(x), that is, a function of a function of x. Let x become x+62; then u becomes u+édu; ie. ut+du=f(~+6x), and when this change takes place in u, y becomes y+d6y, so that y+éy=d(u+Su) ; and hence when x becomes w+6x, y becomes y+d6y, the change in uw being the same quantity du throughout. Now clearly ee since the quantities concerned are all small finite quantities, and therefore the dw could cancel. dy oy dy du dy du A C:= xr=0 te ==s © r) ae =Htdx.=0 bx he, éu 6x du dx for when 6x=0, ay becomes ie ; also 6u=—0, and therefore ou dx by becomes dy bu du This can ee be extended to any number of functions. Thus if y=f(u), w=d(v), v=x(w), and w=F(x), so that y=fpix(F(2))5] raves dy dy du H dw dz du dv dw dx’ which may also be written f’(a)-’(v)-y’(w):F’(x), a fact which should be noted, drawing attention as it does to the 76 SCHOOL CALCULUS [CHAP. III meaning of the notation /’(w) as the differential coefficient of f(u) with respect to wu, and so on for the others; that is /’(w) d f(u) d f(u) oe Mee means — and not — x EXAMPLE 1, on differentiating the sum or difference of functions. Differentiate y =3x° —2 aes men Caer dy Spe A Here —- = 9x2, + —, Sia poe many Fa EXAMPLE 2, on differentiating the product of two functions. a2 52 “COS & Ys k Here ey — i ‘sin x an GOs & 2 (x sin +2 cos @) ie EXAMPLE 3, on differentiating a quotient. See sree PY Gadi s) oe. Hore dy 3 (—2 sin «)—2 cos x 6a dx oa" __ —2 (x sin «+2 cos 2) is 323 ; Note that the same function was differentiated in Ex. 2, as a product, which is often a convenient method. EXAMPLE 4, on differentiating a function of a function. y =loge (tan 2). Let tan x=z. Then ts =sec? x, dy dy dz 1dz andes ee dx dzdx zdx ] mo sec? a tan x =? eosec 22. 4 CHAP. III] DIFFERENTIATION ce We add one harder example on a combination of these methods. : : ) Examp.e 5. Differentiate y=x" e* *"*, Let sin «=z, and sin x=v so that z=~/y Phen y— 7" -e- dy dz es eerie ore eae) dz dz dv 1 ENC ee --CcOs & dx dv du 24/y eeecos & 24/sin x ; 4 ee Seat: Substituting in (i), and taking out the factor x"~1 e**"* ay Tyne evant {4 _& COS & | fe | 2V/sin a) It will be found that with a little practice, most of these steps can be done mentally, and the result written down at once. EXAMPLES Find a in the following cases: Pru ox + 5. 15. y= 3-7 sin 2. aa 16. y=5 cos x+ 3 tan x. - Y=ax?+ 2bxr+c ae ae Ot: 5d. y= *7Tx? — 1124+ 7 18. y = Te®. 7 on ay 2 19. y=acosec x+ b cot x. a 20 = x Sin x: -Yy=Ve. 21 yw (log, a.— 1) 2/x 9. . 22. y=2? cot x 10. y=+/x+ —.. ry ype ahi a/ x sin x ll. y = are’. 25. y = log. =. a y = Sal5 — Qa-"7, 26. y = cos (ax-+ b). 3. y=(5 — x)?. 27. y =sin 22. 14, y= 23414 +¢, 28. y =2 sin? x. bu? 29. y =sin (27). 78 SCHOOL CALCULUS [CHAP. IIT — gin? (22 i 30. y = sin? (x). MG ope 2+ sin v 31. y= cos oan : Wiig ae 48. y=sec (2%). 32. y = (5 — z)J8. 49. y = cot a 33. y= ¥x+ 1. paste! 34. y = loge (ax + bx + ¢). 50. y= Vat be. 35. y = et log, x. 34 36. y = 5 logig x. Oly = ; 37. y = logio 2a. Vy tee 38. y = log, sin x. eR eae 39. y = logio sin ax. Seria 2 40. y= 44/2 — 2) 53. y = log, tan (x7). a/ % 54. y = (log, tan 2)?. hh pa BB ty ee loge x loge x 42. y ate 56. y = sin (é*). 43. y= cos xe. 57. y = tanz log, sinz. 44. y= log, «-e—*. 58. y = e# cos (bx-+ c). 45. y = cos 2a-e— 32, 59. y = loge (xe*). 46. y =sin 2 cos 32. 60. y = log, (w+ Va2+ a?)- 15. Logarithmic Differentiation. When we have to differ- entiate either a function of the form uw? where both w and v are functions of x, or else to differentiate a product of three or more functions, we take logarithms before differentiating. 1. Let y= (x) o(x)- F(x). [Classi eam Taking logarithms we have log y=log f(x)+log ¢(x) +log F(x). Differentiating both sides with regard to x we get dye glass ] } 1 . y dejan ga) * ey ae and in this form the result is easily remembered, and can obviously be extended to any number of functions, or we may multiply through by y, i.e. by f(x)-¢(x), F(x); so that we get d eal) F(a) f(w) +f @) Fw) 6”) +f) (a) F(a). Whence the rule, Differentiate each function in turn, multiply each of these differential coefficients by all the other functions, and add the results. II. Let. y={f(x)}¢(9 Taking logarithms we get loge y==(x) :loge f(x). CHAP. III] DIFFERENTIATION id: The right-hand side being a product we apply the rule for differentiating it, and we get as before 1 dy 2 1 wy, a8 Wy, ane ie) f(x) +loge f(x) +$’(x) or multiplying up by y, i.e. by {f(x)}*™ OY (a {flay} 9 -f’(x) +loge f(x) : {f(x}? : ’ (a). This result need not be remembered, but the process for arriving at it should be used for each example of the kind. Note. In this section we have twice had to differentiate loge y. This is a function (namely a logarithm) of a function of x (namely y) and therefore (by § 14, V.) its differential coefficient is ey y dx eer . sin 3x ExampLeE. Differentiate y=(cot 3x) 2 . Taking logarithms loge y= Beer loge (cot 3x) ts L4y_aee ee a ee loge (cot 32) es tae 3x! —3 cosec? 32): __ 3x cos 2S oF itaes (cot 32) sae 16. List of Differential Coefficients of Combinations of Functions. 1. bie a“) +P (x) +(x) + OY af (x) +F"(x)+¢"(e)+ ... 2. Am =/(x) —$(2) wy (x) —p’(x) 3. y=f(x)- (2) 7] oY (ar) -$!(@) + (2) +f) 80 SCHOOL CALCULUS [CHAP. III x yal p(x) dy _o(e)-f"(x) —fla)-$/(2) da {p(2)}? 5. y=o(u) where u=}(x) dy dy du. da du dx 6. ie —f(x)-o(a):-F(x) . . . \ Use logarithmic 7. y={ f(x) J differentiation. ri { EXAMPLES se Yen ; Find dx 2 each of the following cases: 1. y = 4x sinzx cos2z. 2¢+ 3 ae 1 10. y=(s> ad) 2. y= (Va? — a?)z- - x Bie 10 4/ le ha 12. y=(1+ 2) (1 — 2x)? (1+ 3a)3. Tle yes (=) 13, y = Da : 14. y = 2(sin x) 008 2. 5.7 =s@ Sinz c*. 15. y = eVsing, 6, af = ersing - eot sa 16. Y= (V1 oi a2)”. “ 17. y = e2?sin2(22), 7. y = (sin «)?*, 18. y = (cot x)32. 8. y = xe—V1e . cot? x. 19. y= a es sin x (1 — x?) (2+ 32) Se 20. 4 = : od atc, Y= | ae 1%. eure the equation to a curve is given in the form x=f(t), y=¢(t) where ¢ is an independent variable, and there- fore i ‘and 2 y both dependent variables. For instance the equation to the parabola y*=4ax may be written «=al?, y=2at, and the equation to the cycloid is most convenient in the form *=a (O+sin @) y =a (l—cos @) #, and @ are called parameters, and the coordinates of any point on the curve are said to be given in terms of a single parameter. CHAP. IIT] DIFFERENTIATION 81 To find es in such cases we proceed as follows : oy dy dy by ot _ dt _ ot) dg lda=0' 5, = Lti=0 da de f'(t) ot dt for when 6%=0, S¢ also = 0. Of course this gives the slope at any point, and to find the slope at a particular point (a1, y1) the proper value t; of ¢, which was used to determine 2; and y;, must be substi- tuted in oe after differentiation. Notice also that ¢’(t) and (4) mean that the differenttations must be performed with ‘regard to ¢, the independent variable. ExampLe 1. Find the tangent to the parabola x=al?, y=2at, at the point for which t=}. Determine also the coordinates of the point. Putting t=4, we get w=s, y=a; and the required point The tangent is therefore a y—a=2(%—7) or 47% —2y-+a=0. EXAMPLE 2. Find at what points on the cycloid given by «=a (@-+sin @), y=a (1—cos @), the tangent makes equal angles with the axes. dy asin@d _ sin@ dx a(l1+cos 6) 1-+cos 6 equal to +1, as the tangents are to make an angle of 45° or of 135° with the axis of wx. sin @ e Pe toa 7 Popa ee “. sin 6=-L(1+ cos @) *, sin 0+ cos O= +1. 6 In this case and this must be 82 SCHOOL CALCULUS [CHAP, III To solve this equation we multiply through by sip and we get ] : 1 ] = gin 04-——. cos: 6 =-- V2 Le = /9 which may be written = ; vin 5 7 : sin @ cos 41098 ? sin qu rsin | or sin (6-3) =-Lsin . ‘, since the upper signs must be taken together, and also the lower signs, we have 3 Tia ° Tt: sin (6 =) == Si) se 4 4 d si t)=—si diet (-7) and sin (0+ 4 sin A sin 4)" ae ae 5G. gy od 5 T and 047 = a ee d= — Hence the points are given by mate a Gl eins | Ey 18. Differentials. Let Dv and Dy be two quantities, small but finite, such that the ratio of Dy to Dx is measured by Z which is a certain quantity obtained by a definite process, namely differentiation, so that we have Dy Gr Oe GLY Dr (which is a fraction) =— (which is a differential coefficient) where the Dy and the Dz are quantities which may be separated, but - must be CHAP. III] DIFFERENTIATION 83 taken as a single quantity indicating the result of a certain operation, dy and dz being inseparable. The quantities Dz and Dy so defined are called Differentials of x and y, and their nature will be further seen from the following figure. Fre. 25. Let the equation to the curve drawn be y=f(x). Then at the point P where x=-145, oY tan SPR. Now, since ae =tan SPR, RS and PR are two small quan- tities whose ratio is equal to ; we may therefore call them Dy and Dx respectively, as they satisfy the above definition of Differentials. So also R’S’ and PR’ might equally be taken for Dx and Dy. Of course RQ and PR are dy and $2, and we see that which equals Zh is also equal to Lt sz —o ce As a matter of fact, these differentials will always be found to appear homogeneously in any equation, and therefore their actual values are of no importance. We shall therefore generally adopt the usual notation of writing them dy and dz. Dy dy N ee ow since 7 =". dy ma Dy = Dx ee (i) 84 SCHOOL CALCULUS [CHAP. II] or with the other notation ey. i teas ne Ce Sota ine For example, if y=sin x we know that oY 00s z. Substituting for @ in (i) we get in the notation of differ- entials, Dy=cos 2: Dz, or as it is usually written dy=cos x-dz, and this notation will frequently be employed in differentiating equations. Example 1. Differentiate the equation 2 tan Ye sin 2. Writing down the differential coefficients of every term, we get dy 2 dy Y dx y2 dx (since y is some function of 2) sec? —322 cos x + 6x sin x ie (se0?y — 9) oY 3 (x cos «+2 sin 2) or since eet dx Dx (sec2y =) dU sey (x cos +2 sin @). Da Now multiplying up by Dz, (sec? i) Dy =3x (x cos «+2 sin %) Da. Hence the ordinary form for differentiating the equation 2 is (sec?y— —) dy=8a (x cos +2 sin x) dz, and we may in future write this form at once. ExampueE 2. Differentiate the equation iy) =$(2). f(y) Dy=$'(x) De, or f(y) dy =¢' (a) dav. Note that /’(y) means the differential coefficient of /(y) with regard to y. We have at once CHAP. IIT] DIFFERENTIATION 85 ExampLeE 3. Differentiate the general equation of a conic, ax? + 2hay +by? + 29x 4-2fy+c=0. We have 2axda + 2h(ady +ydax) + 2bydy +29dx +2fdy =0 or (ax-+-hy+g) Du+(hu+by+f) Dy =0 or (ax+hy+g) da+(ha+by+f)dy=0. 19. To find - when y is given implicitly as a function of x. We have only to differentiate the equation as in the last section, and thence find dy ; as in the following examples. dau EXAMPLE 1, Find dy for the general conic of § 18, Ex. 3, dx at the point (x1, y1). We have found in that example (ax hy +9)Dax + (ha -+by +f)Dy =0. _ Dy axthy+g pe Re he by 4} at any point; and at the particular point (24 41 ) i oe dy aa, +hyi+g Da which is equal to ie (ee eee Note that (21, y1) must be a point on the curve, that is, “1 yi must satisfy the given equation; otherwise we are trying to find the slope of a curve at a point which is not on it. EXaMPLE2. Find the slope of the curve 3x3 + 2x2y —24y24-54 where it crosses the line x=3. To find the required points we must substitute «=3, and solve the resulting equation in y. We have 27+ 18y =24y?+ 54, . 8y7— by —9 =0 Besy- 8) (2y—3)—0. 6°. y= = or = 86 SCHOOL CALCUCUS — [CHAP. III dy dau 927da 4-207dy + 4aydx =48ydy. _ dy _— 9a? +4ay Now differentiating the equation to find guavas gg J See NSBSe Ze SaNNee BEROAANNA BERS aARN ees betinte ls aes! ARRAS ROE PREECE PO HAO 47 Cee A ee pele ele aaa SEN beg | ata Jaane RRZGew See ee Sf fal gots tata a a] Seo HOR aRee Pk bets bac | PaRnRage MAEDA Ow eats elu tee Gist: al BCU Goce ce Pe ls fete she aaa eee] ahaa a SC RRA See Af SENSE | | 7 ia ei | V/V 7.9 GS AR SSaSSe Cocoa eaBane Seale ate Fie. 26. eee 3 Substituting «=3, Ye dy es 2h 3) eee x Mo (ee 0) ea 3 Again substituting 2=3, 2 a0) dy 3(27+6) 11 dx 2(36—9) 6° The curve is drawn in the accompanying figure and the tangents at the two points have been constructed, by drawing CHAP. III] DIFFERENTIATION 87 lines through them with slopes as calculated above; and we see that they touch the curve as they should. 20. Tangent and Normal to any Curve. In § 11 we saw how to write down the tangent and normal to a curve whose equation is of the form y=f(x). We can now find the tangent and normal to any curve. For instance, to find the tangent at the point (3, 5) on the curve 3x° -27?y =24y2+54 which we discussed in Ex. 2 of the last article. We there found that the value of a at this point is a and the equation to the tangent is therefore or lla—6y =24. And in general for any curve f(x, y)=0, we first find eB 2 as in § 19, and the equation of the tangent at any point (a, y;) ; d is y—Y1 =(q4),@-) where ay means the value of eS at the point (#1, 41), i.e. when x; and y; are substituted in it for v and y. Since the normal is perpendicular to the tangent and passes through the point (x1, y1) its equation is (“") (y¥—Yy1) +4 —2 =0. +4 Examece 1. Find the equation of the tangent and normal to the curve zy=11, at the point (k, =) Differentiating the equation we have Roe eee x dy-+-y dx=0. Tee enter ul The tangent is therefore y -> =— ae —k) or llv+k?y =22k 88 SCHOOL CALCULUS [CHAP. IIT k2 or k8x—llky=k4 —121. and the normal is a (v—;) =x«—k EXAMPLE 2. Find the tangent and normal to the curve ey=sin x, at the point where we Taking logarithms to find y at this point we have 7 y=loge sin x=log, sin F loge = —loge 2. Also differentiating to find a x ey -dy=cos x da. - dy COs cos: "cde ev sin x =—cot x. , since eYy=sin a, CYS eae and when w= dg OO 6 = V3. The Hence the equation of the tangent at the point ( @? —loge 2) is y+loge 2=/8 ( -§) and the normal is /3 (y+log2)+-2—%=0. Sometimes numerical values are required in such equations, though usually the above forms would be sufficient. In this case = vie loge 2= +6932, +/3 =1-7320 and ¢ = - 5236, so that the tangent is y+ -6932=1-7320(%— - 5236) or 1-732%—y—1-6000=0 and the normal is 1-7320(y+ - 6932) +a— -5236=0 or «-+1-732y-+ -6771=0. CHAP. IIT] DIFFERENTIATION 89 EXAMPLES Differentiate equations 1-5. LT. w*y = 2x — 3y. 2. 3(v-+ y)? = Tx? — 5. 3. 442+ Qy? — Tay = 2x-+ 3y. 7 4. (x+ y) (2a — 3y) ce’ pes 5. xy? = 5 (9 — 2). x y 4 6. Find the slope of the curve in Ex. 1, at the point (9, al ; 1\ 7. Find the slope of the curve in Ex. 3, at the point (0, 3 ): 8. Find the slope of the curve in Ex. 5, at the point (5, 2). Find the equations to the tangent and normal to the curves in Examples 9-14 at the points stated : 9. y = 3x2 — Tat the point (2, 5). 10. y = 3x3 — 54+ 1 at the point (1, —1). ll. v?+ y? — 2x =4 at the point (2, —2). zx x 12. y=3 (e6+e 6) at the point (x, 1). 13. xy? = 12 at the point (3, 4). 14. 27y? = 4(x — 3)3 at the point (15, 16). 15. Find the equation of the tangent to the curve 22+ y? — 4x+ 6y = 12 at the point (5, —7). What is the curve ? 16. Find the equations of the tangents to the curve 9x2+ 4y?+ 18% — 8y — 39 =0 which are parallel to the line 4y = 9x — 3. 21. Lengths of Tangent, Normal, Subtangent and Sub- normal at any point of a Curve. Let the curve in the figure be that whose equation is f(x, y)=0. Then PT is what is meant by the length of the tangent at_P, PG the length of the normal, 7’N the length of the subtangent, and NG the length of the subnormal. Let PT make an angle ¢ with the axis of x. 90 SCHOOL CALCULUS [CHAP. III Then the tangent P7'=y cosec ¢ Vi +tan? } _ Vi ee tan ¢ Y dy dx The normal PG = = u (4y\" =y sec 6=Yy \ 1+(3%) The subtangent 7N =ycot d= + dx The subnormal NG = y tan NPG dt Dey Hu in each case being of course the value of the differential coefficient of y when «=O. EXAMPLES Find the lengths of the subtangents and subnormals of the following curves at the given points: 1. y2 = 4a at the point es 3% eee) 2. y = 4ax? at the point (—". ia a’ a) 3. 22+ y2 =r? at the point eo Bes 4, 2y+ x2 = 43+ 3 at the point (5 oy \ 5, 42+ 3y2 = 12 when x =1. 2 2 6. as when x = 4. 7. ay = 36 at the points (4, 9) and (9, 4). 8. y= log, x at the point (71, y}). 9. y?(x — 3) = 2a at the point (6, 2). 10. x = y(1+ x?) at the points (2, -4) and (4, -4). 11. Find the distance between the points in which the tangent and normal to the curve 9y? = x°(6 — x) at the point (3, —3) meet the axis of z. 12. How would you find the points on the curve ¢(x, y) = 0 at which the subtangent and subnormal are equal? Give two methods. Also show that the distance between the points where the tangent and normal in such a case cut the axis of x is equal numerically to 2y. CHAP. IIT] DIFFERENTIATION aL 22. Differential Coefficients of the Inverse Trigonometrical Functions. We now return to the differentiation of sin —!z, cos —!x”, etc., referred to in § 12, XI. t sin —'x. -Let y=sin —'z Then sin y= *, cos y dy=da aye eel 1 ] “dx cos y ~ 4/1 —sin2y iver II. cosec—1x. Let y=cosec —z. Then cosec y= -, —cosec y cot y dy=dx eel ae 1 ‘du cosec y cot y 1 1 ~ cosee y1/cosec2y —1 x21, Hi cot x, Let. y=cot —!2. Then cot y= ay. Pies dy =dx . dy 1 a ctaie Coseczy, a5 1 a 1 l+cot2y 14x? IV. Similarly the remaining differential coefficients may be found, and the results are: dy 1 ih ees Ce aha y=Cos x, 7 Trae dy 1 If y=tan-1 ae : ee de 1px? dy 1 If = —1 = = - . YOO Ss y= aay 23. Hyperbolic Functions. The values of the Hyperbolic Functions of x are by definition (from analogy with the exponential values of sin x, cos a, etc.) : Ye om ‘g é he= sinh & 2 92 SCHOOL CALCULUS [CHAP. III Cr Cae cosh «= 2 __ sinh % ete = tanh x ~ cosh x — exte—x 1 and cosech4———.—— ec: sinh 2 The elementary. properties and relations connecting these functions, analogous to the ordinary formulae of Trigonometry, are stated and proved in any text-book on Higher Trigonometry, and will be hereafter quoted and used when necessary. The shapes of the curves y=sinh x, y=cosh 2, etc.; are shown in Fig. 27. It will also be assumed that sinh w=loge (w@++/1-+-2) cosh—! x=loge (4 ++/22— 1) 1+a 1 tanh—1 Poe loge 5 < where x must be < 1 a and coth—! — loge — j Where x must be > 1, The differential Coefficients of the direct and inverse Hyper- bolic Functions will also be required. 24. Differential Coefficients of the Hyperbolic Functions. I. sinh x, Let y=sinh epee BL ign poms Then pe =cosh x. II. tanh x. ; inh 2 Let y=tanh =~ Sate Mire, pals Goat _ dy _cosh a-cosh x—sinh «:sinh a ie cosh2 x hae (since cosh? #—sinh? x=1) =sech? x 93 DIFFERENTIATION CHAP. IIT] i a i | if a ERRER BER OCS One eos SE Seas Bees a | ne | ae a | a a 1s i a fF a a i a | fw |e i | | Fh | Pi ttt tty ft Ett Te kG ea ae i a | ga i a a | | HSB eS BeBe i ae ab a | ae a eS | | a a | | | HERRASA CESARE SREY | fs Fa | a | a a ea as a ee Ff | SSG Bones 5 ae i a Po en a if of [| | PSEA HE 15) oe a sf a ae ae a oe a aa as fi a a i i TT Tt Ty TAANESA [| (em | a a a a i fe) a DORSRES Rae eFC BBGES SER EES CODER Ee BARS sso Beoos SRESSERRES Zee Se el lol bao iet eb oiels lol aig eiae orstal aie ope: Teles eis AVSREASNRBESSh Ses lel fell tee al ia ale Tet cias Pee Telos Te ea seo tale tebe | a | | | a a a ea ee fa a a ea i | a a Pa ae et A ae PE +} EF EERE So RS RoS sees wee EET Ee eer eee ERDSEaR BSE SBeeaeae SP ial ele Seoere dels iol ele sis ero tle) ts edema bets i oT a i] om es | fe |= EEE i fe Sa i | Fi a a Ff | | a oe fa | es | eta | | ek a | wt Ne IN 7 | SL | Ef fe a a Fa a ie a i | [i a is | ie | af sf om 1 i Ne NY tN a i | a a Ui et a il fo [a i i stats rigs 2 (epeean abel eet fe lsiaa eet Nips amg PON lett 12] ofall ee etl cate Lest rial fis bel ba oe kde estes ey Pere el eioballel eel cre aie ie te et eer Leal re Red bt tee hate ees ee ote re ast ae ee ie a | J fe | | fm Pee | |g i | | i | || NY Fe Ff ee | i | | oh a a | fA ad wef el oe fect | a df Oe | OTH UT Np POT TN Sa | i a | i Pn a fa Ye i i Th i Fecpeieen ad legal pe) eel enemies A hens er re AE eens Poteet be eol | bebe ded deh cteds | Lape be ithe bea psi ed at pig EEnEe pat Lt Lt ft td Pt EIN 9 HSH + He H+ fe HE HOS oLo BOOBS AS CROs Behted Bho ERS Soe REE Pepe AL eee HERE REBAR S SSeS sae eae aees tt Rt pF EE tH Sa sen a er ee a in| Game a fas] ce gal oat ae ae af eo a aN Te Fi [| ee caf | ew || | ee |i | ee UDR Rao BeR Ns GRRL ae PORN ReP Soe SOs Loe Sees Reese ese ee ee eee ee eee te epee rr SSESC RS SRR ABD SR RRDRS REE 2U EMO SLE Lae Se aN CORO RN CR eRe tle Cee pled pele esl el oe Tee leet 7 ee eee oie ere Norenerage iat ber) tae et) er tele | ee a aa Fe a es a Fa kr f BER OR CRS SENENOS BS SEe Re ae BRR Ree ibe a |) (Won |e ie a | Fc fae ERM USGL DNASo ae ae ee See eRe OCS eE Bases eenoaeee Eee EE Kea SPe. Gi. Ge SESDUOR EERE RECS EEN 4e ESEES) TEEes NAZIS i ef i fi el | os PAH PCE) eat] a ae a ch Ae ae! HREREECeISas ERR ceases p77 | BRDU AEN Se, SBP REM ORR ADERRPeS Seer ‘am te ee a if a ef ae) re ka Pa ee HUBS Aka HESRUASRL ERS AUR RP RASA BEanenwe \ | Se denn nee eee | SeaboerczsenevaseTataauesetseresesaenactie aT | a NGSEO GE MOMMA AEN RCS cet EEE CCE EEE EEE ml) | PE tt STORM EE RESP ROR SCREAMER E RRMA ss SEE gna SE EEE EEE EEE EEE ECE EEE EE ned Seth keene ee See eRwe PRBS AE se: ea | ' Bra? Similarly the differential Coefficients of the remaining y—cosh & Hyperbolic Functions may be found, and the results are— FEL 94 SCHOOL CALCULUS [CHAP. Ill y =cosech x “Y= —cosech x-coth # y=sech 2 oY —sech x-tanh x y=coth & OY —cosech? x. 25. Differential Coefficients of Inverse Hyperbolic Functions. Ie) smnhe 4x; Let y=smh—! x sinh ya . cosh y dy=dx _dy_ 1 "* dx cosh y ik 1 1 ~ V/1+sinh2 y V1-+x2 II. tanh—! x. Let y=tanh—! x , tanh y=% *, sech? y dy=dx _dy_ 1 ‘*dx sech2 y o 1 ~ 1—tanh2 y edie! ~ 1—x2 III. sech—! x. Let y=sech—! 2 SCC =n 1 ae cosh mes .. sinh y dy=—, dx patie a enue) de. «2 sinh y A 1 a?4/cosh2 y—1 CHAP. 111] DIFFERENTIATION 95 eee x Jt —1 2 a I s xV/1—x2 IV. Similarly the remaining differential Coefficients may be found, and the results are: dy 1 If y=cosh—! zg, eee ‘ dx +/¢2—1 dy 1 COU a: ao Y dx ¢2—] d i! y =cosech—! z, Me dx ar/ 42+] EXAMPLES Differentiate the following functions : 1. sin—! 22. 14. x? sinh 32. 2. 2 cos} 32. 15. (a2?+ 2) sinh x — 2a cosh x. a. tan! 7?. 7 16. : 4. ay cot—1 le : V5 3x+ 1 17. cosh—1 (2a+- 1). 5. sin—! log, x. 1 1 1823 smh) 2s =, 6. tan—l : 3 Var+1 l 7. cos—l e—7, 19. tanh—1 : Le Lh. 4x+3 S) —= gin! : 5 ie ee Sua 4/2 5 ee. 4/3 9. ——_ tan—! : jones 3 21. SS ee sinh—] ~ 10. cot— e cot 4). 2 11. sinh (222). 29. 5.2 sinh (") heal ate 12. V cosh x. ; x 23. cosh—! (cos 6+ sin 6). fee ee beuth of 24. tanh—! (sin x). 26. Asymptotes. Definition. An asymptote to a curve is a line which touches the curve at infinity, but does not itself lie wholly at infinity. Or: An asympote is the limiting position of a tangent 96 SCHOOL CALCULUS [CHAP, III whose point of contact with the curve has moved along the curve to infinity, while the line itself remains at a finite dis- tance from the origin. From this it follows that as we proceed along the curve towards infinity, the distance between it and the asymptote diminishes indefinitely, so that the slope of the curve con- tinually approaches that of the asymptote, though it must be remembered the curve may also cut an asymptote at a finite distance from the origin, just as it may cut any tangent at another point besides the point of contact. It is not proposed to consider the finding of asymptotes in general, as for the curves with which we shall have to do they will generally be obvious from the tabulation and graph of the curve, and by extending the tabulation could always be so found. But there are three points which may be noted as a help to finding the asymptotes of the curve y= ¢(a), which form of equation will cover all cases that we shall require. (1) If the curve has an asymptote parallel to the axis of y, y must become infinite while ~ is still finite (from the definition of an asymptote) and therefore as explained in Chap. I., § 8 d(x) must be fractional, and as there shown, those values of x which make y infinite are found by solving the equation denominator of d(v)=0 .. (i) and each root of this equation gives an asymptote parallel to the axis of y: for instance, if one root of this equation be “=a, x—a=0 is an asympote: for as x approaches the value a indefinitely, y becomes infinitely large, and the point moves off along the curve to infinity, getting closer and closer to the line x=a, which is therefore by definition an asymptote. Similarly each root of the equation (i) gives an asymptote parallel to the y axis. So also, if the curve can be written in the form #=f(y), and f(y) is fractional, we can find asymptotes parallel to the axis of x by equating the denominator of f(y) to zero. (2) If Liz « (x) is finite and equal to a say, this means that y continually approaches the value a as x increases indefinitely : hence y=a is an asymptote parallel to the axis of x. It will be found that this method is equivalent to the preceding one, but perhaps sometimes more convenient. (Ss) ali wigan z is finite and equal to a say, this means that as @ increases indefinitely the slope of the curve con- CHAP. IIT] DIFFERENTIATION 97 tinually approaches the value a, and therefore the curve touches at infinity some line whose slope is a, i.e. touches at infinity the line y=ax-+c, where c has to be determined. These oblique asymptotes can in general be easily drawn in a graph, and as we shall not require more than their directions, we shall not further consider the question of determining c¢, i.e. their exact positions. EXAMPLES Write down the equations to the asymptotes of the following curves, and draw graphs to justify your results. 1. 2y ‘ate 6. y(x — 5) = 2?V2(10 — 2). “ag Sone 5 a) 3y eset Si y(4 mee x)? + uv — QO. ak aide gata A Nala) 3. yV14+ e=2. y2 4. ay = V1+ 22. 9, = Aa. 5. yV5 — & = 5a — 2%, Loe 7 — 12, 27. Undetermined Forms and Limiting Values. It was explained in § 6 how the limiting values of certain simple functions, which assumed the form ; upon directly substituting some particular value of the independent variable, could be determined : it was also stated 0 that the form 9 Was the commonest Ee and most important undetermined 2 a form, and one upon which the others could be made to depend. We will now consider how the limiting values of functions assuming this form may in general be determined ; for though sometimes some simple procedure, such as that of § 6, may be sufficient, this is not in general the case. To find Lt,. Ae where He assumes the form > Let y=f(z), and «=¢(z), and suppose the curve in the figure be that obtained by 7 98 SCHOOL CALCULUS [ CHAP. III plotting corresponding values of 2 and y (cf. § 17): then since «=0 and y=0 when z=a, we require the limiting value y of 3 at the origin O, i.e. in the figure we want the limiting value of ee when both these quantities are very small, that is we require the limiting value of when é6a=0, namely a Now (as explained in § 17) dy _f'(@) dec (2) . the value when 2=0 of CY ee of HAL when z=a. dic $' (2) If then on substituting z=a in the fraction i) only the ’ (2) numerator or denominator vanishes, or neither, we have found the limit required. If both vanish, however, we can, by similar reasoning, differentiate both again, for ih. f(z) 2 en f’'(z) Tis he) daa eae as before: and we can continue the process until we arrive at a fraction whose numerator and denominator do not both vanish on substituting z=a, and which therefore gives us the limit sought. EXAMPLE 1. Find, by this method, the Lt,—4 ae (See Chap.III., § 6, Ex. 2.) Here Lty—1 (ot the form 5) x—l 0 noni =n, the same limit as previously obtained. EXAMPLE 2, Find Litg.-0 aaa (of the form ). ; 1—cos @ 7) This =Lt,—0 a (of the form a) sin @ = Leno > — CHAP. 111] DIFFERENTIATION 99 28. There are other undetermined forms which may be : : 0 made immediately to depend upon this form =. 0 ee For instance 0x oo =0 x aan odeedie els s0 See OO and the procedure in each case is similar to that just explained, though we shall not enter into the proofs. 1 EXAMPLE 1. Find Liz. 3” (27—1). This is of the form cox 0. Putting ar (as in § 6, Ex. 1) 1 ae, 0 Fee ee (TT ts 0 “—— (ot dette ;) ) y ers. 3x 2 et 2 ea 108, 2. EXAMPLE 2. Find Ltz~o loge ‘s cot x This is of the form =. Differentiating we get 1 ts Liz =o pee dye aa (of the form =) cot x —cosec? x CO sin? x (rearranging) =Ltz 9 — (ot the form 4) 2 sin x cos x (differentiating again) —Lt, —9 — ; —() 29. Undetermined forms which upon direct substitution of some particular value of the variable assume the form 0°, c09, 1% may be again made to depend upon the form ; thus: Let $()/@ be such a function, and put y=d(a)@), Then taking logarithms, loge y=f(x) loge (2). 100 SCHOOL CALCULUS [ CHAP. III The right-hand side of this will now be seen in each case to be of the form 0x, hence we can find the limiting value of loge y just as before by differentiation, and hence we can find y. 2 ExameLe. Find Ltz=, (2—2)'~*. This is of the form 1%. 2 Put y=(2—x)'* 2 = 10R. yi e loge (2—2) . Ltgai loge y =Lte <1 — loge (2—2x) (of the form oo x 0) = Dite a1 2 Togs me) (of the form ;) l1—2x 0 —2 Te ae =2 .. y=e, which is the required limit. 30. Many undetermined forms can also be evaluated at once without the help of the Calculus ; but if general methods could be indicated for many cases yet they vary so much for each individual one that it is not worth while considering them. Two examples however are added as hints. 2th EXAMPLE 1. Find Leas 5 . This is the same example as previously given in § 6. Its undetermined form is clearly due to the vanishing of the factor («—3) in the numerator and denominator. Now since in finding the limit we are finding that number to which x2 —9 ane approaches as x approaches 3, and we therefore never make a absolutely equal to 3, we may divide out this factor, and we get Meee Lity=3 Cae x =6 as before. 2 EXAMPLE 2, Find Liz., (2—x)'~*- (Cf. Ex. of § 29 above.) CHAP. II] DIFFERENTIATION 101 Putting l1-~=y we get 2 2 Ltya1 (2—2)'~* =Lty—o (1+y)? 172 =Lty=0 la sy) | =e? (by § 3). But these methods are only occasionally shorter or more convenient, and are just mentioned here for use on such occasions if preferred. EXAMPLES Find the following limits : ee Be an ee tT (a?+ x2) log, 2 305 — bat 2 a— x. LeeLee 6 Pay: 2a? — 5a — 12 ; ui a 2 bets” il Joya emebaaasl oe ~ gin bé 3. pee eS x cos x — log, (1+ 2a — 5V a+ 3 13. Ltz—0 OEE EL 2 ati re y= yo — 1 14, Lty=1 Tate 5 Lyla Pa eee 5 ae loBe 6. a as 3x ; Aap | 16. Ltg—o pote lais Sah verse loge cos 8 a —] vy : logio (« me °) ee v1 V7. Liga = y ° 2 tan iw e9 =a e-9 18. Li * 6 = sin 6 cos 6 6 =< ia oge(l+y)— y fee +e 2 Te I bare (oc es ¢=0 ——353—— y y2 at — >t in 0? 10. Lzag ~~". 20. Lto.o = Je MISCELLANEOUS EXAMPLES Find the differential coefficient of y in the following 50 examples with regard to the variable involved : 3 1 - ¥= 32° — -—.- = 3 Y= Watts 2. ea ee 2 Ve+] Ee? IES 102 SCHOOL CALCULUS [CHAP. TIT 2x loge sin 0 5. Y= e 14 j=" “Opa ] qf es oe ees hoe lo Oe 2) aie Beles by 15. y= | = ast 7. y=4 sin 22 cos 3. 4 8. y? = 32° (7 — 2). 16. y=) 9. y? = 2x? (8— 2). x 10. 9y2 (6 — x) =x (9 — 2x2. 17. y= A tande*®, 3 . c pee ATR ee ys 18, 7a 11. Uae J/9+ a 8 ¥ yi-06° 10 loge x 12. = SS 19. = 2 ss "1 sine I~ cota 13 ey 20 eta -y=anre’, -y=sin-) TT > 2 ees 21. y= Woe ie 2n 008 8 9 variable and n constant. 1+ n?+ 2n cos 6 22, Ditto n variable and @ constant. 37. y = (x logioz)*: 23. y =sin? (364+ 4. ae ie ( +5) 38., log. y = x sin 32. 4-7 (r4— 24) 2 Sey = in - r ; 39. sin y = La" 25. y(5 + x) = a (5 — 2). 40. cos y = 403 — 3b. PAL UI Recah nad Poti 8 2 Vie 4]. tan y= +5) dé sin4@ sin20 36 ere 25. A few important examples to illustrate these and other substitutions are here added. ExamePie 1. fsectx dx=/(1+tan2z) sec2« da. (Put tanz=y, .°. sec? dz =dy) =/(1+y?) dy tan*x The same method will apply to any even power of a secant ; or of a cosecant, by putting cota=y. ExaMPLe 2. fsec’x dx. To find this put tanzsec~=y . dy =(tan2x secx-+secz) dx = {seca (sec2x—1)-+sec3a} dx ==(2 sec3x—secx) dz. .. 2 fsecka dx —/secx dx =y =tanxsecx 1 1. or /sec3x da=5 tanxseca +5 Jsecx dx =; tanx secx +5 log,tan G +e) (§ 29, IT). The general case of /Sec2"+1xz dx will not be considered. ExameLE 3. ftan®x dx —/(sec2a—1) dx =-tanx —x-+C. Examee 4. tan? x dx—/tana(sec2x—1) dx =/tanxsec2x da —/tanx dx _ tan’x 9 —log,sec x-+C. CHAP. Iv] INTEGRATION 129 EXAMPLES Evaluate the following integrals : 1. /sin36 dé. 6 cos®d 2. /sin3dcos6 dé. 9. f chu 3. /sin®6cos?6 dé. oe 10. cos 4. /sindcos?6 dd. en, ainca ag. 5. / sin®écos*é dé. 11. /cos58 dé. 6. f ae. 12, foost9 dé. Bone 13, sind sin3d 44 If | ae ae nog 7. He ma dé at 14. f? cos36 cos S dé. 8. sin?d Vier 15. /sin36 sin4é sin5é dé. 26. Integration by Parts. A product of two Functions of x may sometimes be integrated by the following method : We have © {f(x)-#(2)} =f(a)-$/(x) +f) $(2). Integrating both sides, f(a) h(a) =/f (x) -"(w) dx +/f’(x) and transposing, 3 Jil) $a) de = f(a) -$(«)—/f'(x) h(a) de. If now we write ¢’(x)=F(x), i.e. “$(x)==F(x), so that x) =/F(«) dx, we get /%(x) F(x) dx=t(x)-/F(x) dx—/[i’(x){ /F(x) dx}] dx. This rule is more easily remembered in words: The integral of the product of two functions is equal to (the First function) x (the Integral of the Second function) —the Integral of [(the Differential coefficient of the First function) x (the Integral of the Second)]. Theoretically it is immaterial which function is selected for the first function, but in practice that function should generally be chosen for the first which simplifies on differen- tiation ; and in particular, if one of the functions be a positive integral power of x, it should always be so taken. This process further emphasises the tentative nature of the Integral Calculus, for we can only integrate some products by its means, and that only by taking the functions in the 9 130 SCHOOL CALCULUS [CHAP. IV correct order ; for evenif the second function can be integrated as required in the first term of the result, the success of the method still depends on the second term, namely SUF (LF (x2) dae} | dx being more readily integrable than the original product, which is not always the case. ExamMPLe. Find /x-sin3zx dx. Taking x for the first function as explained above, fesin38« dx=ax /sin3x dx —/[1(/sin3x dx)] dx ___ & COS3H ff | oh 2 ee ( 3 x cos3x sin 3a 4) 9 +C, 27. Some Functions which are not Products may be integrated by the rule of integration by parts, by considering them as a product of the function and 1. In these cases, 1 must always be taken as the second function : for obviously if taken for the first, its differential coefficient, which is zero, makes the last part vanish. EXAMPLE 1. /log,x dx. flogex dx =/logex-1 dx Slogew-x—f— da =2 logex —/da: =x (log,x —1)+C. Similarly SJ logi9X AX =/logioe-log.x da =x loge (logex —1)+C =x (log 49x —log, e) +C. ExaMPLE 2. Integrate by parts /V/a2—2x2 da. As before /Va2—22 dx =.Va2—2x2-1 dx SVG tae eas ((,¥8—2) «as bed x2 2e PLeyy: REN be % Vara ire dz ’ CHAP. Iv] INTEGRATION ee) —(a —7 Ve—a+f~ Wipe ax (Notice this step) Cee a2 : =% Ve—2+/ TI da—{(V a2—x2 dx a= x sg Peak eee: —7 Va2—x2-+a2 sin—1 qe’ —(V a2 — x? dx. Transposing therefore we have Peer ee 2 2 (Va2—a? du =a Va2—x? +a? sin ae ———; eV ae—x? a2. a ee ae gin ~ fva xu? dx 5 T5 sin anes the same result as in § 12. Exampce. 3. Find ftan—'x dx. ftan—=<2 dz=/tan—x-1 dx 1 Prone : stanta-a -f- qe de To find [~~ dx, put x2=y, so that 2a dx=dy. fg emf ry WH 5 Be (140) 1-22 2S 1+y 2 1 ese a5 loge (1+ 2?) =log.V1 +22 -. fan dx=x tan—1x—log,/1-+x?-+C. 131 28. Sometimes integration by parts only partially simplifies the product to be integrated, and a repetition of the process may enable us to find the integral required. EXAMPLE 1. Find /x2-e* dz. Taking x? for the first function [x20 dx =x? fe® dx —f(2x fe* dx) dx =72 -e® —D/x-e* dx. 132 SCHOOL CALCULUS [CHAP. IV ~ Integrating again by parts the simpler product /xe* dz, we get /xet du =x -e*—/1-e* dx =x(er—1)+C so that finally, /x? e* dy =x? et*—2 jx (e*~—1)+0} =e" (y2—-247+2)+C". EXAMPLE 2. Find fe** sinbx dx. Here we take sinbx as the first function, so that 3 4 ear ear Se sinba dx =sinba -— —/bcosba -— da a a etx sinby ob , : = - mae Cosby dt) jaa) Integrating again the product /e*cosbx dx by parts J/e% cosba dx —cosba“" — f° —bsinde =) dx a = 6% COsDs : a fe sinba da. Substituting in (1) IE sinbx eee oe ee sie 4 Sav ORE cose ye eat sinba ie a a a a J ax 2 =5 (asinbe—beosbar) —, fet sinba da. Multiplying through by a?, and transposing (a2 +b?) fe sinba dx =e (asinba —bcosba) eax a? tb? ., fe -sinbs dx = (asinbx —bcosbx) +C the constant being finally added as usual, though omitted for convenience in the intermediate steps. Similarly /Ae—®! sin (Ct+D) dt A : = Seats ak {B sin (Ct+D)+C cos (Ct+D)} +K which should be verified. CHAP. IV] INTEGRATION Additional Useful Integrals. 29.1. fcosec x dx = —— da 2 sin 9 cos 3 on dividing numerator and denominator by cos? 3° Now put tan s=Y 1 2 a My 88h p da =dy sec? 5 Ea dx=f — dy 2 tan 2 =logey+C =log,tan 5+. II. /fsecx dx=/cosec (F+=) da —log, tan G +5)+¢ (by the above) [Or independently thus : 1-++tan2 > secu dx -{ — dx 1 —tan? _ 2 sec? 5 -f sae a 1—tan? a 2 133 134 SCHOOL CALCULUS [CHAP. IV x 1 x pee a ReGe eanhg ee Put tan 5 =Y, 80 that 5 860 5 dx=dy a (3 a ; =) dy, (by partial fractions) =loge (1+-y)—loge (1—y)+€ x 1+tan —~ =loge — Ty +C=loge aC 1—tan — 2 x tan {ttan 5 2 =log, ————————— +C (since tan T=) 1—tan re “tan 5 =log.tan (7 aoe ee x 1 1 —tan2 5 III. Roa AS dx. Put cose= — = + 1+tan?2 9 2 j 1+tan? 9 _S site "= | ued Meee | a+b+(a—6) tan? 5 Now put Va—bd tan SEE ae) 22a a 5Va—b sec? 5 dx=Va-+b-dy, and a-+ b + (a—b)tan25 —(a+b)(1+y), and the integral eae 1 peri REDE GEA becomes 2 \ CHAP. Iv] INTEGRATION 2 1 ee a2 a —1 V/a2—b 1+y? V2 — 0 aaa Yar 2 a—b x ———— tan ——- ta s) C Va? —b? ( np)" EXAMPLES Integrate the following : 1. /xsina dx. 12. Asiré + écosd) dd. 2. /xcosx dx. 13. /sin-le dz. 3. /xet dx. 14. /tan~1z dex. 4. /xe™ dx. 15. /3axsin2xcosx dx. 5. /ecoshx dz. 6. ; 2asinha dx. 7. jxsin2a dx. 8. /26sin36 dé, 9. /e*sina dz. 10. /e-**cos2ax dx. ll. /clogex dx. 30. Areas by Integration. be the graph of y=¢(z), and let BB’ be a fixed ordinate yi, Whose abscissa is x1 : also let PM be the variable or- dinate y whose abscissa is x, let A be the area between the curve, the axis of x, BB’. and PM, so that A also is a variable, whose value at any time increases as x in- creases in this particular case. Let PM in the figure be some position of the moving ordinate, and let P’M’ be 16. /3asin2rsectx dz. 17. /eot—ta dx. 18. /62sind dé. t 19. /t2e 2 dt. 20. ;sec1a dx. 21s. ,CO8: 7x da: 92. _xlogex dz. Let 135 the accompanying figure another ordinate near PM, such that MM’=8é8x, and con- sequently P’M’=y-+éy. Then the area PMM’P’ or 8A lies between PUM ’Q’ and QM M’P’, i.e. between yédx and (y+ éy)dx 6A - Ox lies between y and y+6y me (1) 136 SCHOOL CALCULUS [CHAP. IV Hence in the limit when $a, and therefore also dy, is inde- finitely diminished a, dx =(x) = ae Lt This may be written (as in Chap. III., § 18) dA=¢(x) dx, ordA=ydx .. (ill) whence by integration A=/6(x) dx-+k ahs, or A =/ydx +k=f(x)+k (say). Hence A is a function of 2, and if OC=ze2, then BB’C’C is A when x=x2 ~. BB’C’C or A=f(x%2) +k where the value of k is determined by the fact that A =0 when 7=%} “. k= — f(a) This result is usually expressed thus A aif h(x) dx where A is the area between the curve, the axis of x, and the ordinates at x; and xo. We have confined ourselves here to an area lying in the first quadrant, where there is no difficulty about the signs of any of the quantities, all being positive. Areas lying in other quadrants will be found discussed in § 39 on the Integral Curve. 31. It will now be clear why the integral of ¢(x) with respect to x is written in the form /¢(x) dx, as explained in § 3 of this chapter, and what is the meaning of the “dz.” For we see from (i) in § 25 above that the small increment of area 6A is very nearly a rectangle whose sides are y and 6x respectively, that is to say the area of 6A =(x) dx very nearly, and we can make this statement true within any degree of accuracy we please by taking 6x small enough. In the limit therefore, when 6z is indefinitely diminished, we retain the same form dA =d(x) dx as used in equation (iii) of § 30. 32. Definite Integral. The quantity fs "2 d(x) dx is called a X{ CHAP. Iv] INTEGRATION 137 Definite Integral, and is read “‘ The integral of ¢(x) with regard to x [or of (x) da] between the limits 7; and x2 [or from x1 to x2], and means that after the general integral, say f(x) -+C, of ¢(x) has been found, the values x2 and x are to be substi- tuted in it in turn for x, and the second result subtracted from the first, in which operation the constant of integration disappears ; so that we have if 3 $(x) da =f(a2) +C—f(e1)—C =f(%2) —f(21). This last result is also written | fe) [ice | 1) | means. f(%2) —f(%1). a1 and 2 are called the Limits of Integration, x, being called the lower, and xg the upper, limit. 83. The area between the curve y=¢(x), the axes, and an ordinate at =x» is obviously found by putting OB, i.e. x equal to 0 in § 30 and is therefore f(v2) —/(0), or : “b(a) da. Similarly also the area between the curve y=¢(z), the axis of y, and two perpendiculars to it at y=y1 and y=yp is equal Y . to hs x dy, where we must first express a in terms of y, that 1 is, write the curve in the form x=F'(y). Again the area enclosed between two curves y=¢(z), y=F (x) and two ordinates at x; and 22 is clearly equal to if: P(x) du m/s: (a) dx, if y=F(x) be that curve which lies farther from the x axis than the other ; this may be written a [F(x) —¢(x)] dx, and ot this result could also be obtained directly by the method of § 30. A rough figure should be drawn if this statement is not obvious. 