2 Geometrical Applications of Differentiation aaaaa
2.1 INTRODUCTION Though we have some algebraic results which give useful information about the graph of function and the function rate of change over most if not all of the functions. But to know the complete insight and details about the graph of the curve in space, we need to know first about certain other things like Maxima-Minima problems, Estimating approximation Errors, Intermediate forms, Role’s Theorem, mean value theorem, Taylor’s theorem, concavity, points of inflexion, sign of first derivatives, Asymptotes, etc. and which in turn can be known only with the differential co-efficient/derivative of the function at a point or over a certain change. Derivatives are interpreted as slope of curves and as instantaneous rate of change. We know that the first and second derivatives together tell how the graph of the function is shaped. Second derivative helps in estimating the linear approximation of the function. Collectively all above inference help in sketching the trace of the curve.
2.2 TANGENTS AND NORMALS I. Tangent and its Equation Let P(x, y) and Q(x + δx, y + δy) be neighbouring points on the curve y = f(x) with a supposition that the curve is continuous near P. Equation of any line through P(x, y) is Y – y = m(X – x) …(1) Y where X, Y are the current coordinates of any point on this B line (Fig. 2.1). y = f( x ) T´ Now, as Q → P, the straight line PQ tends in general to a (xxyy + δ , + δ ) Q definite straight line TPT', which is called the tangent to the curve at P(x, y). (,x yP ) Whence the slope of PQ, which in other words known as A Ψ ΨδΨ + ()yyy+δ − δy X gradient of PQ becomes = and in the limiting O T T´´ ()xxx+δ − δ x Fig. 2.1 δydy case when Q → P, lttan===Ψm δ→x0δxdx
97 98 Engineering Mathematics through Applications
Therefore, the eqn. (1) becomes equation of the tangent at P(x, y), and reduces to dy Yy−=() Xx − dx or Y – y = tanΨ(X– x) …(2) Cor. 1: Intercepts of the tangent on x-axis (i.e. OT): Putting Y = 0 in eqn (2), we get y OT()==− X x dy …(3) dx Intercepts of the tangent of y-axis (i.e. OT"): Putting X = 0 in eqn (2), we get dy OT"()==− Y y x … (4) dx ∂f dy f ∂ Cor. 2: =−x =− x If the given equation of the curve be f(x, y) = 0, then we know that ∂ dx fy f and hence eqn (2) reduces to ∂y
−=−fx() − Yy Xx or (X – x)fx + (Y – y)fy = 0 …(5) fy Cor. 3: If the given equation of the curve is in parametric form x = φ(t), y = ψ(t), dy ψ () dy ==dt 't dxdx φ '() t and whence the eqn (2) reduces to dt ψ't() Yt−ψ() =() Xt −φ() φ't() …(6)
Example 1: Find the equation of the tangent at any point (x, y ) to the curve x2/3 + y2/3 = a2/3. Show that the portion of the tangent intercepted between the axes is of constant length.
