1] Curve Tracing 111. The purpose of curve tracing is 9 obtain an approxi- mate shape of a curve, without plotting a large number of points on it. The following rules help in determining the shape of a curve. 112. Procedure for Tracing of Cartesian Curves 1. Symmetry. (a) If even and only even powers of ¥ occur jn the equation of a curve then the curve is symmetrical about x-axis because for a_ given value of x, we obtain two equal and opposite values of y. For example the parabola }*=4ax is symmetrical about x-axis. (b) If even and only even powers of x occur in the equation of acurve then the curve is symmetrical about y-axis, because for a given value of y, we obtain two equal and opposite values of x. For example the parabola x*=4ay is symmetrical about y-axis. (c) If on interchanging x and y, the equation of the curve remains unaltered, the curve is symmetrical about the line y=x. For example the curve x*-+> }? =3axy is symmetrical about the line yp=x, (ad) If on changing x to =xand y to —y, the equation of a curve remains unchanged, the curve is symmetrical in opposite quad- rants. For example the curve xy=k* is symmetrical in opposite quadrants. (e) If on changing x to—y and y to—x, the equation of a curve remains unchanged, the curve is symmetrical about the line Thus the curve x*—y?=3axy is symmetrical about the line y=—x, 2. Origin. (a) Find if the curve passes through the origin. It will pass through the origin if the equation of the curve has no constant term in it. (6) If the curve passes through the origin, find the equations of the tangents at the origin, by equating the lowest degree terms in the equation of the curve to zero. (c) If there are two or more tangents at the origin, then it is called a multiple point. Further the origin is called a node, a cusp or an isolated point, according as the tangents are real and different, real and coincident or imaginary. 3. Intersection with Axes. (a) Find the points where the curve meets the axis of x by substituting y=0, in the equation of the curve. Also find the points where the curve meets the axis of y, by substituting x=0. 265zou () Find the tangents at these points. This can be done easily by shifting the origin to these points of intersection and equating the lowest degree terms in the changed equation to zero. (c) If the curve is symmetrical about the line y=x or y=—x, find the points of intersection of the curve with these lines also, Find the tangents at these points. (d) Find the position of the curve relative to the tangents obtained in steps (6) or (c) whether the curve lies above or below the tangents. This can be done by finding the ordinates of the curve and the tangent, near the origin. 4, Special Points. (a) Solve the equation of the curve for y (or x) if possible. (6) Find fy. and the points on the curve where the tangent is dx =0 or ©. Usually parallel to the x-axis or y-axis, according as a at such points the abscissa or the ordinate of the curve changes its character from increasing to decreasing or vice versa. (c) Find the points of inflexion, if any. 5. Imaginary Values. Find the regions where no part of the curve lies. This region can be found by solving the given equation of the curve for one variable in terms of the other, say yin terms of x and then finding those values of x for which y be- comes imaginary. 6. Asymptctes. Find the asymptotes of the curve if these exist. Usuaily asymptotes parallel to the axes are nceded and these can be found by inspection as explained in the previous chapter. 7. Region. (a) Consider the yariation of one of the variables say y as other say x varies, paying special attention when x jncreases and finally approaches 0. (6) Similarly observe the variation of y as x decreases and finally approaches —2o. Important Note. Make use of only as many steps of the above procedure as would be sufficient to give an approximate shape of the curve. Example 1. Trace the curve y*(a—) 1. Symmetry. The curve is symmetrical about x-axis, as there are even and only even powers of in the equation of the curve. 2. Origin. (@) The curve passes through the origin. (6) The tangents at the origin arc given by y°=0, i.c. y=0, ince the two tangents are real and coincident, therefore the is a cusp.we 3. Intersection with Co-ordinate axes. The curve meets the co-ordinate axes only at the origin. 