34. Change of the Independent Variable in Definite Integrals. We have already seen in § 8 how, by a change of the Inde- pendent Variable x, an integral may often be obtained. In finding a Definite Integral, the limits must also be changed. We must find the values of the new variable 138 SCHOOL CALCULUS [CHAP. IV which correspond to the limits of the ‘original variable, and these become the new limits of Integration. EXAMPLE. Find el ; dy by putting y=siné. J, oy We have dy=cosé dé, and (1 —y2)* =cos30. Also when => O=-, and when y=0, 0=0. Hence the given integral becomes = sind yf costo cosé dé ss af ® secétand dé 0 =| seve |’ (§5, 9) T =sec yom 0 Notes. Two details should be noticed in connection with definite Integrals. (i) Interchanging the limits of integration changes the sign of the result. b For J f'(x) de={(b)—f(a) and f,* j’(x) de=f(a)—f(b) fe f’(x) du=—f * je mane: (ii) It is immaterial in a definite Integral what letter we use to denote the variable ; for if d(x) be any function of z, and ¢$(y) the same function of y, then: J, ¢'(w) dx=$(a)—$00) and /* $'(y) dy=4(a)—$00) . fe p(x) dx =f ¢'(y) dy CHAP. Iv] INTEGRATION 139 These results will also be obvious from the graphical meaning of a definite integral. a EXAMPLES Evaluate the following definite integrals : 2(5a -—- ee 8 1. oo) eee a 1 Br : « dz ive as — dé. 2 & ar dex. Ex 2eoty dé = 1 f 13. Se a. /,* 5sinx dx. hs eon eoe £ 3 af, (1 — 4! dt. 4. f ST oe eee Bato <1 x2 ey ie ———_ dy, ue 5 V/ cosa 15. de 2 4sin3@cosé dé. ‘An 2 [4a $ ie ee oe heate 16: Jk 3 sin26sin30 dé. fie hk ee ‘sa 95 le 5 dx. 17. ve cos4d dd. 4-5 , 1 4g? da. 18. / wer dx. 11 l EE 9. —— ag, 19. 2 (sind+ Acosé) dé. ee a . / ea osd} 3°5 t 10. / aV x2 +1 dx. 20. / te dt. ‘v do. An Area as the Limit of the Sum of a very Large Number of very Small Areas. It was stated in § 3, Note, that “/,” the symbol of integration, was really a form of the first letter of the word “sum’”’; and it was also proved in § 30 that the area bounded by a curve, two ordinates at x, and x2, and the axis of x is equal to the value of the definite integral 96 is ydx, or of. * (a) dx, where y=¢(2) is the equation to the 71 antl curve in question. It will now be proved that : (i) Such an area can be expressed as the sum of a very large number of small elements of area of such a form that they can conveniently be added up. (ii) When these small elements of area are added up their sum will be ie pe ydx, independently of the former proof in § 30. x 1 This point of view of a definite integral as the sum of a 140 SCHOOL CALCULUS [CHAP. IV very large number of very small elements is of great import- ance in the finding of Volumes, Centroids, Moments of Inertia, etc.: for though the method of treatment in § 30 gives true and intelligible results, the other aspect of a definite integral as a sum is more immediately intelligible as a process, and more obvious in its applications. 36. The Division of an Area into Small Elements which can be conveniently summed. Let ABC be part of a curve, BD i SO GERRY EEE EEE PT EP EE eee SOp 7) SURRSREGRROSRME SRR CCC, cafe Be a ST 5 0 A Cs PECL LECCE CCE eh eel eels eee A PEC EEEE EEE 9, LOLRRS | AT BERR RRBERPESREESAeNANeAaP A LS EERE EE Ea SERRA mPa A EEE SSS ral oa CECE Pea PETTITT ls Te ee CECE ie Coe, Pe TT @ ee | eee CECE eet oe Po ee CECE le eer Cea CEC ere, Ce COAT BES ee ES ese SESCZ GREGG BGGRGREBEERER Si Or Cons as OT II OO OO 8 SS EASESR REGO PR OM CeCe el te ee BRERA NORA. SEBERRGDERBERISSRE ERE COC DCM, CIMT Melati eC Hp) Gj HHA HE Ic |— Sh a EE HiG.f ol. and CH two ordinates, and let the area considered be that between the curve, the two ordinates and the axis of x. Draw intermediate ordinates P,;M,, PoMo, ... P5Ms at equal intervals of -2 units, say ; through B, P 1, Po, ... Ps, C draw lines BRy, Q1P1Ro, .. . QeC parallel to the axis of 2, to meet the ordinates as in the figure. Then we see the area required is greater than the sum of CHAP. IV] INTEGRATION 141 the rectangles BM,+Pi1Me2+ ...+PsH, but less than the sum of the rectangles Q1M1+Q2Me+ ... +Qc6H. Now the sum of the first set of rectangles DM,-DB4+™,Me-M1P14+ ...+MsH-MsPs5 -2(-57+-65-4--74+-85+ -98+1-17) in this figure -2x 4°96 -992 units of area. And the sum of the second set =DM,-M1P1,1+M1Moe:-MoPot....+MsH-HC = -2(-65+ -74-+ -85-+ -98-4-1-17+1-42) = 0° S 1 =1-162 units of area, ll ll lI and the true area lies between these two sums, the error between the true area and the first sum being the areas of little triangles inside the curve such as BR,P, etc.; and the error between the true area and the second sum being the areas of the little triangles outside the curve such as BQ,1P1 etc. Let all these little triangles be moved parallel to the axis of x into position on the longest strip Qe, the triangles from inside the curve being shaded in the figure. Then the two sums obtained above differ from each other by the area Qchs (i.e. by -17 units of area) and each differs from the true area required by about half that amount. If now the interval between the ordinates BD, P;M, etc., be taken much smaller than -2 units of length, say -001 units, the area of the last strip becomes much smaller, and therefore also the difference between the two sums as obtained above; and an the limit if this interval be indefinitely diminished the two sums become equal, and therefore the true area required, which lies between them, becomes equal to either of them. 3/7. The Summation of these Small Elements of Area when the Curve is given by y=¢(x). In the last figure, let y=¢(x) be the equation to the curve: let OD 7, OH =a and DM,=M,Mo2= ens =U Then DB= ¢(#1) and M,P1=¢(2#1+ 621), MoP2=$(a1+4+ 262) ete.; and therefore as in the last section the area BDEC lies between 61 {h(%1) + h(%1 +821) +(%1 +2621) 4- . . . +6(%1+5621)} and 621 {b(%1 +621) + (41 + 28a1) + ee + (21 +6621)}. Now we have seen that the smaller Sx, becomes, and the more intermediate ordinates we consequently take, the more 142 SCHOOL CALCULUS [CHAP. IV nearly does either of these sums approach the true area. If therefore we take n ordinates, where m is very large, 624 becomes very small, and the true area is equal to Litsx,=06%1 { (#1) +(%1 +624) + (#4 +2624) + vie ae +(x1 +n —1-621)} .. (i) choosing the first of the above two sums. In the limit this sum, which only differs from Litsx,-=0 6x%3{ (21) +h(21 +624) -+ EO +$(11-+n—1-8a1)+h(21+7n-8x3)}.. (ii) by the additional term [621-$(%1-+n-621)], which is added in (ii), is itself equal to the value of (ii), since the added term [which may also be written 621-¢(%2)] represents the area of the last strip such as QeH in the figure, and this strip be- comes indefinitely small, i.e. vanishes, in the limit when 6z1=:0, provided that (x2) or CH itself remains finite. The area required is therefore represented by (ii), a sym- metrical expression which may be written Ltse=0 7 o(x)-8a .. (ili) by which is meant: ‘“‘ The sum of all terms like ¢(x)-dx obtained by writing 21, 21+6%1, 71+262%}, etc., in turn instead of x, until x has thus assumed in turn all the values from x to x2 inclusive, passing by increments 6x1 continuously from one to the other,” that is in other words (ili) is a convenient way of expressing the sum (ii). We have now to prove that the limiting value of the sum (iii) is the definite integral previously obtained. 388. An Area as a Definite Integral. Let f(x) be a function whose differential coefficient is d(x); i.e. let d(x) =/f/(2). Then by definition (21) STi +624) —f(%1) O24 . Ltdz,-0 $(41) 6% =Litdz,-0 if(u+6x1) —f(#1)} Similarly Litsx,=0 P(%1 +621)6x%1 =Ltaz,=09 i f(%1 +26a1) —f(a1 +821)} Lt3x,=0 $(%1 + 2621) 641 =Ltsx,=9 {f(%1+382x1) —f(x1 +26%1)} and so on: } and finally Ltsz,=0 $(%1 +n —1-6a)dx4=Lsz, 0 if(a1+nda1) —f(v1 +n —1 +824); CHAP. Iv] INTEGRATION 143 Hence by addition : Tite, Br (ples) +H(en +86) + ple +2BE) + -- +6lEr+n— A Bey 7 (i =Ltge,-0 {f(x1-+-nd21) —f(x1)} =f(x2)—f(a1) [since 11-+ndx1=22] s Wh is (a) da the meaning of JE *: d(x) dx being asin § 30, so that it is simply | another way of writing f(%2)—/f(a1) the real result required. Hence the Limit of the sum (i) in the last section, which x was otherwise expressed as Ltjz-0 re $(x) dz, is thus proved 1 to be fe ya p(x) dz, and the meaning of a definite integral as za the sum of an infinite number of infinitesimally small terms is thus clear. It will also now be more obvious why the ‘dz’ appears when writing an integral such as /6(x) dx; for in integrating we are really finding the sum of an infinite number of rect- angles each of which has one infinitesimally small side dz, while the other side is of variable length ¢(xz), and therefore continually changing in value as 2 passes from one limit to the other. This method of summation can clearly be applied to all kinds of variables; for when it is said above that we are finding the value of a particular area, we are in reality only expressing a series of very small products graphically, and it is only the numerical values of those small areas that we are adding up, and the results are therefore quite independent of the units whose number was originally expressed as (2). It is on this account that the aspect of an Integral as a. Sum is of such importance, as the process can be applied to the finding of Volumes, Surfaces, Moments of Inertia, etc., in a manner similar to that employed for finding an area, all these quantities being in fact capable of being represented. graphically by areas. EXAMPLES Figures should be drawn in the following examples. 1. Find the area between the parabola y = 8?, the axis of x and the ordinate: at2=2. Also the area between the parabola, the axis of y and the line y = 4. 144 SCHOOL CALCULUS [CHAP. IV 2. Find the area of the circle 72+ y? = 4. sae the axes and the ordinate 3. Find the area between the curve y = at x= 2-5. 4, Find the area between the same curve and the lines y= 1 and y = 2:5. 5. Find the area between the curve y =3 , the axis of 2, and the x (2 ordinates at x=1-°5 and x= 5. 6. Find the area between the line x = 2 and the curve (1+2?)y? = 42?. 7. Find the area between the curve y = the axis of x and the ordinates 2 — 4’ at 2=3 and x= 4. 9— 72 8. Find the area of the curve y = a Sad 39. Integral Curve. It has already been proved that the area A between a curve, the positive directions of the axes, and an ordinate at =a, is given by A= -[ 4 ydx. HW y=¢d(2) 0 be the equation to the curve and f’(x)=¢(z), this may be written A=/f(x1)—f(0), so that f(0) is a constant whatever be the numerical value cf 71. Hence the equation A =f(x) —f(0) gives the area between the curve, the axes, and the general ordinate y, and of course d=0 when z=0. If therefore we draw the Integral Curve A =/(x)—/(0) this will pass through the origin, and ordinates to this curve will represent by their lengths areas between the original curve and the axes up to an ordinate whose abscissa is the same as that of the ordinate in question to the integral curve. Now ordinates to the integral curve will not be always positive, that is to say some areas will be represented as positive quantities and some as negative, and we can show that areas lying between the curve, the axes, and an ordinate in the Ist or 3rd quadrants will be represented positively and in the 2nd or 4th quadrants negatively. For, as proved in § 30, dA=ydzx always, and if the small change da of x be always taken towards the right, i.e. if x always increase slightly, dx will always be positive. In this case however dA in the 2nd and 3rd quadrants will be a slight decrease in A (as can be easily seen by drawing figures similar to Fig. 30, but in the other quadrants) and will therefore be negative, while in the lst and 4th quadrants dA will be positive. Also in CHAP. IV] INTEGRATION 145 the 2nd and 8rd quadrants y is negative. The results in the four quadrants are therefore Inthe lst dA=ydz » >», 2nd —dA =ydx ~.dA=—ydz fa » » ord —dA=—ydx ..dA=yda es » » 4th dA=—ydz where dA, y, dx are numerical quantities and independent of their signs. From this we see that areas between the axes, a curve, and an ordinate in the Ist Quadrant give a positive branch of the Integral Curve 2nd 99 oe) negative CH) Mr) 9 oy) (8) 3rd 29 we) positive i) oy) / 5) 99 . 4th + », hegative ,, . se > Again the area between a curve, the axis of x, and two v2 F 6 ordinates at x1 and x2, which isf yda, is equal to thedifference vy between the two ordinates to the integral curve which give the area from x=0 up to x=x2 and x=, respectively (for the constant /(0) disappears on subtraction): therefore such areas in the lst quadrant are given by (A2—A,) (if Ag and A, be the ordinates to the integral curve at x2 and x;) where Ag and A, are both positive and Ay, is >A, therefore the area given will be positive. Similarly in the 2nd quadrant i ydx is equal to (42—A1) where Azgand A, are both negative but Ao is =< Ai, and the result is therefore again positive. In the 3rd quadrant ifs % ydx=(Az2—A1) where Ag and A, are both positive but Az is < Ai, and therefore the result is negative. Lastly in the 4th quadrant IE e ydx =(A2—A}) where Az and A, are both negative but 42 > A, and the result is therefore negative. Hence we may summarise these results thus : HY Je * ydx is positive in the Ist quadrant | ad | ee OSILIVG = ond 0, eters. (y) » » negative ,, 3rd * | » », negative ,, 4th _ 10) 146 SCHOOL CALCULUS [CHAP. IV In finding an area therefore which lies partly in one quadrant and partly in another, we must be careful not to obtain merely the algebraical sum of the separate parts: if the parts lying in different quadrants would on integration be of different signs, we must find them separately and add the numerical results. The following figure, which is hypothetical but reasonable, will make these explanations easier to follow: the dotted | Sooaeae BSEBEMRRREREMANELOS BRRRRBRERERRERROOE SURMERERERESRRBURRS ST BERERERSERRRREREOR mat BERR aso eas FONG EEE HERS BRED SSN BABER ap \ PRCCEE Bits N a WAN iS q CUULUYL be st branch] [TT V2 Bo ol ei shal 4 eps He es Z4YCLGLALLEZS cpa tel a a F eu eke riceiziata HOCRRERC Ee TA. Aah] State or u ZA7EN WLiLAA| S| > _| Net te tat 4 fe S17 dade tas Poet Abed CIS eleys Eaten a [i pete} ft pees] || | 5 & ia iG ki a G30 iG ware | es i & Zi Taw Ey Be eh be fe Pee cota ete Siete | Fre.32., curve is the integral curve, and the number of the branch indicates in which quadrant the area lies to which it refers : the shaded areas are those represented by the difference of the ordinates to the integral between which they lie. Reference should also be made to the Integral Curves of the following chapter. 40. Infinite Limits of Integration and Infinite Ordinates. Hitherto we have assumed that the limits x; and x2 inthe ~ydae were et finite, and also that the ordinate y was always finite between those limits. CHAP. IV] INTEGRATION 147 (i) Now it is obvious that if v2 be indefinitely increased, so that we are finding the area between the curve, the axis of x, the ordinate at «=a, and an ordinate at infinity, then if y remains finite or itself becomes infinite, the area sought must also be infinite and therefore cannot be found numeri- cally, and Je. yd« will merely give the result co. ay If however the axis of x is an asymptote to the curve, so that as x becomes infinitely large y becomes indefinitely small, the area sought may be finite. In this case the area will be correctly given by ifs ydx, though the area even then i may be finite or infinite, as will be seen from the two examples - which follow: so that in every case if x2 become infinite, vA yd«x gives us correctly the numerical value of the area, 1 provided y does not become infinite for some value of x between “, and infinity, but either remains finite, or diminishes in- definitely, or increases indefinitely as x increases indefinitely. (ii) If y becomes infinite as x approaches either of the limits 21 or 2% (but does not become infinite for some value of x lying between those limits) the area between the ordinates . Xo at x; and 22 will still be correctly given by ifs “ydx, at any rate cy in all the cases we shall require, though as before this area may turn out to be either finite or infinite. (Gii) If lastly y becomes infinite for some value x of x between x, and a2, we must find the areas from x; to ao and from a to x2 separately and add the results, and again the whole area so found may be finite or infinite, for the statements contained in (ii) now apply to each part of the area so found. The proofs of these statements and explanations are not here given as being beyond our scope: but we can convince ourselves of their truth sufficiently well by noticing that when x2 becomes infinite but y remains finite as explained in (i), there is no violation of the assumptions made in finding, “ wv as we did, the result ye “ydz, and we may therefore suppose a1 the results are correct even when x2 is indefinitely increased, v.e€. With an Infinite Limit of Integration. 148 SCHOOL CALCULUS [CHAP. IV If y become infinite as in (ii), so that wz=a, or w=2 18 an asymptote to the curve, we have only to change the origin so that this asymptote becomes the new axis of y, and the area under consideration can now be found as ( Jf “xdy + a finite yy rectangle), as will be obvious from a figure. But “xdy is Y of the same form as ydx, so that the case of Infinite Yl Crdinates is reduced to that of Infinite Limits just discussed, and is similarly valid. A consideration of the ordinates to the Integral Curve (see last §) will also further explain these points. 3 e2+4e+5 and the positive directions of the axes; also the whole area between the curve and the axis of x. ~\\ Exampte 1. Find the area between the curve y= | 4 aaee tee EREVAaReS BRVSRURBERESRRERS BEECH IEE PCE CECE EPs gene 2] BCC ret ae ee PEER CE eee Fia. 33. The form of the curve is that given in Fig. 33, and the 0 00 3 = wes | Glee ee d the axis of xi area OC'D f ydx =f. ay eanT? dx, and the axis of a is an asymptote to the curve. oo 3 oe 1 N oy Ae »§ 18 ow f oa Bes sf, me: a7” a Sf Te 7p de (putting «+2=2) 2 (for when #=0, 2==2 and dxv—dz) INTEGRATION 149 .=3[ ten" |, (by § 5, 13) =3 (tan—!,. —tan—}2) ia —= zy ain) @ since 1-11 is the circular measure Olitaine zs CHAP. IV] —3 (1-57—1-11) =—3 x°46 —]-38 units of area. Also the whole area between the curve and the axis of x is equal to + 0 3 39 ore u2+ 4e+5 da [see § 39 (y)] a3 [ tan—t2 | EP ) — 37 —9-42 units of area. EXAMPLE 2. Find the area between the curve v2y=«x—y and the positive direction of the axis of «. per NY pemteeretteeeee cebel eb sbaeeat ee a ee ee eee CU (oS Tao ibe = Se eee Reese [0 PREORDERS een Peete Pitot ttt ine eee CaSaceean Sao Basie OO 0 papreseeererore eer Pssoicion es teil eles) Per Per err Fia. 34. The equation may also be written y= eae and that branch of the curve given by positive values of x is drawn in Fig. 34. 150 SCHOOL CALCULUS [CHAP. IV The area required = af ‘ ydx =[ aes dx 0 lta? et ir ac es ay = /, 1 +-a? (the numerator now being the differential coefficient of the denominator) = [ 108. +22) | =) (loge co —loge 1) =~. Note. Comparing these two examples we see that though the axis of x is an asymptote in each case, and y finally diminishes indefinitely in both, yet in the second curve y does not diminish sufficiently rapidly to give a finite area, though it does in the first ; and as previously stated though Ui * yd does always give the correct area, yet this result may be either finite or infinite. EXAMPLES 1. Find the area between the curve (1+ x?)y? = x, the axis of y, and the line y= 1. : 2. Find the area between the curve y= a. » and the axis of 2, from x = 1 1 to z= 3. 3. Find the area between the same curve, the axis of x and the ordinates at v= 3 and x7 = 00. 4, Find thearea between the axes on the positive side, and the curve y= See 5. Find the area between 3y (x? — 1) = 2, the axis of x, and the lines x = 1 and # = 5. 6. Find the area between the same curve, and the axis of x from x= 2 to x= ©. 41. Differential Equations. It is not here proposed to enter into the general theory of Differential Equations, whose solutions when possible are usuallw difficult : but it is useful CHAP. IV] INTEGRATION 151 and necessary to understand their meaning and origin, and there are some forms which are easily solved. If we differentiate the equation y=sin X meee (a) dy the result is dog 008 ® aoe tit) Ody =COsc. ax se) (11) The solution of this latter equation, in whichever form it be written, means the obtaining of equation (i) from which it arose by differentiation. From this it will be clear : (i) That all the difficulties arising from the tentative nature of integration are still present; and there are in addition others peculiar to these equations. (ii) That to return to the exact original equation requires some further data; for the solution of equation (ii) gives us y=sin «+C eel) where C is an arbitrary constant which would disappear again on differentiation. If we differentiate equation (ii) again we get GU en 5 Pa eae 56 (v). Now if this equation were given us, on integrating once we should have d . 4008 x+C, ween iE) which contains an arbitrary constant which did not appear in (ii); and integrating equation (vi) again y=sin «+C,x+C2 .. (vil) which contains two arbitrary constants which did not appear in (i). Such constants will clearly always appear in the solution (i.e. integration) of any differential equation, and there will be one of them if the equation contain @ only, when the equation is said to be of the first order; two if the equation 152 SCHOOL CALCULUS [CHAP. IV f me when it is said to be of the second order : n and in general n such constants if the equation contain ary dan and lower differential coefficients but no higher ones, when it is said to be of the mth order. These constants must be determined from some given conditions in the usual way (cf. § 2) to enable us to return to the exact original equation. The only type of equation it is intended to consider is that in which the variables are separable, which means that it is possible either directly, or by some simple substitution, to collect dy and all the terms in y on one side, and dx and all the terms in x on the other. Such equations are represented by the general forms p(y) dy=f(x) dx and f(x) dy=(y) dx since this latter equation may be written | dy ta dx p(y) -- f(x) To solve either of these forms we have merely to integrate both sides separately, though of course even now the inte- grations may be difficult. also contain EXAMPLE 1. Solve the equation (2y4+1) OY ype. Separating the variables we get ] (2y a) dy =e* dx and integrating this 1 Pd sath Enea AH 9 me 2y? a ae which is the complete solution of the given equation, though in any particular problem we should now require some con- dition to be given us to determine C. EXAMPLE 2. Solve the equation xsin2y dx=ysin2x dy. On separating the variables we get acosec2a dx =ycosec2y dy. CHAP. IV] INTEGRATION 153: Integrating by parts, and taking x and y for the first functions. respectively —axcotx+/cotx dx = —ycoty +-/coty dy *, —xcotx +log-sinz = —ycoty +log,siny-+C Inv ; , sl which may be written —— =A e*ot?—yeoty siny where A =e“, dy ExamMp_Le 3. Solve (%+y) 5 eee (i) The variables here are not immediately separable, but become so on writing x+y =z so that dx-+dy =dz A (ii) *, zdy=a dz, by substituting in (i) =a(dz—dy), from (ii) aie y=a loge (z+a)+C =@ loge (wt+y+a)+C which will be seen to be equivalent to y eC x=Aea—y—a, where A=e 4a, EXAMPLES Solve the following Differential Equations : dy , bx dy ae eee tes a = QO. 6. ‘ 2 b —— =v. : Sepiay vi by oe sin 2) + 1 f 2, = ote geen) dye (cl sain Nde. ES .: 8. (1+ 22) dy = 2y dx. 3. 7 = — 9. ycosx dz +esinz dy = 0, 4, cdy= Vc? — y? dz. 10. ny WY ¥ —2 (32+ 4). 5. 4ydx+ (4 — x?)dy = 0. dx a#+4 State the following as Differential Equations, and solve the equations which result. 11. The rate at which the distance (s ft.) of a body from a fixed point is. increasing with the time (¢ secs.) is constant. 12. The rate of increase of s in the last question varies as the time. 154 SCHOOL CALCULUS [CHAP. IV 13. The rate of increase of s, again as before, varies as the distance already ‘traversed. 14. The rate at which the velocity (v ft. per sec.) of a body is increasing with the time (é secs.) is constant and equal to f ft. per sec. per sec. 15. The rate of increase of v in the last question is partly constant and partly varies as the velocity at the moment. ° 16. The rate of increase of the velocity (v metres per sec.) with the distance {s metres) varies inversely as the velocity at the moment. 17. The rate of cooling of a body (i.e. the decrease of its temperature T° C.) with the time (¢ mins.) varies as the excess of its temperature above the temperature 7° C. of its surroundings. 18. The rate at which the area (A sq. in.) of a curve is increasing with the abscissa (x in.) varies as the ordinate (y in.) at the point. 19. The rate at which the depth (h cms.) of the water in a tank is de- creasing with the time (¢ secs.) varies as the square root of the depth at the moment. 20. The rate at which the depth (h cms.) of liquid filtering through a conical filter paper is decreasing with the time (¢ mins.) varies as the surface of paper covered by the liquid at the moment. 42. An Important Case : ae Ky. This may be written Kae ~. loge y= Kax+C -. y=Aek”, where Ae" This equation states that the rate of increase of y with sespect to x is proportional to y itself, or if AK is negative the decrease of y is proportional to itself: e.g. (1) The rate at which the pressure of the atmosphere decreases at any height is proportional to the pressure at that height, which is stated by the equation dP pee les (2) The rate at which the temperature of a body decreases if left to cool is proportional to the excess of its temperature above that of its surroundings at any moment. (3) The rate at which under certain circumstances an electric current in a circuit dies away if the impressed electro- motive force be suddenly removed is proportional to the current passing at any moment. (4) The rate at which a sum of money increases at compound interest is proportional to the sum itself at any time, sup- posing the interest to be payable at indefinitely short intervals. CHAP. Iv] INTEGRATION 155 From this last example the law of such rate of increase is often termed the Compound Interest Law. 43. We add the solution of one differential equation, which is not of the type so. far considered with variables separable, as it is required in a pie es namely a een iye=0 aly, To solve this put y=e*, X being a constant whose value we shall find. Then dy =e dx dy d Sei 2 er an dx nr *, (i) becomes on substitution e* (A2+h4rX+/) = which will be satisfied if \ is chosen so that r2-+-krx+1=0 ie (ii) i.e. if X is either root of this quadratic. There are however three forms of the solution according as (ii) has real, equal, or imaginary roots, i.e. according as k2—4l is positive, zero, or negative. Case l. k2—A4l positive. The equation (ii) has then two real and different roots, say A1 and ro. Then either e*:* or e2* is a value of y, and it is easily seen by differentiation, and this should be verified, that y=Ae**+ Bex. ene ir) also satisfies equation (i), where A and B are arbitrary con- stants. Since the constants are two in number this is the complete solution. Case Il. k?—4l negative. The equation (ii) has now two imaginary roots of the type Ay +tdkeg where i=*/—1, so that as before y=C} ea, +t At 1 Oy eA, —tA,)u is the general solution. In this form it is inconvenient and requires some alteration. In this case it can be shown that the above form is equiva- lent to y =Ae** cos (A,xX-+B) .. ads Peg 156 SCHOOL CALCULUS [CHAP. IV and it should again be verified by differentiation that this is actually a solution of equation (i). Case LII. k?—4l=0. In this case equation (ii) has two equal roots, A1 say, so that we get y=C'7 eit +C'5 poted = Ae", where 4 =C1+Co. This contains only one arbitrary constant A, and will not therefore do for a complete solution; but it can be shown by differentiation that y=(A+Bx)e*.* .. = =o Ley) satisfies equation (i), and this contains the necessary two arbitrary constants. Note. A and B are not of course the same in each of these results, being arbitrary constants to be settled in any par- ticular problem by given conditions. MISCELLANEOUS EXAMPLES Find the values of: te 2 2 2 1. (ie dg Be Put 22+ 444+ 5=y. 43 2. f- : dz. Put lova= y: SS eae dx. POC eae J x logex 6. /sin®é dé. sa Slaton dx. 7. SV x2 + 2x dx. Put l=. SF ean aeeaten Vi ut 2+ y - /————_. te Pai SF Veit 4e —5 9. feVat+ xf dx. Put 22 = y. ae 10. ase dx, Put 2? — a? = y?. un. /r/ 7 dx. Putxz—a= y?. 15. (1 + x)5 da. x—a Crk 16. feVa+adx. Puta+a2=y? 12. /cos? ssind dé. sind id. Pp } B t tané = 2. 13. fp dx. i cosof aid a 1 at Awe ie! ti et+ e-2% agra eabaeeecd ace 18. fang cos%6 a 19. / (sind + cos6) dé. Put e®sind = z. CHAP. IV] INTEGRATION. 157 1 Esra a i a1. /-———;—_. : aa (a? 1) (x? 4 4) dx. Use partial fractions 99 x 28. /cos®é dé. @an@+a x3 eee a" 2 = y, a3. f- Ne dy 29. frp Publy amy e4/ Tee Ge — 72 Ay? 1 30. /——— dy 2 —————— lepy Va — 4a 8" 25. /x%e8* dx. Integrate by parts. 31. aa dy. 26, /2tan?dsecd dd. Veli pao.) 60 0 27 osin-ssin 5) dé. 32. Stas az. 33. *—— dz. Can be reduced to Ex. 26. V1 + 22 2 = x (% — 5) du. 34. far ap dz. Put x=tané. . 36. [V2 (a — 5) dx a seo 37, (V1 —4y? oe: 35. a dx. Putlta=y. ve Wie dy. Put 2y=siné. 5 ie 38. /- 22 V5 —adx. Put 5 —x=y7. CHAPTER V COMMON CURVES 1. In this chapter the following curves of common occur- ence will be considered : i. Straight line. § 2. li. Circle. § 3. iii. Parabola. § 4. iv. Ellipse. § 5. v. Hyperbola. § 6. vi. Rectangular Hyperbola, referred to its asymptotes. §7. Vil AGUDIC == R se vill. Logarithmic Curve; y=logiox. § 9. ix. Exponential Curve; y =107. §10. x. Cycloid. §11. xi. Catenary. § 12. A special numerical case has been chosen to illustrate each of these curves, and in connection with each curve will be givenits equation and figure, also the figures of the Differential and Integral Curves with their equations : the original curve in each case is shown by a continuous line, the Differential Curve by dashes, and the Integral Curve by dots. A specimen tangent is drawn to each original curve at a point marked P, the corresponding ordinate of the Differential Curve which measures the slope of this tangent being lettered AB; similarly a specimen area of the original curve is shaded in each case, and the corresponding ordinate of the Integral Curve which measures this area is lettered CD. In each figure the same scale is used for all three curves, except in the Hyperbola (Nos. v and vi) and the Catenary, where the small numbers on the y axis indicate the scale of ordinates to the Integral Curve, of which otherwise only a very small portion could have been drawn. The Integral Curves are the graphs of A =/ydx+C, C being 158 CHAP. V] COMMON CURVES 159 chosen (as explained in Ch. iv. § 39) so that A =0 when x=0 ; i.e. the Integral Curves pass through the origin, except where: otherwise noted in the text: for it must be remembered that the position of the Integral Curve depends upon the particular value assigned to the constant of integration, which is arbitrary when not fixed by some special condition (Ch. iv. § 2). Saas spe eleie oP lee eels. pach ice | mini i eieiees Binlelciwiaisls Leite) feta- ie . SUSE eae 12 2000 SSR eee tater elt ieee La EEE EEE CEE CECE eee nile awl lelotele pets )slapet tefl ldate |e | BEET Sie alee eile paisielelal=lots Spe apes | spars) =| fo Pee eee Ce elsiabeleia els] =| mmlelstatsie ie falsisiniatals Seodeg 2090 SER EDS B06 UMBRO Boe an = Psa © miaele Dmaieeieistiaisisie aie COOe es CeCe S20 S080 0000000000 0888085117 408 SSS SES e555 SCo See elena an Be EEE Ee SECO ra aga Hit emia cee eli EaretSi cle Bee eee BEEECEEEEEEEEEERN ECE 7000050 Pee eS SAsseeunpgeannRene ERLE mee ae silat siti.) GRRRENECCL pagalslaiatal aa isietetel RARRALO Seiiepienicbn ti. NSS =] Ve bueeee NNNS ONE =EEERE ERE 7 REC PEER LL ttt 6 eZLARANARENE EELS COSTE ETT W7TTRSSSSSN mice esi CCS? FP Sauce reenaey rauca SACS SN nap ININ IN N Beh Fol H41 |e SANs SSH a TOU20 S90 4002 ERR ARREo ese SEH ve Sue Choe Leo ae _ SEB a oed oe OL Go eeGoeomaa ee CCE CCC eee SC Rgeo CGwG Se ee ee ee aie eatel ay. elects sista stateielaietoteisietatcieie eetaleimisicelsisticel beni Sustor cay pavatasronsetaesonsote f | Reel CEE EH HH ++ eeyaralialetele tele laiatafelepobels (ooh toie tot Tato Fie. 35. (For the straight se in § 2.) Since for the most part the units chosen along the two axes. in this chapter are different, the apparent shapes of the curves. and slopes of the tangents are not in general the correct ones ; and the slopes of any lines must be determined by measuring 160 SCHOOL CALCULUS [CHAP. V the ordinates and abscissae on their respective scales, and cannot be directly found by a protractor. Similarly when calculating areas by counting the squares, the number of squares in one unit of area must be first determined in each case. 2. The Straight Line. SJ ydec =I - 2502-1 5a. Since the slope of a straight line is the same at every point (in this case 2-5) no particular point P could here be reason- ably selected. The shaded area, from 7=2 to x=3:-5 is given by CD, the difference of the ordinates to the Integral Curve at those points, and is by calculation 7-69 units of area nearly, which agrees with the length of CD. The ordinate to the parabola (the Integral Curve) at any point where x=z,, say, includes the negative area lying between the line and the axes in the fourth quadrant; e.g. when x=1-4 the ordinate is 0, whereas the area is not, and allowance for this fact must therefore be made in calculating any required area (cf. Ch. iv., § 39). With reference to the position of the Integral Curve, see the remarks at the end of §9 on that Integral Curve which apply here also. 3. The Circle. Pie te ees an 4/9 —22 2. Sf yde= =r Ssin— i eae EN cf Only one quarter of this ee and its corresponding curves has been drawn; but the general shape and direction of the remainders are similar to those drawn more fully for the ellipse. The slope of the tangent at P, where x=2:-5, is by calcula- dion —1-51, and is shown graphically by AB. The shaded area from 0 to 1-6 is by calculation 4-56 units of area, shown graphically by CD. 161 CURVES COMMON CHAP. V] The ordinates of the Integral Curve are (as explained in Ch. iv., §39) negative in the third and fourth quadrants PERSE EEE CE FEE Ptr cis selaal | SuSeSSSSSHSREGRr/ Zh 1 Leola at eesateestesssetteenteenterteets oe Seen ee ( on PEC a a Do RMa ana NEES E SesEeos Coast ceateneteecsters a=, Seroneeesecel iat bia Fria, 36. The Integral Curve is discontinuous, becoming imaginary if x is numerically greater than 3 as we should expect, as corresponding respectively to the second and fourth quadrants of the circle. Li [CHAP. V SCHOOL CALCULUS 162 there is no part of the circular area lying outside the limits r=+3. 4. The Parabola. rr ee eet eg te ree eh tele eerie ere bot eee 4 eee Pere ete hers ep apere LILI PSO SAAB Ae ae AR RESGEERERES ete ly tet a a ell af iet SC te pe Paella ae te | | BRIGRRRRRRSRLS KOBE Re ae TOC | Se jemi tes [res 5. The Ellipse. 16—2x2 8 a 3a ANE 6sin—1 Syde : j ; ' | ie Sos all Bae! { ’ BE || [esi a a a a ie i a es a a BRE CP Pel as wa Peeper creer er er Bean Segeen Bosh lS keene se st eeeeceeeees beget ae a | a fe Tt ef a ee A Fei i a Na iF ef ce se ete el at | Be paste Liha Une Ra Seh Bee RE RERPeESEERa sear ache SSeS peas Fe el BONAR SS REE Be ARBs ERE A eo. ARR ASO Reonoee SALRCASECE RRR RRR aRRUSSRRRNERIAP RARER AA eee SARS Iss ee vf ee as Sh | Ye Ff | a |e | Te |Z ea wa i a aN i po bE ER np Ys Ne Hott hp Ep a pb Ly | SSERMURDSPeaVWERe see es 02 a a af Fa | oe ia 71 | A ale ie fa ed a fe a aN a a a ae BREE ATS eR SeenseAo fa ed | ti fF a a i | re i a a a A) i CER REE Bea Sa neeEee SY ee |e ea 2 a fe ie |e PST Ry app ee ERE eNews ewe AeA SR SREY LVN e Nea OING OVS i Tee ah co [cl if se Sl ce REARS SEe oS Roe ses S ea eS Rae EN SP ND We G4 CLM ews Owe, As AoE NeanSnss a ee al ee AACA ALA St eee Tel ae eee BUDRSBREE DOERR See Oe ESA SWE AGGWE CNA GAG OCCA ees Sees BERS eee SEER VAG Ise GNSS IAG Cie Ove 7 BSB Bee eWweeawoo BETES RS RAs SSeS ESRAAD OAs IINE ONE Glas od DARA Bese Ss BOERS RE RSS ree S ROSE Se AeA SACO Ns OS Cvs LOR RR Sees ff sf ln a ae a es =e TY DIVAIW ASL LL VP A/VAVEL TT TT TL odeg ey TTT TT TT TT t+ tt eg} +-+++ PET TT TTT lee TT AV le VAST PRAT A eT bet eT | = ya Ltt BRSESeS EBRERSRaS GREER Pt VALS VALLE FRx eD Sena Mes | ARRAS ae ease AeA Rasa SNS Teese SRACS MEAs Beans i ef Ts a Li a a ae pr a | Tg af |S Sa a Fie TT as A a i EBRERSEEE ERE Ss Sooke BASE Chae Roe e eH VeReeeese 44 tt NT geet} pe tt pes a ep hep ELEN AP CB ESRERARARE Seer eenee BERREPARRSSLee Oo SOPs eee EAS Bee eee, sees ERERESES SRR Oe ees Teo oes CaS Cases SSR ERs Nees BERS eS enoe es ee Ae ae PtP TP eee TT TN tt eg Ht tt SRE SRE RABE RAR SehSoR owas BRAN Rees. DaVTREeeaes SEe tae Pen eA ee MDAROs Ce REAR SNe BRERA TERRES anne BRO Be Rese he Rennes BERDRERSRERAC VOR er Neue SP OES IRA SE eae tte FA REESE eS sae AOS aw eeas Spe lo ke, pobre gle ke [aps od abate NERA ORAS eee senses MER S ARS eas Pe yer A | a ae Frac fa | Sa a ene i Oe i Fe] Ln | ep (et fe NN ee ele Nhe BRPUGRSRNSSASSUA Se SIENA Aa eR SRE RRS RARER’ aa SO ee er ¥ SRR SR PRRs eee Rea ae SR RE eRa EO EES ARIES PaP OBR eee Reese SOREN Res De eNBR RARER Bees SeRESARP EAR OABENAOReS ee SERRA ERRACEES SE KERRER EMRE EAeYeeeS I \ Fia. 38. 164 SCHOOL CALCULUS [CHAP. V The slope of the tangent at P, where x= —2-6, is by calcu- lation -642, and the area from 0 to 2 is similarly 5-74 units of area. | | It will be seen that the branches in the first quadrant of the Differential and Integral Curves are very similar to the corresponding graphs for the circle: the general shapes there- fore of the remainders of the associated curves for the circle may be seen from this figure. The four branches of the Differential Curve are fairly com- plete, though this curve is not discontinuous, but has the lines a= +4 for asymptotes. The branches of this curve in the fourth, second, third and first quadrants correspond to those parts of the ellipse in the first, second, third, and fourth quadrants respectively. The branches of the Integral Curve are completed absolutely in its first and second quadrants (which correspond to the first and third quadrants of the ellipse) and the branches in the third and fourth quadrants, which are not completed, are similar: for here again the Integral Curve is discontinuous, becoming imaginary when x is numerically greater than +4. as there is no part of the ellipse lying beyond these limits. 6. The Hyperbola. ch gd 9 4 dy _ 2x dx 3/x2—9 _ tVa2—9 a +V/x2—9 [yd 7 —3log. 5 ; The slope of the tangent at P, where «=4-98 is by calcula- tion —-835, and the shaded area from 3 to 5-6 is similarly 5-11 units of area. Only the right-hand branch of the hyperbola has been given in the figure, and the branches of the Differential and Integral Curves drawn correspond to those parts of the hyper- bola in the same quadrants as themselves. The left-hand side of the figure is exactly symmetrical about the axis of y, but those branches of the Differential and Integral Curves which lie in the third and second quadrants correspond re- spectively to those parts of the hyperbola in the second and third. 165 CURVES COMMON CHAP. V| Bee we aot @)@) es 2 © o O° S&S ~! ee S z CARE NERC ETE EEE EE 2883 ao _— fe Be susczivtctasozes eves taseaant stat fentensesestatester7 7 a oe ae ne FGDs CoS RS PR isbtalaabsteh ced tl See ° eye a Sree GE LL aera} RUAD > ASRS RR EARS Sas ae o 8 qo ite ate a BEC EEE HEE EE EEE EEE EEE EEA Pm Go 8 & SS SG088 908 \OSSSS SSERs 2 napa TS o oergiced = i FECEREE CECE EEEC HEE EEE EEC HEHE S 8 —s- uSe SGRReanS NEOPA SE eT es Er EA anGnnE ee geile zs SS SSSSRSSSR68\005 35058 050088 TSieteh Loe fag es Ve Act ie o fo.ert Ss HENLE PECs aoa EGE RUGema PaMEaD a HS @ n 2 © eso et tan LLL be reel TAT YT a 16, 8 CaS, 2S BS088 SESS SEAMS eB EDERDSSSCARGy AB4Newue = ~ 2 i ARE SENT TARSAOeMeRET A | a 5 oo Bird sone SOSG0 SG OS0 SR K0RN URE SO Cees EAC Eero a 8 .& 2 sae ERM OE aS CAA CLEA CUT /LOqMEPAEESAae HARE EEE ° Bons 2,9 EEE EEN ee eat AHH HH Seen ct Soo) 0 ce eae Co YZ MOUDEH LEE amet a ACA 32 scl by 2? Biba ish wih ial RCNA EAE PR he es DOSS ER RE ERER GER C02101/70/7, 00087 ae af ff a Bees, 8.2 9 5 BEE EEE EE SLE RASA SHE PAT oN g HP cn SD 2 ol © Bist tert hore oink foto ister ore ar AIT N ial Ll) | A Zee HH dq 3 oe : ' ebsieteted BS = ead sae PRECINCT Poe oe oe Sa 7S greg gees EERE EEE TNC oo el dom omen ae BEERS ata CCE Re CO eee ~™ 5 2:a 8 : eat bf WET CDAY TS HH An 2 ea CSR SnG se ee PSA eet Ht tt Ses ae Oe Se Soe gS Rk Bd oP Be SaSGeuOmemnmaane ames cris C2208 87 WS Ss ae BO Oe ee ee > n° PEEEEECE EERE EERE EEE EOE EEE EEE Bip eh ee eee S0000000000000000085RR 0000 2 PU a Oa eee = be S06 SOS0 8 SOS 88 SSE ONES Sey oy a Ye © pe a ene an PERE PACE EE EE EEE EEE cee tee gS Bao SUGGGE0GaREGCS S00 000 S000 00085 48008 2S 7 Oia a ee oper BREE ECEE EE N EAE EEE oe eet 2 PREECE eee A st io 8 ES © ie te ce aber ot eaten pitt IN TA HAE ECE PREECE a a oor on ro) AB CM Mo Ree EBSn 23 weeeeeee tA TL BNA RAS Co = o5 © BasS sooo SERA RAST P AS y oY pee pt a Peay at Sere eer es SS0S0 000000500008 SS888 SER 4e See eto ciae hia HS Gk cee | Seta Pri 20\ 0000/00 Soe oe a 2556 Soe a SaeSn EE poor: PAECERNE CCE See = bei 57 Secearee ol TAA a v BEER rans LL . = e7 "8 Pett ee eE eee a et SB sigs ee 2 essa aenna SSEbSRS0 00008) 60 eR Ran Corrie To ie +~ s B oo > EERE EOE EEE HA AH REE ECEE a a Peete : PEL eer 2495 Ren eae e ee aoe ao o > Pe Scan HOE Cs a Se Say Sanam AKRee ce RESaL . 2 ee: SR Sad brere rcmedeeen Seasdenteferfaitostza - ~ £282 } | | : n (eB) | se | SeSNSE SRE Ree ae HH Bese w ee 2 fF oss Nia S 4 & ohs BAe SBS - AS S 166 SCHOOL CALCULUS [CHAP. V finitely in the other directions. The constant of integration in this curve has been so chosen that the integral curve passes through a vertex of the hyperbola, and not through the origin, so that ordinates to it measure the area between the hyperbola and the axis of x up to any point. 