Solution: The given curve is x2/3 + y2/3 = a 2/3. 1/3 22−−dy dy=− y On differentiation with respect to x, we get xy1/3+= 1/3 0or . 33dx dx x Now equation of the tangent at (x, y) is Y – y = m(X – x), i.e. (Y – y) = –(y/x)1/3 (X – x) …(1) Further, intercept on x-axis, i.e. Y = 0 implies
y12 2 2 1 2 1 Xx=− =+ xxy3333333 =() x + yx = ax dy …(2) dx Intercept on y-axis, i.e. X = 0 implies dy y 1/3 Yyx=− =+⋅ yx =+ yx2/3 y 1/3 dx x Geometrical Applications of Differentiation 99
= (x2/3 + y2/3)y1/3 = a2/3 y1/3 …(3)
2222 The square root of sum of square of the two intercepts =+()axy333() ==aa4/3 2/3 a Thus, the portion of the tangent intercepted between the axis is a which is a constant length. x y Example 2: Prove that +=1 touches the curve y = be–x/a at the ‘point’ where the curve ab crosses the axis of y. Solution: The curve y = be–x/a shall cross the y-axis at a point where x = 0, i.e. y = be–0/a = b Now at the point (0, b), dy − −11bbb ==−=−=−be xa/ 0 dx()0,b a()0,b aa a a e ()0,b
∴ Tangent at (0, b) is b (Yb−=−−) () X0or aY + bX – ab = 0 a x y or +=1 (Because relation is true for all x, y). ab Example 3: Find the equation of the tangent to the curve y(x – 2)(x – 3) – x + 7 = 0 at the point where it cuts the x-axis. Solution: The curve cuts the x-axis at y = 0. But y = 0 implies –x + 7 = 0 or x = 7 Thereby means tangent is taken at the point (7, 0) The equation of the tangent at (7, 0), however, is given by
()−=−fx () − Yy Xx, where fx = y(2x – 5) – 1 and fy = (x – 2)(x – 3) fy (7,0) 1 or ()YX−=07() −, i.e. X – 20Y – 7 = 0 is the desired equation of tangent at (7, 0). 20 Example 4: Find the equation of the tangent at any point ‘θ’ on the curve whose equations are x = a(θ + sinθ); y = a(1 – cosθ).
Solution: As the given equation is in parametric form whence the equation of tangent at any point ‘θ’ is given by y'()θ Yy−θ=() () Xx −θ() x'()θ 100 Engineering Mathematics through Applications
dx dy Here =+θa()1cos , =θasin dθ dθ asinθ ∴Ya−−θ=()1cos{} Xa −θ+θ() sin a()1cos+θ θθ θ2sincosa Ya−=2sin2 22{} Xaa −θ−θ sin or θ 2 2cosa2 2 θθ θ or Ya−=2 sin2 tan() Xaa −θ−θ⋅ sin tan 22 2 θθ θ 22 Ya−=2sin tan ( Xa −θ− ) 2sin a 22 2 θθ or YXa=−θ()tanor()tan yxa =−θ 22 (on changing the current coordinates to general coordinates). m m x y Example 5: If the straight line p = x cosα + y sinα touch the curve +=1 , prove ab m m m that ()abpcosαm–1 +() sinαm–1 = m–1 .
xm ym Solution: The given equation of the curve is +=1 …(1) abmm − xm−1 ydym 1 On differentiating with respect to x, we get mm+=0 abdxmm
− m m1 dy bx or Slope of the tangent at ()xy,==− . dx a y ∴Equation of the tangent at (x, y) becomes
− m m1 bx (Yy−=−) () Xx − ay
− xxmm−1 yymm1 or XY+=+=1, (using (1)) …(2) ababmmmm Now if the line xcos α + y sin α = p touch the given curve then it should be identical to equation (2). cosαα sin p Whence == mm−−11 xy1 m abm Geometrical Applications of Differentiation 101
m−1 ααm−1 xa==cosy b sin ⇒ and ap bp
mm Y ααmm−−11mm N abxcos+=+= sin y ∴ 1 N ppab o rm T´ whence the result. al P II Normal and its Equation