4. Special Points. From the equation of the curve, a (g—x) te (a-x) O when x=0 or x=3a/2 « Rejecting the value x=3a/2, because y» is imaginary when x=3a/2. The tangent at x=0 is parallel to x-axis. “4 y’ Fig. 101 5. Imaginary values. If x<0, v* becomes negative and is aginary. Hence no part of the curve lies in the sccond and the third quadrants. Also if .v>a, » becom curve lics beyond the point x s imaginary, Hence no part of the268 ENGINEERING MATHEMATICS 6. Asymptotes, Equating to zero the coefficient of y*, the highest degree term in y, the asymptote parallel to y-axis is x—a=9 or x=a. There are no asymptotes parallel to the x-axis. 7. Region. As x increases y also increases and when xa; yo. Also when x=0, y=O Hence the shape of the curve is as shown in the figure, The curve is known as cissoid. Example 2. Trace the curve 9ay?=x(x—3a)*. 1. Symmetry. The curve is symmetrical about x-axis, as there are even and only even powers of y in the equation of the curve. 2. Origin. (a) The curve passes through the origin. (b) The tangents at the origin are given by 9a2x=O or x=0, 3. Intersection with axes. The curve meets the x-axis at (0, 0) and at (3a,9) and the y-axis at the origin only. We shall find tangents at the point (3a, 0). To shift the origin to the point (3a, 0), let X=x—3a, Y=y. The equation of the curve becomes 9aY*=(X+3a) X? Now tangents at the new origin are obtained by equating the Jowest degree terms to zero, thus tangents are 9a¥?—3aX?=0 or yettex or yet JF (x—-3a). The tangents at (3a, 0) are inclined at an angle of tan-#( dr) i.e., 30° to the x-axis. The ordinate of the curve in neighbourhood of the point (34,0) is greater than the ordinate of tangent at this point. (sce figure). 4, Special Points. From the cquation. of the curve, a5 (x—3a). Wx (taking positive value only) ay (x8—3 axl?)269 dy —0 at x=a, thus the tangent a: x=a is parallel to Now Gx anit: =o, “2. becomes infinite and f | vatso when X=O> Spe ies infinite and hence the tangent jgin is parallel to )~axis f., the y axis itself is the tangent to at the OME” s the origin. e 2 ‘ the curv! i ginary Values. If x<0, becomes negative and ) 5. Hence no part of the curve lies in second and third is imaginary quad Asymptotes. The curve has no asymptotes parallel to eS. . . the a Region. As x increases from 0 to a, y also increases ically) and when x increases from a to 3a, j* decreases to zero. (oumers ‘Mcreases beyond 3a, y also goes on increasing and as x—- or, when oO. yr" lso when = x=0, y=0. Hence the shape of the curve is as shown in figure. $a? (x#+ 4a) 1, Symmetry. The curve is symmetrical about y-axis as there are even and only even powers of x in the equation of the curve. Uxample 3. Trace she curve ¥ 2. Origin, The curve does not pass through the origin.ab ENGINEERING 4, ATH Mate, “y 3. Intersection with axes. The curve meets the the point (0, 2a) which is obtained by putting x=0, in the: 2% a of curve and it docs not meet the x-axis. We shall find the (tution to the curve at the point (0, 2a). ‘angents To shift the origin to the point (0, 2a), Ict X=x, Y=y—2a. The equation of the curve becomes, (Y4-2a)(X? + 4a”) = 8a". The tangents at the new origin are obtained b. lowest degree terms to zcro. Thus the tangent is 4a? Y: or Yy=0 oe y—2a= ry equating the The tangent at the point (0, 2a) is a line parallel to x-axj distance 2a from it. P © axis ata 4. Special Points. From the equation of the curve, __ 8a* Yaa) : ay x(t 4g?) = 16a°x . Ge = —16a%x(x* 4a") = — Now “<0 when x=0 and, therefore, y=2a (from the equa. tion of the curve). Hence the tangent is parallel to x-axis at the point (0, 2a), a fact already established in (3). ? 5. Imaginary values. Solving the given equation of the curve for x. x=2a A) "2a—Y (taking positive sign only) y Ify<0 or if y>2a, x is imaginary. Thus no part of the curve lies below the x-axis or above the line y=2a. 6. Asymptotes. Equating to zero, the co-efficient of x1, the highest degree term in x, we get y=0. Hence x-axis is an asymptote of the curve. The curve has no real asymptotes paraflel to y-axis. 7, Region, As x increases stcadily from 0 to ©, » decreases from 2a to 0. Also when x=0, »=2a. Hence the shape of the curve is as shown in the figure.CURVE TRACING x’ x Example 4. Trace the curve y"(a?-+x°)= x*(a?— x’). 1, Symmetry. The curve is symmetrical about both the axes, as there are cven and onlyeven powers of both x and y, in the equation of the curve, 2. Origin. (a) The curve passes through the origin. (b) Tangent at the origin are y?