7. The Rectangular Hyperbola Referred to its Asymptotes. 12 wy=12 ean xy ONY ae dy 12 dx 2. Suds = lO are The Differential and Integral Curves have been again drawn for one branch only of the hyperbola, that in the first quad- rant: they are both symmetrical about the axis of y, that is, the other branches would be obtained by folding the figure over along the axis of y. The slope of the tangent at P, where x=2, is —3, which also agrees with the length of AB in the graph. With reference to the graph of the Integral Curve : Let us find the area between the curve, the axes, and the ordinate at x=, say A x, 12 Then Ay={ : — de 0 e “4 = 12 [ logex |’ =12 (loger; —log,0) Hence the area from 0 to a, is always infinite, and the Integral Curve which would give all such areas would lie wholly at infinity being of the form y=12log.x-+o. The constant has therefore been omitted in this case, and the graph of y=12log.~ has been drawn. From this we can find the area between the curve, the axis of x and any two ordinates at x; and x2 say: e.g. in the figure the shaded area is represented by CV —FM, i.e. by CD which is very nearly 9 units long: by calculation the shaded area (from « =3-2 to «=6:8) is 9-04 units of area. Similarly the area between the curve, the axis of 2, and the ordinates at any two points is equal to the difference of CHAP. V] COMMON CURVES 167 R' Ge & fa be | iz Pal a | a ae ae BHAw 5 | a a a 4 Boe ak | a0 [3G BE Pale Mrmr le a he IS a Se teeter eb Meoperapsiete i hlel ie perp Znoane Hp LLB REE ReRe Beee GUI _C RR SSR RERBRSSREReB\ SERRE SESE eeEe ee oie erie Se ae tae eee [aetep re ee SE EEE EERE Se ee ee eee NoPE meee Pott | —h- ] HOARD U SSS CRBS VSR EER aaah: st Lash fa eee tele ieee pete plete Joe tft _ = JU SSEORERa GER eae eN RANG eon Chct20 0 2 i be ata shalt te AN SAREE Sereletai aia “5S Beeo aoe CCCETECONNRRENS NAN CSE 12 SOSHESSR ME Rie Aa RNS WARS LLL eee eee AN | HH eden Ho HEN 4 RSS fis TER aaeee ee Aniotelone Tl holed bee l= manent cee CH SEE EET SOUS EPSON eee BREA Ree fee eee eet ee eee tt JUS SED RRR eae eee peer Pee soars A tet te eee eee EET ag an doe I foe Vest] BLM pO ieee | fe CEB az HO oe pe eee ee HORA: +L 4 merece iel [Shetapepee oe isl) Sage h' WR DEG IU Ree oe et ETN ero rye) ieee efeletslatelal tela|s| a2 80g |_| gee eee let Sieh] Fie eo] a] Sih Te lobe Te eho _ SRS. UR Rae Ae rere else WNiniel intel ii tee by LEE EEE eee eee eee at ele ise lola slate bebe boljele | be) i DORR Ve? Pee Beene LOR SGR BER Rae eee pepe rais eis) e hal alata ile Dal p S| eee ee eee lalla eis letelet elo beat te fy fe JLB EeEER EG Se ee 00S Vee Pee ee ease ite eo lei ietsieielete| tele sl biel. _ _. (ADRS Ree ae ne eee emer i iey Veieiefey salable fel eler Tee ry . (UU BED Ne Bae eee Fie. 40. the corresponding ordinates of the Integral Curve drawn, attention being paid also to the sign of those ordinates. 8. The Cubic Curve Us ats [CHAP. V SCHOOL CALCULUS The Differential Curve is of course a Parabola, and shows The slope of the tangent at P, where v= —2-4 is by calcu- lation 17-28; and the shaded area is by calculation 29-4 units that the slope of any tangent to this cubic is positive. of area, and both measurements agree with the ordinates to the associated curves. 168 | SSGKRSSACHRBR TSN SEED HDS ee Cet eet ial SOMEUNan ean bole | meres t+ EH ee eR SS eRe eee PEEP EER Se CEST new ee HE SORES RE SRESSREEES a SSeS Rae Neeees BCOPLEESOERR RD UOREA RR AASCUKGSH DURES OREM Oe oTie eter been. ee eee ed eae eae roa gan ecded epee toi HRRRRER ESS SOSH RAS Ree KO ReReees SNe oeNe Haase | BEDSRARARDES Vea s Basse ¢ UR ASSaSReAe ee BRSEUSR ARCELOR aes PEPE Ete es (47007 RR RRR Ree eee BERBERS RSee tease REE PSkReee ERR EERE CEE CECE EEE ECE EEE EEE EEE SX ee ad Pe DS beh ee eel elas eee ered PebehopM tater te) lat tw ON ye dol betes lalate alates SERRERNNSN VBR SAL eee Oe ER CRe RARER BRERA eK ahaa ewes PE Ee cee pale bol eget toe ep ecgebotes tte iE tat clot de tel ateled aelet loiete ls bal eist op EEE EEE EEE EEE EEE EEE 2° EEE R EARS bole PET be) ToL bd Lt ae i eet te Sea le thle SS MeRLeD AW ag 0008 20008 S000 0S8 See ee | Oe Be SEARSS SESE BBO AME HN ha SESAME eh 4 it tt BERE rTittis SECS EEE Pp fe BRL CSSS hao (EEE LEEAS en Ree ASCE CEE Se eeO re biel pare iol ta icq seteisete te SBN CS SSRs AERP RASA ELAR RREs PORES hee VIBES RAMS Bee SUMMER ERASERS RRR Pe SeM SRA eae Sk080 ESE BRRGRER PEASE eV ERP A Se ARR NEMS RRR EVEN SY SORE OSES HHH BOSS 5 ppt Ne ane 228 ERASRRR SAR RSERE A RERURUStSS PERSE S SOMO MN Ses wee ame ne ae ERR ee HE NA ERBEOie see A EPL EE A NAPE ree PREG RAE DERG ORR SERRA AR EG SCR SERRA PRORM EA ORAS OR Hes DSLRs ARPANSA Sa MSRM PR ARRMSSEEe pte bE SiVSaReS Pek eAs SeCRLAR RRP e AAA ptt ft | Se BEEBE Ealtelaia3 it el de teeb ae ia Senate ie SRA SRS RENAE BARRERA RRA ARE AREA PIRES RSS ease REBESES CASPER ASRS SPARE ARES BSS ARCA ERARENES LN GeE SR HA HHH -H--A--89 & ax, of the types y Fia. 41. ' Cis The three curves drawn again afford an example of the curvatures near the origin of curves y =b23, and y 169 x (logi9% — ; 4343). -173. CURVES COMMON x (logi9v —log1o0e) da Jude This curve can be used to find cubes and cube The tangent at P, where 7=2-5 is by calculation roots (cf. § 4). Also the shaded area is by calculation 1-419 units of area. 9. The Logarithmic Curve. CHAP. V] Note. |_| | | | a B | gue ueeseenssee CenseuaUUEEGEEEeT ofsveezaee BEEEEEEEEEE ERE EEE EEE EEE EMM ARD EERE RER EEO Lea DSO TS BARR RAROREANASRaREe Oh _ | CBSBEnRaeo Be ee YT es | ESnsUrTHhORaGGOe Sees r |_| DERE ERERoweeess ae SS ens ne BEAGR WERE EB CHEER BAaRRaMse PREECE CEPT Ey COENEN CCC Ee ac ERR RES Rae 2 VATAI Os vAValae wave @.04 so nS FCCC EEE ELINA AVA BERRA ase COCO CCS AVP SS BRR ERR eM RA CASRBeaMR ae VAN ALAA A tt POP ANA VV TAMA A sl sielal se sini SRRERDESDEREALST OR BAE 6 PoC MGOM ALY ALS Ste Ss 2 Sen ee Pe ee aie leh to NVAAL TPL AAP CYA WC Pe MAS Aa AA tt tt ete ease a NA A ea [a a eee (ea NALA on fi ZA tail LE SER DRE oeo PET TTT QV SAS TT TT ET ER RRR SEA TR ROR SERA 4eeUe RSRREaIiaS BERBERS NAA REDE a AAR BRE Shella shlt Seesaw eAaSsese MAA) HUM MACRAele ta esse R2IEeeee. Cnee ERBEEESISE Sea Resse eee Dee ANW46 7 GREER BERELSIAO ARS SOAK SRR RRSP Use, es ROR SMa Ree BRABOIS ss eee Aceh eh] LRU aeae) Bes fF a fs | 0 et 9 el a lo-fi TNS de ine of BARI MNASARRM SCE Se PP ee DOOR IRARA SAGs SSeS BERRA. a PI a a pe tet led Nd er al SE CER RRS Se OR ROE R Ee DROS ROM ee SERSRR PCC TES Soe ees ' ' ( for though both curves approach Fie. 42. In the figure this is given by CD which is the Algebraic differ- ence of the ordinates at 1 and 4:5, i.e. their numerical sum. -4343. The Integral Curve should be compared with that of the Rectangular Hyperbola (§ 7) : The Differential Curve is of course the Rectangular Hyperbola the axis of y as an asymptote, this one approaches it much more closely, and the whole area in the fourth quadrant ay 170 SCHOOL CALCULUS [CHAP. V between the curve and the axes is finite, and measured by the ordinate to the Integral Curve at w=1. This area is also equal to the positive area from x =1 to x=2-72 (i.e. x=e), as we see from the fact that the Integral Curve crosses the axis of x at that point. Notice also that when «=0, «a(logi9x— +4343) is indeter- minate, being of the form 0x (—0oo): but by plotting points near the origin we see that the shape of the curve is as drawn, and that the Lt, —o x (logi9% — : 4343) is 0, as may also be proved by the method of Ch. III., §27. The Integral Curve also becomes imaginary when x is =< 0, but continues to infinity in the other direction. It must be noticed that ordinates to this Integral Curve from x=1 to v=2-72 are negative, though they apply to a part of the original curve lying in the first quadrant (see Ch. IV., § 39). The reason for this is that the area between the curve and the axis of x vanishes for the first quadrant part of the curve at x=1, and for the fourth quadrant part at x=0, and for convenience we have so chosen the constant of integration as to make the Integral Curve cut the axis of x at the latter point, instead of at the former. The same ex- planation applies to the Integral Curve of § 2. Note. From this curve logarithms and antilogarithms can be read off (cf. § 4). 10. The Exponential Curve. y= 10% dy _ 7, = 10 logel0 =2-3026 (10*) 4b ] SJyte= i, 10 e (10% —10°) = -4343 (10*—1). The slope of the tangent at P, where x= -6 is by calculation 9-18, and the shaded area from 0 to -9 is similarly 3-016 units of area. The constant of integration is not zero in this case if the Integral Curve is to give us areas such as that shaded: but it is easily determined, since the area from x=0 to =a is Ls Shy pea nes. dfs tee loge ah = +4343 (101—10°) | = +4343 x 10°— +4343 _ a, now becoming variable, 171 COMMON CURVES CHAP. V| The required equation and curve are therefore those given at the beginning of this § and in the figure respectively. PULLS DRAB RS Reese GEMS ete BBO BANA RROBRaBECe): SO EE EEE EEE ECE EEE EEE EEE HS ERS TAA EOS CULE ORANS ASRS ALAN SBAERAAIE SESS EEDORER LAS MOD eee ke aR R ES BR ABR ele (elegy oreo tel y erie erat ie TR aS eels Peds ee hehe eit fore aiiie tele ay le ela eS Paes | RCS Seo Rae ess al eet retebe i elors eh olel Nelailole ho BRRSERSRGRa pote el eal ote ia siet tatiana dN otete ib) fap SLE ahh sn Kena eee HRSBRReoeAey BREESE ARROBRLes. SEEe EShalcaaanea let ame ree iey (ote [elie BRSSMROeRee Sac See eee SUSsRSERbeae ita sf ep a |) a | a ee 2 i a RSGsa Sra Sopa eet se bbe ist tal a SO Oa Coan. ee Reisen cea PIV TAA TPT ARIGheissS CERNE AE Rae SOL LGA LAG AH Press oe echrastsehete SO TVLAINNEO UNAS Ee CRGSGeO TERA ae RASS HERA RED RsRse ewe Raee ee eieicieisiet ale tal etic ee elated igs Mae PAA ake ig | Ys | fa a fie Sa eT [fT She NZ AZ eee is | 4 | i a itn | aa Si i LN eet ee cL isi botak 1 lel betel telat Rae | arf) a ef | ef am a Ad Ssrriceitrreniseetrcr incepta (2 DEBS aa we Is seeseesenseneens 7 20H Sos bot 20 sn a SE RUARSSPORE PARA SRRARS NAH STA SCALARS eee Be ERR ERNS PHOEBE BO ERROR PRES ANE aR pt fy RORRROR AES RAR BRE BERR REA VAAE SS CORRES aes FEEEEEEEEEESEE EEE EEE EEE SEE ER eer bee ee ete be eet SRLS RR ANOS RRA TERR SSSk SARE Res Ht Fia. 43. The three curves are all exactly the same shape, the Differ- ential Curve being the Exponential Curve moved -362 units to the left, and the Integral Curve the same one moved -362 units to the right, and -4343 units downwards. This curve is really the same as the last with 2 and y interchanged, and can similarly be used for reading off Note. 2 = is: — oo i] fan] ao (o) | ee | eh ars Sec oO S > —o So Bs a ey rie S fa) In this figure the axes are still axes of « and y, corresponding values of a and y for different values of @ having been calcu- [CHAP. V Seviicie = (me loa alee Sis Bete lelel zetale isisteN LAV ATA ABGeeS22ER S500 =taareae aseSCuuaqom an ae eee iets p err Tee eo ae ERECT ESSE ESE EE AUIGS RWS Faas Wee Bee SMe REMEBER SRR NER Lee aRease ARR EERSUBE Ska TE PAR hea eA eRe SUVLSIOATSRS SEUSS LI SER NARS ee CREM eT aT eT ERAT AAV ALA Tite TS] BRUBS EARS SESSA e BD rie we CFs. SEBR ee ee ree ee ee OBSSURERE ee Se RRS RAE RAR ean ree Soe re AV ett Ld COEF aX 7779 FO 8 5 a i eee fs es eA Vd Re DBRS SSSR Oe SESS REeRoE Way sasne QU BUEERESOUUE LEER RSE GEROSSSReAY Caeeeee GU PSOE SEER BESS EL ERA ROR eee 25 A SA EN ngs] sf as] FPP aT [Ts lew HH i a a a | we a a P| a Ps [Ff es 5 =1-038 1 Curve 3°8 and @ -hand branch SCHOOL CALCULUS not terminate as in the figure, , the period being given by one Fia. 44. 172 SI UREES batceecececescececseaenc! Fine EERE PARE ROPE eT Elsie Si tessrecessceaseeee EH HeaREEReRe eS ESEee EERE EH a SEES ES EEEITCEEETESEESIEEESI pane BURGE RRBkSoaReeweee G SUGESTSIEEES Poe Sa eeatectest astestastettaem fatten aoe SESCGREESIGES Saaneeeee Surectesrtartects ARSEESEEE sedaesatvesseveessareee N HHH 0 to 6=1-502) is similarly 2-87 units. GE 6 Ok a aos Preiibigeiert tat Peete ¢ ALEC EEE ADSRECSe aGe BLE See Pa ister CEC EEE GEC eer rr a LAL ee eS A ee EEE EEEERE EEE EEEEEEC EERE EEE Deeb CIuS tesa w NP Ae one ASRONS. ueo deen eae EEC BEE ae ee @ gn> S005 88 Che LH The Differential Curve has asymptotes x=+6-28 since the tangents to the Cycloid when x==-+6-28 are parallel to The Differential and Integral Curves are again drawn for the axis of y. the right-hand branch only of the Cycloid, but the other The slope of the tangent at P where x All three curves are periodic lated in each case and the graphs then drawn to show the The Integral Curve does but only that part of it drawn applies to the right relations between 2 and y, and abscissae do not represent values of the parameter 0. branches are clear as the curves are symmetrical about the axis of y. ating circle continued to roll away to the right and another Cycloid were traced out. is not, its general shape and direction being however obvious is by calculation -572, and the shaded area from «=0 to x would apply to the corresponding areas obtained if the gener- revolution of the generating circle, i.e. while @ increases by a period 27; the Cycloid and its Differential Curve are re- peated exactly during each period, but the Integral Curve if we consider the areas to which it refers, of the Cycloid drawn: the continuation of the Integra (i.e. from @ CHAP. V] COMMON CURVES LTS Another figure of the Cycloid is also here given to explain how it is generated, and to show its geometrical properties. B Fia. 45. The Cycloid is described by the point P on the circum- ference of a circle rolling along a straight line B’B :* when the circle touches the straight line at A, P is at O, and when the circle has rolled along the line to & the tracing point is at P as marked. (i) Since the circle is momentarily turning about #, P is moving at this moment at right angles to RP: hence PT is the tangent to the cycloid at P, and therefore PR the normal. (ii) To find the equation to the curve let OCQ be @, and ‘a’ the radius of the circle: the arc TP or OQ is equal to the distance the circle has rolled, that is AR. Therefore eel Ee =NQ+QP =NQ+AR =NQ-+are OQ =asind +a =a (@+siné) also y=ON =O0C —CN =a (1—cos@) (iii) Angle PT x=QOT =QAO in the alternate segment Ee 2 dy Le, 4-=tan g Pte iS Par ¢ 174 SCHOOL CALCULUS (CHAP. V (iv) To find the length of any arc s. ds? =dax? +dy? —[a? (1+cos6)2-+a? sin2] dé? = 2a? (1+cos0)d62 == 492¢ ott dg2 hes 2acos, dé. ay g d Coe dé PAG re =3 asin, there being no constant of integration, since s and @ vanish together at 0. But 0Q=2asin, *. the are OP =-2 x chord OQ. 12. The Catenary. y =5cosh— aS) esate de =sinh 5 Sua = 26sinh;. The Catenary or cosh curve, and the Differential and Integral Curves which are both sinh curves, should be com- pared with the similar curves in Chap. III., § 23: and the whole of this figure should be compared with that of the parabola in § 4. (i) Since y=c when x=0, c is the height of the vertex of the catenary y=c cosh above the origin, and the axis of x is called the Directrix of the Catenary. (ii) The length of the arc measured from the vertex to any point P is equal to the length PR cut off from the tangent at P by the perpendicular AR upon it from the foot of the ordinate PA, and AR is constant, being 5 units Me e. c) in this case: and s=PR. 175 1 CURVES COMMON CHAP. V] i NEONS ToRGOs a RRR CHa OSU Roema ss BUEN a RM EL LE IS Pept ast (eta ial-( feletel se) iaterenet, Pela foley poss Pl ded sd ate bafolicy tele [ck erates (ale tt SHEP EO CeR SASS cESa SsEMaGece ta) SOP SSSGaeGhCcomrcceccn ae aie eitats ed to peelotsial ieteicr ieteieiseebiere rs eb lol cl teeter ea pele 000000 ee Ee pet ete doled: ieee i bao tee intal toler Tee eel peer ete) paps berens foe Pictaletetoted cla] ce einer ale sieieblalons mie latsloboie| [ole lsian loads bets nk | lapse opal os SUCRE ts BEE EEE EEE EEE EEE CEERI EE EEE EEE EEL CE RCL 2 sw celttieiee sie eel Iasi eh iter Mise bse eter ert Dbl dete, | re a [alcisiaiat alts iste |eisiaieial cist: | lol (ot tetera cho ied Wor intel ete ie Ped ee 2000S SSS Se ees Pees eee Pe CEE COCO CCC N a = REECE ELE PEELLCLELULLCLL LLEVA ee © EECCCC EEE EEE RA ; SEEGER EERE ERE Ea a aaa. ) Per eset ee ee bist Ie) io lel del NAA Ararat ob k bake ~ 2 PERS ELEECEEEL eel eh ecb Leer reer errr TQ Pe CCC ihe CLL ole LELEbEni: erage ec COR FCCC ECCS COCs Peay TOC he rr SECCCC CC SEEEEEE See ape ee ep nN DRS ta ec A one eS Ue a — SCEPC EEE ee heer >|,8 EECEEEEEECECCEEEE EERE EEE EEE ARE Oper i tel lcloh rs is BREET R SET cer here — eds bape deed ct else oleh leh ok talc Op Ziclebot C1 ert Cc er at eg Poisistelel Wolee ek | tote 1) y eRe 6 Wee Se SAAC Pee AA AN — CONTE RSENS see Ae IN LLL te) OS A AY COPECO Ns CCC 996 AY US Cet eh ee ee et ea < SSGCHERER SND weeoM Ans ZeCC FSCO CCE EN ACE EEEEEEEEEEE EEE ee | PEEEEE ERECT ee SEN Eee | 8 fe Shoe CePA ARREARS eR BR ook Me Rees Ie ris SU DECUSEO SSR ER SSSA o GEN EMER O RE OSG Nea OES ee Oe Pi ta et bie feb icl ie ieboattal tee tl eeiers yo ie my OC eet CS eet ert ti © MCC bia Lee ie tt eee oat ieee g tt Tk Lot eto Ey Pais oe eee ht hil Sis rt let ee tr er rt te Bee ee OO ee aia ae eel eee eee & Wo ro : 2 oD) S Se) ie) ae Spo : ae ae a a iz 176 SCHOOL CALCULUS Also AR=PA cos PAR =y cos PTO numerically isi Seba Bea VAM V1+tan2PTO = Ds aeaES for tan PTO aoe wl 1 @y dx Beosh~ 5 f AP Nes ae 7 Lor Te nhs cosh; - =5, and is therefore constant. Again PR=V/PA?—AR? BY Ae == D Us cosh —] ent ==5sinh— sinh. =s from equation (i). [CHAP. V So that if the line PR be wrapped along the curve, A will coincide with the vertex. Similarly in general for the catenary x y= ecosh - OY ey ae oF sinh ‘ PR=s=c sinh = _, dy ° dx and AR=c. The importance of the catenary is due to the fact that it is the curve in which a uniform string or chain hangs if suspended by two points. Now we see that the catenary bears a strong resemblance CHAP. Vv] COMMON CURVES 177 to a parabola, and Fig. 47 shows that this agreement is very close near their vertices provided a suitable parabola be selected. Problems frequently occur concerning wires, etc., which are somewhat tightly stretched, and which therefore form part of a large flat catenary, namely a relatively small part of the curve near its vertex. 18. The Parabola and Catenary may be compared from Fig. 47, which gives the graphs of the catenary y=4 cosh ; (lowered four units so as to pass through the origin) and the a2 parabola Ere jy Ce | | "es Cy | eee Fia. 47. If the catenary be supposed to represent a stretched wire with a 400 foot span, and the scale be one unit of length on each axis to 100 ft., then the wire will be represented by that part of the curve from ~=—2 to x=2, the sag at the middle being about 50 ft.: and we see from the figure that even in this case, where the sag is about 4th of the span, the wire would also be fairly accurately represented by the parabola 12 178 SCHOOL CALCULUS [CHAP. V instead, which from x=—2 to w=2 agrees closely with the catenary. This may also be proved from their equations, for in the catenary (which it must be remembered is displaced in ie figure four units downwards) when «=2 a y =4e osh7 1 =4cosh- cosh; =4-5104 and when x=0, y=4 *, the sag = -5104 units of length =51-04 ft. on our scale. Also the length of the wire =2s=2 (4 sinh 2) =2{4 sinh(-5)} —=4-1688 units =416-88 ft. Now taking the parabola when w«=2, poe = +5 .. the sag=-5 units of length =50 ft. and the Jength of the wire e=Taeee ee +2 log ga (Chap. VI., § 19) —V5+4 logel-618 —=2-236-+1-924 =4-16 units of length =416 ft. Hence even in this case the errors from assuming the wire to hang in a parabola instead of in a catenary are not large (in the length of the wire it is very small), and these errors would of course be still less for a more tightly stretched wire. In general if a wire hang in the curve x y =c cosh— c CHAP. V| COMMON CURVES igs) this may be written 2h ee mara (a C a =5 4! tat oat Got at igedice ro nll oo oa ieee, a =e) 1+508 ae eee and if = be small, that is near the vertex x2 YO ae the fe pare (BS 4 omitting oe and terms involving still higher powers of =. 2 Now (i) is the equation to the parabola I=5, raised through c units of length, so that its vertex coincides with that of the 2 original catenary: hence the parabola Y=5 represents approximately the shape of the wire, provided that etc., oa 24c%’ are very small, and the degree of accuracy can be judged from the case just considered, in which c was 4, so that the 2 catenary was y=4 cosh ; and the parabola y= EXAMPLES 1. A point P moves uniformly reund a circle whose centre is O, PN is the ordinate and PT the tangent cutting the axis of x at 7’._ Prove that the rate of increase of PN is proportional to ON, and hence is greatest when PN = 0. 2. Prove also that ON- NT = PN?. 3. Prove that the tangents at the points (aé?, 2at) and -=) to the parabola y? = 4ax are at right angles to one another: also that the line joining these two points always cuts the axis of x in the same point (i.e. the focus). Hence state a property of the parabola. 4. Prove that the subtangent to the same parabola is bisected by the curve, 180 SCHOOL CALCULUS [CHAP. V 5. Prove that the subnormal of the same parabola is constant. | 2 2 6. If C is the centre of the ellipse “+ - = 1, PN an ordinate and PT a the tangent, prove that CN CT =a’. . he 1 at the ends of any diameter are parallel. (Take y = mz for the diameter, and by differentiating _ the equation to the curve show that the slopes of the tangents depend only upon m.) 5 fae 7. Show that the tangents to the ellipse + Z 8. Prove the same property is true of the hyperbola rons - — 9. Find the slope of this hyperbola when 2 is indefinitely increased, and b hence prove that the slope of the asymptotes is -+- a 10. PN is the ordinate of any point P on the rectangular hyperbola xy =c, PT the tangent cuts the axis of x in 7’, and O is the origin. Prove that NV bisects O7'. 11. Find the equation to the curve passing through the point (1, 4) whose subtangent at any point is one third of the abscissa of that point. 12. Find the equation to the curve whose subtangent is equal to x log.x, and passes through the point (e, 2). 13. Find the curve whose subtangent is equal to — 4 and passes through the point (0, 3). 14. Find the equation to the curve whose normal is of constant length a, and passes through the point (0, a). 15. Find the equation to the curve passing through the point (2, 5) whose subnormal at any point is four times the abscissa of that point. What would the curve become if it passed through the origin instead ? 16. A string hangs from two points in a horizontal line 90 feet apart, and the slope at the ends is 9°. Assuming its form to be a parabola, find its sag at the middle. 17. A telegraph wire hangs in the curve y = 500 cosh on a and y being expressed in feet. If the points of suspension be 80 feet apart, find its length and the sag at the middle. Also find the tension at each point of suspension if the wire weigh 2 oz. per foot run. 18. Find the length of the rope hanging between two points 180 yards apart in the form of the curve y = 100 cosh saa’ x and y denoting yards: also its sag at the middle point. CHAPTER VI PLANE AREAS AND LENGTHS OF CURVES I, PLANE AREAS 1. In this chapter we shall assume a knowledge of the usual methods of finding areas of rectilinear figures, and con- sider only figures with one or more curved sides. 2. The Circle. (first Method). Divide the area up into a large number of equal triangles, such as OPQ in the figure. Then when their number is indefinitely increased, the base ‘ome Bie. 48: PQ of each will be so small as not to differ sensibly from a straight line, and the area of any one of them is $PQ x height, which height is ultimately the radius r. Hence the area is Rhy , and if s denote the length of the whole circumference of the circle so that 6s denotes a very 181 182 SCHOOL CALCULUS [ CHAP. VI small element of the arc, the area of the whole circle, being the sum of the areas of the little triangles, is equal to Tie fee =" fis mx 8 3 : Now by the definition of 7, s=27r the area of the circle =5 Qrr =r, 8. The Circle (Second Method). Let the radius of the circle Pp Fie, 49. be r. Then its equation referred to axes through eae centre is 72-+-y2=712'; y=V 12 —22. Take a narrow strip PP’Q’Q of width dz. Then the area of this strip is 2ydw=2V/712—22 da v a Hence the area from 7 to 2 1s af” 24/7292 da x 1 a v2 —22 Des E Aye iB (Ch. IV., § 12) ja Le r eras tS ae To find the whole area, we must take limits r to —r, and CHAP. VI| PLANE AREAS 183 we get r2 sin—11 —r?2 sin—! (—1) (the second term disappearing at both limits) =r2 (as in § 2). To find the area of a segment of height h we must take limits r—h and r, and we get ee) vk y2 { sin—1] —sin—1 = r VR r—h_ (r—h) / 2rh |. =4 + cos! - r r 4. The Parabola. The equation referred to its vertex is y? =4ax. Fiaq@. 50. Hence as before, the area from A to any double ordinate PNP’ where AN =, 18 Tv 1 ans = 2 ale 8 sete if 2yda—4 f Vax-dx=4V a E F =3 Van This may be written SAV au, 24 DA did 2 ar ee: EP. ‘AN=,% rectangle PK K’P’. Also by a property of the curve, the tangents at P, P’, 184 SCHOOL CALCULUS [CHAP. VI meet at 7’ on the axis, where AJ =AN. Hence the area of the triangle 7’PP’ a+ DN -~P Ries Ave re =area of rectangle PKK’P’. Hence the shaded area is two thirds of the triangle 7’PP’, and the area 7'PAP’ outside the curve is one third of that triangle. It follows also that the parabolic area PAN is two-thirds of the rectangle PK AN. 5. The Ellipse. Let the axes of the ellipse be 2a, 26. Then P p’ the equation referred to the centre, is y2 b2 b ° SN Ge or y= Ve —a The area from x; to %2 is Wes Lit be a *4/ a2 — x2 da eee ae nit use| (Ch. IV., § 12). a 2 Ly Comparing this with § 3, we see that the area is : times the area of a circle of radius a between the same values of 2, i.e. of the auxiliary circle; a useful fact to note, as areas of various parts of an ellipse can now be at once obtained from the corresponding areas of the circle. Thus by its means (or substituting a and —a as limits in the integral) the whole area of the ellipse is rab, CHAP. VI] PLANE AREAS 185 Similarly the area of a segment of height h cut off by a double ordinate is { -1a—h (a—h)V2ah—h?| abicos = —— a at a dy 6. The Hyperbola. This is of course an unclosed curve, and as a whole has no definite area. Fia. 52. The area of a segment cut off by PQ, where CN= 1 is easily found. The see to the curve referred to its axes, ay is 3 i=l , Or y= V2 —a?. Hence since PQ =2y, the shaded area is oS. Vx? —a?- da (2 Pe a a (Ch. IV., § 13) sea bay A/ 1242 —q a b / 242 —a? a —ab loge (#1-+WVa12—a?). Since =PN=y, say, this may be written xy, —ab loge @ + us) 186 SCHOOL CALCULUS [CHAP. VI Clearly the area of the triangle CPQ is ay,. Hence the hyperbolic sector CPAQ has for area ab loge (= +H) and area of cap=© loge (2 +%). Let the area a ts be called S. y1\_ 28 Then loge (= - a 3 © 25 or eee VAGer ee (i) a 6b 4? 2 es But from the curve 5 Br = ee $e Sp 3 : vy Y1 a Ate *, dividing (ii) by (i), pees adding and subtracting (i) and (iii), 28 28 ne Ue ieee h 2k 2 Memo geass! oa 2g 38 a__¢ Mere eae ay eet and y1=b a) eaebane sinh ab We now see the reason for the names hyperbolic sine, hyperbolic cosine, etc. Dh yr Just as a point on the ellipse “atl can be expressed in the form #=acosd, y=bsingd, where ¢ is the eccentric angle of the point, so a point on the hyperbola can be expressed in the form x=acoshu, y=a sinhu, where uae, and uw has therefore a definite geometrical connection with the hyperbola being the ratio of the areas CAP and CAB in Fig. 52. 7 The Rectangular Hyperbola referred to its Asymptotes. The equation to the curve is xy=c?. The area PMNQ where CM =2;1, CN =o, is fs y dee =) ee 2 C =| #logex | : ra 2) =c2log.— Bea CHAP. VI] PLANE AREAS 187 Fig. 53. It will be noticed that this area depends only on the ratio of x2 to #,, and is the same for any pair of abscissae with the same ratio. Fig. 54. If we require a similar area between the curve and the y axis, we have only to interchange x and y, which, since the equation is symmetrical in them, gives the same result namely loge where CM’ =yo, CN’=y, Yo 188 SCHOOL CALCULUS [CHAP. VI 8. There is another curve somewhat similar to the rect- angular hyperbola which we must consider, namely yx’=c where y is some constant, on whose value the precise shape of the curve of course depends. When y=1, the curve becomes the rectangular hyperbola. When y is=>1, it will be found that the curve approaches rather closer to the x-axis than the y-axis. | eas RASBo mee Sta ttt ty pe bh tp pe ehh] epee Ht tt BRR IR ES Bs BSReaelasekea BOOCMORERCRGCREMGSEOeco Coe COE BSeIRG Emi ee BRET ERER SSM Pfft EAL LE 7 a = vn SEE ele | | {| NAAN H MAS raya ats iet aia a SHO Bom EE H MRED ESSSSUeeRoke DRERARaR SOR NSMas apa "a SSSSEG8 nanSeoee | REREE saree Fia. 55. The area of the shaded part in Fig. 55, in which y=1-408 wg and c=100, is y dx, where OM =21, ON =x2 4 | 189 PLANE AREAS CHAP. VI] This result may be written where 41, y2 are MP and NQ. a, yx’=C, xy 9. To find the area between the four curves d. xy =b, yx’ i , f | eh etoile) ebe po Te Teele he aes eta ate ACURA RR OA hE RES Omi oe DURE RS See ALeP SEEN Eee Soin Aan aieleelulslalel(arnlstatalelelalehelalgialal hgrael es ROEM ANRaRe Shee eEE oe ens My BSR EE RASS EERE. Seas BAAR OR SRO aARaeaaRey, | gH SCE Sash CoRR SSS Orne polmie is Simi stage betel vierebelele ere lelete | fl aii re tal oe Seo SRee Ane? | KoRn SHAR SSR Re Ress ARE Sees =e Hg SHES RRERERSERSRSRRAe imo AR USERS SOA SARRRReNS il AGRE RARS DOERR SAR MM ee id oon BOVE SReanBoonS ee A —|—| — | i ole (Stet tol: beleietelcl ein fs als 71M) el elate nme t-te REBORN RES UESAeS NOUR aes BR REREERERSRAM AMA AAnP Aeneas ER BOB RBG O Res S40 i/o ee es RE BRMBMMERE VOC a2 BERBER BRSEVSa Er Ns FaSEGEER( RR RRRRARAR Sioa toma BREA BNO Wl as AER soars Se y BB i Sadho se ene Resee ERRRa> = atGROSCeaA eee as =~ aah RRO eesnAenn +++ 4 hes Ve ete | ol el ay te a] ea Le a esi ye ee thee el le ee ed he rire lel ate bf pave eels (ete | eel ils eae reese etal Fia. 56. Let PQ, QR, RS, SP, represent these four curves respec- tively. bscissae, namely OH, OK, —RLMS—SMHP. OL, OM. Call them 2, x2, x3 and a4 respectively. PH, QK, RL, SM. Then the required We have first to find the four a Draw the ordinates area is equal to PHKQ+QKLR d d, we have y;= a 1 a, and yx’ = Since P lies on xy 190 SCHOOL CALCULUS [CHAP. VI b c i » oo 8, me =p u=(3)™ Now area PHKQ=a loge (;;) (§ 7 x ~ ay — l — Y = a 1l0ge : 1 a ert (‘). Similarly RLMS =b log, (") mp loge AR Also QKLR=——(13'7 —xg'Y) (§ 8) a) f Similarly SMHP =, d (5-3) no b—a Hence the required area is equal to In the figure the curves drawn are as follows: PQ is xy =25, QR is yx! © —90, RS is xy=18, SP is yar =50, 26-18 50 . Area enclosed is pene loge (55) be lela Ps (;) Ty 66 ee oN 7 anne -5878 =6:234 units of area. 191 27 27 —40 14 12 v3 + 3x2—8xa—16 ble : side of the y axis. —16 | —24 | —32 —16 | —20) —12 ans =¢ (See Chap. IX., § 28.) AREAS 0 —8 4 8 area lies in its connection with the PLANE 16 —4°5 —91-125 | —64 | —27 —10-375 10. Find the area between the curve y To plot the curve we have the following ta and the axes on the negative The importance of this work done by an expanding gas. —8x—16 and the curve is as in Fig. 57. CHAP. VI] x 3x? neice le ainiate ee alaieieinial efoista sista morale eta te Ted oben stostnd nln eos ecto fe nf St [clobel | ae todo len inpe as bel sl alates ote ee een el ={a[al she lela le le ke el ee a ee ete s le he bel sha ae ee DL de bb fo [emt | feed be Foe oR ta Ff hg fat Poets ad fe pct hats Pale [else belabet eters bale lal a tel thers nana BOSRE RTS SESE SeOPEV Eee BeonUe ET Pied bb. eld ce|mals is bode) apa o lsat anae peel ealeel a alate RABE SCPRSS Pees eMame sero eee latslelefpolsietels|s|slelets palate SRRGR 000008 G0005 00000 RRR eee An ett + be ls HH HF fe tt Els ++ [sis CTT A eT Mist hse VAAN AAV LAL eee Lopletetetele lal |e VA AAZVYAALV AA 2 ee elatlehip oY AAS AZ ae ae es el cele ale TAZ VA Ao le eee ria ehetaerate it alti | PAA lol stale te [otal ane ant epee sib ett Vt AAT beet] ct ale aarp et CE eB OR RS ACBRRRESSDESREELECL oe iairimmiete lle 1 lec | alll a) ata be eal ee a isle peal pear A /A/IN [ll | tole la] eran ele a raner ane ye simile VAA// (8-4 |e tele eee ahaa eat AMMA AMZ elo lel: [cle eee eee AAA AAT ZoReeaTor Regen ACER Biase H+4 Ce ARS Ren ea Seer ee +4 Bs 0 ed oe Sk ee Fiq@. 57. 192 SCHOOL CALCULUS [CHAP. VI 0 0 ired i — 3 2 gy oe The area required is yh ; yd if : (x3 + 342 —8x—16) dx B 3 Ay? 162 | | = mime om Ry —64+64—64464 0. The meaning of this result will be understood by reference to Chap. IV., § 39, and we see that the areas in the second and third quadrants (i.e. above and below the axis of x) are numerically equal, and opposite in sign. To find them separately, we must find accurately where the curve crosses the axis (at about —1-5), and integrate from there to «=0, which would give one part, the other being of course equal to it. To do this, two methods are at our disposal. We might solve the equation 23-+3z2—8x—16=0, and so find all the points where the curve cuts the axis. In this case this is easily done, since we know that —4 is one root, and therefore x+4 a factor of the expression. Dividing it out, the other roots are given by 22—x—4=0. 1 a/ 17 ees 2 2 whose roots are , and the negative root required is —1-561 nearly. Hence half the shaded area is given by et 0 E +43 — 442 — 162 | = — PEST 4 3-804-49-748—24-976 —1°561 — —12-91 or 12-91 units of area approx. In case the equation cannot be solved we may approximate to the root required, by plotting that part of the curve on an enlarged scale. We have the following table : A ae ee en Bier | et 1:5". | sae Pea ie ee a 5 4.096 3x2 Sahel *o5 ba ee Bee eee —(8x-+ 16) ope aire, BCE Py mre CHAP. VI] PLANE AREAS 193 es |_| 0 a | a ae ial oi |_| B je a re Ns and we see that the curve crosses the axis at —1-56l, which agrees with the result already obtained. 11. Sometimes (as in Chap. III., § 17) a curve is given by its coordinates each in terms of a third variable, as x=f(t), y=¢(t), t being the independent variable, and x and y both dependent on it. Differentiating «=f(t) we now get dx=/f'(t)-dt, and the area becomes /ydx=/y f'(t)-dt=/$(t)-f(t)-dt between suitable limits, f(t) meaning of course the aca co- efficient of f(#) with respect to 7: 12. The Cycloid has for equation x=a (O+siné) y =a (1—cosé@) a being the radius of the generating circle, and the axes being the tangent and normal at the lowest point. HEH pf) | pore BERS e BOC UEaoee Cees BA ft ae ian Seo aa Ree ildateie te) aan cenee & a a a a aS a Fane é SACRE (Bo ZReVaseeee EEA 6= 0 gives the point O, 0=7 the point A. 13 194 SCHOOL CALCULUS (CHAP. VI Hence since dv=a (1+ cos@) dé, the area WE yd becomes v1 94 if a? (1—cos20) d0, where 61, 02, are the values of @ corre- Oy sponding to 2; and #2. 6 =a? f)* sin?9 do =e sin20 7% 2 eae I, calle {42-815 (sin202 — sin20,)}. 2 2 Though the equation «=a (6+sin@) cannot be solved for 6, when x is given, by any simple method, yet the required values of 6 for any given abscissae can be read off from a figure such as Fig. 59 by drawing PQN and measuring the angle 0, or values of a (9+sin@) can be plotted and a graph drawn, from which the equation a (@+sin@) =a; can be solved. In this case, where the radius ‘“‘a” is 1-5 units, when x=3, 9=63°-21"—1-1054 radians approximately, and there- fore the shaded area is 5 4: 1054—3(- 8021) | = - 7926 units of area. : . To find the whole area outside the curve put 0=7, and we 2 eee SMES But from the way the curve is generated, the circle rolls half round from B to A: ... BA=7a, BO=2a, and .. area of rect. OBAC =2ra? 2 “.area AOB inside the curve a and the whole area AOA’ =37a? =three times the area of the rolling circle. Examp.LE 1. Obtain this last result directly by taking BA, BO for axes. {x will be unchanged, and for y we must write 2a—y i.e. a (1+cos6).} | EXAMPLE 2. Obtain the same result by keeping O for origin, and finding /xdy. CHAP, VI] PLANE AREAS 195 EXAMPLE 3. Find the area inside the curve from O to 6,. o ExampLe 4, Find the area inside from 6, to 6g. 13. The equation of a parabola being given by x=at?, y=2at, find the area enclosed by the curve, the axis, and the ordinate at t=1,. We have y= 2at, du =2at -dt. : t -, fydaf* 2at-2at dt =402 f t2 dt 0 _ 404143 mes This may be written Sats? 2aty = rectangle ANPK. (See Fig. 50, § 4.) EXAMPLES Figures should be drawn for all the following examples. 1. Find the area of the curve 2y = 2V16 — 2?. 2. A field is enclosed by a straight hedge and a circular railway line of radius one quarter of a mile, forming a segment whose height is 100 yards. Find its area to the nearest tenth of an acre. 3. Find the area common to two circles of radii 10 and 20 yards, whose centres are 25 yards apart. 196 SCHOOL CALCULUS [CHAP. VI , the axes, and the ordinate 4, Find the area between the curve y = j 4 5 2x at x=4; also prove that the whole area between the curve and the axis of x is the same as that of a circle of radius 2. 5. Find the area between the same curve and the line y = 2. 6. Find the area between the two lowest parabolas and the axis of x in Fig. 28, the unit of length on each axis being 2 cms. 7. Find the area between the parabola y? = 12%, the axis of x, and the ordinates at the points (3¢,2, 6¢;) and (3t2?, 6t2). What does this become when 4 = 4 and =a? 8. Find the area between the curve y = soa and the axis of x (i) from z= 2to2=3 and (ii) from z= —2 tov=8. 9. Find the area between the rectangular hyperbola xy = 1785 and the y axis from y = 85 to y = 223 (approximately half the area of a section of the Skerryvore lighthouse), the units being feet on both axes. 10. Find the area between the curve 3y? (4+ 2?) = 8 and the lines x = 0 and x = 6. 11. Find the area of the loop of the curve y = 2% V5 — a. 12. A table top 3 feet square and 2” thick has four pieces symmetrically cut out of its corners, in the form of rectangular hyperbolas whose asymptotes pass through the centre of the table parallel to the edges, the curve of the hyperbolas being xy = (with 1 foot units); find the weight of the table top, if the wood weighs 40 lbs per cubic foot. 13. Find the area between the curve (7? — 1) y= 6, the lme y= 2 and (i) the axis of a (ii) the line x=1 (both these latter being asymptotes). N.B. y =.x cuts the curve at the point (2, 2). 14. The curves zy =a and yx’ =b pass through the point (30, 150) and two more like them through (100, 44); if y = 1-06 find the area enclosed between them. 15. Find the area between the curve y = x® — 5%+ 12 and the axes in the quadrant where it cuts both of them. 16. Show that the curve x* — a*x — y = 0 has two finite parts above and below the a axis such that the areas between them and that axis are equal. 17. Find the area enclosed between the curve y = x2? — 6x2+ 32+ 10 and the axis of x. 18. Find the area between the curve y*+ y? = x? and the line y = 2. 19. Find the area enclosed between the curves y = x and y = 82%. 20. Find the area between the curve y = logiov°, the axis of x and the ordinates at 2 = 1 and 2 = 10. 21. Find the area between the curve y = logiox and the y axis from y = 0 LO 4 =) 30. Find the area between the curve 2xy?-+ x — 3=0 and the line x = 1, 23. Find the whole area between the x axis and the curve given by x = 7, 3 ome Tg 24. Find the area between the same curve, the axes and (i) «= 10 and (ii) 2 = 30. 25. The curve y = x? — b? is required to be drawn so that the whole area between itself and the axis of x is 36 units of area: find 0. CHAP. VI| PLANE AREAS 197 TRAPEZOIDAL RULE AND SIMPSON’S RULE FOR AREAS 14. Trapezoidal Rule. The area between a curve, the axis of a, and two ordinates may be approximately obtained by the following graphical methods. COIS S Se aaa Mma JIE 8 a OD wos SES8D7. . <4 SSeRSeeee Pee DOS 2eG0 See SEAo Gees Be Dect ORES SSeS Zk Ses SSS Ne Se ECCECC RSE ERE BORER 2S So5 SAU ee Poe fe HEN pd ie Se SI SS ee SS me S CEECECERSERES ERS SSC EEE poe FCCC TRRS SENSES eT CCCCCCORSREREEESRES EERE FCCC RES ESES ESAS ESSER PECL & RE SER S SS8 Swen UL Sebid Sao Sa eee Wee CSS sae TLOCCCCRSSRESESSESERSSSRRSERSS ZJEREESSSS PR SSSSSSSSy SESSSE HO FATES EESSSSSSSER SESS TL L MAIN & Rete se eee sb FFA Ba Glsiaiale reali Fig. 60. Let AHSL be the area required. Divide LS into any number of equal parts LM, MN,....RS (seven in the figure) the length of each being A. (The more there are of these divisions the greater will be the accuracy, but the labour involved will of course be greater). Through each point of division draw ordinates to the curve wl On SH. Wet their lengths be y1, yo,.... yg. Then neglecting the small portions of area inside or outside the chords AB, BC,....GH, the area is approximately the sum of’ the trapezia ABML, BCNM,....GHSR. These areas are hx WES hx Lae ES Ber ued [foe ae Adding these we get ° h (“440+ re + +%) and a similar formula clearly applies to an area divided up into any number of strips all of the same breadth, 198 SCHOOL CALCULUS [CHAP. VI Hence we have the Rule: Divide the required area into any number of strips by ordinates at equal intervals. Add together the lengths of half the first and last ordinates and the whole of all the other ordinates and multiply this sum by the common interval. Or, if there be » ordinates and the common interval between them be h units of length, the approximate area between y, and y,, the first and last ordinates, may be expressed by A=h (21180 yap ys Se +%n-1): If the curve is partly convex to the axis and partly concave, as in the figure, the result will tend to be more accurate than when the curvature is all one way, as the small portions omitted or included to some extent neutralise one another. In the ihe the area by this rule is— 3 (-84+1-62+1-39-+1-02+4-76+- 66+-° 69+ - 40) 3 x 7°38 =p. 214 units of area. By counting the squares, the area is found to be about 2-22 units. 