Normal to a curve y = f (x) at any point P(x, y) is the straight Ψ line (NP, say) through P(x, y ) but it is perpendicular to the X O T 1 tangent at P. For m to be the slope of the tangent, − (say m’) Fig. 2.2 m will be the slope of the normal (since we know that for two perpendicular lines, slope of one is the negative reciprocal of the another). With this notion, it becomes easy to write the equation of the normal, if the equation of the curve is given, i.e. 1 dx (Yy−=−−) () Xx or ()Yy−=−() Xx − …(1) m dy dy f Cor. 1: If the equation of the curve is f(x, y) = 0, then =− x so that equation of the normal dx fy reduces to (Yy−) ()Xx− ( −=−) 1() − or = …(2) Yy Xx −f ffyx x fy Cor. 2: If the equation of the curve is in the parametric form x = φ(t), y = ψ(t) so that dψ dy ψ't() ==dt ,the equation of the normal becomes dxdφφ '() t dt 1 ()Yy−=−() Xx − or [X – φ(t)] φ'(t) + [Y – ψ(t)]ψ'(t) = 0 …(3) ψ't() φ't()
III Lengths of Tangent, Normal, Subtangent, Subnormal, etc. Let y = f(x) be the equation of the curve with PTT' as tangent at point P(x, y) meeting the x-axis and y-axis at T and T' respectively. Let RPN be the normal at point P(x, y) meeting the x-axis at N and PM be the ordinate, OQ be the perpendicular from the origin to the normal (See dy Fig. 2.3). Let ∠MTP be Ψ so that tan Ψ= =y and ∠MPN = Ψ. Then we have the following dx 1 geometrical results: 102 Engineering Mathematics through Applications
1. Length of the tangent Y 2 =Ψ=+Ψ=+2 dx ()TP MPcosec y1 cot y 1 R dy Px(, y) (i.e. portion of the tangent between the curve Y Ψ
and x-axis). 9 – 0
°
2. Length of normal Ψ dy 2 y ()NP=Ψ=+Ψ=+ MPsec y 1 tan2 y 1 dx Ψ Y (i.e. portion of the normal between the curve X O T M N and x-axis) Q dx 3. Length of subtangent ()TM=Ψ= ycot y . T´ dy Fig. 2.3 dy 4. Length of subnormal ()tanMN=Ψ= y y . dx y 5. Length of intercept (of the tangent) on x-axis ()OT=− x (see cor. 1 on tangents). dy dx dy 6. Length of intercept (of the tangent) on y-axis ()OT'=− y x (see cor. 1 on tangent). dx xy− y 7. Length of perpendicular from (0, 0) on tangent ()OQ = 1 + 2 1y1
(since the equation of the tangent (Y – y) = y1(X – x) may be rewritten as y1X – Y – (xy1 – y) = 0 comparable to aX + bY + c = 0. Therefore, length of perpendicular
xy− y c from origin on the tangent become OQ = 1 comparable to + 2 22+ 1 y1 ab xyy+ 8. Length of perpendicular from (0, 0) on the normal ()OR = 1 + 2 1y1 dx (since on rewriting the equation of the normal (Yy−=−) () Xx − as X + y Y – (x + dy 1
yy1) = 0 which is comparable to aX + bY + c = 0, the perpendicular distance from (0, 0) xyy+ c becomes OR = 1 comparable to . + 2 22+ 1 y1 ab
dy Observations: To usual conventions, y, both are positive and T is taken to the left of M and further sign of dx TM is positive, otherwise, negative. Similar interpretations can be given for other values.
Example 6: Find the equation of the normal at any point t to the curve x = a(cost + t sint), y = a(sint – t cost). Verify that these normals touch a circle with its centre at the origin and whose radius is constant. Geometrical Applications of Differentiation 103
1 Solution: Equation of the normal at any point ‘t’ is given by (Yy−=−) () Xx − where y' x' dashes represents derivatives with respect to t. or [X – x(t)] x'(t) + [Y – y(t)] y'(t) = 0 …(1) dx Here x'() t==−+ a()sin t sin t + t cos t = at cos t dt dy and yt´()== a()cos t − cos t + t sin t = att sin dt ??? Hence the equation of the normal at ‘t’ becomes som [][]Xa−+(cos tt sin tatt ) cos +−− Ya (sin tt cos tatt ) sin = 0 mi Xcost + Ysint – a(cos2t + sin2t) = 0 or xcost + y sint = a …(2) (on replacing current coordinates to general coordinates) Now the perpendicular distance of this normal from (0, 0) (or in Fig. 2.3) is given by xyy+ atttatttt()()cos++ sin sin − cos tan 1==a +22+ 1y11tant sin2 t atacos + ==costa, which is a constant sect Hence it touches a circle of radius ‘a’ having its centre at (0, 0).