=x? or y=ex. Since the tangents are real and different, the origin is a node. 3. Intersection with the axes. The curve meets the x-axis at(a,0) and (—a,0). To find tangents at (a, 0), shift origin to this point, let X=x-a,Y=y The equation of the curve becomes ¥?La*+ (X+4)*] =(X+a)*"[a*—(X+ a)*] or ¥?(2a?+2aX +X7) = —(X+a)*(2aX +X"). The tangent at the new origin are obtained by equating the Jowest degree terms to zero. Thus the tangent at (a, 0) is —2a°X=0 or x=0 a x--a=C. Thus the tangents at (a, 0) isa line parallel to y-axis at a distance a from it. By symmetry tangent at (—a, 0) is also parallel to y-axis. 4, Special points Here _ 2a® x?—x! dx (at#—x* F(a? + 2x") - ay . dx =0; when at—2a*x?—272 ENGINEERING MATHEMATICS sta 2a? Vda +4, or @(—1+-V2) fe xackaV(V 2-1) (Rejecting the other value for which ~x is imagin: Ty) Thus the tangents are parallel to x-axis at “x=taWW2—1) Also =o, at = cha. Thus tangents are parallel to y-axis at (a, 0). 5. Imaginary value. From the equation of the curve we have yet af e+ ue ota, ie. | x | >a, then y is imaginary. Hence no part of the curve lics beyond the lines v=-ba 6. Asymptotes. The curve has no asymptotes. 7. Region. From the equation of the curve, we have yoxa/ (Taking +ve value only) When x=0, 0. As x increases from 0 to a\/(/2—1), » goes on increasing. At xeaV 2-1) 2 =0 Further as x increases from aV¥G/2—1) to a, y decreases ultimately to zero. ‘The shape of the curve is as shown in the figure.273 (pTRAGING x\r, fy \P ample 5. Trace the curve (2) +(2) =1. Es : symmetry. The curve is symmetrical about both the axcs, 1. 5¥™even and ouly even powers of both x and y in the thers, of the curve. Origin. The curve does not pass through the origi mtersection with the axes. The curve meets x- 85 ati us ae 3, (i, 0) and y-axis (put x=0) at (0, -£8). yoy Special points. From the equation of the curve, . eye -(2 279 + ($)"=1-(4) “ 2 zed cn 4y"2 ~ ole b dx 3\a a ay Bye _ (y" of dx a (xjaj? ~~ Vaex 'y qd Now “i? when y=. From (1), when y= 0, | x=sha Hence the tangents are parallel to x-axis at the points (a, 9) oO, when x=0. i Also a From (1), when x=0, y=tkb Hence the tangents’ are parallel to y-axis at the points (0, +8).ENGINEERING MATHEMATICS Imaginary values. From (1), we have a 1 yp yes oo en |x! >a, (¥ ) <0, thus » is imaginary. Hence no part of the curve lies beyond the lines x= cha. Similarly we can show that no part irve lies beyon a n b part of the cul yond yeu. 6. Asymptetes, The curve has no asymptotes. Region. From (1) above, when x=0, youd and when y=0, ska Also as x increases from 0 to a, 3: decreases from b to 0, in the first quadrant. The shape of the curve is as shown in the figure, This curve is known as hypocycloid. xttl | xT Example 6. Trace the curve y a] about y-axis ds there |, Symmetry. The curve is symmeti are even and only even powers of x. 2. Origin. The curve does not pass through origin. 3. Intersection with the axes. The curve does not meet the x-axis but meets y-axis at the point (0, —1). To find the tangents at the new ori (0, -1. . we shift the origin to Let x=x year+ The equation of the curve becomes “+1 1 Y¥- or (KY -DCXt - DH XA4+1 or XY —2N?— Y=0. Tangent at the new origin is obtained by equating the lowest degree terms to zcro- Thus the tangent at (0, —1) is Yy=0 or y+1=0. Hence the tangent at (0, —1) isa line parallel to x-axis at a distance —1 from it. 4. Special points, From the equation of the curve, we haveI f curve TRACING._ . 275 Gl) Now From (1) when x= oO. yer Hence tangent is parallel t x-axis at (0, — 1). Also ce oo, whenx=+1 But when x=<-i, from (1), we have y> oo. Thus the tangents are parallel to ‘y-axis at ©, tence v=+1 will be asymptotes. 5, Imaginary values. From (1), we have yt y-l when »lics between —1 and 1, x* is negative and: hence x is imaginary. ‘ Thus no portion of the curve lies between the lines y=-L1. 6. Asymptotes. (i) Equating to zero the coeflicient of the highest degree term in x, i.e. x*, we have y—1=0, | the asymptote is parallel to x-axis. (ii) Equating ‘to zero the coefficient of the highest degree term in y, 7.e. }, we have x*—-1=0 or x=+1 as the asymptotes parallel to y-axis. 7. Region. “Solving the equation of the curve for x, we have a/ vr <1 In first quadrant as } decreases from © to 1,x increases from 1 to co. Also for the portion of the curve in fourth quadrant, as y decreases from —1 to —o, x increases from 0 to | As the curve is symmetrical about y-axis, we have the shape asshown in the figure. x276 ENGINEERING MATHREy, ATIC, 3 EXERCISE 11 (a) Trace the following curves : ayax, 2 a@y=x*(2a—x). 