15. Simpson’s Rule. A closer approximation may be obtained by supposing the curve joining any three consecutive points such as A, B, C in Fig. 60, to be a parabola with axis parallel to the axis of y. Thus suppose ABC to be a parabola, and taking the origin at M, the foot of the middle ordinate, assume the equation to the curve to be y=a+bxa+cx? where a, 6, c, are unknown constants. Then putting x equal to —h, o, h, in turn, AL=y aL alata BM =y2=a aT) CN =y3=a+bh 4+ ch2| Then the area ACNL = f- ydx for this parabola =f (a+bx-+cx?) da ba2 case ]h =| aet+—> +3 |, =2(ah *) =2h (a +5) CHAP. VI] PLANE AREAS 199 But from equations (i) yi t+y3=2 (a+ch2), and yo=a hai ty¥3_ 4 Yi tys—2y2 2 2 2 — aoe (a +2 )=2n (yo4% ue “ley h Ss (y1+4y2+yYs). Similarly the area CHON is 5 (ys +4y4-+y5), and so on. Taking then an even number of intervals and therefore an odd number of ordinates, so that we can pair off the in- tervals in this manner we get for the area in the case of six intervals (yr +4y2+y3+y3+4y4t+ys +y5 +4y6+y7) [LS wla = (yi +4y2+2y3 + 4y4+2ys +4y6 +Y7) and the result can clearly be extended as before to any even number of intervals. Hence we have Simpson’s Rule for Areas. Draw an odd number of ordinates at equal intervals; add together the lengths of the first and last ordinates, twice the alternate intermediate ordinates (the third, fifth,....) and four times the remaining ordinates (the second, fourth,....), and multiply the sum by one-third of the common interval. If there be 2n-+1 ordinates, and the distance between each be A units of length the area between y; and ye2,;; may be expressed : h A=ly1 ce depie (3 4-Yoo et Yn 1) 4+ Foye .. ---Y2n)) EXAMPLE. Find the area of a quadrant of a circle by each of the foregoing rules. ! Let OAB be the quadrant, the radius being 1-2”, and let the ordinates be taken at every tenth of an inch. An odd number of ordinates has been taken, so that the 200 SCHOOL CALCULUS [CHAP. VI same measurements will serve for both cases. The lengths of the successive ordinates are, 1-20, 1-20, 1-18, 1-16, 1-13, COBRA REVRSSZweo ms he Paella eile ete ere tsa eet eel steep! SERS SREechEnease i Sy ees errs bagted 1-09, 1-04, -98, -90, -80, -66, -48, 0, and the common in- terval is -1. By the Trapezoidal Rule we get for the area -1 (-60+1-20+1-18+1-16+1-13+1-09+1-04-+ -98 +-90+-80+-66+-48-+0), = 1x 11-22—1-122'80. inches: By Simpson’s Rule we get 4x -+1(1-20+4-80+2-364+4-64+42-26+4-36+2-08 +3-92+1-80+3-20+1-32+1-92-+0) Se -129 sq. inches. .Q)2 The true area is mC 2Y = 1-131 sq. inches so that the second rule gives a very good approximation. The first result is also fairly close, owing to the large number of ordinates taken. EXAMPLES Find the areas between the curves through the points given in the following examples, by the methods of §§ 14, 15 between the first and last ordinates given. Where ordinates are omitted or are not equally spaced they must be interpolated by measurement from a careful drawing. 1. #.0M8,, <1. 0 3 6 “9 Hcl 2s l-o ais yoms... | 3°5 | 2-6 | 1-92) 1-42/1-05| -78 | -577 Use the Trapezoidal Rule. CHAP. VI] PLANE AREAS 201 % xft. ..| 0 | 5 |1-0| 1-5] 2-0 Vite ese een eelO Ow) | 14 Use both rules. ‘ pa Mere RS) Sie nel) un pei Pam a. eC 14 | 16 | 12 8 10 | 24 Use both rules. ; tinsel) Leo Leb 176) 2-0 yft. .. | °825)1-41 |2-42|4-13| 7-08 Use Simpson’s Rule. oi Cie pee 45 19°55) --6 Fe- 7 | 85 | 1-0 Us oe 410 -427 428 -423| -4 | -376/ -362 | -333 | -289 | -247 Use Simpson’s Rule. ‘i ems... | 10 20 40 50 yoms... | 27:36 | 34-47 43-4 46-79 Use Simpson’s Rule. 16. Evaluation of Definite Integrals by Means of Areas. We have seen that the area between the curve y=f(x), the x-axis, and the ordinates at 71, x2, is equal to fe a f(x) dx. It follows conversely that the value of any such definite integral can be obtained by drawing the curve y=/(x), and finding the area below it by Simpson’s rule, or by counting squares. This method is a most useful one in cases where the Integral is difficult to obtain by any ordinary method of Integration. EXAMPLE 1. It is shown in § 23 of this chapter that the are of an ellipse of axes 2a and 2b is a /V 1 —-e*cos* -dd V a2 —b2 where e (the eecentricity) =——— Now, as is there stated, this integral cannot be found by any simple method, 202 SCHOOL CALCULUS [CHAP, VI By plotting however the curve y=V1—e*cos*@ for any particular numerical case, in which ¢ is taken as the indepen- dent variable, the arc is easily determined with quite sufficient accuracy. hus: let a@=s4eeb =o. (This is the curve drawn in Ch. V., § 5.) 16—9 7 2——__ —-__—. Then e 16 ié 4375. Tabulate results as follows : g@ degrees .. a 0 15° | 30° | 465° 60° 75° 90° cos .. 4. | 1 | +9659! -8660| +7071} -5 "2588 | 9 e*cos*h Clix. =. 4375 -4081 -3281/| -2187| -1094| -0293|) 0 Th e2cos%p -.. | 5625 | 5919 | -6719|-7813| -8906| -9707) 1 VI —ecos®p .. 15 | 7694 -8195|-8838| -9437| -9852| 1 ¢ radians . .. | 0 |-2618) -5286] -7854 | 1-0472 | 1-3090 | 1-5708 The limits 0 to 90° for ¢ will give us a quarter of the length of the ellipse. Now plotting the last two lines of the table we have Fig. 61 in which however only the upper part which includes the curve is shown, as indicated by the numbers on the vertical axis. If the area is found by counting squares, the lower part whose area is -7x 1-5708, or 1-:0996 units of area, must be added on. As however we already have an odd number of ordinates at equal distances -2618 apart, we may with advantage use Simpson’s Rule, taking the calculated ordinates direct from the table, and not from the curve. We get ae (:75-+3-0776+1 - 6390 +3 -5352 + 1-8874-+3-9408 +1) _ +2618 SUE SUUIIS) iv This has to be multiplied by a, which equals 4, to give the length of a quarter of the ellipse. Hence the whole perimeter of the ellipse=16x 1-3814 = 22-102 units of length, PLANE AREAS 203 CHAP. VI] | EEE REESE fe are ters ears Pere INL me BAUS IAA NEE EEE CSRERS HEME RRERRRERLESEO et israel tee Ne EERE EEE Son BE S@SERS ANGERS HASSE. 60 -1406 90 0 02637 55 - 1482 -007478 20 -01209 50 -1423 -004186 2 sin4@cos26 dé. -1250 T 10 - 1002 -0008824 Fira. 61. 07264 70 | -0000572 65 -04688 0 0 by intervals of 5° is very quickly plotted from the 7 2 tables of Logarithmic sines and cosines, but 6 is strictly ex- The table of values for this function while @ increases from EXAMPLE 2. To find value of ve pressed in radians, and when calculating the area between the curve and the axis of @ from the figure or by Simpson’s Rule, it must be remembered that the horizontal distances or abscissae must be expressed in radians. 6 (in degrees) sin*6 cos? é (in degrees) | sin*@ cos?26 6 (in degrees) = sin*@ cos? 0 to [CHAP. VI CALCULUS SCHOOL 204 The curve is that shown in Fig. 62, and by “ counting the squares ’’ the area will be found to be -031157, remembering EEBSBBRSROReEEL BERRA SaUMbeS popes beeie (stele iolo bt by beats] SEE SAR SERRORE SAS sek Sas ERRCSABRRBDRS 2a eloas ERRRUSSSEP Taos RRGRERS Aah BESBRUNSe ae eUeiaseunney BSREMS ZAR BRL ARS aR See ERC 2ERDNEEERAREAReE oes BBoPS Bee 3 BHR PAUBES SERBS BEERS NBBEeS; HBL SGRAe sess BARRENS ES BSR Seeeeee SURES SENRORSeeeeo es) ak | | BOSE RBUBCA CREB DBEQEho8 SSSaRSR Fia. 62. 5 Units long, so that 1125 small that the base of the figure is squares represent an area of - 17. By Simpson’s Rule the area is -0312597, while the correct -031257 exactly. Similarly we can evaluate other definite integrals of the form /sin"@ dé, /cos"@ dé, and /sin™@cos"@d@ when m and n are even; see Ch. IV., §§ 22 note and 24, ii. value of this Integral is Tr a : : ye sin2@cost@ d@, as is obvious Note. ye ante dae ade 8) 0 so that KIN 0) and we integrate from 0 to 2 the curve would merely be reversed from right to left. T since sing =cos ( = —dd. od, so that dé T Or thus: CHAP. VI] PLANE AREAS 205 Tv 0 Then / * sintécos29d0=: —/, costsin2d dp J, S = dh 2 sin2pcos*d df | 0 7 = i 2 sin2@cos*@ dé 0 Ch. IV., § 34, Notes. EXAMPLES 1. Find graphically the area of the loop of the curve 4y? (3 + x) = #7(3 — x). 2. Find the value of af 2 /25 — 16cos*p dp, which gives half the : perimeter of an ellipse of major axis 10 and eccentricity sf 3. Find the whole area of the closed curve 10y = tiv 4 — a 4, It is shown in § 19 that the length of the arc of the parabola y? = 4ax . vy ae y? . ° from y=0 to y= y is ws 1+ a dy. Evaluate this graphically when 0 a a=1 and yj = 2, and verify that the result is 2:29 approximately as there found. 5. Find the area of the curve 4y? = 4x3 — af, Find graphically the following defintte integrals : id pale Ce Did uceMees: aa 3 as 6. / Va? —7 dx. 9 fev sin36 dé. ul 2 1 Cn aera 10. ——_ dy, 7. f° V1 = sind ae, ¥. Ala Chale or 35 | a renee ne 2 Ee /it Saare at ll. Vi sin2dcos2d dé. 12. bh * sin®@ cos?é d@; and show that the area of Example 3 may be evaluated by means of this integral by putting x = 4sin?0. us r 13. ye prema dé. 14, ie eaorsa de: 0 206 SCHOOL CALCULUS [CHAP. VI II. LENGTHS OF CURVES 17. The length of any arc of a curve can only be found by Integration. There is one apparent exception, the circle ; but it should be noticed that we define 7 as the ratio circumference diameter | circles. Hence the length of the circumference is 2rr, but this does not really determine this length until the numerical value of 7 is known. Taking some other curve of fixed shape, say the cycloid, we might take as definition length of arc radius of rolling circle whence the length of the arc is or. o has now however to be determined, before the arc in question can be supposed known. It so happens that the quantity 7 appears in many places in mathematics besides in its connection with the circle, such as continued fractions, trigonometrical series, etc., from which its value has been calculated. Such methods are not however available for other curves. 18. Consider any curve y=f(x) and take two points on it , Since it is easily shown to be the same for all =some constant oa, Oo M N Fig. 63. near together, P and Q. If s denote the arc measured from some fixed point A on the curve, the are PQ=8s, PR=6éa, RQ =Ssy. Then the chord PQ and the arc PQ are very nearly equal, and become more nearly so as P and Q approach one another CHAP. VI] LENGTHS OF CURVES 207 indefinitely, until the difference between them ultimately is very small compared with either. Hence in the limit we can write ds? —dx? +-dy? 2 =de2+-(54-az) dx / ‘.ds= Ni +((¢) ae and s =f, Vite Pe -dx between suitable limits. We may also, of course, if convenient, write dst (F dy) ay =a?) +(ay) | and .°. af, /\ +(f) -av. 19. Arc of a Parabola from the vertex to any point P. J 6 Fia@. 64. It is more convenient here to take y as the independent variable. The curve being y2=4axz we have aay dy 2a and of ee 1 +5, dy, where PN =y, =f f° V 4a2+-y?-dy eh ene 4q2+y2 ay Wes oe is eee +y | (Ch. IV., § 14) ee hee YitV 4a2+-y32. - 4a a aloge 2a 208 SCHOOL CALCULUS [CHAP. VI If we take the arc as far as R, the end of the Latus Rectum, we have y;==2a, and we find for the arc the value za 2a + V80? pel 2a Sena (1+ -V2)} =2-2956a. 20. Sometimes a small portion of arc near the vertex of a large parabola is required. This may be found approximately as follows : Fia. 65. Taking the axis of the curve vertical so that the equation becomes x?=4ay (and a is a large quantity), we have, by interchanging x and y in the last article af fits ‘da =i (ee =) dx nearly (by the Binomial Theorem, or by taking the square root to two terms, since further terms involve still higher powers of the small quantity =, and may be neglected for our present purpose), = + 5-5 (i). This shows that to a first approximation when 2 is small the length of the arc is equal to x, the difference between them 3 being the small quantity Saga" se Result (i) may be written x (+555 sea z (1 +2), a form which is sometimes useful, when y is known at the end of the are. EXAMPLE. A wire between two poles 60 yards apart CHAP. VI] LENGTHS OF CURVES 209 droops 2 feet in the middle when the temperature is 6° C., and 3 feet when it is 30° C.; find the coefficient of expansion of the wire, assuming the curve in which-it hangs to be a parabola (which is very nearly the case). If PAP’ (see Fig. 65) represent the wire, we have in the first case AM =30 yds., AN = yd. If the curve be x?=4ay, PN2=4a-AN, or 2 900 =4a- —, ee 3 :3. ae lyds. Hence for the half span Y — ]+— 8 x ( +e) 2 3 8 Be teraeaael <0 (bozant) 8 =30 yds. +5 inch, phew rs 64 Hence the wire is longer than the 60 yard span by 90 °F ey beinch. Similarly in the second case the wire will be found to be 1-6 inches longer than the 60 yards. Hence the wire has expanded 1-6—-711 inches=-889 inches in 60 yards with a difference of 24° C. of temperature. The coefficient of expansion is therefore nee 60x 36x 24 — -0000171 approx. Note. That the assumption as to the shape of the curve is a reasonable one may be seen by reference to Ch. V., § 13, where the degree of approximation of a catenary and parabola near the vertex is explained. 21. For the length of the arc of a catenary see Ch. V., § 12. 14 210 SCHOOL CALCULUS [oHaP. vi 22. Arc of the Curve x = f(t), y= d(t). If we are given a curve in terms of a single parameter as above, we have __| (dx dy snip (a Piet al da\? ml 7 1 /G + (i) # =Vif OP +{¢O}? dt and s= /V{f(®)}?+(¢ (0)? dt between suitable limits, f(t) and ¢’(t) denoting as usual the differential coefficients of f(t), d(t), with regard to ¢. 23. Arc of an Ellipse. The ellipse being given by the equations x=acosd, y=bsind we have dx= —asing dd dy =bcos¢ dd .8=/V ae a*sin* +b2c0s*h dd =/Va? — (a2 —b?)cos* dd =a/V 1 —e2cos2h —— dd since a4—b*=—a?e?, where e is the eccentricity, and to find the length of a quadrant, the proper limits of integration are 7 from 0 to 5° This quantity is known as an Elliptic Integral, and there is no simple method of finding it. A numerical example has however been worked out graphically in § 16, and the method there employed is very useful for this and other cases where the direct integration is difficult or impossible. 24, Arc of a Cycloid. The cycloid being given by the equations x=a (0+sin@), y=a (1—cos@) we have dx =a (1+cos6) dé dy =asin6 dé 6 Hence s=a f mVAGl +cos6@)?+sin26 dé = ee V2-+2cos6 dé 0 0 @ = 1 —_ 2a f COS5 dé 6 6, ° 04 =| dasin5 | = fosin-S CHAP. VI] LENGTHS OF CURVES 211 which gives the are from the vertex to the point given by 44. Putting 6,;=7 for half the arc, and doubling, we get the value 8a for the whole length of the curve, i.e. four times the dia- meter of the rolling circle. EXAMPLES 1. Express the arc of the parabola y? = 4ax from x = 0 to x = 2] in terms of v1. 2. A stone is thrown with velocity 80 f.p.s. at an angle of 60°. Find the whole length of its path taking it to be a true parabola. ; prove ‘ at at? 2 2 — j _— ns SS 3. The circle z?+ y?=ay may be written x ite?’ ie that the arc may be written fi a dt. Show also by drawing the curve that the line y = x cuts off a quarter of the circumference, and hence that we get this length by integrating from t=0 to #=1. Deduce that the whole cir- ce of the circle = za. 4. Shew the length of the loop of the curve 2y = 2V1 — 4x is given by 4 we is vi he 72) 101— 188 a 18z) 5 CHAPTER VII VOLUMES AND SURFACES I. VOLUMES 1, Let the figure represent a solid, whose volume is required Fia. 66. between the parallel planes FG, HK. Let Ox, any convenient line, perpendicular to these planes, and cutting them at L and M, be taken for axis. Take a thin slice PP’Q’Q parallel to the boundary planes, and cutting the axis at NN’. Let its area be A, and its thickness 5x : then its volume lies between Aéxz and (A+6A) 8&2, and as 6% diminishes indefinitely, this ultimately becomes A dx, and the whole volume required is the sum of all these indefinitely thin slices between FG and HK, i.e. the volume is Ltsz-0 5a A $x, where OL =21, OM=x2. If now A is | given in terms of x, so that its value can be found at any point along Ox, say by the equation A=d(x) the volume becomes Xo x, . fs Adz = fe d(x) dx, and can therefore be found, if the 41 a | integration can be performed. 212 CHAP. VIT] VOLUMES 213 The principal case we shall have to consider will be that of Solids of Revolution, that is, solids generated by revolving a curve about some line in its plane, and that line is called the axis of the solid. Then taking this axis for x-axis, the section by any plane perpendicular to it becomes a circle of radius y, whose areais7ry?. If then the equation to the gener- ating curve be y=/(x), the volume is if wy2dx or wf f(a)}? dx between suitable limits. 2. Volume of Right Circular Cone. The cone is formed Fie. 67. by revolving the line y=mz about the x-axis, where m =tan a, 2a being the vertical angle of the cone. Hence the volume of’ a slice PP’Q’Q at distance x from the centre, and of thickness dx, is given by dV = cy? dx=rx"tan2a-dx a] .. V=nrtan2 af x2 dx 3 =Ttan2a - Zs where 2, is the height, OM. ; J 1 This result may be written =-7 (a,tana)?-a = Y1" 1 3 and my 2 is the area of the base CD. 1 Hence the volume is 3 (area of base) x (height). Frustum of a Cone. If ABCD (Fig. 67) be a frustum of a cone, its volume can be expressed in terms of its height h, and the areas of the cross sections, A;, As, of its ends, as follows: Volume of the frustum=volume of whole cone— volume of the part OAB 214 SCHOOL CALCULUS [CHAP. VII ] a mtan2a (%23 —%43) where OM =a, OL=x1 1 a4 amtan2a (%2—2%1) (%o2-+%2%1 +2?) iE = 9(%2—*1) (TY? +-myoy1 +7Y17). But TYyo*—=Ao, TYy17 =A}, and TY2Y1 = V AoA while z2—27,=LM =h. *, volume of the frustum = (Ag+WVA241+4}). 3. Volume of a Sphere. Let the radius of the sphere be r. Fie. 68. Then the sphere is generated by revolving the circle 2? +-y?=r2 about the axis of w. .°. y2=r2—2? and the volume of a slice, aV =n (r*—a22) du... Van (r2—x) dx, on integrating for 34° 3 half the sphere = [ 22 LL FADS. 0 3 47r3 Hence the whole volume of the sphere=—.—. For a cap cut off by a plane at distance h from the centre the limits of integration are h and r. Bes Be .. Volume of the cap =7 | 2x > \ h CHAP. VII| VOLUMES 215 7 oa (2r3 — 3r2h +h3) (r—h) (2r2—rh —h2) wi wy (r—h)? (2r+Ah). 4. Volume of an Ellipsoid of Revolution. There are two ellipsoids or spheroids which can be formed by revolving the ee ellipse “atp =1 about its axes. These are called respectively the Prolate and Oblate Spheroids according as the curve is revolved about the major or the minor axis. Volume of a Prolate Spheroid. As usual the volume of a slice PP’Q’Q perpendicular to the z-axis is given by AV =cry2da ee OP oo 2) d =m — (a°—2*) da. Therefore for half the spheroid, UE (ieee ; (a2 —2x?) dx ute aa”. Gime ape) 208% 2 ae a2 Sit Tae Re ees and the whole volume is arab? 216 SCHOOL CALCULUS [CHAP, VII EXAMPLES 1. Show that the volume of a cap of a prolate spheroid from x=Ah to ub*(a — h)? (2a+ h) [It will be noticed that this result x =a is equal to q 3a? 2 is us times the corresponding volume of a sphere of radius a, given in § 3, a ; ; Die : ; just as areas of an ellipse were — times the corresponding areas of a circle.} a 2. Find the whole volume of an oblate spheroid. 3. Find the volume of a cap of an oblate spheroid from y= k to y= b. 2 5. Hyperboloids of Revolution. Let the hyperbola ia be revolved about the z-axis. There is of course no part of the solid formed between x=a, A 4 IN ' : i] } ws 2 : Fria. 70. and x= —a. We shall therefore find the volume from a to x1 (v41> a). : sah This volume is wy*dx a2 ab2 [xs “1 ah j ms Bh daly eae 32 9A2 3 seralts a x |, 32 (71° —3a2x14 +203) 2 = (ay —a@) (%12-+a%1—2a7) arb* Sagt (v1 —a)” (x4+2a). CHAP. VII] VOLUMES 217 Fre. 71. EXAMPLE If a hyperbola be revolved about the conjugate axis, show that the volume as shown in the figure from C to M, where CM is yj is 2 “pr tw) 6. The two hyperboloids mentioned in the last section are called respectively hyperboloids of two sheets and one sheet, for as we see in the figures the first one consists of two separate parts with vertices at A, and A’, and no part of the figure lies between them: while the second is one continuous figure through the centre. The transverse axis is the axis of symmetry of the first, and the conjugate axis of the second, and both figures stretch to infinity along their axes. 7. There is one other hyperboloid we shall need, namely that formed by revolving a rectangular hyperbola about an asymptote. Fig. 72 shows only the part above the z-axis, there being of course a similar part below. The equation to the curve is xy=c?. 2 U . = and the volume as shown from yj to y2 is ye ” murdy | i 218 SCHOOL CALCULUS [CHAP. VII act This may be written —— (y2—y1) Y1y2 2 Cc =17X14X2 (Y2—Y1), since 4; =, etc. 1 EXAMPLES 1. Find the volume of a cone 2 feet high, whose base is of diameter 10 inches. 2. A beer mug is in the form of a frustum of a cone 6” deep, 3-6” diameter at the top and 4-2” at the bottom: find its contents in pints when full to a depth of 5 inches, if 1 cubic foot = 50 pints. 3. Show that the volume of a cap of a sphere of radius r inches can be ; nel H written either (i) . (387 — #) or (ii) 3 (A+ rH), where H inches is the height and A sq. inches the area of the plane face of the cap. 4. If a cone with the same plane base be placed in the above cap and its vertex is at the highest point of the cap, prove that the volume between the wH?r cap and the cone is aN cubic inches. 5. A spherical basin contains water to a depth of 4”, its surface being a circle 12” in diameter. If water weighs 62-5 lbs. per cubic foot, find the weight of water in the basin. 6. Show that the volume of a paraboloid of revolution of any given depth is half the volume of the circumscribing cyclinder. | CHAP. VIT] VOLUMES ) 219 7. Show that the ‘aca a a cap of an oblate spheroid of semi-axes a a? H and b may be written (i) ~ fae * (3b — Hf) or (ii) 3p2 (A+ 7bH), where H and A are as in Example 3. Find the volumes formed by the revolution of the curves in Examples 8-14 between the planes indicated, and in each case make a rough freehand drawing of the volume found. 8. y? = 4-18z about the x axis from z= 1 to x= 2. 9. The same curve about the y axis from y = 3 to y = 3. 10. xy? = 150 about the axis of x from x= 3 to x= 6. 11. The whole volume generated by revolving the curve 9y? = x? (9 — 2?) about the x axis. 12. (1+ 2?) y= 12-5 about the x axis from x= —3 to x= 3. (To integ- rate put x = tané.) 13. The same curve about the y axis from y = -7 to y =12°5. 14. The volume obtained by revolving the same curve and the ordinates at x= — 2 and x = 2 as one whole figure about the axis of y. 15. An iron pedestal is in the form of the solid made by revolving the curve (1+ 2”) y? = 40 about the x axis from += 0 to x= 12, and a hemi- spherical hollow of diameter 34 is left in the base. Find the weight of the whole, if the iron weighs 450 lbs. per cubic foot, the units throughout being inches. 16. Find the difference between the volume of the solid formed by the revolution of 10-8y? = x3 (6 — x) about the a axis, and a sphere of the same diameter, i.e. 6. Find the volumes formed by revolving the loops of the following curves about the axis of 2: 17. 2y = 2/6 — zx. 18. 4y = 42/6 — ax. (Put 6 — x = 622.) 19. y= Vx (6 — 2). 2aV'6 — x V6+ 2° 21. A cylindrical bottle, 34” internal diameter, has a projection inside at the bottom in the form of a paraboloid of revolution extending to the sides and 3” high: find the amount of liquid inside when filled to a depth of 6 inches ; when filled to a depth of 2 inches. 22. A lighthouse is of the shape formed by revolving a rectangular hyper- bola about an asymptote (as in Fig. 72). If its height be 110 feet, and the diameters top and bottom are 13 and 38 feet respectively, find its volume in cubic feet. 23. Show that the volume formed by the revolution of a rectangular hyperbola about an asymptote is equal to that of a cyclinder of the same height whose diameter is the geometric mean between the greatest and least diameters of the hyperboloid. 24, It can be shown that if a liquid be revolving steadily in a vessel formed by a surface of revolution, the surface assumes the form of a paraboloid of 20. y= 2 revolution (neglecting friction) whose latus rectum is aye where w is the @ angular velocity. Find at how many revolutions per minute the liquid 220 SCHOOL CALCULUS [CHAP. VIT must revolve in a cylindrical vessel of 18” diameter and full to within an inch of the top, that the liquid may just rise to the top without spilling. 25. As in the last example, but the vessel is now a cone of depth 15’, diameter at the top 1 foot, and half full (i.e. half the volume, not half the depth). 8. Guldin’s Theorem on Volumes. The volume of the solid generated by revolving an area about any line in its own plane which does not cut it, as axis, is equal to the product of that area, and the length of the path traced out by the centre of gravity of the area. Bia. 73. To prove this, let any small part dA of the area be at distance y from the axis. Then if y be the distance of the centre of gravity of the area from the axis, and A the whole area, y= He (Chap. IX., § 2). But the element dA traces out a ring whose radius is y, and its length is therefore 27y : thus dA generates the element of volume 27ryd.A, as shown in the figure. .°. the whole volume generated = 2r/ydA =27y-A, from the above value of y. But A is the whole area, and 27ry the length of the path traced out by its centre of gravity : hence the theorem is proved. EXAMPLE. Find the volume of a curtain ring of external diameter 5” and whose cross section is a circle of 3” diameter. The ring is generated by revolving a circle of radius 3” about an axis in its plane at a distance from its centre of 2}”. CHAP. VIT] VOLUMES 921 Fia. 74. The centre of the circle is its c.g., and its path is shown by the dotted circle. The area of the circle is 7:(?)? and the length of the path of its ¢.g. is 27-24. The required volume is therefore Boe hLeairs tha —~_— =5- b ORG 5-898 a ic inches EXAMPLES 1. Verify by Guldin’s Theorem that the volume of a cone formed by re- volving a right-angled triangle about a side containing the right angle is one third of the circumscribing cylinder of the same height. ) 2. A square, whose side is 14”, is revolved about a line in its plane parallel to a diagonal of the square and 2” from that diagonal: find the volume of the ring generated. 3. An interior of a motor tyre is formed by revolving a circle of 3-54” diameter about an axis distant 14-18” from its centre: find its volume. If it be filled with a substance weighing 125 Ibs. per cubic foot, what weight is required to fill it ? 4, Find the centroid of a quadrant of a circle of radius r by revolving it about one of the bounding radii (for its area and the volume of the hemi- sphere generated are known). Deduce the position of the centroid of a semi-circular area. In Examples 5 and 6 draw figures to illustrate the solids formed. 5. Find the area of one loop of the curve 5y = 2/5 — «®. Its centroid is at a distance we from the axis of y: hence find the volume formed by revolving the loop about that axis. 6. Find the volume formed by the revolution of the loop of the curve 4y? = a? (9 — x) about the axis of y. Its centroid is at a distance 5+ from the origin. 222 SCHOOL CALCULUS [CHAP. VII 9. Determination of Volumes by Simpson’s Rule. Suppose any solid to be cut by a series of planes perpendicular to the axis of x say, and let the intervals between these planes be small and each equal to h. If then we are given by a table the area A of each cross section made by these planes, or if those areas are given by an equation of the form A=¢d(z) where x is the distance of any plane from the origin, we can draw a graph exhibiting the relation between A and the distances of the planes from the origin graphically, in which the ordinates will represent the areas of the cross sections ; and therefore the area between this curve, the axis of x and the end ordinates will represent the volume required, for this area represents }Ah or }Aézxz, and therefore forms an approximation to the volume required. We may therefore apply Simpson’s Rule (as in Chap. VI., § 15) to find this area, and so find the required volume. If the cross sections are given by a table, and the relation between A and x cannot be found, some such method is the only possible one; and even if they are given by a relation such as A=d(x), so that the volume could be expressed as /Adz, this integral is sometimes difficult or impossible to evaluate, and we again fall back upon the above method, or some equivalent, which is practically that of Chap. VL., § 16. EXAMPLES In Examples 1-5 find the volumes of the solids whose areas A are given at distances x or y along an axis in the following tables, interpolating when necessary by drawing a graph : i; xin. 1 1-5 2 2:5 3 3°5 4 A sq. in. | 11-4] 11-3 | 11-8] 13 | 15-1 | 17-8 | 21-9 2. | ecm. sa ap 3-0 31 3-2 3°3 3°4 3°5 Asq.cm. .. | 39-38 | 38-99 | 38-31 | 37-01 | 35-48 | 33-8 eon hy | 3-6 3°7 38 3-9 4-0 A sq. Sed pears 31-63 31-29 31-31 | 31-82 CHAP. VII] AREAS OF SURFACES 223 3. | wyds. .. 0 3 *45 | -75 °Oe | 1-00 1) 12 A sq. yds. 0 10-4 | 14-3 | 18-1 | 17-5 | 13°7 0 4, yom. .. 0 “5 1-5 2 3 3-5 4 Asq.dm. | 0 | 14-7 | 22-7 | 22-5 | 17-2 | 14-6 | 14:7 5. y in. Patan ie5. wl 7s) 12 Avagattr ©. 1 -34 | -56 | 1-28 In Examples 6-8 the diameter (D) or the radius (R) of the cross sections of certain solids of revolution at distances x along an axis are given. Cal- culate their volumes. Beeeeign ee | 4251/0604) 6:5 | 6-0'| 6-5 | 7 | 7-5 Dfeet .. | 2-02 | 2-58 | 2-96 | 3-06 | 2-92 | 2-2 | 0 7 7 2 cm. RleearesiG | 30: |) 40 | 80°) 60 eegaes nt |e 10, ) 16 | 23°9.|25-1 | 23-7| 10 8. OF ei eerie cg er ery ea aa y cm. .. | 25 | 2-8 | 8-4 | 3-7 | 4-3 Dom. .. | -212 | -148 | -071 | -046 | -026 Find, by Simpson’s Rule the two following volumes : 9. That formed by the revolution of the loop of the curve (1+ 2) y?= «2 (1 —2) about the x axis. 10. That formed by revolving the loop of the curve y (y — 5)? = 2? about the y axis. II. AREAS OF SURFACES. 10. Surface of a Cone. The area is the sum of all the little strips such as PP’Q’Q round the cone, cut off by two parallel planes at distance dx apart. If OP=s, and PP’ =ds, and PN =y, this strip is ultimately 224 SCHOOL CALCULUS [CHAP. VII a band of length 277y and breadth ds, and its area is 27ryds. y Hence the total area of the cone is anf ; yds, where CM =y 0 and we have now only to express y as a function of s, and Fie. 75. perform the integration, with a suitable change in the limits. In this case y=s-sina and when y=y1, s=OC=l say, and I the area is 2msinaf sds=Tl?sina which may be written, 0) . a: lgina:l=rrl. 11. Area of any Surface of Revolution. Consider the sur- face generated by revolving any polygon about an axis in its plane. Let AB, BC, CD, DE, ... . be consecutive sides of / re Fia. 76. the polygon. Then any one of these, say CD, generates a small portion of a conical surface CDD’C’, whose area as in CHAP. VIT| AREAS OF SURFACES 225 the last article is 27yds., where CD=ds, and y=CN or DN’, which are equal in length when the sides CD, etc., are inde- finitely short. In that case also the polygon becomes a con- tinuous curve, and the area of the surface generated by the revolution of that curve is ultimately 27/yds. between suit- able limits for s. If y is given in terms of s, this can be integrated immediately. If not we may (as in Chap. VI., § 18) write ds in either of the Wey is ae forms a/ 1 +(ZY dx, or 4 1 +(F) dy, and use whichever of these gives the simpler quantity to integrate ; so that a surface of revolution about the axis of x is given by S =2nfy Jf oe ey dx or of y ve +(e) dy. Similarly if the curve revolve about the y axis, the surface is 27r/xds, which may be put in either of the forms mf x a ‘ | Gal dx or oof a J +) dy. 12. Surface of a Sphere. Let the radius of the sphere be r, so that it is generated by revolving the circle 2-+-y?=r? about a diameter. Then the surface of the cap cut off by a plane at distance h from the centre is 27/yds., between suitable limits. Now differentiating «2+y2=r? we have adx +-ydy =0 CB ee dx y ‘, yds=rdx, 15 226 SCHOOL CALCULUS [cHaP. vit and 27/yds. = 2nr " de (the limits now being easily assigned) h =2rr (r—h). To find the area of the surface of a hemisphere, put h=0 .. Surface of hemisphere =2rr2 and the whole surface of the sphere =4z7r?. EXAMPLES 1. Show that the area of the zone of a sphere cut off by two parallel planes at distance H apart is equal to 27rH. 2. If a cylinder be drawn of radius r to touch the sphere, show that any two parallel planes perpendicular to its axis cut off equal areas from the sphere and the cylinder. 13. Surface of a Paraboloid of Revolution. The equation -to the generating parabola is y?=4az. . ydy =2adz, Hie. 7 7 and the surface from A to N =2n/yds we mV 402 -+y? res =2nf no ae ydy. Put y2=z. .°. 2ydy=dz CHAP. vit] AREAS OF SURFACES 297 a Be 212 -dz,21 bei | == =3,/, V/ 4a2-++2z-dz,21 being the value of z when y=y Be (i277 98) ob ol 2 2 | ~ Qa EG at) ieee acral 3 LR er 2) 8q3\ 35 (4@ U1) 8a, 1, 14. Surface of a Prolate Spheroid. Let the ellipse — = a be revolved about its major axis to produce the EA Then the surface required =27:/yds. B Fia. 78. Now y = Ve ae. Als fee + UY 0 =0, dy _ be da ary le FAe Ss b472 3 ds= Ale ip dx (b422 + ay? +aty? 5 2 /b4a:2 +-a2b2 (a2 —?) sm arty? n a/ ab? (a2 —x?) Oe 4 fy_2 = h2 Russe a (a? b*) a2 ap a2 (a — 7?) 228 SCHOOL CALCULUS [oHaP. vn =" V/ a4 —a2e2x2 dx, since a2—b2=a?e? — oP — ea? a? —erx? dx. .. Area of surface of a cap from 4 to a =n V/ a2 —e2x2-da. Putex=asingd. .'. edx=acosd dd a Xi , __2Qarb acosd ,, 2%abs 5 go LOE ere ee tag 3 _ mab To +4sinzg |=" | sin as seeks = || a2 > a2 — e242 =7O sine tev 1—e%—sin—! a oa Putting 7;=0, the area of half the spheroid =700 (om +V/1 12). Notice if e=0, the spheroid becomes a sphere ; in that case gine Lite—0 =1, and V1—e2=1 ‘. the surface of the half spheroid becomes aab (1-+1) =2rab—2rr2 (if a and 6 which are now equal, are each written 7)=the area of a hemisphere. 15. Guldin’s Theorem on Surfaces. The surface of the figure generated by revolving a curve about any line in its own plane which does not cut it, as axis, 1s equal to the length of the curve multiplied by the length of the path traced out by the centroid of the curve (not the centroid of the area it contains, if closed). To prove this let ds be any small element of the curve (whose whole length is S), at distance y from the axis, which is taken for axis of x. Then as in $11, the whole surface generated =27/yds. But if y be the distance from the axis of the centroid of the curve (noé of its area) gal (Chap. IX., § 2). CHAP. VIT] AREAS OF SURFACES 229 ', Qr/yds =2ryS: and 27y is the length of the path traced out by the centroid of the curve. Hence the theorem is proved. ExampuLeE. Find the whole surface of the curtain ring in the example of § 8. As before the length of the path traced out by the centroid of the curve (which in this case is the same as the centroid of the area) is 277 x 24”; and the length of the tracing curve My is the circumference of the revolving circle, namely a 3 17 Hence the required surface = * = oh -45 sq. in. nearly. EXAMPLES 1. Find by integration the curved surface of a frustum of a cone whose height is # inches, and the radii of its ends rg and rj inches, respectively, of which rg is the greater. 2. Find the surface of a reflector in the shape of a paraboloid whose depth is 12 inches and diameter of the front 32 inches. 3. One third of an arc of a circle, whose radius is 2”, revolves about the tangent at its middle point. Find the surface of the reel so formed. 4, Find the area in the same way if a parabolic arc cut off by the latus rectum revolve about the tangent at its vertex, the length of the latus rectum being 6 inches, CHAPTER VIII MAXIMA AND MINIMA 1. We have already seen that if y is equal to any function of x, i.e. if y = f(x), y is as likely to decrease while x increases steadily as it is to increase, and that as a rule, while x increases from the smallest value of which f(%) admits up to the largest value it can assume, y is sometimes increasing and sometimes decreasing. 2. The problem now before us is to examine those values of y, or of f(x), with a view to determining the maximum and minimum values assumed by y as a increases steadily from its smallest to its largest value. In some cases this will mean finding actually the largest or smallest values which the function under consideration ever assumes. If for instance we draw the graph of the function 3x2-+ 6x+ 8, i.e. the curve y= 322+ 62+ 8, we see from the figure that y, i.e. 3x2-+ 6x+ 8 has absolutely its least value when «= —1, and that this minimum value is 5. aaN PEPER PEE Ca ae WEP LEE CE TE Say eS PER REE EERE EEE EEE PEEL oer A EE tt ett tt A FEES EEEEE EEE ! iN 4 ep SHEE +R A HE PETAR EEE TE LT Tee - PTET EE CELT e226 a FECEEE ERR EEE EEE EAE EEE BEERS EEE EEE fe P| et STS | SNS | apt BP4RERRS PTL ONDE EEE ee Peep RTP ry TT EAA Te eee REREARRERERRRohs. =a UCU Pee Tee PED Eee EE CS Sa Se SEECCOPSCC Eee MES bias DOG RROMEREECOwneeode 2 ee CHAP. VIIT] MAXIMA AND MINIMA 231 3. But the problem of finding the maxima and minima of any function of a variable (and we shall only consider functions of one variable) is a much wider one, as will be seen from the following definition, and the example which follows it. Definition. If while x increases gradually and continuously through a particular value 21, ¢(x) increases to the value ¢(%) and then decreases (both increase and decrease being through a finite interval) that value ¢(a1) is said to be a maximum value of ¢(x): and if under the same conditions ¢(a) decreases to the value ¢(%1) and then increases (both decrease and increase being through a finite interval) that value (21) is said to be @ minimum value of ¢(z). It must be carefully noticed that if ¢(x1) is to be amaximum value of ¢(x), must increase through the value 21, not merely up to the value x,; and that ¢(x) must first increase up to the value $(a%1) and afterwards decrease, all the changes of value considered being through finite (i.e. possibly small, but not infinitely small) intervals ; and similarly for a minimum. 4. Example. Consider the function of x: 32° — 4x + 5. The curve y = 32° — 4x%+ 5 is that drawn with a continuous line in Fig. 81 (p. 232), and the table of values is as follows : | —2 ‘67 3°22 1 4 15 9:12 2 21 —15 88 —1 6 — *67 6°78 6°62 — 33 6:22 33 | 5 378 | 3-37 et 5 (The dotted curve is the graph of its differential coefficient, that is to say is its first derived curve, to which we shall need to refer later, and the straight line is the second derived curve.) Now we see that as x increases from say — 1 to — g Y or 3a% — 4% -+ 5 is increasing, but that when x further increases ane 2 y decreases. Therefore by our definition when «= — 3 we have a@ maximum value of y or of 3a — 4%-+ 5, and its value is 6°78. Now it is clear from the figure that this is not at all the largest value that 3”° — 4%-+ 5 can have, as it is always pales 3 greater than this whenever x is greater than, for instance, 3° but this is the largest value of the function in the neighbour- hood of that value, which is practically in a few words what our definition means, | [CHAP. VIII SCHOOL CALCULUS 232 AIAG Shee AER RAREAB URES AS GORA PUNE RARER Bane aang BRRER oN SRUREERS RS CREMP RRRSE ERAS VEAL SHKARKO lA BABAUMS SS CRE RREERERSREE SARTRE ASR aes BESS GRREAREE RES ARBRE SONA ADE DCRR ES BORD R CARR Aes UPSVSSRRRR RES SURED BERRAREEA REPAY BSAA CAO wEeRe “82% 2 VO00 S82 s 2288 PSEA RON eRe eon Cee ees PSEC eC PCs CLCLE MSc TLL Pid ee, else ier tcl Sho. Re SSo hl ee ORD. NSE RSE ERGER SAS. BOSE cane ERUGR SURER EER“ RRR R Se AR BPRERNEDERS WRU eee ete alee eee ad ee ee ee ere | T) SOSaa nse Oe eek ha Jo EE SSeh eis eeR es RAED ReS Sk RARER BER ESENEERee ote eee Teh ses a i Sees epee sett ot NEE ela [Te haul ae te eh lectin RBESS Ee SSS URES EBLE RRO REA REED MBP CERO AT Eee SE Nee. Sana EEE Een Deen Sanaa nhsee | a | | a a g a | a | | a | 7k uo a | IA | V Wi a a aa a ay a! a | a a | | asf | A_| Ht a a oa | 5 z a Bb if a aie LUGSE WEARS ROOCROE CHAM SRe a contaeda beet sleet Hath BARRE AMBRE EER SCR ISN ot SERRE SSF SS SSSR SSE RS8 Ss ES Ss Bees) bee Seek Sees eee SERMS SRR E Ee Pees eee eee eee Nee OSG Rd SRS PASSIONS KAS RR OE ZS SBR SSO es RSENS 2S Ses PRPS SER KNEE eS ROOT RAs Seer mari irr hi beet Ke etre h hr minions det ar ESSe2 CBA SR RR RAEE LORE TUE NAMES SE SERRE PRR Eee EEE PEN Eee pire eee e le wets eed ats Dad er el cet per olerals eae Teepe | ater DSS 0 SSSaSh Bese BES ee RA RR RAR “SERRE See Eee ee ee ee Pee eter ott tei etek. i eee oe bale lapel persia i fee eS ee eee eee ee en see aise tee lesa | teen et eee tet Ls aero relat iy lola iat ata| fe bal een ee edie | cet ia ne BAGGP 2 ho RO CRE RES 2Us Oe e eee CEs ee ph) tape est lt bed Pel oleleeh Perch Tet (ores Peter el el ere tate poiedl ee et te ee ee ere ie else ee tetoere ey 2S tek ey Eta Ser ee Eel eds dete et ee etsy: be ae pone eed el etal ait: | BUSS 2S URANO RoR SE eee Rd PRESS Eaeee SARS eee ae ees SeMGSRRons bees eee Sees eae Fie. 81. 30° — 4a + 5 Lae 3 3 ¥Y OF Similarly if x increases from say 2 is decreasing: but if # continue to increase beyond 3 ins , 343 — 4a4+ 5 Z 3 Therefore by definition when x creases. It would seem natural to suppose that the maximum and minimum values of 32° — 4x%-+ 5 were respectively +-oco and —oo (since those are the values of the function when z@ is +oo and —oo), and these are respectively larger and smaller has a minimum value, viz. 3°22; and this as before is merely than any other values assumed by the function. the smallest value of the function in that neighbourhood, and by no means the smallest value it can ever assume. —oo are not maximum and But the values -+-oco and CHAP, VIII] MAXIMA AND MINIMA 233 minimum values at all of 372— 4¢-+ 5 in the sense we are considering: for the essence of the definition is that during some finite increase of x the function itself should first in- crease or decrease through a finite amount and then decrease or increase through another finite amount, so that the maximum or minimum values should be the largest or smallest of the values of the function in that neighbourhood on each side of the maximum or minimum itself. Jf we consider the. value of 323 — 4x -+ 5 when z is oo, 7.e. when & increases indefinitely, we can only say that 323— 4x%-+ 5 continues to increase indefinitely also, and when x becomes oo we cannot find any value of 323 — 4a+ 5 greater than all other values in that neighbourhood on each side of the one we wish to select : similarly when a decreases indefinitely or is equal to —o, we can only say that 323 — 4x + 5 decreases indefinitely also ; in neither case do the infinite values satisfy our definition of a maximum or minimum. 