Example 7: Find the equations of the tangent and normals, the lengths of the tangent and subtangents, length of the normal and subnormal for the ellipse x = a cos t, y = b sint, at a π fixed point (x1, y1) for which t = /4. Solution: The given equation of the ellipse is x = a cost, y = b sint …(1) dxdy dy b dy b =−atsin , = bt cos , =− cot t , =− Thus, we get π …(2) dt dt dx a dxt= a 4 See the geometry, here, PT is the tangent, MT is the subtangent, PN is the normal, NM is the subnormal. Now, equation of the tangent at P(x , y ) is 1 1 Y (Y – y1) = m(X – x1) Px(, y ) dy 11 −=−ππ π or Ybsin t== Xa cos t /4 ttπ 44dx t= X 4 ONM T −=−−bb a i.e. YX 22a ⇒ bx+− ay ab 20 = …(3) Fig. 2.4 104 Engineering Mathematics through Applications
Equation of the normal at P(x1, y 1) is dx a (Yy−=−−) () Xx or Yb−=+−sin tπ Xa cos tπ 11t= t= dy 4 b 4 ()22− −=−ba a −+ba = i.e. YXor (ax by) 0 …(4) 22b 2
2 2 dx a b a2 ab22+ Length of the tangent PT=+ y1sin11 =() b t +−= + = dy b2 b2 2 (at t = π/4) …(5) dx a a Length of subtangent MT ==ybt( sin ) π −=− …(6) dyt= b 2 (at t = π/4) 4
2 2 π+dy bba22 b Length of the normal, PNat t== y 1 + = ( b sin t )π 1 +−= t= 4a2dx4 a …(7) πdy bb2 Length of subnormal, MNat t== y = ( b sin) π ⋅−=− …(8) π t= 42dxt= 4 a a 4 IV Angle of Intersection of Two Curves By angle of intersection of two curves, we mean the angle between the tangents at common point of intersection. Let us suppose that the equation of tangent of the two curves AB and CD be
y = m1x + c1 and y = m2x + c2 T1 T4 with m1 and m2 being the slopes respectively. Then the angle of intersection of the above two curves will be D − A θ= mm12 tan + …(1) P 1 mm12 θ When m m = –1, θ becomes 90°, i.e. the two curves cut 1 2 T θ θ 3 orthogonally whereas if m1 = m2 then tan = 0, i.e = 0 means the two tangents will be a common one. C T B2 Example 8: Find the angle of intersection of the circle xy22+=2 a 2 and the rectangular hyperbola x2 – y2 = a2. Fig. 2.5
Solution: Intersection means a common point viz. take the sum of the two equations, implying a 1 221xa22=+()⇒x=±()21 + 2 2 a2 a 1 For this value of x, we get y2=−()21 or y=±()21 − 2 2 2 aa1/2 1/2 So that common point are ±+±−()21 ,() 21 22 Geometrical Applications of Differentiation 105
For the given curves fxy(,)=+− x22 y 2 a 2 = 0 and φ(x, y) = x2 – y2 – a2 = 0 the equation gradients will be φ f 2xx x 2xx m=−x =− =− and m=− =− − = 1 2 φ fyyy 2 y 2yy
xx −− mm− yy 2xy ∴ tanθ=12 = = = 1 (with above obtained x and y) +− 22 1 mm12 xx x y 1+− yy π whence θ= . 4 1111 Example 9: The curves ax2 + by2 = 1 and cx2 + dy2 = 1 shall cut orthogonally if –=–. ab cd Solution: Let P(h, k) be the point of intersection for given equations ax2 + by2 = 1 …(1) cx2 + dy2 = 1 …(2) Means this common point must lie on both the equations i.e., ah2 + bk2 = 1 and ch2 + dk2 = 1
hk221 On solving for (h, k), we get == −+bd −+ ca adcb − hdbadcb2=−()()/ − or …(3) kacadcb2=−()/() − Differentiating equation (1) with respect to ‘x’ dy ax ()=−ah =− or mP1at …(4) dx by bk Likewise differentiating equation (2) with respect to x dy cx ch =− or m =− …(5) dx dy 2 dk
For orthogonal intersection, we must have m1 × m2 = –1 −−ah ch i.e., ×=−1 or ac·h2 + bd·k2 = 0 bk dk db−− ac db−− ac or ac⋅+ bd =0 or +=0 ad−− cb ad cb bd ac