4. y*(2a— xt, 6. ay?=x7(x—a). 8. y=(e—2)(e+1)*. V(atx)=x(3a—x). 10. xy®+(x+a)*(x+2a)=0. 11. xy?=a"(a—x). 12." y?=(2—2)*(x—5). © I—x*)=x (4x7). 14. ytx=xt, *—a)G*—b*) = a7. 17. =y'(2a-y). 113, Procedure for Tracing of Parametric Curves Equation of 2 curve in the from x =/(r) and y=¢ (1) is known as parametric equations of the curve, with / as a parameter. To trace such curves the following methods are employed. Method 1. If possible, eliminate the parameter 1 between x=f(t) and y=¢(1) to chtain the corresponding ca:tesian equation of the curve which then can be traced as explained earlier. Example 1. Trace the following curves : (a) x=a cos t, ya sint (Circle) (b) x=a cost, y=b sint (Ellipse) (c) x=a cos? 1, y=b sin*t (Hypocycloidy i) x=a sin? wo ie issoi (@) x=a sin* 1, vre st (Cissoid) 2 SpeenURY ~~ soe , : gol. (a) Eliminating the parameter ¢ from the given yations> we have a x*-+y*= a" (cos? t+sin! =a’, xb yt= ai, . a ich is a circle with centre at the origin and radius a and hence so pe traced easily. (b) Here x=a cost y=b sint “ x o moos t a (1) aod -pssint wee(2) Squaring and adding, corresponaing sides of (1) and (2), we have a 2 x +45 = cos? t+sin® r=1, 5 ve ot gah which is the standard equation of an ellipse with centre at-the origin and the co-ordinate axes as the axes of the ellipse. The length of the semi-major axis is a and that of the semi-minor axis is 6. The shape of the curve can be obtained by the procedure ex- plained earlier. The shape of the curve is shown in the figure. Y278 ENGINEERING MATHEMATICg (c) Here x=acos*t y=bsin't . x ya o (*) ™cos t (1) » \ Aya and (+) essin f +e(2) Squaring and adding corresponding sides of (1) and (2), we get ays fy VRS ( a ) +( rt) =cos* f-+-sin? f= 1 x \3 ( y \a xyr4( ee yrs or ( @ ) o ) 1. (which has been traced in example (5), page 273) (d) We have x=a sin® ¢ a’. sin? rt! ye cost sin’ ¢ cos* ft ved, or Wa-y =X which is the cartesian equation of the curve and has been traced in example (1) (page 266.) 1:4. Method IT This method is used when the parameter ¢ cannot be easily climinaied from the given parametric equations x=f (1) and y= (1). The procedure is given as under. 1. Symmetry. (a) Ifx=f(r) is an even function off and vy (1) is an odd fiinction of ¢ then the curve is symmetrical about x-axis, because for every point (x, )’) on the curve there exists a point (x,—y) on it. (b) If x=f(t) is an odd function of and y=¢ (f) is an even function of f, then the curve is symmetrical about y-axis, because for every point (x, y) on the curve there exists a point (—x, y) on it. 2. Origin. If fora rea? value of f both x and y are zero, the curve passes through the origin. 3. Intersection with the axes, (a) Find the points where the curve meets the x-axis. This can be done by putting y=0 and obtaining the value ot f and then finding the value of x for this value oft. (b) Similarly find the points where the curve meets y-axis by putting x=0.CURVE TRACING 279 4, Limitation of the curve, If possible find the least and the greatest values of x andy and hence the lines parallel to tho. axes within which the curve | 5S. Asymptotes, Find the asymptotes of the curve if these exist. 6. Special points, Find points on the curve for which dy =0 or © i.e. where the tangents are parallel to the co-ordin: 7. Region. (a) id the regions where no part of the curve lies by studying the imaginary values of x or y. ’ , () Consider the values of x,» and -42 for suitable values ate UXe of f. (ce) If x=f (1) and y=¢(1) are periodic functions of t having ncommon period, the ‘curve is to be tracedifor one period only, Further values of 7 will repeat the same curve over and over again and no new branch of the curve is traced. 115. Some Important Curves 1, Cycloid. The cycloid is the locus ofa point om the circumference of a circle which roils, without sliding, along a horizontal linc. The horizontal line on which the circle rolls iscalled the base of this cycloid. One complete revolution of the circle, along the horizontal line generates one arch of the cycloid. The curve consists of an infinite number of congruent arches on both sides of is and hence an infinite number of cusps (see figure in example 1). The arc of the curve between two consecutive cusps is known as one arch of the cycloid. The point on the curve at the greatest distance from the base is called the vertex of cycloid. 