5. Maxima and Minima Graphically. ‘The following figure, which we will suppose to be the graph of y= ¢(x), where (a) is some continuous function of one variable 2, will further explain the nature of the problem of finding maxima and minima. There are many other possible cases besides res ieier \eleie betes | TE Pret Pee LE LEE BEC EEEE EEC a= io SE REPLIES Beene et tp tt mip LH yf rep sye lst ALEC A iy Pie ela pe eee eis ele lee PEL Pele E sep bene s beat mer ereress el ehels fell. ie fal let ale eee a eae ee ogg hal eb SEE de @u ) ROR Rae AS ARSR SRV NORM Oaee _ (LUE SER SSRN GERRI REE ELBA 2 Sy Be BEEPS CURE EEE ARE Ao NE EEE EEE EEE ENE HLA HA tat acl HOUSsRRRE SANS Re DaR lee. se 24a AES SBE? ee EEE eH Seepeeer er ae rene: Taceereees (eat Teeueee LOS 1211 ZO} 0 W701. Has, U ete BLE tf tt B&VVaaaees ttt AE tt Pt ede | EE nieve ttoleqalsiedei ich bi peb fa D4 MAS ee ejeleyyienis sao rape Ret Ate y | 1 toe ee ba Ef (34 JURE SORE SO Ee IRS Rees eeeee ERIN ReE Sepa rere te ropey 16) fe) Sint e{ 2) sable pelt a eee bg rere Sheree Neiet toe lenis) leteiolie ete te) foe ead EEE HEE HE HE Hg FEE BEER EEE REECE EERE EERE EEE EEE EE EEE ee ore aeeiicecnie secs piece tLe 3 ie es Fie. 82. 234 SCHOOL CALCULUS [CHAP. VIII those illustrated in the figure; but they are mostly of rare occurrence and will not enter into the graphs and problems we shall consider. The ordinates at K and P are asymptotes to the curve. Now as 2 increases from —12 through the value 0 up to 70, that is to say as the ordinate y travels from left to right, y is continually altering in value, and it is obvious that at A, C, and R we have maximum values of y, that is, values of y larger than any in the immediate neighbourhood of those values: and similarly at 6, Ff, and LZ we have minimum values of y. The following points should now be noticed : (1) We are only considering the maxima and minima of continuous functions, that is, functions whose graphs have no gaps or sudden changes of direction; and it is only in connection with such functions that the definition of § 3 need at present be considered. (2) The same function may have several maxima values and several minima values, and as in § 4 there may be many other values of the function both larger and smaller than any of these maxima and minima. (3) Between any two equal values of the function, i.e. of y, there must be at least one maximum or minimum value, provided y does not become infinite and change sign between those values, as at P and Q; for if as y passes through a particular value, say 50, it is increasing or decreasing, it must either decrease or increase respectively to again arrive at the value 50, and so pass through either a maximum value or a minimum. (4) Similar reasoning shows that at least one maximum must lie between two minima, and one minimum between two maxima: hence we see in the figure that at K we have another maximum of ¢(x) between the minima at F and L, as is also clear from the fact that the ordinate at K is the largest in its neighbourhood: but at P or Q we have neither a maximum nor minimum, as the function increases indefinitely up to its value at P, and again continues to increase from its value at Q up to its value at R; while in addition the ordinate PQ lies between a minimum (at ZL) and a maximum (at &), not between two maxima or two minima. (5) The points on a curve at which maxima or minima values of the ordinate occur are often called Turning Points, and the values of the ordinate or dependent variable at those CHAP. VIIT] MAXIMA AND MINIMA 235 points are called Critical Values of the variable. The signi- ficance of the terms is obvious. 6. Maxima and Minima by Differentiation. Although in some cases it may be easy to find very approximately the maxima and minima values of a function, and the values of the variable which correspond to them, by drawing the graph of the function, yet this method is at best not exact, and it is often tedious, though a rough graph is generally useful. Now it is obvious from Fig. 82 that at such points as A, B, C, F, and K where maxima and minima occur, the tangents to the curve are parallel or perpendicular to the axis of x. If therefore the graph be that of the curve y= ¢(z), = 0 or oy 00, le. d'(v) = 0 or oo. Hence we see that the values of x at the points A, B, C, etc., may be found by solving the equations $’(x) = 0 or $’(x) = oo. (In practice it is hardly ever necessary to consider the equation ¢’(%) =°00, i.e. ae = 0, as the corresponding values of y or of ¢(%) are, as at K, merely infinite ordinates, and such maxima are not in general of any useful meaning in the solution of problems in which maxima or minima are being sought: we shall therefore neglect this case, unless in any particular question it is clearly necessary to consider it, a matter which is easily decided.) But the solution of the equations ¢’(~%) = 0 and ¢’(%) = co is not of itself sufficient to determine whether we have a Maximum or minimum, or whether indeed we have either. For consider the points D and # in the figure: at D, gy coo and at #, zu 0, though it is clear that at these points there are pothes maxima nor minima. There are three methods of deciding what results are given by the solutions of oY = 0 and wo =e (1) A rough graph is usually a fairly quick and satisfactory guide. (2) Consider the function ¢’(%) itself: at such points as A, C, K, R, where maxima occur, oe (i.e. the tangent of the angle made with the axis of # by the tangent to the curve) is diminishing from a positive value through 0 to a negative 236 SCHOOL CALCULUS [CHAP. VIII value: at such points as B, F, and L, where minima occur, ze is increasing from a negative value through 0 to a positive value: while at such points as D and #, eH either increases up to 0 and then decreases again, or decreases down to 0 and then increases again. (These results may be easily verified by drawing a rough graph like parts of the given figure, and dy da y moves along from left to right: a specimen is given in considering the values of =~ as x increases, i.e. as the ordinate - Fig. 81, where the dotted curve is the graph of ey for the continuous curve.) Hence we see that if 2 = a is one solution of the equation ’(x) = 0, so that we expect a maximum or minimum value (v1) of the function d(x): if ¢’(a) is positive when @& is a little less than a, and negative when « is a little greater than a1, then ¢(#1) is a maximum: if under the same conditions d’(x) is first negative and then positive, $(x%1) is a minimum ;: but if ¢’(x) does not change sign when « is first a little less than x; and then a little greater, then (x1) is neither a maximum nor a minimum, and we have a point such as D or £. (Notice that it is better to avoid the notation 2 — h and a, + h for the values to be assigned to x in testing the signs of ¢’(x), because we have always considered h to be a very small quantity ultimately diminished indefinitely, whereas in questions of maxima and minima, as before stated, all changes considered both of « and ¢(x) are to be finite, how- ever small they may be taken.) (3) Just as we can use dy or ¢’(%) to determine whether y dx or $(%) is increasing or decreasing, i.e. to determine the slope 3 2 of the curve y = ¢(x), So we can use day or ¢’’(x) to determine dx whether 4 or $’(x) is increasing or decreasing. If therefore, as before, x = 2; is one solution of the equation ’(x)= 0; then if ¢’’(x) is negative, $’(x) is diminishing when 2 = #1, and therefore is as explained above $(%1) is a maximum ; and if (x1) is positive, p’(x) is increasing when x= a, and CHAP. VIII] MAXIMA AND MINIMA 237 we have therefore a minimum value ¢(%) of the function: but if (x1) is also 0, then we cannot decide by this means, and it is best to fall back on the graph itself of the function. 7. Hence the rules for discovering where maxima and minima values of ¢(x) occur are : 1. Solve the equation ¢$’(x) = 0. 2. If «=x, be any one of the solutions obtained, decide whether a maximum or minimum value is found (or whether there is neither) by : (a) Drawing the graph; or (b) Testing whether ¢’(z) is decreasing or increasing, or doing neither, when 2 passes through the value a1 ; or (c) Finding the value of ¢’(a,): if this is negative ’(#) is decreasing and ¢(%1) is a maximum: if it is positive ’(x) is increasing and ¢(a) is a minimum: if (#1) = 0 draw the graph any way, even if it has not been drawn before. The different methods will be explained by returning to the Example of § 4. EXAMPLE 1. Find the maxima and minima of the function 303 — 4x -+ 5. The graphs of y= 3x3— 4%4+ 5 and of 9a2—4 (ic. oY) dx and of 18x (i. a) are given in Fig, 81, dan? To consider now the different methods of §7: 1. Solving the equation oY 0, 1.e. 942—4=—0 gives us t= ze or 2 = z eS. 2. (a) From the graph we can see that when “as we have a maximum value of 303 — 4% + 5, viz. 6°78; when t= -; we have a minimum value of the function, viz. 3°22. Or (6) To test whether a is increasing or decreasing when t= Z or — = 3 3 238 SCHOOL CALCULUS [CHAP. VIII 2 : If w is less than — 3° © Il = ae aate 4] which is +ve. : 2 dy If xis greater than — 3&8: if r=— ‘6, 7 ‘76 which 1s —Vve. re - is decreasing, and we have a maximum value of 3x3 — 4a + 5 as before. Similarly in the neighbourhood of a= : we find wi is increasing, and therefore 3x° — 4x + 5 is there a minimum, Or wae dy = a dy (c) When «= — $4 7am 18%—= — 12; therefore Fe d7y\ . (whose slope or rate of increase is measured by Ta) is decreasing, and therefore at that point 323 — 4%+ 5 is a i ogee 3) daa therefore # is increasing, and therefore at that point we Bi maximum. Similarly when 4x 183 ="128 have a minimum. (As explained in § 4, when « = +00 we do not get maxima or minima, and there is therefore no need to consider the d equation at co; and we can see from the graph that we dx have now found all the maxima and minima of the function.) The graphs in §4 should be carefully studied, and the numerical values of the ordinates of each curve considered in relation to the other curves, as also their slopes at different points, and the points where they intersect the axis of @. 5 3 ; uy = 0 and therefore dy is @ minimum (as dx? dx shown by the figure), and therefore at this moment the rate of change of 3x3 — 4% -+ 5 compared with z is least. EXAMPLE 2, Find the maxima and minima of the function 43 + 22+ 8 w—4 © K.g. when «= 0 CHAP. VIIt] MAXIMA AND MINIMA 239 34 242+ 8 = bp To draw the curve y = i ate a we have the following 4 — table of values: a=|—4|—3/—2%5) 2 |/—1) 0 1 2| 25 Ay Ne Can ts alee) y= | —2|— 2] 217 | wo |— 3)/— 2|— 3°67) 16-06 |10-6 8°67 |8°71 |9°25 | Jel Es Ge eres sere ei erae | pei fey tr erie eta pe EEE EY EEE ttt ee ie al eisai ciel labeiitety (isin isl iter per) eer — SIDS CRUE Bee eee eae eee re ee ener ere tcneme terme tel ett LA tere Lo Td rere meee el etiiet, ele ep tA eee te TT ere Pre lee ier oem mre tN ete bel aie tel tT eer ela ieee tee sre et elit TT SI 2S0 RSIS eaNS CBRN Eee eae pe eie ieee) Pe ekotorateteie (oP | tee eer ty pe pe Lt Ld ft tei bo pp SRESE pre ermiominpel eter ee CSR GO OOS ee eee Petar tare Yee si fe pet See esto ett : eee PPT TT 4 pie eee Gr | a a =} EEE AA A Hh pad ft td ae Ona RRE RRR AS HA HY eoanenonee memeeeretie ete) epee reer te ie) CT rr ee tt +H pd EE EE A sy et tH Et py fe JRE See eee ia tt th ty TT one Peleleie ets JES USE ee Cee bee BHR Ree SEE EEE EEE ECE EEE EEE =F ret ti Sete a Jab SSO Saka SRA ewes Fie. 83. and the graph is approximately that drawn with a continuous line in the figure. ose) a (Notes for quickly drawing a rough curve: y is infinite when « = + 2, giving us two asymptotes ; also the curve may 4(x-+ 4 SORE be written y= 2+ 2+ soe on dividing by x?— 4, a form which makes the calculations of the ordinates shorter, 4(x%-+ 4) and also shows that y = «+ 2 is another asymptote as TEE decreases indefinitely as ~ increases or decreases indefinitely, so that a rough curve is very quickly obtainable.) dy x(a? — 12% — 32) ar du (22 — 4)? 240 SCHOOL CALCULUS [CHAP. VIit (The graph of this has been approximately drawn with a dotted line, though not required for the solution of the question, in order that the relations of the two curves to one another, and the meanings of the ordinates of the graph of 2 may be again studied.) To find the values of x which give maxima and minima of : : ed the given function we must solve the equation ee 0; ies dx x(a? — 124 — 32) = 0. Whence x = 0, or #3 — 124 — 32 = 0. This second equation (as may be fairly quickly found by drawing the graph of y= x? — 12%— 32, which should be done) has only one real root, x = 4°36 nearly. We can now see at once from the graph of od = * that : i. When 2 = 0 we have a maximum of ¢(x), viz. — 2. ii. When a = 4°36 we have a minimum of ¢(#), viz. 8°68 approx. iii. There are no other maxima or minima, as may be seen by referring to the consideration in §5 of the ordinates at K, P, and Q in that figure. Notice that the minimum value of ¢(x) is here greater than the maximum. Also that a rough graph of the function under consideration should always be drawn, and in this question it is practically necessary and certainly very much quicker : for the value of dy makes it inconvenient and tedious dx to calculate its changes in the neighbourhood of «= 0 and .. & a2 = 4:36, and to determine these changes by calculating a becomes even more complicated, whereas the graph of ¢(2), when we have noted its asymptotes and general directions, is very quickly drawn well enough to determine the general position of the maximum and minimum values; for of course nothing even as accurate as the figure given is at all necessary. CHAP. viTT| MAXIMA AND MINIMA 24] EXAMPLES Find the maxima and minima of the following functions, distinguishing between them, and stating for what values of the variable they occur. 1. 5 — 6r+ 7x2 Anes 3 14 2°77 2. Qa+ ‘a < 9:3 3. 323 — a+ 8. | 4. 403+ 9x? — 1204+ 13. ‘(i 5. 22(4 — 2). N14 40° 32? — Qn + 3 io, V#8(6 — 2) 6. Bx é 1 te ees eg, = Sas 3: LC y. i 510 x, 34 By ee ee i Sct © J/24+ By? 7 (In Examples 15-19 no angle is to be greater than 7.) 15. a sin x+ b cos @. 16. cos x — 2 cos 2z. 17. sin 6 sin ¢ when 6+ ¢ is constant, and equal to a say. 18. sin 6 sin ¢ when 6 — d=a. 19. cos 6 cos @ when 6 — g=a. 20. Find the value of 6 (supposed less than a for which acos26-+ 2hcosésiné + bsin?4 is a maximum. 22+ Ta+ 4 Qn? —Ta+4 22. Find the maximum and minimum ordinates of the curve et 22 — 5 -— eS 21. Find the maximum and minimum values of 8. Practical Examples. EXAMPLE 1. A railway crosses over another at right angles, and 2 trains are moving along them with velocities 10 yds. per sec. and 20 yds. per sec., respectively ; the positions of the trains are such that a passenger on the first train will just be on the bridge when the second train has cleared it by 30 yds. When will the second train subtend a maximum angle at the eye of the said passenger if it be 100 yds. long, the breadth of the second train and the difference in altitudes between the two lines being neglected ? 16 942 SCHOOL CALCULUS [CHAP. VITI Fira. 84. Let P be the passenger, and AB the second train; ¢ the time in seconds reckoned from the moment P is on the bridge O; and let OPA and OPB be the angles 6; and 62, respectively. Then OP = 10t yds. OA = (20¢ + 130) yds. OB = (20t + 30) yds. 20¢-+ 130 24+ 13° SPER We p= LO ae 20+ 30 2t+3 ee OE (ot 13)h = (21-8 tee . tan (01 ~ 62) = 9 97-113) (Qt-+3) BP +324 +39 HOY. Differentiating to find those values of ¢ which give maxima and minima we have : df(t) 10: —-10¢ (10¢ + 32) dt ~ 5t2+ 32t+ 39 (5t + 32t-+ 39)2? Z oe = 0 when 10[5¢2 + 32¢-+ 39 — t(10¢+ 32)]= 0 ‘ie. when—5i2 + 39 = 0; t= 2 A/* — + /78= + 2°79, approx. 243 before the 5 whose denominator is positive (5t2 + 32¢-+ 39)2 is increasing, so that ¢(t) has P(t) d t MAXIMA AND MINIMA 10(39 — #2) When ¢ = — 2°8, 39— # is negative ; When ¢= — 2°7, 39— # is positive ; dt for all values of t. d o(t) We. -, when ¢=— 2°79 gd Therefore maxima or minima occur 2°79 secs. O first train reaches the bridge and 2°79 secs. after. N there a minimum value, CHAP. VII] | S25 So RARE S Ree RENO M OI SOee ees reNe DEER ES SSeaR ERM ERESRRERnO Rae ee Reeee | SShhsseeee seo SRO SSA RESO RePeoo AMSA BARS Ie tells | eter shee ss era sae at ie | eam et aian] cle) a] aie ale ate aia tats tet ST Shen ade) he bel bathe bl hs bel tte Ls tae) Ppa Po feel Sate ae gee A Pek ge lea ge de ret Leta te Tne Je Pe Tc bes oe (rt Vg as Ud a to EBERSAR Se SRO aaa ARP OMAR Nees HRRARSSOOM MARRS ARR Re ARBs eea seen Sad OPIS SE ROR BORSA Mae Seka ae Fig. 85. 244 SCHOOL CALCULUS [CHAP. VIII d $(t) dt we find ¢(¢) has there a maximum. Hence it would appear that 2°79 secs. after the first train crosses the bridge is the required and only solution to the problem. But consider Figs. 85 and 86: Similarly, by testing as ¢ passes through the value 2°79 10¢ BP Bat 39 OF PLM) for which the table of values is as follows: The first (Fig. 85) is the graph of :-.. ..| =8 | —7 | 26 25 (Se 10¢°:. . ..| =80| —70 | = 60 | =O |-140 ee tee eer 52+ 32+39 | 103 | 60 27 4 | —9|— 11-75 Soles g(t) .. .. |—°776|— 1-165 |— 2-221 |— 12-5] 4:34] 2-080 | 25 | 2-561 t .. .. |429l-16) 21 | 5901-5 1 ee 10 .. .. |~ 20-15} —10| —&10| 5 |\10 | 200) sue 52+ 321+ 39 |—5/ 225 | 12 | 2425] |56-25] 76 | 123 | 180 | 247 | 324 pit) .. .. | 4 |—6-67|— -834|— -206) 0 |-0905/-1303)-1626)-1666|-1620|-1542 The second (Fig. 86) represents the positions of the passenger and the second train for each two seconds from 6 secs. before until 6 secs. after the passenger has crossed the bridge: numbers on the vertical axis represent the positions of the Fie. 86. CHAP. VIII] MAXIMA AND MINIMA 245 passenger at the times they indicate ; numbers on the hori- zontal axis represent corresponding positions of the rear end of the second train; and the angles as indicated are the various values of 0; — 0. From this we see that the actual angle subtended by the second train at the eye of the passenger before he crosses the bridge is 360° — (0; — @2), so that when t=— 2°79, 0; — 02 being a minimum, the angle we are considering is a maximum. Hence the angle subtended at the passenger’s eye is a maximum 2°79 secs. before he crosses the bridge and also 2°79 secs. after. [In the figure giving the graph of ¢(¢) the branch AB is the graph of tan (6; — @2) when- @;— 62 is an angle in the 4th quadrant: the branch CDE gives tangents of (@; — @2) when that angle has passed into the 3rd quadrant; for the part GK (6, — @) passes back into the 4th, and for the part KD into the Ist quadrant. These results should be considered and verified by the help of the last of the three figures. | EXAMPLES 1. At what time or times in the above question does the second train subtend a right angle at the passenger’s eye? (Solve this graphically and theoretically. ) j 2. Find the times at which the second train subtends a maximum angle at the passenger’s eye, if the first move with velocity 20 yds. per sec., the second train with velocity 10 yds. per sec., and if the second train clear the bridge by 40 yds. when the passenger is on it. (The figure for this case is approximately given in Fig. 87 on page 246.) ExamMeLeE 2. The diagonals of a parallelogram cut at a fixed acute angle 0, and their lengths are in the ratio n: 1, where n is variable. Find the ratio of the sides, and deduce the maximum and minimum values of this ratio for different values of n. Let ABCD be the parallelogram whose diagonals cut at O. Let OA=y, OB=ny, AD=2z, AB=mz, and AOB= 90. Then AOD = 7— @, and the ratio of the sides is m: 1. Now m2? = y?(1 + n? — 2n cos @) (i) and #2 = y?(1 + n2 — 2n cos 7 — @) = y?(1-+ n2-+ 2n cos @) (ii) From (i) and (ii), by division, Palen — 2700s 0 Tage ieee ey SY a 1+ n2-+ 2n cos 0 [CHAP. VIII SCHOOL CALCULUS 246 | Ht eg ft tt SAU DEN Bee RAR RSRSMeE OR eaAS eee BRVGSE HRlERS ARERR EREU Renee eo eae a Op CY Bed) MS FP SO ee ot a GEina hii yk ae Rass EeSeRke eevee ee A ee ee ee a a eae {+ a tt et ita pel Tete eee erect el fl see a eer 1 Bisehup Mes i lehman ace tayateemise cl) ae Tee pe ele rel PAP erode a ey ie pol rey pes ie aaa erie | tet Tabs eve ne yy aaee a Soa RESRRRRER AOSD oLoes tH Pe eel eee ead bl. hr al etet ge ie) oa SREER RARE SALBe Lee eee en aca YH -H+-H1 te Seer ime Peete ety ede seated ed | es ECEER EH BoPe dice. --san88 Sram S28 S8RERSe SUSGBGRI eS Soe Su aueaenaTsegens C Bee EE g | | Bane RSehesS EOeeeeooeden any ueeeeeee EGE EERE REECE Pee 4 SRR AR re ae Ppt L | OO ae 245. Fic. 87, FOR AN EXAMPLE ON P. We have now to find the maximum and minimum values of this quantity. Differentiating with respect to n, we get 2m ss (1+ n2 + 2n cos 6) (1+-n?-+ 2n cos 0) (2n—2 cos #)—(1+-n?—2n cos @) (2n-+2 cos 8) and equating the numerator to zero gives those values of 0, since m cannot vanish. dm dn n which make - 0 i.e. 4n? cos 6— 2 cos O(1 + n?) “.n—1=0, 1 (it cannot equal — 1). n= CHAP. VIII] MAXIMA AND MINIMA 247 Substituting this value of n Seowly 2 fs Tie ncep 2 sin pte 2 aunt Gata i 2° Feo 8 4 / cout 8 SecA If, however, we had taken m as the ratio AD” then AD ss i = ta g a b AB m7 tah g a8 above, or m= cot — sf Hence the maximum and minimum values are tan : and Ga : cot 5, i.e. m lies between these values, and it depends on the 2’ magnitude of 8, which is the maximum and which the mini- 0 : mum. If dis> = then tan 3 » cot y i.e. tan : is the maximum | @ rife : and cot. 9 the minimum value of m. If 0dis< bs these must be interchanged. EXAMPLES 1. Find the maximum area of a rectangle whose perimeter is 100 feet. 2. A cone is such that the height and the diameter of the base are together equal to five feet; find the height when the volume is a maximum. 3. By the postal regulations the length and girth combined of a parcel must not exceed 6 feet. Find the volume of the maximum cylinder that can be sent by post. 4. The stiffness of a rectangular beam varies as its breadth and the cube of its depth. Find the measurements of a cross section of the stiffest beam which can be cut out of a cylindrical log one foot in diameter, and compare its stiffness with that of the largest beam of square cross section which can be cut from the same log. 5. Find the volume of the greatest cone which can be made from a sphere containing one cubic foot. 6. Find the height of the cone (i) whose curved surface is a maximum (ii) whose total surface including the base is a maximum, which could be cut out of a sphere of radius one foot. 7. Find the volume of the tray of greatest capacity which can be made out of a sheet of metal 18” by 12” by cutting squares out of the corners and turning up the edges, 248 SCHOOL CALCULUS [ CHAP. VIII 8. A river is 2 miles broad, and a person on one bank wishes to reach a place 6 miles down the other bank. If he can row 3 miles an hour, and walk 4 miles an hour, how far from the place should he land on the opposite bank to reach it in the shortest time ? 9. Show that the volume of the largest cylinder which can be inscribed in a paraboloid of revolution is half the volume of the paraboloid. 10. The expenses of a a at speed v knots per hour are equivalent to the cost of 60+ 20+ = Pig of coal per hour. Find the most economical speed of steaming ms a current running 6 knots per hour. 11. A rod one metre long rests on two supports each distant x cms. from its middle point and in the same horizontalline. When weights of 750 grms. are hung from each end the displacement y cms. of the middle point of the x2(50 — a) 9850 : find for what rod above the line of supports is given by y= value of x y is a maximum. 12. C is the centre of a circle of radius 6 in., D is a point on a fixed dia- meter one inch from C, and DA perpendicular to CD cuts the circle at A. PQ is any chord perpendicular to CD; find the maximum area of the tri- angle APQ. 13. Find the volume of the greatest cylinder which can be cut from a cone whose height is one foot, and semivertical angle 55° 42’. 10 xu? — 4¢+ 5 find the position of these ordinates when the area enclosed between them and the curve and the axis of x is a maximum. 15. Find the position of two ordinates two units apart to the curve y? = 5x2? — x when the area between them is a maximum. Explain the other critical value you obtain by means of a figure. 16. If the curve x? = 2y(y — 6)? be revolved about the axis of y, find the distances from the origin of two planes perpendicular to that axis and two units apart which intercept a maximum volume from the solid so formed. 17. Two trains are moving with velocities 30 and 40 miles per hour, re- spectively, along two lines at right angles to one another. If the trains are at some moment 500 and 200 yards distant respectively from the point where the lines cross, find when the distance between them is a minimum and how far apart they then are. 18. A man six feet high is approaching a tower 70 ft. high on which is a 12-foot flagstaff. Find how far he is from the foot of the tower when the flagstaff subtends the maximum angle at his eye. 19. The sum of the axes of an ellipse is constant. Find the ratio of the axes to one another when the difference between the volumes of the two ellipsoids generated by revolving it about its axes is a maximum. 20. Find the area of the maximum triangle which can be inscribed in an ellipse whose axes are 2a and 2b, with one side parallel to the major axis. 21. Water issues horizontally from a small hole in the side of a tank which is kept filled to a height h ft. If the velocity of outflow at depth z {t. below the surface be »/2ga ft. per sec., find how high the hole must be that the jet may hit the ground at the greatest distance from the tank which is standing on the ground, 14. Two ordinates are drawn to the curve y = 3 units apart ; CHAP. VIII] MAXIMA AND MINIMA 249 22. In the figure, a supposed indicator diagram (cf. chap. ix., § 29), the effect of the inertia of the moving parts of the engine causes an apparent loss of pressure approximately represented by measuring the pressure up to the curve from the sloping line HK. Find at what point in the stroke the apparent pressure is least, and its value there. Given that OD = 30’; AB=6"; AO=80 lbs. per sq. in.; OH = DK = 36 lbs. per sq. in.: also the law of expansion is pv =c, so that BPC is a rectangular hyperbola. 9. Points of Inflexion. It was mentioned in §6 of this chapter that at the points D and # in the figure to § 5 critical values of y, i.e. of $(x), are apparently obtained from solving OY ce or oY 0, since at those points the tangents to the graph of $(x) are respectively perpendicular and parallel to the axis of «. But it is obvious from the figure (and in any particular case it would also be clear from any tests applied for maxima or minima at those points) that at these points y is neither a maximum nor minimum. The points are in fact Points of Inflexion, and the points marked a, B, y in the figure of §5 are also points of inflexion, though the tangents at these points are neither perpendicular nor parallel to the axis of 2. Such points are, in theory, easily found in all such cases as dy dx either (i) increases in value before the point of inflexion up to some value (0, or a finite value, or o), and then decreases again from that value after the point of inflexion, or else (ii) decreases to some value before the point of inflexion and afterwards increases, In the first case therefore at the point the equations we shall require : for it is clear from the figure that =~ always 250 SCHOOL CALCULUS [CHAP. VIII ne Oe dy . ? } of inflexion “2 ig a maximum, and in the second case a dx minimum. Hence to find the points of inflexion on the graph of y = f (a), we have to find the maxima and minima values of au and 2 these in turn can be found by solving the equation = Y 0 2 (and sometimes also FY =<), and the methods of this chapter for deciding the maxima and minima values of any function can now be applied as usual. EXAMPLE 1. Find the point of inflexion on the graph of = 34° — 4x + 5. The figure is given in § 4. We see there that the graph of ou (the dotted curve) has a minimum value when x= 0, as is also clear from the fact that the value of oe (whose graph is the straight line) is also 0whena=0. Hence the original curve has a point of inflexion where «= 0, and therefore y = 5. Otherwise : dy : eee ae, AE 9x Critical values of this are given by d7y ; a 0, i.e. by 18% = 0 so that x = 0. d3y dy ae ae = 18, a positive quantity, so that when x= 0, fe has a minimum value (for it is necessary to decide that aw really has a maximum or minimum value, which would not be d°y decided if a3 had come to 0). Therefore the given curve has a point of inflexion when it ean) CHAP. VIII] MAXIMA AND MINIMA 251 ExampLe 2, Find the points of inflexion on the curve %—3 Y= Jo at 3 (3+ #)#(3— a)? Differentiating, oY Differentiating again, d2y 3(3 — 2x) da. (3+ a“)? (3— a)? This is equal to 0 when # = S and from a graph (which can be quickly plotted) we see this value of 2 does give points of J inflexion, and y is then equal to + _ re seis. The points required are therefore (1°5, °577) and (1°5, — °577). EXAMPLES Find the points of inflexion on the following curves : . ¥ = 323 — «+ 8. 5. (a2+ x?)y = 63. . ¥y = 403+ 9x? — 1244 13. 6. (16+ 2?)y = 5a. y= x4 — x). 7. Oy? = «3(6 — =). . §x2y = 3x2 — 2x+ 3. 8. xy? = 25(5 — 2). . Show that y =a sin n 6 has a point of inflexion each time it crosses the axis of 6. 10. Find those values of x not greater than 180° which give points of inflexion on the curve y = sin r+ 4 sin 32 (see Fig. 13). Co rWN CHAPTER: 1: APPLICATIONS TO MECHANICS I. CENTRES OF GRAVITY AND CENTROIDS 1. Let A, B, C, D, EH, etc., be any number of particles in a plane whose masses are m1, M2, m3, etc., respectively, and let M be their total mass, so that M=mi+metm3+.... Let the distances of these particles from any,line Oz in their plane be y1, y2, y3, etc., and let y be the distance of their centre of gravity from this line: then by taking moments about Ox we have My =myitmo2ye2tmsy3t+ .... MiY1 TMY2TM3Y3F oss M1 + M2+Ms3 + Ae et Similarly the distance x of the centre of gravity from any other line Oy is given by nes M141 +M2%e+msr%3+.... m,+m2+m3+ oo oo where x1, %2, x3, etc., the distances of the particles from that line. Taking these lines for rectangular axes in a plane, and a third Oz at right angles to them both if we are dealing with masses in space, we can determine the coordinates of the centre of gravity with reference to these axes. These formulae are usually written (Mx) od yar) " &(m) S(m)’ in which, as they are only abbreviations of those given above, no cancelling can be done until the numerators and denomi- nators have been evaluated separately. 2. Now a continuous body of mass M may be considered to be made up of a very large number of very small parts, 252 t= CHAP. Ix] APPLICATIONS TO MECHANICS 253 the mass of one of these parts being dM, and the centre of gravity of these parts is that of the whole body. (Note that aM is not the same for every part, as is seen below.) The shapes of these small parts vary in each particular-case, and are so chosen as to make the necessary summations as simple as possible ; and'in general we have | S(dM-x) fedM _fxdM SGM) ~ am ~ MM” It frequently happens that the body has an axis of sym- . metry, which must therefore contain the centre of gravity, and we choose this line as axis of a. Let a plane perpendicular ta this axis, at distance x, from the origin cut the body in a t= Fig. 88. section of area A; (Fig. 88). Then the volume of the body cut off between this plane and a parallel plane through 21+ 624 lies between A,5a,; and (4;+6A1)5%1; so that in the limit when 6x; and therefore 6A, are equal to 0, its volume is ulti- mately A,dx, and if p be its density, its mass is pAida, and its moment about O is pA1dxz1:21. Therefore taking the sum of all these moments we get X(pAdu:x) /pAxdx S(pAdx) = fpAdax’ where A must first be expressed in terms of 2 from the known form of the body, and both integrals then separately eva- luated between suitable limits for 2, — | CHAP. IX SCHOOL CALCULUS 204 If, as is usually the case, p is constant, this may be written x p/Axdx /Axdx [Ads ° To find the centre of gravity of a semicircular lamina of uniform density. ~ pfAda EXAMPLE 1. Weseooee 32 ee, Boe FP BD BE LA BF AP «Fs EP BS SB SE EA FAD 2. AP GB A EAA Aw ae ONO Pea wee * ne TNA OOOH 25 VO a 6 TE a RE tabetha f= EL Sis P| MeL meal sll sa Pater e ae Fia. 89. Let the equation to the circle be x2+-y?=r?. The centre of gravity of course lies on the radius which bisects the figure, and we therefore take this line for axis of 2. Let m be the mass of unit area, then the mass of the whole then the x, and MM’=62;: Let OM mass of the strip PP’Q’Q in the figure lies between 2my16x1 and 2m (yi+6y1)5%1, and is therefore in the limit equal to T~mr2 2 lamina is 2 dx. Q2mMV r2—24 Now take moments about O, and let 2 be the distance from O of the required centre of gravity. 2my 1x4 OHAP. 1x] APPLICATIONS TO MECHANICS O55 Tmmr2 _ xr 5 eae WMV r2—22-xdx 0 =2m a 22)" |? r3 Be ye a8 —. of : KS aa 425r approximately. If the radius of the circle be 16 cms., let x be the distance from O of the centre of gravity of the strip cut off by two double ordinates distant 7 cms. and 13 cms, from O. If M be the mass of the strip fa Poh Aa M-z=2m f 4/162 —22- ada 7 oo i eo and M=2m f” V/ 162 —22-da 7 1 3-113 =5| (162-22) | 7 _7= a (I, 162 =r yr. TY ele eemeeeei6.> |. 2 : 3{ (162-72)' (102-132) | TG gig 7), BVI TVIP TP 2 16 16) * 9 2 722-5 Sey =9-81 cms. nearly. The centre of gravity is therefore 2-81 cms. from the longer edge of the strip on the axis of symmetry. Notes. 1. It will be seen from this example what was meant when it was said above that (i) The centre of gravity will generally lie on a known line. (ii) The parts, into which we suppose the body to be divided, are so chosen as to make the summations or integrations as simple as possible. 256 SCHOOL CALCULUS [omap. Ix (iii) The areas of these small elements must be ex- pressed as a function of x in order to effect the integration at all. 2. The Centre of Gravity is also frequently called the Centroid, especially in relation to a surface, whose thickness is neglected, so that it has not strictly any mass and the term Centre of Gravity is not strictly applicable. ExamMPLeE 2. Centre of gravity of a hemisphere of uniform density. The volume of the hemisphere is equal to e ars (Ch. Vile oo Fie. 90. 2 i a™pr “ 3 § 3), so that its.-mass is therefore 2 a™Pr, where p is the density. Let the equation to the generating circle be x2+y2=r2, and take that radius in which by symmetry the centre of gravity must lie for axis of x. Then (asin Ch. VIL., § 3) the volume of the element PP’Q’Q, at distance x, from O, is my 1?da%1 so that its mass is wpy12da. Taking moments about O, we have as before =p f° (v2 —a?)adax = ilrp om ————— —— 0 oer. 8 To find the centre of gravity of a portion of a sphere of radius 5”, cut off by two parallel planes 33” and 1” from the centre and on opposite sides of it. CHAP. IX] APPLICATIONS TO MECHANICS 257 3+5 mwp(25—x?)xdx _ =—] # Here x~= inches 3°5 Je mp(25—x?)dx -1 E eos leoe elt os ee a - 43 3°5 252 — — | a 8 my 103-4 : Boga -049 inches. .. the centre of gravity lies 1-049” from the centre. 3. When the Density is Variable. EXAMPLE 1. Find the centre of gravity of a right circular cone, whose height is 8 cms., and in which the density of any element varies as its distance from the vertex measured along the axis. Let the cone be formed by the revolution of the line y=mx about the axis of x. The volume of the small element PP’Q’Q (see Ch. VIL., § 2) is equal to wy2duz=7m?x2dx. Now the density of this element oc x and is therefore equal to kx say, where k is some constant, whose value we shall not as a matter of fact require. Therefore the moment of this element about O is mmx? -kx-adx, and its mass 7m2x2-ka-dax. Taking moments for the whole cone .. the centre of gravity lies on the axis 6-4 cm. from the vertex: for the density of any such element as PP’Q’Q is 17 258 SCHOOL CALCULUS [CHAP. IX uniform since every part is the same distance from the vertex measur | along the axis, so that its centre of gravity is on the axis, and therefore the centre of gravity of the whole cone also lies on the axis. EXAMPLE 2, A vessel in the form of a paraboloid of revolu- tion is filled with a substance whose density varies as the depth. If the diameter of the surface is 3 ft. 6 in. and the total depth 3 ft., find how far the centre of gravity of the substance lies below the surface. Fia. 91. Let AOB in Fig. 91 represent a section of the vessel through the axis, and let the equation to this generating parabola be y=ax2, Since this passes through the point (1-75, 3) there- fore 3=a(1-75)? 16 48 a=3X —>=— 49 49 Y—- 95x The volume of PP’Q’Q is 7242dy1 aE .. 1t8 mass is equal to guew k(3—y1) myidy1 -98 i ( and its moment about O = 3—Yy1)Y1 for the density raises as the depth, i.e. as 3—y1, and therefore the density of PP’Q’Q=k (3—y1), where k is some constant. CHAP. Ix] APPLICATIONS TO MECHANICS 259 3 kor 3 De Ys gi oe 2 4,3 2 if. Te tee wy J, (3y° —y? dy 3 k 2 “og YR—y)dy i (By —y?)dy 443 Es Bile bs 4 lo fo 6" a ee 2 le .. the depth below the surface is also 1 ft. 6 in. Notice that the diameter of the surface does not affect the result. 4, General Results for Variable Density. As before this result may be written more generally as follows. Let x be the distance from a fixed line of the centre of gravity of a body of which dV, is a small element of volume whose distance from the line is 7: and let the density of this element of volume be given as equal to (21). Then the mass of the element =¢(21)dV1 and the moment of the element about the line=2 -¢(x1)dV1. fe: d(a)dV Jo(x)dV © Now dV must be expressed as a function of 2, say F(x)dx. Hence y%= Eee (2) (2 da ' Then x= [eb(ae) F («yd numerator and denominator being integrated separately between suitable limits for x. If now the centre of gravity lies on a known line, this one calculation will determine its position: if not, y, and if necessary z, must-be similarly determined. Note. The Centres of Gravity of irregular figures, which are not defined by any equations, can be found by graphical methods which do not involve any Calculus and will there- fore not be considered. EXAMPLES 1. A square is inscribed in a circle of radius 5 cms. Find the distance from the centre of the centroid of one of the segments outside it. 2. Find the centroid of a semicircular arc. 3. Show that the centroid of a semi-ellipse cut off by the minor axis is at a distance = from the centre. a 260 SCHOOL CALCULUS [CHAP. IX 4. Show that the centroid of the area cut off from the parabola y? = cx by the line x = }, is at a distance = from the vertex. 5. Find the centroid of the area cut off from the hyperbola 4x? — 9y? = 36 by the line x = 5. 6. Find the centroid of the area between the rectangular hyperbola zy = c?, the axis of x, and the ordinates at x = 2, and 2 = 2». 7. Find the height from the ground of the centre of gravity of the Skerry- vore lighthouse, assuming the measurements given in Ch. VI., § 13, Ex. 9. 8. Show that the centre of gravity of a paraboloid of revolution is two- thirds of its height from the vertex. 9. Find the coordinates of the venue of a quadrant of the ellipse y2 a ee 52> 10. A hemisphere is filled with liquid whose density varies as the depth. Find the depth of its centre of gravity if its radius is 5 feet. Find the centroids of the areas given in Examples 11-15. 11. One loop of the curve 4y? = 5a? (16 — 2?). 12. The area between the curve (9+ x?)y? = x and the line a = 4. 13. The loop of the curve 8y? = x? (7 — 2). (For the integrations put 7 — x = z?.) 14. The loop of the curve 2y? = x* (6 — 2). 15. The loop of the curve (10+ x)y? = x? (10 — 2). Find the centres of gravity of the volumes formed by the revolutions of the curves as given in Examples 16-20. . 16. (16+ w?)y? = 27, from x =0 to x = 4, about the axis of . 17. (27+ «?)y = 122 about the axis of x, from x= 0 to x =3.- 18. 7y = 2aV'7-+ x? about the axis of 2, from x =0 to = 10. 19. 5y? = 2% (2 — x) about the axis of 2. 20. 3y? = x (16+ y?) about the axis of y from y = 0 to y= 4. 21. Verify by means of Guldin’s Theorem that the centroid of the area between the parabola y? = 4az, the axis of x and an ordinate at x = 2 is at a distance from the axis equal to 3 of that ordinate. 22. Similarly find the y ordinate of the centroid of that part of the area of the curve 2y = «V/10 — a? which lies above the axis of a. II. MOMENTS OF INERTIA do. The kinetic energy of a particle of mass m lbs. moving ina pet line with constant velocity v ft. per sec. is equal to ee foot-poundals, in other words this is the amount of work it can do before coming to rest. If now a rigid body be rotating about any line as axis, and at any moment the velocity of any particle of it whose mass 18 m4 be v; feet per sec., the kinetic energy of the particle CHAP. Ix] APPLICATIONS TO MECHANICS 261 M4012 is foot-poundalsat that moment: and since particles which are at different distances from the axis of rotation have different velocities, the kinetic energy of the body, which is equal to the sum of the kinetic energies of all the particles composing M401 Sm 1202 it, is equal to This is the same as -— ee where 7, feet is the distance of the particle m, from the axis, and w radians per sec. the angular velocity of the whole body at that moment as it is also the angular velocity of every particle composing the body. ‘Therefore the kinetic energy of the © 2 body is = Smr2, and the quantity S'mr? is called the Moment of Inertia of the body about the axis of rotation, and is generally denoted by JL. If now m, the small element of the mass, can be expressed as a function of the variable 7, say m=d(r)dr Then [=X (r)dr x r2= /r2-h(r)-dr between suitable limits. If M pounds be the mass of the whole body and we write >mr2= M kA then the length & so determined is called the Radius of Gyration, and it is clear that the whole mass might be supposed con- centrated at a distance k feet from the axis of rotation: for in that case each particle would be k feet from the axis, and mr? would become >mk2 =k25'm (k being now a constant) =hk2M. EXAMPLE. Find the moment of inertia of a triangle about any side. Take this side for axis of y, and its middle point for origin. Let the length of this side be 2a units, and the coordinates of the opposite vertex (b, c). Then in Fig. 92, the whole of the small element PP’Q’Q is the same distance x from the axis of 262 SCHOOL CALCULUS [ CHAP. IX rotation, so that its moment of inertia is 6M -x2, where $M is its mass. Nae PM _MA me a OA OL b of the lamina, , 80 that if p be the density Mae ee 6x approximately. edt eee = dx 2oa -” 7 I=" (6—x)x? da _ 2pa al Sb Spe Sak eee Ie AG Cae? b2 =pab - - 2 — Me ee where M is the mass of the whole triangle. b Hence also k =V6 = -408b. 6. There are three important propositions by means of which the finding of moments of inertia is much simplified. (1) Theorem of Parallel Axes. If the moment of inertia of a body of mass M about an axis through its centre of gravity be J,, then its moment of inertia J, about a parallel axis at distance d from the former is given by I=I,+Md?. Take the axis through the centre of gravity for axis of y, and let the distance of any element dM of the body from this axis be x; then its distance from the new axis is «+d, and its moment of inertia about that axis is dM(x#+d)?. *,[=f(x+d)*dM = f(x? +2ad +d7)dM =/x2dM +2d/xdM +d2/dMmM. CHAP. Ix] APPLICATIONS TO MECHANICS 263 Now /x2dM =I, ; and d?/dM =Md? ; while /xdM =0, for if x be the distance of the centre of gravity from the axis of y, fred M M ae I=I,+Md?. ExampiLeE. In the example of §5 find the moment of in- ertia of the triangle about an axis through the centre of gravity of the triangle parallel to the side. r= (see § 2), and in this case, x=0. : : lh Mb2 In this case the distance between the axes is 3 and I arc M -b? b2 Oc yaa +M-o Mb RS oto which is the required moment of inertia. 7. (2) If the moment of a lamina about any two axes Ox and Oy at right angles to one another in its plane be J, and J,, its moment of inertia J, about an axis Oz perpendicular to its plane is given by L=],+\. Let any small element of the lamina of mass dM be at distance 7 from O, and let its coordinates with reference to Ox and Oy be x and y; Then [,=/r?dM =/(x?-+y?)dM ==/1?d M +-/y?2dM ae I he ExamMeLE 1. Find the moment of inertia of an elliptic Fig. 93. disc of axes 18 cms. and 8 cms. about an axis perpendicular to its plane through a focus. The equation to the ellipse is DO ayy a eae, SL To find the moment of inertia about the major axis Ad’, 264 SCHOOL CALCULUS [CHAP. 1X the mass of the small element PP’Q’Q is 2pxdy, where p is the mass of unit area, so that its moment of inertia about AA’ is 2pady - y? eal 20x -4>-dy. For convenience we will take the equation to the ellipse as g2 yd (ua b cl 45 f yp OV R—ydy. (Put y=bsind .".dy=b cos6 dd.) adel” b2sin20 -bcosé -bcosé dé Tv —pab3 [ 74 sin26 cos26 dé where JM is the mass of the lamina. 2 Similarly the moment of inertia about BB’ is M < and 2 therefore about the latus rectum is ue 5 Mae [$ GO) =i 2 2 Hence J, (=/,4+/,)=M (g i +02), 2— 52 In this case a=9, b=4, and eeu OED . ar 81 at (2418 465) =M xX 89-25. Hence for the radius of gyration k—=V89-25—9-446 cms. CHAP. IX] APPLICATIONS TO MECHANICS 265 EXAMPLE 2. Find the moment of Inertia of a circular plate of radius a, (i) about a perpendicular to its plane through its centre J, (ii) about a diameter, Io. Fie. 94. (i) We here divide the plate into elements each part of which is the same distance from the centre, i.e. into narrow concentric circular rings. Consider one such ring, at distance « from the centre, and breadth dz. Then its mass is 2rpudx where p is the mass of unit area cut out of the plate. Hence its moment of inertia is 27px-dx-x2=dTy, say. a Ly=2mp f. x3 dx for mpa2=M, the whole mass of the plate. (ii) Since the moments of Inertia about a diameter are clearly all equal by symmetry, calling them I2, we have by Ma, Ma? the theorem of this section 2/5= 2 4 266 SCHOOL CALCULUS [ CHAP. IX 8. (3) If a lamina have two axes of symmetry Ox and Oy at right angles to one another, and the moments of inertia about them are J, and J, respectively, then the moment of inertia J about a line OO’ through O making an angle @ with Ox is given by I=I, cos?@-+-I, sin?0. Let dM be a small element of the lamina at a point P, whose coordinates referred to Ox and Oy as axes are (%, y). Fia. 95. Let PL be the perpendicular on the new axis of rotation 0O"~ Then in the figure PL=PK —LK =PK—HN ~ =y cosé—~x. sind. The moment of inertia of the element dM about OO’ is dl =PL?-dM =(y cos@—-x sind)? dM “.L=/(y cosé—x sind)? dM =cos20 fy? dM —2 siné cosé fay dM +sin20 fx? dM —=cos?6-J,+sin?6-J,, and the middle term disappears. For xydM may be written xx ydM or yx ad M and the integral of either of these is the sum of all the products of the moment of each element about one axis multiplied by its distance from the other axis, and therefore vanishes if the body is symmetrical about both axes whichever meaning be assigned to f/xydM. : EXAMPLE 1. Find the moment of inertia of the ellipse CHAP. IX] APPLICATIONS TO MECHANICS 267 2 +5 =1 about a tangent inclined at an angle of 30° to the axis of x. If M be the mass of the ellipse, its moment of inertia I, about the major axis is (asin the Example of § 7) ee : similarly ol 7s Draw a diameter inclined at an angle of 30° to the axis of a, L,= Fia. 96. that is, parallel to the given axis of rotation. Then J, the moment of rotation about this axis as just 3 cos230°-M 5 sin230°-M proved is given by J,=-— rl rl ; To find the perpendicular distance between this axis and the given tangent, the equation to the tangent is y=« tan30°+/5 tan230°+3 (i.e. y=ma+Va2m?2 +62, the general equation to the tangent g2 y2 to the ellipse — ~ stea=1 making an angle tan~1m with the axis of 2). The perpendicular on this tangent from the origin is the V5 tan230°-+3 /1-+tan230° — Hence the required moment of inertia about the given tangent is (see § 6) given by distance between the axes required and equals 268 SCHOOL CALCULUS [CHAP. IX 3 cos?30°- M 5 sin?30°- M 5 tan230°+3 f= 4 4. sec230° M _M (2 Bca0 eee cos?30°) 4 4 _5M (: ?) eee ria 4 Hence the radius of gyration is V4-375=2-092, the unit of course depending upon that of the given axes. EXAMPLE 2. As an example of the moment of Inertia of a body in three dimensions let us find the moments of Inertia of a Right Circular Cone (i) about its axis (/;1, say), (ii) about a line through the vertex parallel to the base (Ja, say). Fie. 97. (i) Consider the cone as divided into a very large number of thin slices such as PP’Q’Q perpendicular to the axis. Let the height of the cone be h, the radius of the base b, and the semivertical angle a. Then b=Atana. Then if the section PP’Q’Q be at distance x from the vertex and of thickness dz, its volume is wx?tan2a-dx, and therefore since it is a thin circular plate, its moment of inertia about the axis is 2tan2 mpxtanta ss 4 da AGL, (UX 51 and we want the sum of all these elements of moments of Inertia. 4 h Hence J; = Pee =f at dx oO a ee =|; 10 0 CHAP. 1x] APPLICATIONS TO MECHANICS 1) lop) te) __tptanta -h® 10 mp(htana)* -h na 10 & mphb* bhp 3b? 10 3 10 2 and since the whole mass M SER *s h=5 Mob. (ii) Notice that the theorem of §7 does not apply to a solid body (though it does apply to a lamina of any thickness provided that its cross section is always the same, e.g. a right circular cylinder) so that we cannot here apply the method of the latter part of Ex. 2 of that section, to find the moment of Inertia of the cone about a line through the vertex parallel to the base. We must proceed as follows. The mass as before of an elementary slice is mpx*tan2a-dzx,=—dM say. 2, Its moment of inertia about a diameter is (aM (Ex. 2, § 7) xv*tan2a where 7 is its radius, i.e. dM and therefore about the 2tan2 axis through the vertex is dM (eee: +22) (§ 5). .. calling this elementary moment d/o, 2tan2 dlzg=7px?tan2a (* yon “ +22) da ‘an2 I .. [g=Tptan2a (1 +r) [oa da O tan2a\ h° Se (1 3) Re mptan-a | 1-+- Hh 5 Bhat = (ietanza+" =e s) 270 SCHOOL CALCULUS (CHAP. Ix 4 =" (ns02 es) since 6=Atana, =P! (i a Beis Caer BUT Posie ( 2 “) moan a (EN 3 b « [2 ee Pha de B=3u(e+5) 2 ee ese the mass of the cone, since EXAMPLES Find the following moments of inertia : 1. A uniform rod of length 2a (i) about its middle point, (ii) about its end, and find the radius of gyration in the second case if the rod is 3 feet long. 2. A square, whose side is of length 2a, about one side. 3. A rectangle of sides 2a and 2b (i) about a parallel to the side 2a through the centre. (ii) about a diagonal. (iii) about a perpendicular to its plane through one corner. [Use § 8 for (ii), and for (iii) use §§ 6 and 7.] 4, An equilateral triangle of side h (i) about a line parallel to the base through the centroid. (ii) about a line parallel to the base through the opposite vertex. (iii) about a median. (iv) about a line perpendicular to its plane through a vertex. Also check (11) independently by direct integration, and find the radius of gyration in (iv) if the length of one side is 2 feet. 5. A circular lamina of radius r about a tangent, and write down the radius of gyration if r is 20 cms. 6. A circular cylinder of radius r about a generator. 7. A right circular cone of height 2 with base of radius b, about a generator. 8. An ellipse of semi-axes a and b (i) about the major axis. (ii) about the tangent at a vertex. 9. A triangle of sides a, b, and c (i) about the side a. (ii) about a line parallel to this side through the centroid. (iii) about a line parallel to this through the opposite vertex, 10. A hollow circular cylinder of internal and external radii r; and ro. 11. The area cut off from the parabola y? = 4ax by a double ordinate at x= 2] (i) about the axis of x. (ii) about the tangent at the vertex. 12. Verify the example of § 6 by direct integration. cHaP.1x] APPLICATIONS TO MECHANICS 271 13. A paraboloid of depth 9 cms. revolves about its axis. Find the radius of gyration. 14. The letter I with dimensions as shown in the figure revolves about the dotted line through its centre. Find the radius of gyration. Il. FORMULAE OF MOTION BY INTEGRATION 9. (i) When the Force is Constant. Let a force P poundals act on amass m pounds. Then if the acceleration of the body be f feet per sec. per sec., by Newton’s second law of motion. Pm}. 1% x cee and is therefore itself constant. Let now a particular starting point A be selected. Suppose the body had there an initial velocity of wu feet per sec., and that after moving s feet from A in ¢ secs. it has acquired a velocity of v feet per sec. By definition, the velocity at any moment is the rate at which its distance from A is increasing compared with the time, that is (peter Chane FEE 3) 8) Similarly the acceleration at any moment is the rate at which the velocity is increasing with the time, that is dv : ds ed eee or from (i) [a2 emit) 272 SCHOOL CALCULUS [cHAP; 1X Since f is a constant, we have by integrating (i1), d ee yates ee ee saeeent TEL i.e. v=ft+C,. To determine C; we know that at A, when ¢=0, v=uw. Substituting these values u=0+C, AB aes C1=u. | whence v=u-+ft. Equation (ili) now becomes ds ; Seale e-3 mn ey the Integrating again s=ut+4ft?+Co. To determine Co, we know that at A, when t=0, s=0 substituting these 0=C2 whence s—ut-t 5ft?. These equations for v and s are true for any period of time, long or short, during which P and therefore f do not vary, for we have integrated equations (ii) and (iv) on the assump- tion that f is constant. 10. (ii) When the Force is Variable. If the variable force be P poundals at any moment, the acceleration f at that moment is equal to = feet per sec. per sec., and is therefore itself variable. ds Now the velocity at any moment is always given by v= FF by re a and the meee at any moment is similarly o b Sa th t = given by {j= at is by f= oe fee i ee Hence we have Ae ae and if we can now integrate this equation once we find the value of es at the moment required: and if we can integrate it again, we obtain the value of s, or the distance travelled up to that moment. These steps are of course the same as those when the force that is we get the velocity CHAP. IX] APPLICATIONS TO MECHANICS 273 was constant, but the integrations are not so simple: and as the force P is now variable and may be expressed as a function of the time, or distance, or velocity, we do not obtain any general formulae covering all cases corresponding to those obtained when P was constant. 11. (a) Let P=¢(t), so that P varies continuously as the time increases, not necessarily either always increasing or always decreasing. Now the acceleration fae at any time. he seme pale whence we obtain v by ascribing the required limits to #, if we can effect the integration. If the general solution of this be o—— W(t) +C4, this may be written Ta HFCs or ds= (* y()+Cr) ae and if we can integrate this again we get s= (5 we) +01) at =| = F(t) +0st-+0s | say, between suitable limits for ¢. Thus the acceleration, velocity and distance travelled at any time have all been found. ExaMPLeE. Two buckets, each weighing 2 lbs., are con- nected by a light string passing over a smooth pulley, one of them having 1 lb. of shot placed in it. The buckets are then allowed to move. Assuming the shot escapes uniformly at 1 oz. a second through a hole, find the final velocity of the buckets. Since the shot is assumed to escape uniformly, it will be 18 274 SCHOOL CALCULUS [CHAP. IX all gone after 16 seconds, after which the velocity will remain constant. : é t Now the force acting on the buckets after ¢ secs. is (1 —14)9 poundals, and the mass being moved is then (5—,,) pounds. fe : = Sea dv (16-bg ~ » Cb6=He de 80 vie ee ie (PES am vf an dt feet per sec. 16 =9f (4) di 16 —32 | #-+64 log. (80—4) | ,. —32 (16+64 log. 3) 32 (16—14-2812) : =—55 feet per sec. approximately. If the space described by the buckets at this moment be required, equation (i) may be written ds soli ted log. (80-1) | +0 it al and since v=0 when t=0, C= —32~ 64 log, 80 pds ie is t \) 1G, = 82 a loge aah Lega: 16 t | = 32/' {t-+64 log. (1—J5) dt feet = 32 [5 +64 (t—80)!1o 0-z Jat 2 lee eer 5 (Cf. Chap. IV., § 27, Ex. 1). =575 feet approximately. Saar. 1X] APPLICATIONS TO MECHANICS 275 12. (8) Let P=d¢(s), so that the force changes as the dis- tance travelled by the body increases. Then we have v mes dt _ds_ dv Mido edi ds sap f an du fads =. _ Ply) a) ds (since j="), Integrating this are fs) ds, between suitable limits for s, giving v in terms of s, say T= ie ONO Again v= ae by DN H(s)+0) dt or OT rapa er a ds / 2 AJ V+ and when this is integrated we get ¢ in terms of s, say i—f'(s) +C2 from which, theoretically at least, we can find s in terms of ?. Harmonic motion, which is a particular case of this, will be treated separately in § 14 on account of its importance. ExampLeE. A 10 kilogram weight is on a smooth table at a distance of 1 metre from a magnet, which attracts it with a force varying inversely as the square of the distance between them. If the attractive force at the start be equal to 30,000 dynes, find the velocity of the body when it is (i) 40 cms. and (ii) 10 cms. from the magnet. Also find the time in (i). If the magnet exert a force of P dynes at distance s cms., k then since Pox te. s Ss- 276 SCHOOL CALCULUS Now P=30,000 when s=100 eke Sea Le Also as just proved “nie ds _ 2k mM. ds ‘. (i) The first velocity is i, by 2X3 x 10° 7s Oe uae -{ 5 ds 104 100 8? 40 =-§ x 104 | —= | S_1100 ipa pa, ee eae ee eae (00 0) 6 x 104 x 60 4,000 — —900. [ CHAP. IX This negative result for v2 is due to the fact that s is diminish- ing, and that therefore ds is negative. ee OD *.v=30 cms. per sec. » : Dao elle (ii) For the second velocity v? = 104 —5,400 0= Fort] CMs. To find the time after which the weight is 40 cms. from the magnet, we have 26° 107 - +C (numerically) 1 ey) ety es Le .0==6:10 192 TY so that C= —6-102 ATOR YS (=) ds _ M100 =5 CHAP. Ix] APPLICATIONS TO MECHANICS 277 / PE LOAOdes tay eS vs Nie as 104/Be fie ge oe ; Ne per Na 100—s Put 100—s=100sin20. .°.ds= —200'sin@ cosé dé _ s=40 and 101/6 t= =f) ay Cost O00 sin@ cosé dé s<100 10 sing s=100 =100 (1-+cos26) dé 40 s= S=100 —100 | o+4 _ and amplitude a+- “ (Put p A-+ ps = x for the integration.) % 10. A body moves under the action of a force pv a per unit mass, towards a fixed point in its path, starting from rest at distance a from that point. Show that the velocity at distance x; from the point is equal to papomiie p ye) 11. A mass of 12 lbs. moving with velocity 20 feet per sec. is acted upon by a retarding force of 1-5v? poundals, where v is the velocity at any moment in feet per sec. Find its velocity after one minute, and the distance travelled. 12. A body of mass m lbs. is moving with velocity wu ft. per sec. If it be acted upon by a retarding force Kv? poundals, find the velocity after t; seconds. Also show that, though the velocity ultimately vanishes, the distance travelled increases indefinitely with ¢. 13. A mass of 1,000 lbs. travelling 20 ft. per sec. is acted upon by a re- tarding force of 100v+ v? poundals, v being the velocity in ft. per sec. at any moment. Find the velocity after 10 secs. 14. A body falls from rest under the action of gravity, against a resistance -02v? poundals per unit mass. Show that the acceleration vanishes when v is 40 ft. per sec. so that the velocity can never exceed this amount. Find also the velocity after 5 secs. 15. A body starts with velocity 1,000 ft. per sec., and has a retardation equal to -05v ft. per sec. per sec. Find the velocity after 25 secs. and the distance then described. 16. A chain 100 ft. long weighing 4 lb. per foot hangs in equilibrium over a smooth pulley ; if it be slightly displaced and runs over the pulley by its own weight, find its velocity when it leaves the pulley. | 17. A steamer of 5,000 tons displacement is advancing 15 m.p.h. relative to the land against a current running 3 m.p.h. whose retarding force is equal to 112 (v+ 4-4) poundals per ton, where v ft. per sec. is the velocity of the steamer relative to the land. Find the velocity of the steamer after 15 secs., also after what time it comes to rest, the velocities being reckoned as before relative to the land. 18. A captive balloon of mass 1,500 lbs. when loaded, has a lifting force equal to the weight of 100 lbs., and is attached to a rope of indefinite length weighing 3? lb. per yard. Neglecting the resistance of the air, find its velocity after rising 300 feet. 19. If in question 16 there be a constant frictional force at the pulley producing a retarding force on the chain equal to the weight of 2 lbs., find how far the chain must be displaced before it will start, and the velocity with which it leaves the pulley. If it be originally displaced one foot more than this find the time taken. 20. A light ball is thrown horizontally with a velocity of 75 ft. per sec. If the resistance to the horizontal motion be +0001 v? poundals per unit mass, where v is the velocity in feet per sec., find the horizontal distance travelled in 10 seconds. 21. Show that when a body is oscillating with damped harmonic motion the magnitudes of the successive oscillations on each side of the position of 292 SCHOOL CALCULUS [CHAP. IX equilibrium are in Geometric Progression. Assuming the small oscillations of a balance are of this type, and that the pointer of such a balance swings slowly over 9, 7, and 5 divisions on the scale past the zero mark on alternate sides in three successive swings, determine its position of equilibrium on the scale. 22. Show that if the damping force which acts on a body moving with har- monic motion is Kv poundals per unit mass, where v is the velocity in feet per sec., and the amplitudes of two successive oscillations are a and b, then K =3 logep where 7’ secs. is the period. 23. A body placed in a resisting medium is moving with damped harmonic motion. The body starts 3 ft. from its position of equilibrium and the force towards this point is then 1 poundal per unit mass, while the resistance of the medium is equal to oe poundals per unit mass when the velocity is v ft. persec. Find the distance moved by the body and its velocity after 3 seconds. IV. ACCELERATIONS, VELOCITIES AND SPACES DESCRIBED, GRAPHICALLY 23. (1) Suppose we have tabulated s1, so, s3, ete., the spaces described by a moving body in times #1, ts, ¢3, ete. If a graph be drawn showing the times as abscissae and the spaces or distances as ordinates the result is called a Space Curve. The slope of the tangent to this curve at any point P whose i ds . abscissa OM is t is of course ap te: the velocity at time t. If the intervals between the times 4, fo, t3, etc., are small, CHAP. Ix] APPLICATIONS TO MECHANICS 293 we can obtain fair approximations to the velocities at times half way between them without drawing a figure: for the slope of the tangent at a point whose ordinate is half way between Q;,N, and @e2No2 in the figure is approximately the same as that of the chord Q1Q2, which is given by oe Sep 144 raed which gives us therefore the velocity at a time half way between ¢; and ta: similarly we may find the velocity at a time half way between ¢t2 and tz. Applying this method again to the velocities so found, we can calculate approximately the ac- celeration at time fo. (ii) If a body has given velocities v1, v2, v3, etc., at times a7, Fie. 102. t1, to, tz, etc., we can similarly construct a Velocity Curve. dv Since / er the slope of the tangent at any point P gives us the acceleration at that moment; and since ds=vdt so that s=/vdt, therefore the space described between times ¢4 and tz is given by the area included between the curve, the axis of t and the ordinates at ¢, and fo. (iii) Similarly if a body has given accelerations f;, fs, 3, etc., at given times #4, to, t3, etc., we can construct its Acceleration curve. Then since dv=fdt, v=/fdt and therefore the velocity ac- quired by the body from time f, to time ft is given by the 294 SCHOOL CALCULUS [CHAP. IX Fie. 103. area between the curve, the axis of ¢, and the ordinates at ty and to. Notes. (1) These curves are often more exactly named Time-Space, Time-Velocity, and Time-Acceleration Curves to distinguish them from similar curves showing graphically the relations between Space and Velocity or Space and Accelera- tion of a body. (2) If will be seen that the Velocity Curve in any given case is the Differential Curve of the Space Curve, and the Integral Curve of the Acceleration Curve: and other similar relations among these various curves of motion will be obvious. (8) These curves can of course also be constructed when the equations to the velocity or acceleration of a body are given, and often form a convenient check of the results of a long calculation when they can be quickly constructed. EXAMPLE Draw the acceleration curve for the Example of § 11, hence find the velocities after 2, 4, 6, etc. secs: draw the velocity curve and so check the result found for s by calculation. (4) For finding the space described in any given time from a number of given velocities at given times or from a Velocity Curve, Simpson’s Rule for finding areas may sometimes be useful: and also similarly for finding a velocity from an Acceleration Curve. (5) In calculating these areas, attention must be paid to the units of length taken along the axes to represent units of time, space, etc. EXAMPLE 1. The following table gives the velocities of a CHAP. Ix] APPLICATIONS TO MECHANICS 295 body in centimetres per sec. at intervals of 2 secs. Draw the acceleration and space curves, and thence determine the approximate accelerations and distances covered after (i) 7 secs. and (ii) 11 secs: et Pee cheep | 8 | 10 [12 | | | - pmkaRy Ae 23-2 f meres lei Nel ite bf EEE EE mi Mee See | | SVOEMoOee Ue et! Has PEELE Ht ttt \ EGS oeeG Rae Gia ae ney | : |_| be BESS he epee NS bets Ts ROERE CA oeh na aes SH4 emai (speateial mastitis os feist a eherte| miele @ipwieie tele r aie lars Piso te ee he eb Eno PCCP RE a og lit | tT | SSR mmri gs non PEER isis sablavallele Jehn ise ctealwatiaiei a eit side PEPER San ee bead oh] ca et Qe Aa ac pe Lies tee | | | te || |42 couiaMa eae Guow mf an eal 1 faa an Vie. 104. 296 SCHOOL CALCULUS [CHAP, IX The small numbers on the vertical axis refer to the space curve, and the large numbers to the other two curves. The acceleration curve is obtained by drawing tangents to the velocity curve at convenient points, plotting their slopes and drawing a fair curve through the points so obtained. The space curve is obtained from the velocity curve by calculating the area between it and the time-axis up to con- venient ordinates (by counting the squares), plotting the results and drawing a curve through the resulting points. From the graphs we see : (i) After 7 secs. the acceleration= —5-5 cms. per sec. per sec. and the distance =228 cms. And (ii) after 11 secs. the acceleration=1-2 cms. per sec. per sec. and the distance =337 cms. e It was not of course necessary to draw the acceleration and space curves to obtain these individual results, but they exhibit graphically the changes in acceleration and distances described throughout the given interval of time. EXAMPLE 2. Find the accelerations at the same times in the last example by the method explained at the end of § 23 (1). To find the acceleration after 7 secs., 23164 dle see 10 f Sine me Oa —5-2 cms. per sec. per sec. and after 11 seconds we have ;_25—23-2_ 1:8 1220 a so that the results thus obtained are not very accurate, as we might expect, the reason being that the intervals of time are too large. =-9 cms per sec. per Ssec., EXAMPLES 1. The distances s feet travelled by a body in ¢ secs. are as follows: bee ene 2-2 2-4 | 2-6 | 2-8 3 8 | 6-7294 | 7-3009 | 7-8559 | 8 -3956 | 89214 | 9-4341 Find the velocities at times 2-1, 2:3, .... 2-9 secs., and thence the accelera- CHAP. IX] APPLICATIONS TO MECHANICS 297 tions at 2:2, 2-4, 2-6, 2-8 secs. Compare with the true values given by the formula » = 7 ormula v= 5 2. The velocities and times of a moving part in a machine are as follows : peer pe te Lie 6 uet-8 | 2. | 2-2 | 2-4, | 2-6 v ft. per sec. | 3-381 4 | 4-589 |5-142 | 5-657 | 6-128 6-553 6-928 7°251 Draw the velocity curve: find the distance travelled from ¢=1 to t= 2:6, and the accelerations at times 1-3, 1-9, 2-5 secs. 3. A circular cam with centre C and radius 1 inch is pivoted at a point O half an inch from C and revolves uniformly in a vertical plane at 500 revs. per minute. A rod AB slides vertically between supports in a line passing through O, and rests on the edge of the cam. Plot a curve showing displace- ments of the rod for one revolution of the cam: hence plot the velocity curve, and find the inclination of OC to the vertical when the velocity is a maximum, and the velocity at that moment. 4. If in the last question the rod AB be fastened to a horizontal cross piece DBF which rests on the cam, plot the displacement curve for the rod AB as before, and thence find its velocities when OC is (i) horizontal, and (ii) inclined at an angle of 20° to the vertical, with C below O. 5. A crank arm BC of length 1 foot turns uniformly about C at 100 revolu- tions per minute, and a connecting rod AB of length 3 ft. is attached to it so that A moves in a straight line through C. If in any position AB is pro- duced to meet a line through C perpendicular to AC in 7, it can be shown that CT’ measures the velocity of A on the same scale that CB measures the velocity of B in its circle. Plot the positions of crank and connecting rod ° for every 15° in a half revolution, and by measuring the lengths of C7’ con- struct a velocity-time curve for 4: from this read off the displacement of A from its starting position after CB has turned through 45° and 75°, and by comparing these results with the plotted displacements of A in the first figure, verify that CT’ correctly represents the velocity of A as stated. V. WORK AND POWER 24. If a constant force of P poundals move its point of application through a distance s feet, then W foot-poundals of work are done, where Weeks If this work be done in # seconds, the power or rate of working f is given by R= foot-poundals per sec., to (where v is the velocity in feet per sec., if that velocity is con- stant). 298 SCHOOL CALCULUS [CHAP. IX Since 33,0009 foot-poundals per minute, or 550g foot-poundals per second (the unit of power or of rate of working generally employed), is called one Horse-power Naples os Re ¢:550g 550g" 25. Work done by a variable force. If a variable force be equal to P; poundals after moving its point of application through s; feet, then the additional work done by this force in moving its point of application through any small distance ds; feet lies between P;-ds, and (P,+dP )dsj, ake dW, a | “S84. Hence generally dW =P-ds .. W=/P-ds foot-poundals, the integration as usual between suitable limits for s. (i) If P be given as a function of s, say P=¢(s), we have immediately W =/(s)ds foot-poundals. (ii) Otherwise from the data of the question as to the mass moved and the acting force we must express P in terms of s, and so find W as before by integration. Example. A mass of 200 lbs. is dragged along a rough horizontal surface (the coefficient of friction ~ being -4) by a rope passing over a smooth pulley 10 ft. from the ground. If the mass move slowly so that its kinetic energy is neglected, and start 15 ft. from QO, the foot of the vertical through the pulley, find the work done in moving it 10 ft. B 0 | | y 200 lbs. Fic. 105. Since the mass moves slowly, the tension of the rope and the friction against the ground do equal and opposite amounts CHAP. Ix] APPLICATIONS TO MECHANICS 299 of work, and we might calculate the work done by either ; but as the direction ag well as the magnitude of the tension are constantly changing, it is simpler to determine the friction, and the work done against it. Resolving horizontally and vertically, Tcosd=pR, TsinO =2009—R $e Pe eee ea from the figure. ph 200gx 200ug2 Sana R= 10 oa cane ane L0n-b-x S0gn |. = ? aad. rae since pw Now dW =phkdz. 15 80gx alee VW . Caress da =80yf~ (1+) dee aid 5 a+4 == 80g E —4log-(a+ 4) | =80g(10—4x +7473) =g x 560-9 foot-poundals = 560-9 foot-pounds. 26. Rate of Working when the Force is Variable. The Rate of working or power is, as we saw in § 24, equal to the product of the force acting and the velocity of the body moved. If therefore the force be variable, the Rate of working at any moment, if the force then is P poundals and the velocity v feet per sec., is given by 15 a) R=P-v ft. poundals per sec. since v may be considered momentarily constant, and at the moment required can be found as explained in §§ 11-13. The corresponding results for C.G.S. units are exactly the same, the force being measured in dynes, the velocity in cms. per sec., and the rate of working in ergs per sec. 300 SCHOOL CALCULUS [CHAP. Ix ExaMP_LeE. If the mass of the train in the Example to § 13 be 200 tons we have v P=(2+7 509) x200 x 2,240 poundals. v ipa (2+7500) foot-poundals t where from equation (i) v=(2,088e »”—2,000) feet per sec. Hence we can find # at any time in foot-poundals and so find the H.P. exerted against the train at that moment. EXAMPLES 1. An iron chain 30 ft. long, whose specific gravity is 7-8, hangs in water with the upper end in the surface. Find the work required to lift it just clear of the water if the total weight be 702 lbs. 2. A chain 50 ft. long, and weighing 25 lbs. per foot run, lies coiled up at the bottom of a rough inclined plane whose slope is 15° to the horizontal and coefficient of friction -5. Find the work done in drawing the chain slowly up by one end until it is just straightened out. 3. An engine of 20 H.P.is employed to pump water out of a circular well 10 feet across and 300 feet deep. Find the rate at which the level of the water is falling (i) when the surface is 100 feet below the ground (ii) after half an hour, if the well were originally full and water weigh 1,000 oz. per cubic foot. 4. A cubic decimetre of cork, of specific gravity -25, in the form of a cube floats in water with its sides vertical. Find the work done in pushing it down just level with the water. 5. A large mooring buoy floats in water with the part immersed cylindrica] and of parabolic cross section: its length is 13 ft., diameter at the water surface 64 ft., and the depth immersed 3} ft. Find the work required to raise it just clear of the water, supposing its plane surface to remain hori- zontal throughout, and a cubic foot of sea water to weigh 64 lbs. 6. The revolutions of the motor of a car going up a hill and using the same gear throughout fall off uniformly with the time in 6 minutes from 1,000 to 500 revolutions a min. At 1,000 revolutions per minute the engine develops 20 H.P. Find the work done in the 6 minutes, the pull being reckoned constant. 7. A water-cart weighs } ton and holds } ton of water besides. It is drawn at 4 miles per hour along a level road, the water all leaking out in one hour. Taking the coefficient of friction for the cart at -05, find the work done in foot-tons, and the average H.P. required to draw it. 8. If in the last example the water leaks out at a rate which varies as the square root of the depth still left at any moment, show that after ¢ hours (ee the cart has gone 4¢ miles and the water left is tons. Hence find as before the total work done and the average H.P. required, the water still taking one hour to leak out. CHAP. Ix] APPLICATIONS TO MECHANICS 301 9. A point moves with simple harmonic motion of amplitude 2a. Find the points in its path at which the work being done upon it is a maximum. 10. A vertical shaft bearing a weight W lbs. revolves in a loosely fitting cylindrical bearing with a flat base, and it is found experimentally that the coefficient of friction at any point on the flat end of the shaft which is moving Vv with velocity vft. per sec. is equal to , p being the normal pressure in lbs. at that point and ¢ a constant. Assuming p uniform everywhere, and neglecting friction round the sides of the bearing, show that the H.P. ex- One as pended in friction is independent of W and equal to ee where o is the angular velocity and a the radius in feet of the shaft. 27. Work done by any Force, graphically. If from an equation of the form P=¢(s) or from independent data, we know that the values of a force acting on a body are P;, Po, Fie. 106. P3, etc. poundals when it is at distances s1, so, 83, etc. feet from a fixed point O, by plotting the distances. as abscisse and the values of P as ordinates we obtain what may be called a Space-Force Curve. Compare with this Fig. 108. Since, as shown before, dW =P-ds, the work done by the force in moving the body from distance s, to distance se feet from O is given by reef P-ds foot-poundals that is by the area between the curve, the axis of s, and the ordinates at s; and so. This area may be obtained by counting the squares on ruled paper, or if suitable by Simpson’s Rule for areas, 302 SCHOOL CALCULUS [CHAP. 1x attention being paid always to the units of length employed along the axes to represent units of force and of distance. An Important Case: P varies directly as s. Let P=ks, where & is some constant, and let it move its point of appli- cation from s1 to So. Then dW =Pds=ksds Sy k . Wea ksds => (So? —84?). This may be written 22851 (So—8}). But ks2 is the final value of the force, say Pz and ks; the initial value, P;; while so—s, is the distance moved. Hence W =(mean of initial and final forces) x (distance moved). We may also obtain this result graphically as follows : Let OS;=81, OS2=S82, SR 1, SoRe the values of the force at S41, So; RyS1_ RaSo Then since P=ks — =k. —=O)S, “Oss R, Ry QO S; Ss, Fie. 107. Hence Of Re is a straight line, and ordinates up to this line represent the value of P at any point, i.e. Fig. 107 is the Space Force Curve. Hence the area S8;h,R2S2 represents the work done: and the area of this trapezium =(mean height) (breadth) _Sifi+S2Ro GHGE 9 =(mean force) (distance moved). ad If the force start from a zero value at the point O, the work done is measured by the area of the triangle OR?2S2 RoSo 5 OS» =(half the final force) (distance moved), CHAP. 1x] APPLICATIONS TO MECHANICS 303 which follows from the preceding, since the initial force is then zero. There are many practical cases of work done by a force which is proportional to the distance moved. No substances are absolutely rigid, though they are often assumed to be so, and when acted upon by forces, usually undergo some change of shape. Such forces are often called Stresses, and the change of shape is called a Strain. For instance any substance can be compressed or elongated to a greater or less extent, de- pending on its nature, by a direct push or pull in the direction of its length ; it will bend under a force in a perpendicular direction, or twist under the action of a couple. The amount of strain in each case is proportional to the stress by Hooke’s law, provided it does not exceed a certain amount, and the s7e® is called the Modulus of Elasticity of the substance for that particular kind of strain. For a simple extension or compression, this ratio is known as Young’s Modulus, and denoted by £. Its value varies greatly for different substances from about 480,000 for some kinds of soft wood to 29,000,000 for wrought iron, the stress being expressed in lbs. per square inch. This means that a tensile force of 29,000,000 lbs. would stretch an iron bar whose cross section is 1 sq. in. to double its length, supposing it would stretch so far before it broke. Similarly for a twisting strain, the ratio is called the Coefficient of Rigidity ratio and denoted by C. Its value for most metals is about 2. In such cases if a force P produce an elongation of amount 2, or a twist through an angle 6, we have P=Hx or P=C@, so that the force is proportional to its displacement, and the work done in such a strain is found as explained at the be- ginning of this section. EXAMPLES 1. A motor-car weighing 8 cwt. starts from rest and the effective pull of — its engine in pounds weight after moving for ¢ secs. is as follows: | | | | | | | | Beeeofi}2) 3 )4) 5 }6[7/8| 9 |10/11| 12 | 13 |14 | nee | | | | _P .. 15 21 28 35-5 45 55-5 GL 68 69 66-5 62 57 53-5 51-5 50- t Show that | if ' Pdt = mv, where m is the mass of the car and v its velocity at the end of ¢; secs. Hence find graphically the velocity after 14 secs. 304 SCHOOL CALCULUS [CHAP. Ix 2. A steam cylinder is arranged with rope tackle to lift a weight of 600 Ibs. through 12 feet in a single stroke, and the total effective steam pressures P in the cylinder at different heights h feet are as follows : hteot ../ 0 | 2°) 3 | 4 | 5 | 6) Salome | | | Pibs. .. 1,000 990 940 790 640 530 400 820. 270 Find graphically the total work done on the weight and deduce its velocity at the top (g = 32:2). . 28. Work Done in Compressing a Gas. Suppose a volume of gas, initially at pressure p;, to be compressed from a volume v; to a volume vz: also since the shape of the containing vessel is immaterial, let us for convenience suppose the gas contained in a smooth cylindrical vessel with a close fitting piston which can move freely in the cylinder, the surface of this piston being A units of area. At any stage of the com- pression let the pressure be p and the volume v, and suppose the piston now to be pushed inwards an additional distance 6a units against the pressure of the gas, which will meanwhile increase to p+ép. Then the acting force les between p-A and (p-+6p)A, and acts through a distance 62: therefore the work done is in the limit pAdz. But Adz is the small change dv in the volume of the gas, dW =pdv | and w=f- pdv. 1 To integrate this p must be expressed in terms of v from the law which connects p and v for the gas under consideration. This is usually one of two laws : (i) pv=constant, when the compression is Lsothermal. (ii) pv’ constant, where y is some constant quantity ; S and y has the particular value h when the compression D is Adiabatic, where S, is the specific heat at constant pressure, and S, is the specific heat at constant volume for the particular gas. Notes. (1) Isothermal compression or expansion means — that the temperature of the gas is kept constant throughout. Adiabatic compression or expansion means that no heat enters or leaves the gas, so that its temperature rises during compression and falls during expansion. CHAP. IX] APPLICATIONS TO MECHANICS 305 (2) It is clear that the work done by an expanding gas is obtained by the same equation as that for the work in com- pressing a gas, since each stage of the operation is immediately reversible, i.e. if the gas be allowed to expand slightly it will do an amount of work pdv. : (3) It is clear that the work done depends solely on the volume of the gas (besides the pressure) since the gas will distribute itself uniformly throughcut any containing vessel, and therefore is independent of the shape of the volume considered. (4) The common values of y for adiabatic expansion are y=1-408 for dry air, and most of the permanent gases. =) for saturated steam. ==-1-66 for mercury vapour. Work done during Isothermal Compression or Expansion. D We have W=/ ? pdv Vy and pu=c which may be written VW Coe =~pv log.r where r is the ratio of compression or expansion (cf. Chap. VI., § 7). Work done during Adiabatic Expansion. We have now w=f es pdv VD and pv’ =c : C ee oe % dy Vgl-F —vu,1-« aaee6 — = ene v vr ay 20 306 SCHOOL CALCULUS [CHAP. 1X 2—P1%1 |. which may again be written in the form Ws Chap. VI., § 8). 