2. Gatenary. The curve in which a heavy perfectly flexible string hangs when suspended between two points is called catenary or chainette. . Example 1. Trace the cycloid x= a (0+sin 9), y=a (I+c0s 0). 1. Symmetry. Here x=a (8+sin 6) is an odd func- tion of 6 and =a (1-+cos 8) is an even function of 8, therefore, the curve is symmetrical about y-axis. 2. Origin. Putting x=0, we have (@+sin &)=0 or 8=0, Now when 0, y=2a (0). Hence the curve does not pass through the origin. 3. Imtersection with the axes. (a) The curve meets the x-axis where , i.e. when 1+cos 6=0 or 9=+7. When 9=+2, g have ashe, hence the curve meets the x-axis at the point tan, 0).ou 7 ~~ (6) The curve meets the y-axis where x=0, i.e. when 6+sin §=0 or 80. When 8=0, we have y=2a, hence the curye meets y-axis at the point (0, 2a). 4, Limitations of the curve. The least value of yiso corresponding to @=7 and the greatest value is 2a when 6=9, Hence the curve lies between the lines y==O0 and y = 2a. 5. Asymptotes. The curve has no asymptotes “because when x0 corresponding to 8-00, y does not tend to a finite limit. Also » cannot approach infinity because | cos @| <1. 6, Special Points. From the equation of the curve, we have . aa (1+cos 4 Gana sin @ . dy asin 9 . et dx ~~ a (i-+cos 9) 2 sin $ cos >- ~~ o 2 cos* 5 stan 2 an =0 when 6=0 which gives x= and and y=2q, Hence the tangent to the curve at the point (0, 2a) is parallel to x-axis. Also 24 +00, when 0=:= which gives x=-:am and y=0, Hence the tangent to the curve at the point (+a, 0) is parallel to y-axis. 7. Region, (a) We have from the equation of the curvey=a (1+c08 8) 6 =2a cos* z Toa e oe cos pat Wy cas . gy, Hence y cannot be negative otherwise cos zr is imaginary, Thus no part of the curve lies below the x-axis. a=0 € 3 z ds v3 The table gives the points on the curve when x>0._ The curve for the negative values of xis obtained by symmetry. If we give further values io it no new branch of the curve is obtained We observe as xX increases from 0 to az, 3 decreases from 2a to 9 : . 2 put if x increases further from a= to 2az, » increases from 0 to aa andso on. Usually we trace one arch of the cycloid for 8 bev —z to zorOto 25. The shape of the curve is as shown in figure. Note, The four standard cquations of the cycloid are a (@+sin 8), v=a (1+cos 8). Example 2. Trace the curve ! : x=a ( cos t+ + Jog tan” +) =a sint. an 5, . ya 1. Symmetry. Here x is an even function of 1 and .. odd function of ¢, therefore, the curve 1s symmetrical about 2-4 °. tet . 17% 2. Origin. Putting y=0 we have r=0, Now when x0. Hence the curve dees not pass through the origin. 3, Intersection with the axes. (i) The curve docs 1 the x-axis because when y=0, 7.c., when 1=0, 1-7 ®- t-mect - aD (ii) The curve meets the y-axis where 2 when f 0, i.e. and this gives282 ENGINEERING MATHEMATICS roan +] or yoke. Thus the curve mects y-axis at the points (0, +a). 4, Limitations of the curve. The Icast value of y is —a and greatest value isa, because sin f always lics between —1 and 1. Hence the curve lies between the lines y=-ba. 5. Asymptotes. When t=0, x00 and y=0. Thus y=0, i.e. X-axis is an asymptote of the curve. 6. Special points From the equation of the curve, we have t . tan = tan*—> =a/—sin 14—1___ 2sin cos , 2 . 1 l=sin’ rt ) a( sin (+7 ) e sin f a cos* t cos ft sin ¢ + =tan ¢ cos* f Now ~~ =0, when t=0 which gives x00, y=0. This shows x-axis is an asymptote. oo, when t=- 5, which gives x=0, Hence the tangents at (0, =a), are parallel to y-axis. 7. Region, Some points on the curve 1=9 x= o 08 |CURVE TRACING 283 The shape of the curve is as shown in the figure y’ EXERCISE 11 () Trace the following curves : lox (t—sin 1), y=a (I1—cos £). 2, x=a (t—sin 1), y=a (l+cos f). 3. x=a (t+sin 1), y=a (l1—cos 1). 4, x=a [log (1+-cos 8)—cos 6], y=a sin 0 5. x cos* 4, y= sin® 7. 116, Procedure for Tracing of Polar Curves The equation of a curve in the form r=f (8) orf (r, 6)=0 is known as polar equation of the curve. The procedure for tracing ofsuch curves is fundamentally the same as for the tracing of cartesian curves with slight modifications. The procedure for tracing of polar curves is given as under. 