29. The Steam Engine Indicator Diagram. By suitable apparatus connected to the cylinder of a steam engine in action, it is possible to make the engine draw its own work diagram. The approximate shape of such a diagram is that shown in A B cD K 3 Wy Wy) se) KR R, Mae Y ) ee | | | D H Stroke K Fie. 108. the figure, the sequence of operations being as follows. At F the slide valve opens and admits steam when the pressure rises to HA, and remains constant while the piston moves to B on the forward or effective stroke. The valve then closes and the pressure falls along the adiabatic line to KC where the exhaust opens and the pressure falls to KD, KD representing the pressure in the condenser, which is usually considerably less, in modern engines, than the external atmospheric pressure as represented by the line XY. The piston returns to H sweeping out most of the steam, when the exhaust valve closes, and the pressure of the remainder rises to HF at the beginning of the stroke, when the operations are repeated. If the engine be double acting a similar diagram of course applies to the other end of the cylinder. During the forward stroke an amount of work repre- sented by the area ABCKH is done by the steam (see § 27), and during the return stroke an amount DHFHK is done against it. Hence the net amount of work gained per stroke is repre- sented by the area of the diagram ABCDEHF ; and when the CHAP. Ix] APPLICATIONS TO MECHANICS 307 scale of the drawing is known, HK representing the stroke of the piston in feet, and HA the maximum pressure in lbs. per sq. inch, usually taken to be that of the boiler, the actual work done in ft. lbs. per stroke can be calculated. EXAMPLES 1. Find the work done by 2 cubic feet of saturated steam at pressure 120 lbs. per square inch expanding to three times its volume, the law of expansion being pu = c¢. 2. Find the work in the last example if the law of expansion is put’ = C, 3. A given quantity of dry air is compressed to one quarter of its volume. Find the ratio of the works done according as the compression is (i) Isothermal (pv = ¢) (ii) Adiabatic (pv!'48 = c). 4. Find the ratio of the works as in the last example if the air be now compressed to four times the original pressure. VI. VIRTUAL WORK 30. It is proved in Text Books on Statics that a body or system of bodies acted upon by any forces is in equilibrium if the sum of the virtual works done by the forces, when the system undergoes a small imaginary displacement consistent with the geometrical conditions of the system, is equal to zero. The displacements of the bodies, and consequently of the points of application of the forces, being very small, can be conveniently treated by means of the Differential Calculus. For if the positions and lengths of the various parts of the system can be expressed in terms of one length x (or one angle @) in the figure, we imagine this variable to become xz+dxz (or 8+d6), and so calculate the displacements of the points of application of the forces along their lines of action as small quantities all involving dx (or d@). Hence the virtual work done by each force can be found, and equating their algebraic sum to zero, we have an equation from which dz (or d@) can be divided out, and the value of one unknown thereby obtained. Any motion of the points of application of the forces (re- sulting from the supposed small displacement of the bodies), which is at right angles to the lines of action of the forces must be neglected, as the forces can do no work in lines at right angles to their own lines of action. ExAMpLE. Four equal uniform rods each of weight W are smoothly hinged to form a rhombus ABCD. B and D are kept apart by a rod of weight w, the angle at A being 2a, 308 SCHOOL CALCULUS [CHAP. IX and the whole is suspended from A. Find the thrust in the rod BD. | Bia. 109: Suppose the rod BD to lengthen slightly, so that the angle a becomes a+da: then supposing A to remain fixed it is clear that all the centres of gravity of the rods rise slightly. Now the length of the rod BD and the depths of the centres of gravity below A can all be expressed as functions of a. Let the 4 rods be each of length 2a: then BD=4asina, the depths of G and ZL below A are each acosa, those of H and K being each 3acosa. Now since BD or I equals 4asina C o =4acosa giving us the ratio of the change of / to the change of a .. dl=4acosa da giving us the actual small increment of J. Similarly the changes in the depths below A of G and H are their differentials, namely —asinada and —3asinada re- spectively, the negative signs showing that the depths diminish, and therefore G and H rise. Also AM =2acosa. .. MW rises by —2asina da. CHAP. Ix] APPLICATIONS TO MECHANICS 309 Hence equating the sum of the Virtual Works to zero 2W(—asina da) +2W( —3asina da) +-w(—2asina da) +7 (4acosa da) =O where 7 is the required thrust in the rod BD. .. 4aT' cosa =(8aW +2aw) sina — ne t ana Note. The dotted lines in Fig. 109 represent the rhombus in its displaced position. Now the change of position of G, for instance, is partly vertical and partly horizontal. But the weight which acts through G can do no work horizontally, and therefore this part of G’s motion must be neglected, as explained above. Similarly the rod BD both lengthens and rises, but both these displacements have to be taken into account: the lengthening gives us the work done by the thrust in the rod which acts along it, but is neglected as far as its weight is concerned; while the amount by which it rises gives us the work done by (or in this case, against) the weight, but is neglected in considering the work of the thrust. 31. It will be noticed that in the last example the reactions at the hinges were neglected ; for the reaction at any smooth hinge consists of two equal and opposite forces which do equal and opposite amounts of work for any displacement. Similarly normal pressures between smooth surfaces may be neglected, since the point of application moves, if at all, in a direction at right angles to their line of action, and they therefore do no work. The usual types of forces whose virtual work has to be found in such problems are : Weights, whose distance and displacement above or below. some fixed level must be obtained. Tensions in strings and thrusts in rods, when the whole length of the string or rod, and its extension or contraction must be found. Friction, for which we have to find the displacement of the point of contact at which the friction acts. ExampLE 1. A heavy uniform rod of length 2a inches rests partly within and partly without a fixed smooth hemi- spherical bowl of radius 7 inches whose rim is horizontal. Find the angle @ between the rod and the horizon in the position of equilibrium. 310 SCHOOL CALCULUS [CHAP. IX The only forces acting on the rod are the reactions # and S at A and C, and the weight W at G. Sas Fic. 110. Suppose the rod to slip slightly in the bowl, so that @ be- comes 0—dé. Then as noted above R and S do no work. Also the total virtual work is zero, so that the virtual work of W also vanishes. Let GC =z inches, and GH =y inches Then 2 =2rcos@ —a and y=xsind , =(2rceos@—a)sin@, which is the distance of G below the fixed level CD. Hence dy ={cos@ (2rcos@ —a) —siné x 2rsin@} dé = {2r(cos?@ —sin?0) —acos0} dé = {2rcos26—acos@} dé . W(2rc0s2@ —acosé) =0 .. 2rcos20=acosé, which determines 0. Note. If W does no work, this means that G moves hori- zontally or that y does not alter in length. ExampLe 2. A symmetrical pair of steps weighs 50 Ibs., and has the middle points of its legs, which are 8} feet long, joined by a cord 2? feet long. The steps are placed on a polished floor, the coefficient of friction being -1, and a man weighing 12 stone goes 6 feet up one side of the ladder. Find the tension in the cord. CHAP, Ix] APPLICATIONS TO MECHANICS 31] A lies 1D Let ABC represent the ladder, and M the man, so that CM =6 feet. : Let & and S be the reactions at B and C. Let the semi- vertical angle be a, and suppose both feet to slip a small distance, so that a becomes a+da. Then JM is 6cosa feet above the ground *. its displacement = —6sinada ft. Similarly the displacements of D and # are each —4}4sina da ft. Again BO =:CO =8isina *, the displacements of 6 and C, where the frictions a and a act are each 84cosa da feet. And ED is 84sina, so that its extension is also 84cosa da ae Hence the equation of Virtual Work is Iigag ae! —168g x 6sina da — 2x 25g x 44sina da+(* + a) 8icosa da +7'x 8icosa da=0 where 7’ is the required tension of the cord in poundals ANN ie ae Te 10 poundals. 312 SCHOOL CALCULUS [orraP. IX Also resolving vertically R-+S =(168-+50) g=218g and since ED =23 feet, sina=5,= "3285 whence tana =: 3419. rer C=. eee) Uy 10 _=gxX 27-3 poundals, nearly =27:3 pounds weight. EXAMPLES 1. Four uniform rods each of weight W lbs. are jointed together to form arhombus: 2 corners are kept apart by a light rod, and the whole is suspended bv one of the other corners, at which the angle is 2a. Find the thrust in the diagonal rod. 2. Four uniform rods each 2 feet long and weighing 4 lb. are jointed to form a rhombus ABCD. A string of length 3 ft. joins AC, and the whole stands in a vertical plane with AD on the ground. Find the tension in the string. 3. Six equal and uniform rods each of weight W are jointed together, and the middle points of two opposite sides are connected by a string of such a length that when the whole is suspended by one of these sides which is held in a horizontal position it hangs in the form of a regular hexagon. Find the tension in the string. 4, If the hexagon in the last question be suspended as before, but is now maintained in the same shape by a light rod joining the middle points of the two upper sloping sides, find the thrust in that rod. 5. A rhombus ABCD of four freely jointed rods each weighing 10 kgms. has AD held in a vertical position, and the middle point of AB is joined to the lowest corner C by a string of such a length that the angle BAD is 60°. Find the tension in the string. 6. A ladder weighing 150 lbs. rests on a smooth floor and against a smooth wall, and is kept from slipping by a cord fastened to its foot and to the corner of the floor and wall. Find by Virtual Work the tension of the cord when the inclination of the ladder to the ground is (i) 70° (ii) 20°. 7. In the example of § 29 show that the same result would be obtained by supposing the rod BD to contract slightly, so that a becomes a — da. 8. A rod of length 2/ rests with one end against a smooth vertical wall and over a smooth peg at distance b from the wall. Show that the height of its middle point above the level of the peg is equal to Jcosa — beota, where a is its inclination to the wall. Hence find by virtual work the value of a when the rod is in equilibrium. 9, A rod 8 inches long rests with one end against a smooth vertical wall and the other end fastened by a string one foot long to a point in the wall vertically above the first end. Find the inclination of the rod to the wall when in equilibrium. 10. A rod rests in a vertical plane with one end on a smooth plane which is inclined at an angle of 50° to the horizon, and the other end on a smooth Hoor, making an angle of 20° with it: also this end is fastened by a string to CHAP. IX] APPLICATIONS TO MECHANICS 313 the junction of the floor and the inclined plane. Find the tension in the string if the rod weighs 30 lbs. 11. A string is slung over two pulleys in a horizontal line 10 feet apart and carries weights of 10 lbs. at each end, and 12 Ibs. between the pulleys. Find by the principle of virtual work the depth of the 12 lb. weight below the line of the pulleys. 12. If in example 6 the ladder is 25 feet long, and the cord is fastened to a rung five feet from the lower end instead of to the end of the ladder, find the tensions as before. 13. An elastic ring of natural radius r and modulus three times its weight rests in equilibrium on a smooth cone whose vertex is upwards, its height being h and radius of its base a. Show that its distance below the vertex of the cone is ee If a=h and r= 4 in., what does this become. 14. Three rods each of length 2/ and weight 14 lbs. are freely jointed to- gether at one end to form a tripod. Their middle points are joined by 3 strings each of length J, and the whole rests on a smooth floor. Find the tension in each string. Find also the tension if a weight of 6 lbs. be placed at the vertex of the tripod. 15. If in the last example there be no weight on top, and the strings are elastic of natural length / and modulus 14 lbs., show that the angle a between any pair of rods in the position of sae tee by, is given by 2sin5= V 3 (2sin D./3 — 4sin?5 16. A square Feren is hung up by one corner, and a rod of one quarter oi its weight, sliding in a smooth vertical guide at a horizontal distance from this corner equal to one eighth of.a side of the square, presses on one edge, there being no friction between them. Show that this edge rests at an angle 6 to the vertical where 16sin26(cos@ — siné)=1. Find graphically the approximate value of 6. VII. CENTRES OF PRESSURE 32. If a plane surface be immersed in a liquid, the whole pressure acting upon one side of it is composed of a large number of parallel forces each acting on a small element of the surface. These parallel forces have a resultant acting at a definite point in the surface, which is called the Centre of Pressure. (i) If the surface be placed horizontally in the liquid at a depth x feet and contain S sq. feet, and the density of the liquid be p pounds per cubic foot, the pressure on the surface is equal to gpxS pounds weight, and the centre of pressure is the centroid of S : if in addition there is a pressure » pounds per square foot at the free surface of the liquid (e.g. atmospheric pressure) the whole pressure upon S becomes (p-+gpz)S Ibs., while the position of the centre of pressure is unaltered. 314 SCHOOL CALCULUS [CHAP. IX (ii) If the plane surface be immersed at an angle @ to the horizontal, the pressures upon different parts of the area S which are at different depths are no longer equal, but the whole pressure upon one side of the surface is equal to the sum of all the different pressures upon the small elements of area composing the surface, and the centre of pressure is that point in the surface at which this whole pressure may be considered to act. In Fig. 112 let the plane of the immersed surface KQLQ’ Fie. 112. meet the surface of the liquid in the line AB, and let LK (which for convenience is here supposed an axis of symmetry) meet AB at D at right angles. Let the surface be inclined at an angle @ to the horizontal, and let & and / be the depths of the highest and lowest points respectively. Let QQ’ be an elementary strip of area dS at depth x, then the pressure on this element is equal to the weight of the superincumbent liquid, so that if P be the whole pressure sought, dP =gpadS where p is the density of the liquid. A l eee JE gpxdS, and dS must now be expressed in: , xa=k ; terms of x and the integration effected between the limits. k to l for x. Again let C be the centre of pressure, at depth x below the surface; then the moment of the whole pressure about AB is Px CD=Px « cosec@: also the moment of the pressure on any element such as QQ’ about the same line is gpxdS x ND=gpxdS x x cosecd CHAP. Ix] APPLICATIONS TO MECHANICS 315 and since the sum of the moments of these elements of pressure must be equal to that of the whole pressure ral .. P+x cosec@ = osec@ 2d 8 gp cosectf "x i x= l “P-z=gof — vd S Notes. (1) This result is apparently independent of 40, and it is proved in any elementary Text Book on the subject that if a surface immersed in a homogeneous liquid be rotated about the line in which its plane cuts the surface of the liquid, its centre of pressure is unaltered. But it is obvious the whole pressure on one side would be altered by such a rotation, for parts, if not the whole, of the figure would be raised or lowered in the liquid, so increasing or decreasing the pressure upon them. (2) Hence in a homogeneous liquid, when the centre of pressure only is required, we may consider the surface rotated into a vertical position, and in our figure DK would then be depth of the highest point and DL that of the lowest. Calling these depths again k and J, and calling KN z (instead of DN), the depth of QQ’ becomes x+& and the pressure upon it becomes gp (x+k)dS x=I—k a I= ° of se (c+k)dS Now if y=¢(x) be the equation to the boundary of the surface, referred to K as origin and KL as axis of a, the area QQ’ or dS becomes 2ydx or 2¢(x)dx Ik sd ate k d sof (w-+k) p(x) de and if further we are given KL =a, this simplifies to if =24f~ (c+k) d(x) da which is now in the ordinary form for integration. 316 SCHOOL CALCULUS [CHAP. IX Exactly similarly we may write a 29pf * (ek)? b(e) dx apf (x-+k) d(x) dx S* (e+ b) par) de fe (c+k) (2) hs This is the procedure in general for finding x which now gives us the position of C in KL, not merely its depth below the surface, but again we give the warning that P as found above is not the same as when the surface was left inclined at an angle 6 to the horizontal, though the position given for C is the same. (8) If one end (or side) of the surface is in the surface of the liquid, we have only to put k=0 in the above results. (4) If the liquid is not homogeneous, but varies in some way at different depths, so that the density p at any depth x iS given by p=wW(x) say, then referring again to Fig. 112 the pressure on any element of area dS such as QQ’ is now given by dP=gw(x)-adS. | Integrating this gives us P. Also the moment of dP about AB is gw(x)-adS x x cosec@ so that we get similarly } —__ fxr rp(u) dS : ae si Wa) dS between suitable limits, and in this case the surface cannot be rotated about AB, for to do so would bring parts (or the whole) of the surface into new strata of the liquid of different densities from the original. (5) If further there be a pressure p at the free surface of the liquid, this will give an additional pressure pdS on the element of surface QQ’, so that dP =(p+g ple) x}dS whence we can find P. Kptg v(x) xed HP +g v(x) &}ds (6) In every case it must be carefully noted that until the Similarly again «= CHAP. Ix] APPLICATIONS TO MECHANICS ole numerator and denominator of the fraction which gives us x is completely integrated between its limits, no cancelling can be effected except of such constants as can be placed before the signs of integration. (7) We have assumed throughout that KZ is an axis of symmetry, and there generally is such an axis. But if this is not the case we can still theoretically find the depth of C or the distance along DZ at which a horizontal through C cuts DL provided the integration necessary is possible; and this as a rule gives us all that we require. It will be seen that the method is in general the same as that employed for finding areas and centroids, or centres of gravity, etc., and its application will be clearer from the following examples. ExampLe 1. A square whose diagonal is 2a cms. is immersed in a liquid whose density is p grms. per c.c., with one vertex in the surface and the diagonal through it vertical. Find (i) the pressure on one side of the square; (il) the position of the centre of pressure. In the figure the area of any small horizontal slice at PP’, Fie. 113. z cms. below the surface, is 2ada sq. cms. since PM=-AM =x. Hence if P; be the pressure in grms. on ABD, dP1=2gp xdu-x —=29p x7dx. 318 SCHOOL CALCULUS [CHAP. 1X ha 29of “aed 0 a3 =29p “3 germs. wt. Again QQ’=2NC =2(2a—2x); so that the area of a small horizontal slice at QQ’ is 2(2a—a)dax, and if Po be the pressure on BCD dP2=2gp (2a—x) dx-x " =2gp «(2a—x) dx. Zo ee =240f- x x( 2a —2) i 2a =29p [ ax = a3 =4gp ‘9 grms. wt. (i) Hence the whole pressure Po on one side is given by Po=P,+Po a? as =29p + 49p =2gp a> grms. wt. (ii) The centre of pressure lies in AC by symmetry ; let its distance from A be a cms. Then the moment of the pressure on the element at PP’ about O is 2gp%*dxx au; and similarly the moment of the element at QQ’ about O is 2gpx(2a—ax)dxx x: and the sum of all such moments is equal to the moment of the whole pressure 2gp a? about O a 2a ‘.2gpae?x x=f/ 2gpxedx+/ _ 2gpx2(2a—x) dx P 0 a a 2a =29p UE widetf- (2a22—z8) de) : ie 2axr> eA 2=51/ 5 I Us —T | poms. CHAP. IX] APPLICATIONS TO MECHANICS 319 EXAMPLE 2. The area cut off from the parabola y2=4cx by the line x=b is immersed in a liquid of density p, in a vertical plane, with its vertex uppermost at a depth a below the surface, and its base horizontal. Find the pressure on one side, and the position of the centre of pressure. Consider the pressure on a narrow strip at a depth xz below the vertex, and of breadth dx. Its area is 2ydx; and there- fore, if P be the whole pressure required, dP, the pressure on this strip, is equal to 2ydx-gp(a+2) “. P=29p/(a+x)ydx ! b A: =t9pve f- (at+a2)Vada (for y2=4czx) - fab” bd’ +2 __ 8gpc° a (5a +35). Also the moment of the element of pressure dP about the line in which the plane of the parabola meets the surface is dP (a+x) which is equal to 2gpy(a+x)?dx and if x be the depth of the centre of pressure 5 sgove f° (a+a)? V xd x Ie 3 5 Z b _ [2a2x" 4ax® Qa” =4gpVe | Te aes i} qe as 89pVe (35025? +4206" 1 156") wr 105 3502 +.42ab +1502 Me : depth. “(5a £36) the required dep 3b(7a-+5b) 7(5a +3b) the depth of the centre of pressure below the vertex. for If we subtract a from this quantity we get 320 SCHOOL CALCULUS [CHAP. IX ~ If the vertex be in the surface, so that a=0, we see that the depth of the centre of pressure below the vertex is = If we put a@=oo, in which case the pressure becomes uniform over the whole surface, the depth of the centre of pressure below the vertex becomes 3b(‘7a-+5b) _ 3b(7-+-5bd) ee Ha ea (5 3h) St a5 abe (putting a=7) 4 3b mie which is the position of the centroid of the area, see Chap. IX.., § 4, Examples 4. EXAMPLE 3. A trapezium has its parallel sides of lengths 2 and 3 feet, and the distance between them is 4 feet. It is immersed in a vertical plane in a liquid whose density varies as the depth, with its parallel sides horizontal, the shorter one being uppermost, and 2 feet below the surface. Find the depth of the centre of pressure. Cc BREAD Fic. 114. In the figure A# is drawn parallel to BC. Let PQR be a strip at depth x below the surface, so that AN =a2—2. Then since DE =}AK, .. by similar triangles PQ=}4AN = 4 —2 CHAP. IX] APPLICATIONS TO MECHANICS 32] Now PQR is at a depth wx ft., and the density of the liquid at that depth is Xx lbs. per cub. ft. say. *. the pressure on the element PQR is gar dx \bs. and the moment of this pressure about the surface is grad" 9 de ft. lbs. D -23(x-+6) da 2 ™ [220 +6) dx 2 Je + 6x3) da 2 Sf + 6x?) dx 0 Ce ot. |” mispes 2 | I> t=] paea0G. 8 aha The centre of pressure is therefore 4-713 feet below the surface, that is to say 2-713 feet below AB. We can now fix its position exactly, since it clearly lies on the line bisecting 4B and CD. If in addition there were a pressure p lbs. per square foot at the surface of the liquid (e.g. the atmospheric pressure) the pressure on the element PQR would be increased by eae alba, So that we should now have carn t= = 4-713 feet dP =(grx2-+p) ele dx Ibs. and the depth of the centre of pressure would be given by 6 Ve (grax? +p) x-(x+6) dx ale feet. So (D2 +p) (040) de 21 322 SCHOOL CALCULUS [CHAP. IX EXAMPLES 1. Find the pressure upon one side of a triangle of base 3 feet and height 2 feet immersed in water in a vertical plane with its vertex in the surface and its base horizontal, if water weighs 624 lbs. per cubic foot (i) neglecting atmospheric pressure (ii) assuming that pressure to be 14-7 lbs. per square inch at the surface of the water. 2. If the same triangle be immersed with its base in the surface in a vertical plane, find the pressure on one side and the depth of the centre of pressure, neglecting any pressure at the free surface. 3. If a triangle whose height is 15 cms. and base 24 cms. be immersed in a liquid of density 1-5 grms. per c.c. with its base 20 cms. below the surface and vertex downwards, and the pressure at the free surface be 1 kgm. per sq. cm., find the pressure on one side, and the depth of the centre of pressure. 4, If a rectangle be immersed in a liquid of density p lbs. per cubic inch, with one side of length b inches in the surface and the other of length c inches, the plane of the rectangle being inclined at an angle of 30° to the horizontal, find the pressure on one side, assuming a pressure at the surface as in Example 1 (ii). 5. If the same rectangle be immersed vertically in a liquid whose density at depth h inches is pA lbs. per cubic inch, find the pressure and centre of pressure (i) when the side 6 is in the surface (ii) when it is at a depth k inches below the surface, neglecting any pressure at the surface. 6. A square is immersed in water with one vertex in the surface, and the diagonal through that vertex inclined at an angle of 75° to the vertical, the other diagonal being horizontal. Find the pressure on one side, if the pressure of the atmosphere be 14-7 lbs. per square inch at the surface of the water, and the diagonal of the square 3 feet. 7. A rectangle whose sides are a and 6 inches long is in a vertical plane in a homogeneous liquid of density p lbs. per cubic foot, with one vertex in the surface and one diagonal horizontal. Find the centre of pressure neg- lecting any pressure at the surface. 8. A lamina in the form of a parabola is immersed vertically in water, with a double ordinate 5” long in the surface, and its vertex 5” below the surface. Find the pressure on one side, and the centre of pressure, neglecting any pressure at the surface of the water. 9. A semicircle of radius a” is immersed vertically in a homogeneous liquid of density p lbs. per cubic foot with its diameter in the surface. Find the depth of the centre of pressure (i) When the pressure at the surface is TT lbs. per square inch. (ii) Neglecting any pressure at the surface. 10. The same circle is similarly immersed, but its diameter is horizontally placed below the surface, and the extremity of the radius at right angles to this diameter is in the surface. Find the depth of the centre of pressure neglecting any pressure at the surface of the liquid. MISCELLANEOUS EXAMPLES * 1. Draw a graph showing the number of criminals con- victed for the years 1896-1905 from the following table : Year i896 | 1897 | 1898 | 1899 | 1900 Number of criminals .. | 11,870 | 12,130 | 12,516 | 11,865 | 11,082 Year 1901 | 1902 | 1903 | 1904 | 1905 Number of criminals .. | 11,924 12,490 13,162 13,736 | 14,160 2. If y=3x2+5, find the increase of y while x increases from 2 to (i) 2-5, (ii) 2:05. Hence find the average increase of y per unit increase of x during the same intervals. 3. Find the slope of the chord joining the points whose abscissae are 1-75 and 2-25 on the curve y=(x—2) (x—3), and explain your result by a graph. 4. Complete the following table oy 4 y 6x 5:85 34°23 5:9 34°81 5:95 35°40 6-0 36 5. Draw the graph which lies evenly among the following points 1-05 4°25 1:6 | 1-82 | 2 2-35 | 71) 1d ae Des | -86 sii | 4°75 | v= and hence find the probable relation between v and t. * The Answers to these Examples are not included in this volume, but may be obtained separately, price 6d. (post free, Td.). 323 324 SCHOOL CALCULUS 6. The following table shows the expectation of life of females : See 35 30-90 25 37-98 20 41-66 10 49-26 5 53°08 Present age Expectation of life in years Draw a graph, and hence find the expectations of life of a girl of 15 and a woman of 30 respectively. 7. Find the values of sy when y=(3—z)?, x=1, and (i) dba=-02, (11) 62=-002. 8. Find the average gradient of the curve 2y=3xz2—5 from x=4 to x=6, and its actual gradient when x=5. . : oS l 9. Differentiate (1) “go (il) < 2.sia cose 10. Complete the following table : t 8 58 8 bt 3°05 23-812 3°06 24-055 3°07 24-294 3:08 24-537 3°09 24-784 11. The following table gives the produce in millions of £ of an income tax of 1d. in the £ for the years stated : Year .. me | 1896 1897 1898 | 1899| 1900| 1901 1902 | 1903 | 1904 | 1905 Amount in mil- lions per ld. tax 2-03 | 2-10) 2-19 2°28 | 2-35 2°48 | 2-53 | 2-54| 2-56| 2-58 Draw a graph, giving reasons for the form with which you draw it. | 12. From the following table find the slopes of the 4 chords joining the five points whose abscissae are 1, 2, 2-5, 2°75, 3 1 2 2°5 2°75 3 0 Y rl —- 1:414 | 2°828 2°281 y= 2°795 Draw the graph passing through these points. MISCELLANEOUS EXAMPLES 325 13. Find the slope at any point on the curve s=3t+32#2 and also at the points given by t=0 and t=3-5. If in this equation s is the distance in feet of a body from a fixed point after ¢ seconds, what do these slopes mean? What would be the meaning of the slope of the chord joining the two points above ? , 4 14. Differentiate (i) eB V3 A/S ee - 1—73 (1) loge 353 15. Complete the following table : oy is Y 62x 3°51 49-91 3°54 50-24 3°57 50°57 3°60 50:9 3°63 51-23 What do you deduce from your results as to the relation between x and y? Find this relation. 16. Find graphically those values of @ less than 360° which satisfy the equations (i) 10°7 sinz=2-6 cosx (ii) 10-7 sinv+2-6 cosr=3. 17. Draw a graph of the English trade done per head of population, from the following table : Year ..| 1897 | 1809 | 1901 1903 1905 ‘Trade per person .. £18 12s. 9d. | £19 19s. 7d. | £20 18s. 8d. | £21 6s. 3d. | £22 10s. 1d. 18. At what points on the curve 2y=—4a3—xz*—4r7+14 are the tangents parallel to the axis of x ? 19. Find the average increase of y per unit increase of x if y =10x —2x? while x increases from (i) 3:5 to 5; (ii) 3-5 to 6-5, Explain your second result by a figure, 326 SCHOOL CALCULUS 20. Complete the following table : 6A 624 Hi = 5D 5D? 5-02 19-7923 5:03 7:90 5:04 7°91 5:05 7°92 The relation between A and D is of the form A=kD". From - your table decide n, and then calculate k. 21. Draw a graph of the expectation of life for males from the following table : Dprocentare oe ee oo | 5 | 10 | 15 25 35°68 35 | = | Expectation of lifein years... | 50-87 | 47-6 | 43-41 28-64 Hence find the expectations of 2 men aged 20 and 30 respec- tively as nearly as you can. 22. The following table gives the gradients of 4 chords joining the first and second, the second and third, and so on of the 5 points whose abscissae are given, Abscissae .. eh ae eas, | 1-5 | 2 | 2°5 | 3 | Gradients of chords... «382s | —-172.| —-502 | —2-236| If the curve passes through the point (-5, 1-118) and through the origin, find the ordinates of the last 4 points, and the equation to the curve. 23. Differentiate (i i) sin +08 - 3 (ii) sin F cos : ; x a Gaye - ) : PA AR f(x) =302—" 47, prove that /’(2)=1-3/"(1) 25. If y=ax?, show that ot Be very nearly if 62 is small: MISCELLANEOUS EXAMPLES 327 hence show that if x be increased from any particular value by one per cent of that value, y correspondingly increases by 2 per cent. 26. Complete the following table, where s is the distance in yards of a body from a fixed point after a time ¢ seconds. és 628 t Bo We . et 582 State, from your results, the acceleration in feet per second per second, and find the relation between s and f. 27. At what points on the curve 2y=523—3xz+7 are the tangents parallel to the axis of x ? 28. Differentiate (i) x sinz ; (ii) log. cosa ; (lines! 29. Find the values of (i) /((372—2zx) dz ; (ii) /a(2a—5) (w7+3)da. 30. In some experiments on friction a block weighing 30 grms. had weights of 100, 200, ... . grms. placed on it, and the force £ (grms. weight) was measured which was just sufficient to move it along a rough surface. Find from the table the relation between the force and the weight. Weivht 130 | 230 330 | 430 | 530 630 Force 55 | 80 | 123 | 167 | 191 | 225 31. Complete the following table: by bty i Sat Sa? “el 103-371 2 145-2 3 151-8 4 158-4 5 328 SCHOOL CALCULUS 32. Differentiate (i) x log.x ; (il) 2% JOCrs. (i11) 107 logioz. 33. If A varies as log.B, prove that the rate of increase of A with regard to B varies inversely as B. 34. Find the equation to the curve passing through the origin whose slope at any point is 4v—7, and find the equation to the tangent at the origin. 35. Show approximately from the following table that if Y=e e also is equal to e*. t oe} 155° | 1-51 1-62) 220) ea a .: ne 44817 4-5267 | 4-5722 | 7-389] | 7-4633 7-5383 36. Complete the following table : t 8 6s 678 Ot ot? oy 1-88 3 3°78 *4 6-32 5 9-5 6 13-32 Hence find the relation between v and t, s being the distance in miles of a body from a fixed point after ¢ hours, and find the initial velocity in feet per sec. | 37. Find ae li May seer (ii) e=sin—1y ; (iil) xy=10. 38. Evaluate (i) /4 sin3a dz ; (ii) /be% dx. 39. If w=2-5x2—5x2, find the rate of change of w when x=}: explain your result from a graph. MISCELLANEOUS EXAMPLES 329 40. Find the equation to the curve passing through the following points : x —5 | -2 | -1 | ~--5| +5 | -75/ 1-5 25 8-75 : |---| y —1-5 | —3-75 —7-5 -15| 15 | 10 | 5 3 | 2 ae : : , ga ge 41. By differentiating the infinite series 1 +2 1g i ak Lee d show that — e*=e*. dx 42. Find 4 when (i) u(1+27)=10 ; (ii) w2(1 +a?) =10. 43. The volume V in cubic oy of a solid up to different heights h feet is given by van 7 (2h2-++/3 3h3). What is the meaning of dee Find the area of a cross section 9 inches dh from the base. 44, If Ee find approximately the relative change in y x corresponding to a small relative increase in x, say an increase of -1 per cent. 45. Draw graphs of y?=23(3—2z), and y2=a(3—2), and find the maximum and minimum values of y in each case. 46. Draw a graph through the following points, and find the relation between x and y. 15 | 1:5 | 35 =3-03| —4- 16 —1-25° , =10) =3 | -1-5 =-5 2 10 —31+25 | | L625 2-08 4-16 12-5 x 625 47. Differentiate (i) log. (az2+ba-+c) ; (ii) x3 loge (tan 32) ; (iii) 107 sin fz. 7x2 —9xy + 2y? x 48. Find Lt,-, 2— 42 330 SCHOOL CALCULUS: 49. If an area A is increasing uniformly at the rate of -05 sq. ft. per sec.: at what rate is V1+2A increasing when A=4:5% 50. Plot the graphs of the functions 3x3 and 7a? with the same axes ; and without calculating any more ordinates solve graphically the equations 323 —7x21+4=—0 373+ 7x2+18=—0. 51, Fill in the following table with the approximate values of ef when @=1. 1-25, 125,5le(oe2U0- dy x Yy dz °75 2°117 1-0 2-718 1°25 3°49 1:5 4-482 1°75 5755 2-0 7-389 2-25 9-488 52. If the acceleration of a body increases uniformly by 2-5 feet per sec. per sec., find the rate at which the velocity is increasing after 5 seconds, if the initial acceleration was 2 feet per sec. per sec. 53. Find a when (i) sinhu=“V/2 ; (ii) e"=~V/cosa. 9. 54. Evaluate (i) Fei dy ; aU, 55. Draw the graph of z2y—y=9. What maximum value has y ? MISCELLANEOUS EXAMPLES 331 56. Find the equation to the curve passing evenly amongst the following points. qe. 6 attr per pa 7:31 20 7°36 100 118 30 | 40 | 50 7-41 7-47 7-52 If J be the length in inches of a spring to which weights W pounds are attached, what is the meaning of ae 57. Differentiate (i) sin log.Vz ; (il) log -sinV'x : (iii) logeV ginz. 58. Find the value of wu if (i) a ae 2 cosV6 : VO ¥ tanV'x i) = 59. If w=202—5xe—12 and v=x2—6x4-8, complete the following table : : U x U Vv = v 4-5 4+] 4-01 3°99 3°9 3:5 242 —5x—12 and hence find approximately Lt, 4 sey eaay & 60. For what value of x is 72\/ 10x—a2 a maximum ? Draw a graph of the function. 61. Integrate the equations (1) an = = “44; (1) ) 2x dy =ydx. 332 SCHOOL CALCULUS 62. If w=3¢+2V2—5 and v=2«—5V2z+3, complete the table ; U CANT Su v 1-5 1-1 1-01 1-001 and hence find Lt, —1 3a + 2Va—5. 22 —5V4%+3 63. From the following table find the relation between P and W. 12 23 38 50 62 75 13-4 | 14-1 | 15-1 | 15-9 | 16-7 | 17-5 (8 If P poundals be the force required to move a weight W Ibs. what is the meaning of the slope at any point of the curve ? 64. Find the values of eo when (i) logex=cos26 ; (11) @=cos—! logyow. 65. If A is the area in square feet of a square whose side is x inches, and if w increases -5 inches per sec., find the rate of increase of A in square feet per min. after a time of one minute has elapsed. 66. Find the relation between A and D from the following table : eee | 0 | 9.5 | 3-21 3-97 | 4-36 lee | A ae 0 4-91 | 8-08 "12-38 14-91 | 18-85 67. Complete the following table Sr ; | 2°36 | 2°38 2°4 i | 2-3 2:82 2-34 y «. | 87-3 | 88-8 | 90-35 | 91-9 93-5 95 LEP eiAe eho. where A is the area between a curve and the axis of x: hence find tho area from 7=2-3 to ~=2-4 approximately. MISCELLANEOUS EXAMPLES 333 68. Differentiate (i) ev? cota ; (ii) ecot vr : (iii) e Yor, 69. ‘The area (A) in square metres of a rectangle whose perimeter is constant and equal to 2k metres is given by A=2x(k—x). Find the values of 6A for a small increase dx of x from (i) z (ii) as (6x)? being negligible. 70. A body of mass m falls freely from rest under the action of gravity. Find the rate of increase of its kinetic energy in any position (i) per unit of distance, (ii) per unit interval of time. 71. The accompanying table gives the distances in feet through which a body has moved in ¢ secs. from a fixed point : 40 3,120 10 20 : 855 | 1,660 30 2,415 50 3,775 Find directly from the table the velocities and accelerations at different times: hence, without a graph, find the relation between s and ¢, for the accelerations so found are as a matter of fact exact, and account for this fact when you have found the equation. 72. Find the relation between V and r from the following table. (Use Chap. II. §7.) “2. -0334 oe 38°75 3 -113 5 -523 1-3 9-2 8 2°13 r V 73. If w=2x2—3x-+1 and v=xz2—1, complete the table | U x U Vv ae v ae Oni dek and hence find Lt; —1 ~ : ea} 334 SCHOOL CALCULUS 74, Find the values of ( rf Tas aa (ii) VE coté dé. 3 75. Find what values of x give (i) maximum, (i1) minimum values of (v+2) (83a—1) (~+5). 76. Construct a table of values of (~—3)? and x?2—9 as a approaches the value 3, and hence find the value of Lt (a —3)? fo 20s 77. Find the equation connecting x and y from the following table. (Use Chap. IT., § 6.) Art 3:4 ‘0185 | -0786 15-3 7-163 20°5 17-23 10-7 2°45 5:2 *281 x us 78. Calculate the ordinates to the curve x?y+y—10=0 for the values 2, 2:1, 2-2, etc., of x, and so find the area between the curve, the axis of x, and the ordinates at x=2, and x=3, (as in Chap. VI., § 15). 79. Write down (i) He 0 (ii) ) [sina cosx da. 80. Find the general equation to the curve whose slope at Tv 9 : cos* 5 * dx: any point is given by oy ae 2) , and the equation to that particular one of the family which passes through the ] point (1-5) 81. Complete the following table : {iss r) ees] (1+ a«)3—1 Sead (1+a)*— wv and hence state approximately Lt; —o MISCELLANEOUS EXAMPLES 335 82. Find the equation to the curve which best satisfies the following values : ‘7°32 5:39 5 6-524 [a .. | 8-51 | 3-98 | 3-09 cy Peeowei-sli) 8°3 (Use Chap. II., § 7.) 83. The velocity of a body in metres per second is given by v=(4-5t+3)2: find the distance travelled in the fourth and fifth seconds, and the acceleration. tr/5 — a2 5 P 84. Find the maxima and minima values of and draw a graph of the function. 85. Calculate the ordinates to the curve y=2xV5—< for the values 1-25, 1-75, 2-25, 2-75, and hence find the area approximately between the curve and the double ordinates at x=1 and z=3. 86. Find the equation which best satisfies the following conditions : x Livo gio. Or | oy. 87. If w=et+e*—2 and v=3z2z2, complete the table U Vv too CeCe S72e oes 88. A man 6 ft. high is walking towards a tower 85 feet high upon which is a 10 ft. 6 in. flagstaff: by the methods of the Calculus find at what distance the man is from the foot and hence find Lt,—9 336 SCHOOL CALCULUS of the tower when the flagstaff subtends the greatest angle at his eye. 89. The following table gives the area in square inches of sections of a solid made by planes perpendicular to the axis of x at distances x inches from the origin. x | 1 4 ed 38 a 4a 5-584 A Peet 2949 | 1-396 | 3-142 8-726 | 12-565 \ Find two limits between which the volume enclosed by the first and last planes lies as in Chap. IV., § 37. Show that their Arithmetic mean gives the true volume nearly, given 2 that A="—. 90. Integrate the equation (1+2?)dy=tan—1z dz, 91, Find the volume of the largest cone that can be made out of a wooden sphere containing one cubic foot. 92. Differentiate (i) etn *x log, (sin 2) : (ii) logio 55 93. Fill in the following table by drawing the curve passing through the given points. 5-6 Hue 10-3 17-6 | 24-8 100 28 rote 7:49 4-96 Test the accuracy of your results by finding the equation to the curve. (Plot E and Ey. w Y 94, Find the values of (i) /x sinxdz ; (ii) fx? sina da. 95. Plot the curve y=3xV/2—a; calculate the ordinates when 2=-25, -5, -75, etc., and so find the area of the loop by the Trapezoidal Rule and also by Simpson’s Rule. 96. Form a table of values of e*~—e-* and 22 when x=1, v__. e— XL -5, -2, -1, and hence find Lt,—o ca oe MISCELLANEOUS EXAMPLES 337 97. Find the equation to the curve passing through the following points : . 12 20-7 | 27-03 40-81 15:3 25-1 (Use Chap. IT., § 7.) 