1. Symmetry. (a) li the cquation of a curve remains unaltered when @ is changed to —@, the curve is symmetrical about the jnitial line. (8) If the equation of a curve remains unchanged when @ is changed to x—8 or when @ is changed to —8 and r to —r, the curve is symmetrical about the line through the pole and perpendicular to the initial line, i.e. about the line a= . (c) If the equation of a curve remains unaltered when 4 is changed to 3 —6, the curve is symmetrical about ‘the line =" (d) If the equation of a curve remains unchanged when 9 is changed to 4 the curve is symmetrical about the line 6=284 ‘ ENGINEERING MATHEMATICS -. (e) Ifthe equation of a curve remains unchanged when r jg replaced by —r, the curve is symmetrical about the pole. wy q.cr Pole. (a) Find if the pole lies on the curve. The will lic on the curve if for some real value of 8, we have r=0, » aie (2), If the pole lies on the curve, the values of @ for which r= give tangents to the carve at the pole. 3. Determination of ¢. (The angle between the radius vector and the tangent to the curve ata point on the curve). Pole , Ge @ Find tan $= a > then ¢ gives the direction of the tan. gent of the curve at a point. (6) Find the points on the curve for which ¢ is 0 or =/2, the tangent being parallel or perpendicular to the initial line. +. Limitations of the Curve. (a) Let the least and the greatest value of r bea@ and 6 respectively, then the curve lies with in a circte of a radius b but outside the circle of radius a. (8) Solve the given equation of the curve for,r in terms of 9 and find for what value of 6. r is imaginary. Let for ¢«<6©, then ei symptotes exist. These can be found by the method explained earlig; 6. Region. (a) Find the variation of r for positive and ive values of @, marking values of @ for which r attains a maximum, minimum or zero value. When r is a periodic function er §. then ve values of 6 need not be considered and curve is traced for one period only. _ _{b) Giving suitable values to 6, find the corresponding valucs ofrto get some points on the curve. Find also ¢ for these values of 6. - The following examples illustrate the use of the above procedure. Example 1. Trace the cardioid r=a (J-+cos 8). 1. Symmetry. The curve is symmetrical about the initial line, because on changing 8 to —9, the cquation of the curve does not change. Pole. The curve passes through the pole, because when ris zero. The line 8 js tangent to the curve at the origin. => Determination of ¢. From the equation of the curve, r=a (1+cos 4) dr . ig =a sin @cuRVE TRACING 7 285 di ___ a(1+cos 8) Hence . ar =~ asin ® __ 2 cos®, $ 2 sin 8/2 cos 8/2 = 9 tan (244 =—cot > =tan ($+ 2 ) 5 * tan ¢=tan (F+3 ) oF Now 5 2 when 6=0, hence r=2a (from the equation of the curve). Thus the tangent to the curve at the point (2a, 0) is at right angle to the radius vector (initial line). 4, Limitations of the curve. The least value of r is 0 (when g@=m) and the greatest value ofr is 2a (when 9=0). Hence the curve lies entirely with in a circle of radius 2a. 5. Asymptotes. The curve.has no asymptotes, because for no value of 4, r tends to infinity. 6. ‘Region. (a) When @ increases from 0 to, r decreases from 2ato0, / be . (b) Some points on the curve 25 ° = ® so > FT FD * r=2a Il'5a a O'Sa 0 c 4 2n 3 5 73s © FC *286 ENGINEERING MATHEMATICS The portion of the curve between = to 2% is traced by sym. metry. The shape of the curve is shown in the figure below. Example 2. Trace the curve r=a+b cos 8 (a > 6). 1. Symmetry. The curve is symmetrical about the initial line, because on changing @ to — 6, the equation of the curve does not change. 2. Pole. When r=0, o=a— 2 cos 6 . | cos 6 |=4> 1 a>b] o b which is not possible. Hence for no value of 9; r equals zero. Therefore the curve does not pass through the pole. 3. Determination of ¢. From the equation of the curve, r=a+b cos 8 Ha—p sino Hence r ge (age cost 8 a tan $= — apb eos? 8 Now for no value of 8, a+b cos 9=0 Hence $0, at any point. Also tan ¢ is 0, when 6=0, i.e. at the point (a+, 0). ‘Thus the tangent is perpendicular to the initial line at (2+, 0). 4, Limitations of the curve, The least value of r is a~b (when @=n) and the greatest value of ris a+b (when @=0). Hence the curve lies entirely within a circle of radius of a+b. 5. Asymptotes. The curve has no asymptotes, because for no value of 4, r tends to infinity. 6. Region. (a) As 4 increases from 0 to =, r decreases from a+b to a—b and for no value of 4, r is zero. . (6) Some points or: the curve. 9=0 Ja n r=atb a—b ¢=-o tan 5) +0 NANCURVE TRACING or Hence the shape of the curve is as shown in the figure. « The curve is known as Lamicon. . Example 3. Trace the curve r?=a" cos 28. 1, Symmetry. The curve is symmetrical about the pole, because on changing r to —r;- the equation of the curve remains unaltered. 2. Pole. When §=+:7» we find r=0. Hence the curve passes through the pole. The tangents at the pole are the lines o=45 3. Determination of $. From the equation of the curve, we have x =a? cos 28 “ or Gi, = —=2a? sin 20 . . d6 a cos 26 ee 0 ap =~ asin 28 008 28 tan #=tan ( 5 +26 ) 5 == 420 or b= +2 When 6=0, $= and r=-ta (from the equation of the curve). ._,., Hence at the points (-ba, 0), the tangents are perpendicular tu initial line.288 ENGINEERING MATHEMATICS 4. Limitations of the curve. The least value r is zero (when 0="/4) and the greatest value of r is a (considering positive values only). Hence the curve lies entirely within circle of radius a. 4. Asymptotes. The curve has no asymptotes because for no value of 6, r tends to infinity. : my 6. Region. (a) As @ increases from 0 to a7 decreases from ato0. When @ lies between = and 37, cos 20 is negative 4 BT and hence r* is negative. Thus between and 3%, r is imaginary, Hence no part of the curve lies between aoe (6) Some points on the curve. We shall consider the points in the first quadrants between 6=0 to 4” as the curve is symmetri- cal about the pole. The curve is known as Lamniscate of Bernoulli. Example 4. Trace the curve r=a sin 38, 1, Symmetry. The curve is symmetrical about 1 line 6=-5» becvuse on changing @ to =~, the equation of the curve does not change.CURVE 1KAUING 289 2x 3° pole lies on the curve and tangents there at are 8=0, 0= = 2. Pole. Here r=0, when 6=0, zz v0 Hence the because other values of 0 give the same tangents. 3. Limitations of the curve. The least value of r js — and the greatest value of r is a. Hence the curve lies entirely within acircle of radius a. 10 4. Asymptotes. The curve has no asymptotes becaus: no value of 6, r tends to infinity. fe for 5. Region. Some points on the curve. = 2 = oe 3s 7 r= 0 a 0 a 0 As @ increases from 0 to = , 7 is positive and incteases from Oto a. When @ increascs from = to > , nis tve and decreases from a to 0. Thus we get a loop between the lines 8=0 and 8: zn . . As 6 further increases from 5 to 2 ,r remains negative and decreases from a to 0 (numerically). Thus we get a second loop 2: to between the lines 9= + Similarly we get another loop . 2: - between the lines On- tom. The shape of the curve is shown in the figure. 8215 e-2%, 7 a=T/3 . , eM a0, y290 ENGINEERING MALMEMATICS The curve consists of three loops and is known as Three Leaved Rose. Example 5. Trace the curve r=a cos 28. J. Symmetry. (@) The curve is symmetrical about the initial line, because on changing @to —9%, the cquation of the curve remains unchanged. (8) The curve is symmetrical about the line 9==/2, because on € ta x—2, the equation of the curve does not change. chang OD % ls e-%, > 2, Pole. Here r=0, when 0 Hence the pole lics on the curve. Tangents there at are 6= 4 4 [The other values of @ give the same tangents] 3. Limitatioss of the curve, The Jeast value of ris —a and the greatest value of r is @. Hence the curve lies entirely with in a circle of radius a. 4, Asymptotes. The curve has no asymptotes, because for no value of 6, r tends to infinity. 5. Region. Some points on the curve 6= 0 r= acURYE TRACING a7L As 8 increases from 0 to 7/4, r decreases fro:a a to.0. When.? jncreases from 7/4 to 7/2, r is negative and increases from 0 toa (numerically). No further values of @ need be considered as the curve is symmetrical about the initial line and the line @= 5/2 The shape is shown in the figure. The curve consists of four loopsand is known as Four Leaved Rose sin n9 consists of » or 2n Note, A curve r=a cos n@ or r=. loops, according as # is odd or even. Example 6. Trace the curve 16 =a. 1. Symmetry. The curve is symmetrical about the line é=n/2, because on changing 9 to—4 and rto —r the cquation of the curve docs not change. 0 2. Pole. The value of r is not zero for any real finite value of 8. . Hence the curve docs not pass through the pole. 3. Asymptote. As 9->0, r+co. Thus the curve has an asymptote. To find the cquation of the asymptote, we write the equation of the curve as 1_6_.. 