98. Show that the tangents to the curves y=d(x) and ele Y~ (x) axes. at their common points are equally inclined to the 99. A symmetrical surface has lines ruled across it at right angles to the axis of symmetry. The areas in sq. cms. cut off by these perpendiculars when at distances J cms. from one end of the axis are given in the following table: | 2 4 6 ot la lll boa tee BE 3:77 | 10-67 | 19-6 | 30-2 | 42-2 | 55-4 | 69-8 | A Draw the graph connecting A and /, and find its equation. What is the meaning of & for this curve? Reproduce graphically the original surface, and find the equation to its curved boundary. 100. Calculate /38xV2—2, and thus find the true area of the loop of the curve in Question 95. x-+5 21)0%+7 (ii) (a +5)V22+10a+7 da. 102. Find the relation between p and v from the following table : 101. Write down the values of (i) Je da 95-04 | 36-25 | 24-34 9 6-78 | 2°75 4°8 6:04 | 10-7 12-6 | 22 338 SCHOOL CALCULUS 103. The following table gives the distances, s yards, over which a body moves in times ¢ secs. : lt 12 : 53°6 | | s Draw the graph through these points, and read off the velo- cities at times 5 secs. and 10 secs. from the start. Also find the relation between s and ¢ (which is of the form s=ae?!—a) and hence find the relation between v and ¢, and test the two results obtained above for v. 104. If a straight line be moved across a circle of radius 5-5 cm. from one end of a diameter to the other, always being kept at right angles to that diameter, find the rate of increase of the area of the segment on one side of the chord, for any position of the moving chord, per unit increase of that part of the diameter on the same side. 105, Find the equation to the curve passing through the point (1, 2) whose subnormal is constant and 4 units in length. 106. Liquid is poured at a uniform rate into a conical vessel, so that one cubic foot would be poured into it in 5 minutes. If V cubic inches be the volume of liquid in the vessel at any moment, ¢ the time in secs, and A the depth of AE ee a One te d liquid in inches, find a and. Me the vertical angle of the cone being 45°. 107. Differentiate (i) e° cos (vie+5) : (ii) loge cos. 108. Find the relation between » and v which satisfies the following conditions : | | v 4-21 | 5-14 | 6-38 | 7:55 12-36 | | | p 102-3 | 83-1 | 66-4 | 55-7 | 45-35 | 33-36 109. A lamina is lowered into water by a string attached to the end of an axis of symmetry. When this axis is immersed to various depths, A inches, the rate R, in square MISCELLANEOUS EXAMPLES 339 inches, at which the area immersed is increasing per unit further increase of h, is given in the following table Nl 3 ‘is eae oe | Pie PP eT. | 422 | Sue ete. he ot | Ye et : es ) | | R .. | 4-472 | 5-657] 6 | 5-657 | 4-472 | 0 Draw the graph and find the relation between Rk andh. Hence find the equation to the boundary of the lamina in its simplest form. 110. Integrate the equation (1—wz?) cosy dy+2dxz=0. 111. If the volume of a solid of revolution be V cubic inches when the length of the axis is / inches, what is the meaning of Je ib ost, find the equation to the revolving curve. 112. Find the equation to the curve passing through the following points : een le Bae | Creer os el ale y «| SB | 3-98 | 4-85 | 5-225 | 5-44 [ (Plot and .) x Y 113. The area of the cross section of a solid distant x inches from the origin is given by A=2a2—7. Find the equation. to the tracing curve, and the volume from x=2 to x=3:-5. 114. If y=cot (x+C) show that dy+ (1+ y?) dz=0. Similarly elimate by differentiation the constant from the equation y= Aewn~, 115 Find the values of W/T= Wen axe x+a — (ii) faV5—2 dx, evaluating the latter ‘“‘ by parts.” 116. Show that the area intercepted between the curve y=¢(z) the axis of x and two ordinates h units of length apart is a maximum when ¢(z%-+/)—¢(x)=9. 340 SCHOOL CALCULUS 117. Prove that if {f)yde= i {(%1—2) dz ; (ii) if ” tw) d= ites dxf a f(x) dx 118. Eliminate, by differentiation, the constant from y= Va2—a2, 119. A weight of. 10 Ibs. is attached by a light inelastic string to a fixed point in a wall, and is held away from the wall by a force P perpendicular to the plane of the wall. Find the minimum force necessary if the string make an angle of 20° with the vertical through the point of suspension. 120. A body moves with a variable velocity given by the equation v=d(t). Show that whenever the velocity is a maximum or minimum the acceleration is zero. 121. Find the maximum area intercepted between the curve y=_V 13—a?2 and two ordinates 1 inch apart. 122. Plot the curve passing evan among the following points, and find its equation : 4 .. | 14-63 | 10-27 | g-08 | 3-91 | 3-68 | 2-95 J) - | 8:92 | 10-8 | 12°37 |) 18-4 | 19*O39e Biss 123. Differentiate (1 evi 10x—22; by LAVA , Va 124, Find graphically the value of the definite integral y- sin®@ cos*é dé. ne 125. The distances of a moving body from a fixed point, t minutes after passing the point, are given by the following table : | ¢ of DAUR ES hea ys ogee i si | | s F194 "136/156 144 100 24 MISCELLANEOUS EXAMPLES 341 s being expressed in yards. Find the velocity graphically or otherwise after 4 minutes. 126. Find graphically the volume of a solid of revolution from the following’ table, which gives the areas in square inches of sections distant / feet from the origin, by plotting the curve, reading off ordinates -1 ft. apart and using Simp- son’s Rule. | |4 .. | 109-7) 83-7 | 62-2 | 24-1 | 12-6 l Sine esate SOI 46.) 81 1-05.) | + dy 1 Die pCOsc. 127. Find re ik y=eerieeD. : 3 128. Evaluate (i) /, aaa iS es Gi) f* sin2@ cos°é dé. 129. Find the maximum velocity attained by a body which starts from rest and after 1 second has a velocity of 2 metres per second, if the acceleration varies as 2i—7, where ¢ is the time in seconds from the start. 130. Find the equation to the curve passing through the following points Pee a | y Pe leet 0 a4 14 e027 PEPER GIT Oca tear a 2:08.00 3)" 'h A | 20-8 (Use Chap. II., § 8.) 131. Integrate, by parts, the following : (i) /sec?@ cosecé dé. (ii) /cosec?@ secé dé. (iii) /((@+sin@) sind dé. 132. Eliminate, by differentiating twice, the constants from the equation y=sin—! be. 342 SCHOOL CALCULUS 133. Find the equation satisfying the following simultaneous values of v and f. gh ke |B SP a a Oa eo v . .. | 2-698 | 1-921 | 1-423 | 1054 | -781 (Use Chap. II. § 8.) 8 siné —sin26 F 135. Liquid is poured into a symmetrical cup and the heights (H) in cm. corresponding to certain volumes (V) in c.c. poured in are given by the following table. 134, Find Lt, =0 ; | | | | Vy .| 1 | 285 | gall zest T0 amen ; | H .. | 1:65 | 2-22 | 2-8 | 3-21 | 3-53 | 4-04 | | | Draw the graph relating V and H, and state the meaning of its slope at any point. Find the relation between V and H, which is of the form v = hH", and hence draw a section of the cup containing its axis. | : 4 136. Solve the equation @ = mnie 64 +-sin26—8 sin@ 137. Find the value of Lt, —o A 138. A flat stone is made in the form of a cylinder with its edges rounded off semicircularly (so that if ordinates were drawn from the ends of a diameter of a semicircle upon a line parallel to the diameter and on the side of the diameter remote from the arc, and the ordinates and semicircle were then revolved as a whole round that line as axis the form of the solid would be of the same shape as the stone). If the dis- tance between the flat sides of the stone be 53” and its dia- meter 11”, find its weight by Guldin’s Theorem if 1 cubic foot of stone weigh 150 Ibs.- (The centroid of a semicircular v7 pears lamina of radius 7” is a from its centre.) T MISCELLANEOUS EXAMPLES 343 139. Find the relation between x and y from the following ata : x eT | | y | (Use Chap. II., § 8.) 140. If the height of a cone be h inches, and the radius of its base r inches, prove that the whole surface (base included) is @ maximum if h=2r-/2, and if the base be excluded when h=rv/2, the volume of the cone being given constant. 25 | 46 | - 100-7 62-2 24-1 81 1-05 1 76 | 2-0 ‘12-6 | 1-84, -96 141. Solve the equation cosy dy=siny (2% —3) dz. : a 6x3 142, The formula y= Ref, 700 44a has been suggested for the outer curve of a cross section of a masonry dam for a reservoir, x ft. being the depth below the water surface, and y {t. the horizontal distance from the inner face of the wall, supposed vertical, at that depth. Find the volume of masonry per foot length of the wall (70 feet high) by Simpson’s Rule. 143. A jar of water at 15° C. is placed in a temperature of —12°C., and its temperature falls 5° in eight minutes. How long will it be before ice begins to form ? 144, An equilateral triangle, of side one foot, is immersed in water with one vertex in the surface, one side horizontal and with its plane inclined at an angle of 25° to the horizontal. Find the pressure on one side (i) neglecting atmospheric pressure (ii) assuming the atmospheric pressure at the sur- face of the water to be 14-7 lbs. per sq. inch, the weight of water being taken as 624 lbs. per cubic foot. 145. Find the centroid of a circular arc subtending an angle 6 at the centre, the radius being r inches. 1800 — -256 sind + 40 sin2@—sin4@ 147. The curve of the sides of a section of a barrel which contains the axis is a parabola, and the dimensions of the 146. Find Lt, 344 - SCHOOL CALCULUS barrel are, height 3’, diameter top and bottom 2’, and diameter at the middle 23’. Find the volume of the cylindrical portion in the centre, and use Guldin’s Theorem to determine the remaining volume outside this, and hence determine the contents of the barrel if 1 cub. ft. is equal to 6-25 gallons. oe 148. Solve the equation xy ee 149. A string 100 cms. long is stretched by a force equal to the weight of 10 kgm. Find the work done by a force perpendicular to the string at its middle point which displaces it 5 mms. neglecting the extra length to which the string stretches. 150. A theoretical indicator diagram is represented by the area enclosed by the curves and lines xy=50, zy=300, y=80, and y=10, in which ordinates denote pressures in lbs. per sq. in. and abscissae distances moved by the piston in inches. Find the work done by the engine in one revolution if the diameter of the piston be 15 inches, and the H.P. at 125 revolutions per minute. 151. The following table gives the velocity in inches per second of a body after ¢ secs. | t 3 | 60h Wes 10-5. 7:8 | v 26 “19-2 14-2 5:8 Find the distance travelled during the whole interval of time (i) graphically, (i) by Simpson’s Rule. 152. A lighthouse is formed by the revolution of the hyper- bola xy =1,785 (foot-units) about an asymptote. Its diameters top and bottom are 16’ and 42’ respectively. Assuming the rooms together to form a cylindrical chamber of 114’ diameter, the lower 30’ of the lighthouse being solid, and the masonry to weigh 156 lbs. per cubic foot, find the weight of masonry to the nearest ton. 153. A cup of coffee at temperature 100°C. is placed in a room whose temperature is 17°, and it cools to 63° in 5 minutes. Assuming that the rate of cooling is proportional to the difference of temperature between it and its surroundings, find the temperature of the coffee after another 5 minutes. MISCELLANEOUS EXAMPLES 345 154. The expenses of a ship are equivalent to 64+ v2 tons of coal burnt per hour, v being the speed of the ship in knots per hour. Find the most economical speeds for a given voyage when steaming (i) in still water, (ii) against a 6 knot current. 155. A rectangular log of wood 12 ft. long, whose cross section is 14”x 18” and specific gravity -62, and shod with 10 lbs. of iron, floats vertically in water (whose weight is 62-5 Ibs. per cubic foot). Find the work done in just lifting it clear of the water, the volume of the iron being neglected. 156. Solve the equation secy dx=V22--a2 dy. 157. Find the work done in twisting one end of wire through 15°, the other end being kept fixed, if one dyne will twist is through 78° when applied at the end of an arm 10 cms. long. 158. Plot a curve showing the volume of a spherical cap of height # inches cut from a sphere of radius 10 inches, for values of h from 0 to 10, and find from the figure the value of # that gives a volume of half the hemisphere. 159. A dock whose length is 120 ft., depth 30 ft., and width at the top 30 feet, has a uniform parabolic cross section. It is completely filled in 80 mins. by water admitted at a uniform rate. When the depth of the water has risen to 20 feet, find the rate at which the water-level is rising. 160. Liquid is filtered through a paper in a conical vessel. Assuming that the rate of passing through is proportional to the depth of liquid left, and that it passes through a small area at the bottom only which remains covered the whole time, if the depth is originally 4 inches, and falls one inch the first minute, how long will it take to fall (i) one more inch, (ii) 2 more inches. 161. A string hangs from 2 points in the same horizontal line 66 yds. apart, and sags 4’ 6” in the middle. Assuming the curve to be a parabola, determine the slope at each end to the nearest minute. Hence find the tension at the end if the string weighs 25 lbs. 162. A liquid at temperature 90° C. cools in 3 mins. to 70° and in three minutes more to 55°. Find the temperature of 346 SCHOOL CALCULUS the room in which it stands, assuming that the rate of cooling is proportional to the excess of the temperature of the liquid above that of the room. 163. A steel rod 3 feet long clamped at one end requires a force of 15 cwt. applied perpendicularly to it at the other end to displace that end half an inch. Find the work done in displacing it another quarter of an inch. Find also the work if the original force of 15 cwt. only displaced the end one quarter of an inch. 164. Solve the equation y tana dx=2 cosx dy. 165. A rope hangs between 2 points in the same horizontal line 50 ft. apart in the curve y=250 cosh sc : find the sag at the middle. If the rope contract so that the sag is reduced to 9 inches and the curve becomes y=c cosh, find c by assuming the new curve to be a parabola. Also find the length of the rope in each case, and the ratio of the tensions at the points of suspension. 166. Find the work done in lifting a weight W pounds just off the floor by an elastic string, of length J feet and modulus of elasticity kW, attached to it. Also show that the weaker the elastic string is (i.e. the less k& is) the greater is the amount of work done. 167. A quoit is formed by revolving an elliptic quadrant about a line parallel to the minor axis and 2 inches from it. If the semiaxes of the ellipse are 14 inches and ? inch, find by Guldin’s Theorem the volume of the quoit, and its weight if made of metal weighing 450 lbs. per cubic foot. 168. Find the area between the curve y=3-7e15*, the axes, and the ordinate at «=4. 169. In the following table A is the area in square inches of cross sections of a solid of revolution by planes perpendicular to the axis of x at distances x inches from the origin. MISCELLANEOUS EXAMPLES 347 Find the relation between A and wz, and hence the equation to the generating curve in its simplest form. 1 2 : : 170. Show that pare Coe decreases steadily as 0 increases He « A ep es from 0 to g Le. has no maximum or minimum between those limits, provided a is not equal to 1. 171. Find the centroid of the area cut off from the hyper- bola gears by the double ordinate at x= a2 62 y Sy { 172. The velocity of a body is given by v=95e 10 in ft. sec. units. Find the distance travelled in 44 secs., and the acceleration after that time. 173. A glass of hot water, standing in a room at 20°C., cools in 20 minutes from 90°C. to 30°C. How long would it take to cool the same amount if placed in a freezing mixture at —10°C. ? 174. A sphere of diameter 4” rests in the bottom of a cylin- drical vessel of diameter 4”, into which water is poured. Plot a curve showing the volume of liquid in the cylinder (the water being supposed to get past the ball) for every }” rise in level up to 4”, and explain the nature of the graph after that. From the graph read off the area of the surface of the liquid at heights 2”, and 1}”, and check your results by calcu- lation. 175. A body slides down a smooth tube in the form of a circular arc, standing in a vertical plane. If the body start at any height h inches above the lowest point of the tube, show by the methods of the Calculus that its velocity at the lowest point is the same as if it had fallen vertically from that height. 176. Water is admitted to a dock whose cross section is of parabolic form, of depth 30 ft. and width at the top 40 ft. Find the pressure on the vertical end when the depth is 20 feet, and the rate at which the pressure is increasing if the water be then rising at 4 inches per minute. 348 SCHOOL CALCULUS 177. Find the cubic contents of a cylindrical gasometer 34 ft. high and of 36 ft. diameter, the top being a spherical cap rising 4 ft. above the sides. 178. A pendulum consisting of a small bob attached to a fine wire swings harmonically with small vibrations, and is found to complete 121 semi-swings per minute in vacuo, and 120 in air. Show that the resistance of the air to the bob is equal to 1-626v poundals per unit mass, v being the velocity of the bob in ft. per sec. 179. A tank is filled to a depth of 4 feet with water which escapes from a small hole at the bottom. If the water level falls 1 foot in 25 minutes, assuming the velocity of outflow to vary as Vh where h feet is the depth remaining at any moment, find the time taken to fall another foot. 2 = 180. The formula Hip ee 1) is sometimes used for the H.P. transmitted by several ropes passing over one pulley, where C is the circumference of a rope in inches, NV the number of ropes, and V the velocity in ft. per min. of the ropes. If the total weight of ropes to be used is given, show that the greater the number of ropes the greater is the power trans- mitted, provided they do not thereby become so thin as to break. 181. Find the maximum velocity of a falling raindrop if the resistance of the air be 2-8v-++ -001v3 dynes per unit mass, where v is the velocity in cms. per sec. at any moment. 182. A tank kept filled with water to a depth of 3 feet has a rectangular slot of breadth 14 inches cut all down one side. If the velocity of outflow be V2gh cubic feet per second at depth h feet, find the amount of water that flows out in a second. 183. Find the acceleration at any point of a particle sliding down a smooth cycloid which is in a vertical plane with its vertex downwards ; deduce that the motion is simple harmonic along the are, and find its period. Also find the velocity at any point, and verify that it is the same as would be acquired by the particle falling freely under gravity through the same vertical height. MISCELLANEOUS EXAMPLES 349 184. Find the distance from the vertex of the centre of gravity of a thin parabolic reflector 12 inches deep and 18 inches across the front. 185. Find the work done in pumping air into a vessel of volume V cubic feet which is already filled with air at pressure p1 lbs. per square inch until the pressure rises to po lbs. per square inch, the external pressure of the atmosphere being IT lbs. per square inch. 186. Find by Guldin’s Theorem the volume of the spindle generated by revolving a segment of a circle about its bounding chord, the radius of the circle being 10 inches and the height of the segment 4 inches. 187. The resistance to the motion of a body is 2-5v+ -001v3 poundals per unit mass, where v is the velocity in feet per sec. If the body is at some moment moving 30 m.p.h., find the rate at which the resistance is decreasing. 188. A string makes 200 complete vibrations per sec., and it is observed that in 2 secs. the amplitude of the vibrations is halved. Assuming the motion of any poy oe it to be damped harmonic motion with the Setione saa + px, find X and p. 189. A smooth stone column of circular cross section is built of 6 stones placed one on top of the other, the volume of each being 4 cubic feet. The height of the stones, beginning with the lowest, is as follows : di? | | INoGeeor stone! 2... ° +: | 1 2 3 4 5 6 2010 “eae LIS TSG 879 | Height in inches ee | 1-98 13°38 | 38-43 | Draw a graph relating heights and volumes of the column and explain the meaning of its slope at any point: hence draw a cross section of the column through its axis to scale, and find the equation to the generating curve of the solid. (The relation between Vols. and Heights for the column is H k= nea) 350 SCHOOL CALCULUS 190. Find the Lt,-ohlog.h. Hence determine the whole area between the curve y=log.% and the axes, in the fourth quadrant. 191. Find the work done in pumping up a tyre of dimensions 810 by 90 mms. from atmospheric pressure (14:7 lbs. per sq. in.) to 70 lbs. more than that, assuming the compression to be isothermal, i.e. pv=c. (The tyre is 810 mm. maximum diameter.) 192. Pure water is poured at the rate of 3 gallons per minute into a vessel containing 100 gallons of liquid of specific gravity 2-75, Find the rate at which the specific gravity of the mixture is decreasing with the time (i) when 30 gallons of water have been poured in ; (ii) at the start; and (iii) after ¢ minutes. 193. A body falls from rest retarded by a force which varies as the velocity and is such that the velocity can never exceed 90 ft. per sec. Find (i) the velocity after 10 secs.; (ii) the distance fallen in 10 secs.; (iii) after what time the velocity is 32 ft. per sec. ; (iv) after what distance the velocity is 45 ft. per sec. | 194. A wire is stretched between 2 points 100 yards apart, the sag at the middle being 8 feet: if it lie in the curve y=C cosh”, find c by assuming the curve to be a parabola. Hence find the length of the wire, and if 10 feet of it weigh one pound, find the tension at either point of suspension. 195. Two equal circular discs are symmetrically mounted on an axis perpendicular to their planes through their centres, and connected by a large number of strings parallel to this axis joining points on their circumferences. If one disc be now turned through any angle the strings being still tight but all oblique to the axis, it can be proved that the form assumed by the surface of the strings, supposed infinite in number, is accurately a hyperboloid of one sheet. Taking the discs to be one foot in diameter and the strings one foot long, find the diminution of volume enclosed if one disc be turned through a right angle. MISCELLANEOUS EXAMPLES 351 196. A tank hf feet high is kept filled with water which flows out through a triangular slot the full height of the tank and with base 0 inches long. If the velocity of outflow at depth x feet be ~/2gx ft. per sec., find how much water flows out in a second (i) when the vertex of the triangular slot is upwards (ii) when the vertex is downwards. 197. Find the time of a small vertical oscillation of (i) a right circular cone of height A floating vertex downwards to a depth h, in a liquid of density p ; (ii) a right circular cylinder ; (iii) a paraboloid of revolution; (iv) a sphere, all under the same conditions, neglecting the action of any waves set up. 198. Show that it is harder work pumping up a tyre from atmospheric pressure to pressure /p lbs. per sq. in. on a mountain height than at sea level in the ratio log, Lf loge | the ITs IT, atmospheric pressures on the mountain and at sea level being respectively IT and IT, lbs. per sq. in. 199. A ball is projected horizontally with a velocity of 5 ft. per sec. from a vertical cliff. The retardation of the air (which may be assumed not to affect the horizontal velocity) varies as the velocity, and is such that the vertical acceleration of the ball after one second is 30 ft. per sec. per sec. If the ball takes 3 seconds to reach the ground at the foot of the cliff, find (i) how much less time it would have taken to fall in vacuo, (ii) how much nearer the cliff it would in that case have struck the ground. 200. A tank of depth 2 ft. 1 in. and uniform cross section 4 sq. ft. 124 sq. in. has two small holes in its side each one half of a square inch in area, one at the bottom, and the other 9 inches above it. Assuming the velocity of outflow at a depth x ft. below the water level to be 2gx ft. per sec., find the time taken for the water to fall to the level of the upper hole. OH 3 . Increasing from 9 to 10. Decreasing from —10 to ANSWERS CHAPTER I Section 4, page 7 . Listed at 5/7, 7/4. . (i) 1-0769, 1-8539. (ii) 1-6010, 2-8834. (iii) 1-4071, 1-9799, 2-7860, Section 12, page 16 . = -33,y= —°68. 3. ©=2-12, y= —4-53. 4. w= 2-69, y = —1-72, Se 30, 4) = 1 39, w= —4:97,y=—-5. x= -92, y= -69. x= —]-14, y= —4-09. w= —1-42, y= -24, Section 18, page 18 3 . Increasing from AG) to 10. Decreasing from —10 to =f 2 o 3 3 . Increasing from —10 to 3 Decreasing from 5 to 10. 2 : 3 3 . Increasing from —10 to 3 Decreasing from —>5 to 10, 2 . = —6°19, —-81. 7. ©=1-+56, —-96. . ©=6°87, —-53. 8. «= —5-92, -60. Section 14, page 20 . Increasing from —oo to —2-67; 3. Increasing from —oco to -28; Decreasing from —2-67 to 0. Decreasing from +28 to 2:39, Increasing from 0 to +0. Increasing from 2-39 to +oo., . Decreasing from —oo to —1; 4, Decreasing from —co to —1-78: Increasing from —1 to +1. Increasing from —1-78 to -25, Decreasing from +1 to +. Decreasing from -25 to +00, x= —-91, Nib 00. = —4-41, -5, 3-91. 8. y= —°3, —1°7, 6°5. 23 353 354 SCHOOL CALCULUS CHAPTER II Section 4, page 36 1. 3-4y+ 5-3x7+ 1°25 = 0. 4, s= 8: 5i?. 2. v=5-+ 2:250. 5. y = °052°, 3. y = 3a24+ 2. 6. D=44T — 2-572, Section 8, page 41 Ly 2a : 4, y=5x 2-714. 2. y= 12-7 Vz. 5. puss = 630. 3. A = b2\/3 (more nearly 1-71b?). 6. v= -096 x 8-6. MISCELLANEOUS EXAMPLES Page 42 Lon Dao a ae 9. pour 48 — 505. JO Re “owe 14-5, 28 3. v? = 6u. 10.18 aR ae 4, HP. = 0125)? 4-31, ll. 56x-+ 25y = 40zy. 5. s = 16-09¢?, f = 32-18. 12:0: 1002-28 6.) P= -19272 412-25. or 6 = 100 x 1-072-* nearly. 7. y= 3503 — 2. 13. (100 — x)r = 2,080. 8. y= 750-3. 14, p = 14-7 x 10~ 2000209, CHAPTER III Section 6, page 51 1 : 3. = let 1. 2 | 3 5. 1 7 6. 1 2 4. ——. Pokal 2/2 6. e Section 7, page 52 1. (i) +0503. (ii) —-006997. 3. (i) -007449; (ii) —-11285. 2. 88 = 1+3864 ft. ; 69-3. () , - Section 9, page 5'7 Fi d 1. (i) 3. (ii) 3. 3. (i) 5. = 24336, (ii) w = +7454, 5 3 veel 2. (1) v= 4. (ii) = 4 4. (i) Onna o (il) 3 io.2) Twn = . (i) Slope = 12. . 15:5. 1; 7-389. . 33 ft. per sec. p 5 5D) . (i) —2, at the pt. (3 ea) (ii) 7, at the pt. (3-5, 0). (il) ANSWERS Section 10, page 63 9. Acceleration = 5-5. 10. (i) 264 m. per sec. 11. 12. (ii) 1 km. per sec. (i) 1-496 tons per min. (ii) 1-45 tons per min. (iii) 34 mins. 9 secs. co dv 7 20- Section 11, page 67 26° 34’. 2. (i) 26°34’. (ii) 19° 39’. (iii) 5° 36’ Section 14, pages '77, 78 1523: 26. —asin(ax+ b).. 2. 14” — 2. 27. 2cos2x. 3. 2(ax-+ b). 28. 2sin2z. 4, —92x?. 29. 2xcos(x?). 5. 1-4 — 5524. 30. 2asin(2x?). 6. pee) : 3] 2 . Qa+ 3 7. —3(1 — x) 7 sin a s =Jk 32. 6(a — 5)>. 1 1 30. > 3,——_— | * 9, Wa 34/(a+ 1)? 1 1 34. wasn 10. Sides ( a) ax?-+ ba+c ou 1 ll. 4a”. 35. ¢ (logx-+ =); 12. 7-5a°%-+ 1-40-17 ahi ea 13. 2(2 — 5). 36. aM oe i 14. 3x77 — ae 37. logi0e 15. 3+7cosx. 35 ee Pio whe E : 16. oe a --- §sine. 39. alogioe-cotaz. sinx Re oe Oe 2 1 i 3 cosec?a. ree (2+ ai Ss Te Ai pe 19. —coseca (acotx«+ beosecz). 4] __loga 20. sina-+ acosz. * ~~“ e(logex)? 21. log. 42, —2xe-*’, ek eens xeosec?2). 43. e%(cosx — sinz). 24, a — cotx) =, 1 ] 7 , 44, e (= — loge). 25. —. 45. —e *“(3cos2a + 2sin2z). 355 356 SCHOOL CALCULUS 46. cosxz cos3a — 3sinz sin3z. 53. 4acosec2xz?, 47. cos’. 54. 4 log, tana cosec2a. 48. 2x secx? tanz?. SGN oe l 1 logex a2a(logex)? 49. x2 cosec? = 56. e*cose?. b au 57. 1+ sec?z log sinz. aye 50. n (a+ bz) 58. e*facos(br-+ c)—bsin(br+ c)}. 3(x — 6) i” ol = 59. —+ 1. 2(a — 3)! xt er —e-—t 1 - 52. ars 60. V 22+ a? Section 16, page 80 1. 4cos2a(sinz-+ xcosx) — 8xsinxsin2a. 1 plogda?—22) 1 3, «(1+ loger). 2, —(Va?—2x?)2 { Dx? Ga agi 1\2+2 a i (logex — 1). 5. sinze*(1 + #+ xcotz). Gens { (sina + xwcosz) cots cosa? |. 7. 2(sinx)** (log-sina+ xcotz). 11 1 —a(1+ 2?) logel0 8. e~ °lcot?a(1— -Ola—4acosec2z). 10%(1+ 22)? sinx De Sen SPOOte — te): 12, 6(1—42—622) (1—2x) (14+3r)?. (2a? -+- 62-+ 29) (2a+3)!® 10. —2:5 5 4H : ia (2+ Tx — 4)* 13. 227 “(1+ 3 log.r): 14. 2(sinx)** (cosa cota — sinx log, sinz). 15, 0S nese 18. 3(cota)** (log.cota—2acosec2z), ' 2V/sinz log t—1. (1 ; x2 19, 2 logex-x. qo \t4sloo Lael 16. (V1 x ) \2 Oge( x ) eh * 6at-+- 6x3 — 21a? — 14x+ 3 17. Daren sine) { sin2(a2) + asin (2a?) }. : (1+ 2x)! Section 20, page 89 1. 2(ay —1) dx+ (a?+ 3) dy=0. 2. (Ta? — Qa — 2y) dx — 2(a+ y) dy=0. 3. (8x — Ty — 2) du+ (18y — Tx — 3) dy =0. 4. (622y — Quy? — 3y° — 2x) dx+ (203 — 2u2y — Oxy? — 7) dy =0. 5. (y2-+ 5) da+ 2ay dy = 0. ANSWERS 357 2 13 9 6. — —. tiga. Sai =. 49 9 20 9. Tangent y = 12x — 19. Normal 12y+ x — 62=0. 10. a y = 4x — 5. » 4y+2+3=0. 1] “A 2y= ax — 6. | Yt ee 2. 12. Piven ty (4 14) 10. ue 12. ” 2( ‘aa Y1) = (e x Normal (e° — e *\(y — yi) + Aa — x1) = 0. . Tangent 3y+ 8%=18. Normal 32y = 12x+ 119. oe) y = 2a — 14. ” r+ 2y = 47. . dv = 4y+4+ 43. . 4y = 9x+ 39. 4y = 9x — 13. Section 21, page 90 . Subtangent = 4-5; Subnormal = 2. ot 32 be 2a ) 99 ee e =— f 3 5 — rV3 99 6 9 99 5} ‘ 99 8 > 99 8 o 4 9 = 2; es aa? 2 29 = 0 ; ” = Oe a 81 (i) 9 4 > “1p = a (ii) ” 7 4 > 99 aoe 9° og) a eeben Staab = ay logy (1 + loge). 1+ logery 1 ae iat 12 > 99 = 3 10 oo > => -048. (i) 5, 5 5 il ae ; 9° — °192. ity e..5 5 6. d . dy y ao = os ty For (1) Yaz ~ dy’ dx or (ii) the tangent makes an angle of 45° with the axis. 358 10. 11. 2 12 sinhx V1 — 4x?" " 2V/ cosh’ z: 6 13. cosechz. : /1 — 9x2" 14. 2asinh3x+ 32?cosh3z. jae 15. x?coshz. lat 16. sinha: e%**, exile oe We 322+ 32+ 2° Vx(1+ x) vee eee eee aV1 — (logex)?’ xt — dx + 10 < x 10. (x2 + IF (a? + 2) eee ca 20. : VE ps V 2x24 Qa+ 3 1 21 etl JB Be = BaP Vara 1 a 22. ras aV «t+ at 6 93 cosé — sind 4sin26 + 9cos?0 ‘/sin26 ° 4xcosh(2x?). 24. seczx. Section 26, page 97 gest 2. 2) 3. py 5. None. 7. «= 4. 9. None. Pa 4, pee eee See ee Section 30, page 101 1 6 i 11. 2a. 16. —2. 5 3 ips. 17. 0. u 1) 2° a3! eee ee = —8. Rak 4 3 e | ive, 19, 3 : oe Plog she Te Go 15. 2. 20. 1. 3 e b: MISCELLANEOUS EXAMPLES Pages 101-103 1 b 9 (22+ =). 3. — Fe i n(a+bx) ” a 1 , 3(1+ a)! 4, da+ 73. SCHOOL CALCULUS Section 25, page 95 ANSWERS 2(a? + 3) Gh ate - (a2 3)?" 13. a e (3—7r). I — 22 : 6. Sea 14. a3 (Acoté — log,sin#). 7. 4(2cos2xcos3x — 3sin2vsin3-2). 15 2(a® — b?) 8 +/32'(35 — 62) (a? — z2)8(b2 — 22)#" ORT Say [pee oe ae 16 — 32 aVa® — x BA Gye on 17. Ae*(sec26+ 3tand). 27 — 184+ 2x2 1-06c 10. 7 Se — sa 323(6 — 2)! ee - o& t ll. — (94 22)F 19. —=+ sec?x loge. ae aa wheel te (1—siné)? (1+ x)V2a+ x? 21 2n(1+ n?) sind " (1+ n2 — 2ncos6)* (1+ n?+ 2ncosé)?* 99 2(n2 — 1) cosé 3sin (60+ A : 4-7( 3724 =) 25 — 5a — 2? "(5 — a)* (5+ x)F V2(3 — 2a) 38V2—<2 V «(x2 — 15a+ 75) ~ Aa —5)2 10 —# ‘ae 22/8 — 2 49 * 27 — x? 2(7 — x2)" 30. Zlogige: cosec2d. a Q(ax-+ b) logi10e . 2logioe: cota. . 10°‘ 3cos(3¢+ 4) — log.10-sin(3¢+ 4)}. " (1+ n? —2ncos6)* (1+ n2+ 2ncosé)?" 34. 35. 36. 37. 38. 39. 40. 41. 42. — 43. — 44, — logiou%:logioe?x. logex — 1 (log.a)” i 3 (a+ 3)V 62+ 2? 2ulogiox:logioex. e*in3z(sin3a+ 3xc0s832) eck 1+ 2? ar, phe ts 1+ v2 x (2 — 22)V1 — 2? xeeint(2 + xcosx) 10 Sy log.10 2V1—a2 - 359 360 46. e*-sin2¢-tan7} ‘| 1+ 2cota — 47. 48, 50. dl. 52. O38. e*™(sinx + acosz). cotz — 2. x SCHOOL CALCULUS (1+ 22) tan-? Tf 49. | cos(log.a). ae V seca esine loge— 11 'a0F +. ¢9 cosz — } | (i) -3. (ii) : (i) 10. (ii) -> (i) 2. (ii) 8V3. (iii) a+ 2y = (iv) e+ 8V/3°y = 30. a loge| 54. (— 370)3 Gs. 57. -O8nr°. 58. -5% 59. (i) 90°. 60.. 22° 31’, (ii) 34° 42’ CHAPTER IV All constants of integration are omitted. Section 7, page 1138 ] n+1 b aE Tay 1 ax* bx ~ gt A) atgeat ace . der, Qe ions} . — 5cosz, . 10tanz. Section 10, page 114 g, (a+ x)* es 18. 20. = Pp w w Section 16, page 120 : oe 4x). pie 5 foe ey mlOgs a ets 6. log{5a? — 122+ 1). . loge 1+ 2x?). : lops(1++ 3x) — 2. 7. 2 logd3-+ 404 52). wo ANSWERS 361 (a+ art? 9, sing es 1 s : 10. etan- Ty “i 3 ( — t)’. 11. —2Vcosz. _— 2V/1 — x. 12. (logex)* 1 3 ~ 6(1-+ 2ax)3" i 1 . /4+ Be. mere 220 72 2c ] é ‘ 8 Sagas lan ea) Section 15, page 118 ro 500836, 9, 3sin”\. sin20 gree POOR ak 10. aV x2 — 25 SEE phe aa 25 sd . 2 2 5 cos 2 1 & Blog. 7 Va = 2 tan46 V3 core 12. tan-1 2x. 6 1 xy/3 . — 9cot =. 135 22 f =f = 3 Por Me ne sec30 , i ie [ee te “fd Sad V3 /3 2sin %4 4 — x Sek ae 15. 5tan-! (x+ 1). in-1% pe 15a ee _ 9sin at V9 — 2 aT ‘as hte ae = 5 108. (at 1+ V22+ 22). aVa2+ 3, 3 xt-++ Veet 3 eee “ae ONY re V3 19. 4/2 tan 5 ve ee tre log. (x+ Vx2+ 1) 362 SCHOOL 1 ae ee 1 2 2, 5 log (2%? — a?+ 5). 1 10. 3 log-(5x3+ 3x2 — 4). 11. 5 log-sec2x. CALCULUS 12. 4 loge sin ©. 13. 4 lowelat+ a‘), 1 14. Qa 3at 15. log (logex). 16. log-tan~'x. 17. log(eA*+ e-A*), 18. log,sin= 1x. 7 lowd2 get 1 19, — 16 | 3 log-(3+ 4x)+ 7 log(3 — 4x) \ : 20. logeV v?+ 4a-+ 8. Section 20, page 124 , li 2x — 3 are ay 9Pr+ 1 Pees CAI a ee V13 V13 Pe 2V/10 Mier ean ell” Be — 6+ VII loge wt et Vx2+ da+ 2 hee V/2 V5, du —14+V5 — Qx-+ 52? log: ; . og = : 6. log Sag 3 e o/6 Section 25, page 129 g ; 4 1. - ee cosé. 8. & o 9 sin46 9. — cot®d. Ty he emg 10. “6. 3 cos®d _ cos?4 egg e Ne 4 p 1l. sind — © sine sin?6 4 sin4d _ sin®d 3 5 ° + . 9 4 6 12. sind? sin2d 30 5 cos’@ _ cos°d 32 4 8 : Aa 13, sind _ sin5d 6. Reve. — log.cosé. a 76 a 56 2 14... 8n = 4S sins 7. cos6-+ secd. 7 BED 2 Log, 2 42+ ViIL+ 40+ 40? aie ee Pte Rl ee V 61 L 4a — 15 8. ———= Ss] -1 es 41/2 sm 257 9, =. tant ee 272 2/2 10, V7 lop 2 "4 + Bet V7 6 weer a Ll. z - loge 3a + 14+V 3x + 2a 5 4 15. 1 coe cos66 cos46 a5 cost), Ee ie AT) 2 10. 11. -~ Ww Wd . sinz — xcosx. . cosx+ xsine. SL Sere Oe ANSWERS 363 Section 29, page 185 12. ésing. 13. wsin-ta-+ V1 — 22. (x — 1)e*. = ems 14. xtan-!x — log, V1+ 22. —~-(mx — 1). 3 m* 15. asin3a+ cosa — 22”, . «sinha — cosha. - 2(vcoshax — sinhz). ie EON log seca. 1 ; . — (2%? — Qasin2x — cos2z). ee 8 17. acot-ta+ logeV1-+ 22. in36 6 6 ae = 18. 26sin — (62 — 2) cosd. t ee (sina — cosz). 19, — 2e° 2(t?-+ 4t-+ 8). < 3 20. asec-1a — log(a+ Va? — 1). tad : 13) CP Aa 21. xcos-tx — V1 — 2. ee Pe — (2 logex — 1). 22. — (logex® — 1). 4 9 Section 34, page 139 *865. 7. 3-°5178. 11. 2-5072. 16. +2597. 2-164. g 7 12. 3:5256. 17. -1485. 2°5. os 13. -O217. 18, 315-1. - 3904. 9, -2194. 14. 6-0461. 19. -1778. 2. 10. 14-124. 16-1. 20. -23. - 7694. Section 38, pages 143, 144 . (i) 21°33. (ii) 3-771. 4, 3-429, 7. °6385 Ar. 5. +4925. 8. 18. . 4°178. 6. 4°944. ¢ * ection 40, page 150 iP en OO;s 3. 3°4655. 4, 6°8. 5. 00. 6.400% Section 41, pages 153, 154 ey _o t= 2 ee a Ob = Cy. ap bare gene p= dart C. 6. By? = cose-+ +4 C. . y=2alogexr+ C. ze 1% 7. 2(y>+ logey*) = 3x2+ 6cosx+ C. eo eet (“+ A): 8. logy=2tan ‘a+ C, Shey Ce Sa sem s+ C. oF : SCHOOL CALCULUS Oe Ok ae ae 16, (i) =". i) ee . 2+ 4de= AVY? — 2, ds v ae ie ge 17. @) = ne ‘ ao é : i 1) (ii) T — Ty = Ce-*. dA . (i) “=ky. (ii) A=kfyda. . (i) e =ks, (ii) s=Ce™. 18.0) ., dh S ., dv € posse =f Cee ree ) Pah Gi) = #+0. 19. (i) 7 =—kVh. (ii) 4h=(C--he)®. BR hae vs ey ee 0. (i dh ___ a9 3) ee . (i) di =a+ bv. (ii) a+bv=Ce™*. 20. (i) Ui . (ii)) wee MISCELLANEOUS EXAMPLES Pages, 156, 157 = 1 wt 24+ Vx2+ 42 — 5 Sr yeceay Pay 4, loge 3 : 1 x2+-44—5 . log-(log.2). 5. ae a. 3log(1—z). 1, («# — 3) g, — 008°, 2cos36 _ ; 9, 08e cea . 5 cosé. (2+ Le 20 oe 1 Nogea+ Tee . Mans Ve, 18. rh logetand. be py. Sa rd 2 a2V/attat at 24. +/at4 at Sr ag ee — 19. e’sind. aVa2? —a2. a2, 24+V2? — a? © ah _ 2+ 3. acme RO © in a eee S a, 1: 2 1 x » V2(% —a)+ clog, 2+ 5 a @i. 3 tan“ 'e — 6 tan 5° Retel 90 Liggett oars cos® 3° 6) eres 1 3 ) 23, 8sin-% = 3 {loge(1+32) + ako oto Ic 2 meee . tan le, OA, loge 24 Wx? — 42+ 8 1 2 , (1 + ar) BY 7 Ob. a (9a? — 6x+ 2). 2 3 y. ; Tica a eee 26. secétand — log,(secé+ tané). 2 = 9 (24 tan%). 27. 2 (4sing — sin4é). 28. 29. 30. 31. 32. 33. awe COIS OR oe . (i) 12-25 sq. ft. . 720 sq. cms. by both Rules. oe eh 2sin36 , sind 3 uy 5 sind — 1 . 5 {1+ x* — loge(1-+ x2)\, 2 {y? — Qy — 34 2 log (1+ y)}. 5 (sin 'y—yV1—#), a —tan~! ; {eVite —logex+ V 12°) }- - ¥ = 423, . y=2logex. yee =, . e+ y?=a?, . (i) y? = 427+ 9, (ii) The lines y = + 2z. (ii) 12 sq. ft. 5. SAA 4y?-+ log, i Cede! OO (1 Pinky, 365 400V 5 21 CHAPTER V Page 180 16. 3-563 ft. 17. (i) 80-09 ft. (iii) 62-63 Ibs. 18. (i) 205-3 yds. CHAPTER VI Section 138, pages 195, 196 42-67. 14. 83-27. 7-87 acres. 15. 38-25. . 62:25 sq. yds. 17. 40-5. 5-304. 18. 6.79. 2(7 — 2). 19. 3414. 45-24 sq. cms. 20. 30-46. . 12(t23 — t13); 28. 21. 3-909. . (i) 1-818. (ii) 8-410. 22, 2-053. leiae. ads tt. ai "5-939. 23, Sovil4 _ 17-63, . 59-62. 5 Ms . 19-05 lbs. 24. (i) 3-660. (ii) 6-505. . (i) 5°296. (ii) ©. 25. b= 3. Section 15, pages 200, 201 . 2-943 sq. cms. 4. 34-905 sq. ins. -3681 sq. metres. 6. 1,548°8 sq. cms. Section 16, page 205 1-9314 7. 245. it. 8. °243. 6-28 9. +929. ; 0. 7854. 5:897 1 Vite 2y : (ii) 1-600 ft. (ii) 43-31 yds. Tats Loe 16° 51 Loge = 256 14, —— 366 —_ — jl mm Oo De Pose aN . 1,157 sq. ins. . Mininum 3° when «=<. | i Minimum 2V6 when x= \/ - Minimum 9° when x= 1 SCHOOL CALCULUS Section 24, page 211 AVA AGL ay bey eo ee 2. 239-05 ft. Va CHAPTER VII Section 4, page 216 Seat, 3 = Bee 3 3b? Section 7, pages 218-220 -3636 cu. ft. 12. 760-4. 19. 367. 1-775 pints. 13. 76-10. 20. 45-67. 9-392 lbs. 14. 63-20. 21. (i) 43-33 cu. ins. 49-23. 15. 45-75 lbs. (ii) 6-43 cu. ins. 1-230 16. 0. 22. 42,678 cu. ft. 326-6. Ly. 2ims 24. 41-6 r.p.m. . 22-61. 18. 97-2m. 25. 98:6 r.p.m. Section 8, page 221 . 28-27 cu. ins. Ar ‘ i= the d ter. . (i) 877 cu. ins. (ii) 63-43 Ibs, 4 oy eee ee 233: 6. 666°57. Section 9, pages 222, 223 . 42-58 cu. ins. 5. 84-96 cu. ins. 8. 34:7 cu. mms. . 04°571 c.c. 6. 1-343 cu. ft. 9. -166. . 14-85 cu. yds. 7. 827 cu. metres. 10. 163-6. . 7,108 c.c. Section 15, page 229 4. 11-88 sq. ins. CHAPTER VIII Section 7, page 241 3 1 3. Maximum g~ when x= —-. 9 3 ¥ : ie 7 al . Maximum —2/V/6 when x = — NA - ME oe 45 when ae 4. Maximum 41 when x= —2. 2 14. 15. 16. We 21. 22. m COD ANSWERS . Minimum 0 when 2 =: 0. 9 Maximum oe when x = re . Maximum oo when x= 0. Minimum = when «= 3. 10 . Maximum 1] and \ = Il. wi tigh ANS a whenz=+V 5. : . Maximum 3Vv2 | and —\ when y=0. Minimum _3v2 | f 2 tS: Cane 3,125V/5 Maximum 2. ——_- and a8 — \ when x = gi Mini 3,125V/5 icy inimum — cake aaa a Maximum Va?2+ 6? when x= tan~*;. Minimum —1 when x=0. 18. Maximum 2 te when x= cos-™. 19 Pot ° a Minimum — ne when 6 = — Maximum sin’) when 6 = > 20. 0= Maximum — ENE, when x = V2, Minimum — oe when x= — V2. Maximum — 3 when x= —4. Minimum — 2 when x= —3. Middle of Section 8, page 245 Section 8, pages 247-249 . 625 sq. ft. 6. . 1 ft. 8 ins. . 2°547 cu. ft. (i) 1 ft. 4 ins. 7. 228-2 cu. ins, . (i) breadth6ins.; depth 10-39ings. _8- 3°731 miles. . 4-762 secs. and 1-638 secs. before crossing the bridge. . 3:347 secs. before and after crossing the bridge. (ii) 367 . Maximum V30 when ¢ _V30 P 6 9 Mininum — ye whenf=— ee . Maximum — 4°65 when v=0. None. . Maximum sea and ~\ when «=4.5. Minimum _27V3 44 None. a 5 : 9 a . Maximum es, when 6 = 5 easy yy) ft. 16 ie 7 10. 9-755 knots per hour. (ii) 8V3: 4. 11. 334 cms. ren fi 12. Two maxima, 22-36 and 13-89 27 sq. ins. 368 SCHOOL CALCULUS 13. 1 cu. ft. 18. 69? ft. 14. Ordinates at x= 4 and x= 3}. 19. 3-732: 1. 15. Maximum between x = 2-230 and - x = 4-230. 99, 8V3ab, Minimum area zero, between 4 a= — -897 and x= 1-103. ye : ft. above the ground. 16. y= 1-085 and 3-085. 2 207 22. (i) 14°14 ins. from the start. CON tia re (ii) 31-77 Ibs. per sq. in. Section 9, page 251 1. (0, 0). é ph ci3 ees 2° 135/ eres: 4 20 a 3b? a 3b? 3. = Ga oe ae a ace PS v — ees Ee 3 457): 2 Ge a and ( 4/3 oe 6. (0, 0) pe 4V3, ov), 7 9—3V3 eae ses NE} _ 3V6V/3 —9 OB VET 2 : 2 2 : (33, 3 5V3) V3) and (3 & en 3 10. 2=0°, 63° 50’, 116° 10’, and 180°. 3 CHAPTER IX Section 4, pages 259, 260 1. a ems, il. ge r 2. = from the centre. 12, £=2-5281 i3snc=4. 5. (4-219, 0). een = 09 a7 LY a Cae 1) 15. = 5-534. Oe a OE 16. x= 1-765. loge 2.1 x2loge— 17. 7 = 2-162 2 a 18. = 8-247. 7. 47-51 ft. Bar, 3m” 3m 20. y = 2-86. 10. 2 ft. 8 ins. 22. y=. Section 8, pages 270, 271 2 2 = 2 EH pee Hipeecaanipiatigns ee 3 3 3 2 22 3, (i) MB, iy 2a aay 4M(a2 + 2) 3 : 3(a2 + b2) (ili) ope 2 2 4 (i) ae (ii) oe ia 2 B. (i) lt (ii) 104/5 ems. 6. 3Mr? 2 re A a 2M 2 Serta) ae (ii) O02 10. M Rome 2 aes... 3ix2 jiygcpy: ees le eee) wD . (i) 40-5 ft.per sec. Mh? ANSWERS 5 Mh2 V15 syle (iv) 3° (v) 3 feet. 11.3612 02) DORE +B? (yee (i) os 2 Cy ae a- 13. 3V3a cms. 14. 3-069 ins. Section 22, pages 290-292 71-14 secs. . b(e” — 1) secs. 124-8 ft. 9-597 cms. per sec. . (i) 126-5 cms, per sec, (ii) 4-18 secs. (ii) 121-5ft. 13. 3-482 ft. per sec. 14. 39-97 ft. per sec. 369 15. (1) 286-5 ft. persec. (ii) 14,270 ft. 16. 40 ft. per sec. 17. (i) 5:48 m.p.h. 18. 28-06 ft. per sec. 19. (i) 38-4 ft. per sec. 20. 3334 ft. (ii) 35-83 secs. (ii) 5-7 secs. 21. : division from the zero, on the side of the second swing. 23. (i) 1-016 ft. Section 23, pages 296, 297 . (i) -1325 ft. persec. (ii) 40-14 ft. nla e 5 ON: Outht+ am m Qu2ke . (i) 8°86 ft. sec. per sec. Section 26, pages . 19,716 ft. lbs. . 23,180 ft. lbs. approx. . (i) 16-14 ins. per min. (ii) 17-98 ins. per min. . 2,812°5 ergs. . 10-20 ft. tons. Section 27, pages . 17°3 miles per hr. 24 (ii) Accelerations are 3. 2°943, 2-357, 1-612 ft. per 4, (LeO 7253: (i) 2-182 ft. per sec. (ii) -746 ft. per sec. 300, 301 6. 1,326 ft. tons. 7. (i) 792:4t, tons. (it) 8. (i) 704 ft. tons. (ii) (ii) -3297 ft. per sec. (ii) 2-45 ft. per sec. -8958 H.P. -7963 H.P. a 9, —= on either side of the centre. V2 303, 304 2. (i) 7,340 ft. lbs. approx. (ii) 3-86 ft. per sec. 370 SCHOOL CALCULUS Section 29, page 307 1. 263-7 ft. Ibs. 2, 254-3 ft. lbs. 3° «7439: 4. 1-148. Section 31, pages 312, 313 . 2W tana. t LAME Naat 9. cosé = Aa 2. V7 ips. we 14 } 10. 12:47 Ibs. Spa ae 11.3: ft. 9 ims: 4. 2WV3. 12. (i) 44-1 lbs. (ii) 275-9 Ibs. 5. 20V7 Kgms. 13. 4-211 ins. 6. (i) 27:3 lbs. (ii) 206-1 lbs. (i) 27 ‘f (i) 2 ; 14, (i) VE ths, (ii) ya! lbs. Shs aera Wp 16. 18° ta Section 32, page 322 1. (i) 250 lbs. wt. 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