1-4-0 Now S()=0 gives 6=0. Also r@-t oat The equation of the asymptote is, | 1 i 8—0)=—— r sin (8—0) FO or r sin 0=a, a line parallel to the initial line at a distance a from it. 4, Region. (a) The equation of the asymptote to the curve is rsin 9=a and of the curve is ro=a or i As sin <4, the value of r for the curve is always less than the value of r for the asymptote. i Therefore the curve lies entirely below the asymptote. (b) As @ increases from 0 to 0, r is positive and decreases from © to 0.292 ENGINE (c) Some points on the curve 0= oO a= Qn on The shape of the curve for negative values of 8, is traced due to symmetry. : ‘The shape of the curve is shown in the figure. The curve is known as Reciprocal spiral. 117. Miscellaneous Examples Example 1. Trace the curve x'+ y= 3axy . (Folium of Descarte's) 1. Symmetry. The curve is symmetrical about the line yx, because on interchanging x and 3 the equation of the curve does not change. 2. Origin. a) The curve passes through the origin. (b) The tangents at the origin are obtained by equating the lowest degree terms in the equation of the curve to zero. Thus the tangents at the origin are xyv=0 or The tangents at the orig is a node. _ 3. Intersection with the axes. (a) The curve meets the co- ordinate axes at the origin only. (4) To find where the curve meets its lines of symmetry, i.c. , putting y=x in the equation of the curve, we get ie., x=0 or 3a/2 or = -3u/2. Hence the curve meets the line y=x at the point (3a/2, 3«'2) reseal and different, hence thet anve TRACING 253 To find the tangents at the point (3a/2, 30/2), we shift the origin 10 this point. To shift the origin to (3a/2, 3a/2), X=x- 3a/2, Yey—3o/2 The equation of the curve becomes (X+ 34/2) + (Y + 34/2)" = 3a ¥ + 3a/2)( ¥ + 34/2). Equating to zcro, the lowest degree terms, we have X+Y=0 or x—3a/2+y—34/2—0, xtys3a, 4. Asymptotes. (a) The curve has no asymptotes parallel tothe axes. (b) To find the oblique asymptotes, put y=m and x=1 in the highest degree terms. of fe b,(m)=1+m* $,(m) = —3am. For asymptotes to cxist, $3(m)=0 or 1+7°=0 or m=—1, the other values being imaginary. gm) —3am _a bam) 3m a c=-a (Co m=—1) Hence the equation of the asymptotes is xtyta=0 (so yo mxte) 5. Region. (a) x and y both cannot be negative simultanc- ously. Hence no part of the curve lies in the third quadrant. (b) To study the variation of y with x, we transform the curve to the polar form by putting x=r cos @ and y=r sin 8, which gives r®(cos® 6 + sin® 8) =3ar* sin 9 cos @ 3a sin 6 cos 0 or cos? 6 = z = 2 35 §= 0 a z 2 37 a _ 3V 2a 6V3a_ —6V3a nO 2 Gt3v5 9 Gv3s-) @ - (=2112a) (=1'51a) (=—2'46a) We observe as 6 increases from 0 to =/2, r first increases and then decteases to zero. Hence loop of the curve is between 6=0 - and 6=n/2. foENGINEERIMG MATHEMAT;: 2 to 35/4, 7 is negative anc numei shown in the figure. Y 4 ary ce cosh xje. e curye may be written es ele 7 7 Symmetry, The curve is symmetrical about the 3-a: _ Origin. cause when X=0, FTC. g¢xto —xthe equation of the curve does Rot The curve does not pass through the origin, be- Intersection with the axes. The curve does not meet the is, but meets y-axis at the point (0, c). 4, Special points. From the equation of the curve, we have, © (geet eH Fie lee terr) (er nF) Now when =0 Hence the tangent at the and 5. Asymptotes. point (0, c) is parallel to x-a) The curve has no asymptotes. 6. Region. “As x increases, y also increases and when xX 0, - yoo,CERVE TRACING 295 The shape of the curve is shown in the figure. c pe - en 0 . X EXERCISE 11 (c) Trace the following curves : sin @ a sin 26 sin 26 r=a (1+sin 6) Py x 9. 10. r=a (sec 8-i-cos 6) i. 12. r=ae™ — (a>0, m0) = asin? é 13. M. r= 2 15. 16. xtty! 17.ENGINEERING MATHEMATICS yeretl, p=0 10. yo x. Exercise 10 (b) (Page 256) woke, yobva 2. > a y=sl 4. x=41,y=41 x= cha, yo tb. Exercise 10 (c) (Page 264) 2r sin @ =a, 20=7 : kz a o— k= y= __4 __ resin ( a )= wees k= rcos 0t+b=a 4. O=9 . =) ._2a rain (0-5) =2 3 at+v2r sin( 64 = )=0. » where « 1s an integer. Exercise 1i (a) (Page 276)ANSWERS 13. M4. 507 f 5. See page 273 (Example 5) Exercise 11 (c) (Page 295)508 ENGINEERING MATHEMATICS. 11. 12. 13. D hes. iM >.6 8 Exercise 12 (a) (Page 301—303) Lo —(P—x+4)[x8+e 2. x-tan? xte 3. 3(°+3 tan™ x) +¢ 4. /2sinx+e 5. —sec x cosec x+e 6. tanx+2 secxte 7. tan x+cot xtc 8. sin x cos x+¢ z % ay 10. 12 11. 2V3+3 12. cosec (ba) log [ sate = Exercise 12 (b) (Page 308—310) 1. Asin x*)+e 2. log log (sin x) +e 3. Asin x)*?+¢ 4. ¥(x+log x)? +e 5. Mlog sec x+tan x)]'+e 6. Ut—4xt+D+¢ 13. e—tan™ (cos x) 4, $x+4 log (sin x—-cos x) +e (VEZ) 6 a x 16, 2 sin x-log tan (43). 2 a 17. - #L log (a+b cos d+apteslt 18. log (e"—e*) +e.