11−=− 11 or bd ac whence the required condition for orthogonality. 106 Engineering Mathematics through Applications
ASSIGNMENT 1
1. At what point is the tangent to the curve y = logx is parallel to the chord joining the points (0, 0) and (0, 1). n n x +=y x +=y 2. The straight line 2 touches the curve 2 for all values of n. Find ab ab the point of contact. n n x n−1 y n−1 3. If p = x cos α + y sin α, touch the curve += 1 , prove that ab pn = (acos α)n + (bsinα)n. 4. Prove that the condition for the line xcos α + y sinα = p to touch the curve xmyn = am + n, is pm + n · mm · nn = (m + n)m + n · am + n cosmα sinnα. 5. Show that sum of the intercepts on the axes of any tangent to the curve xya+= is constant. 6. For the curve x = a sin3θ, y = a cos3θ, find the angle which the perpendicular drawn from the origin to the tangent at the point θ makes with the axis of x. 7. Tangents are drawn from the origin to the curve y = sinx. Prove that their points of contact lie on the curve x2y2 = x2 – y2. 3 3 3 8. If the tangent at (x1, y1) to the curve x + y = a meets the curve again in (x2, y2), show xy that 2 +=−2 1. xy11 9. If the tangent to the curve x1/2 + y1/2 = a1/2 at any point on it cuts the axes OX, OY at P, Q respectively, prove that OP + OQ = a. 10. Show that the tangents at the points where the straight line ax + by = 0 meets the ellipse ax2 + 2hxy + by2 = 1 are parallel to the x-axis, and that the tangent at the points where the straight line hx + by = 0 meets the ellipse are parallel to y-axis. 11. Show that the exponential curve ybe= xa/ , the subtangent is of constant length and the subnormal varies as the square of the ordinate. x x 12. In the catenary yc=cosh , prove that the length of the subtangent is ccosh and that c c 3x of the subnormal is csinh . c 13. Find the length of the tangent, normal, sub-tangent and sub-normal of the cycloid x = a(t + sint), y = a(1 – cost). 14. For the curve x = a cos3θ, y = a sin3θ, show that the portion of the tangent intercepted between the point of contact and the x-axis is y cosec θ. Also find the length of the subnormal. θ 15. For the curve y = a sinθ, xa=−θlogcot cos , find the lengths of the sub-tangent and 2 sub-normals at the point θ = π/4. Geometrical Applications of Differentiation 107
16. Prove that the subnormal to the curve xy = c2 varies as the cube of the ordinate.
17. Find the angle of intersection of the curve x2 – y2 = a2 and xya222+= 2.
− 3 18. Show that the parabolas y2 = 4ax and 2x2 = ay intersect at an angle of tan 1 . 5
x2 y2 x2 y2 19. Prove that the curves +=1 and +=1 will cut orthogonally if a – b = c – d. ab cd
2.3 TANGENTS AND NORMAL FOR CURVES IN POLAR COORDINATES
dè I. Angle between the Radius Vector and Tangent tanφ = r ⋅ dr Let P(r, θ) and Q(r + δr, θ + δθ) be any two neighbouring points on the curve r = f(θ). Let PT be the tangent at P(r, θ) and φ be the angle enclosed by this tangent with the radius vector OP. Join PQ and draw PN ⊥ OQ, then from the right angled triangle ONP, NP = r sinδθ, ON = rcos δθ, B δθ δθ ∴ NQ = OQ – ON Qr( + r, + ) N δ δθ = (r + r) – r cos α = δr + r(1 – cosδθ) δθ Pr(, θ) =δ + 2 rr2sin …(1) r 2 f If ∠NQP = α, then δθ A δθ qθ α=NP = rsin tan δθ O NQ δ+rr2sin2 …(2) 2 p In the limiting case, when Q → P, i.e. δθ → 0, the chord T QP turns about P and becomes tangent at P and thus resulting in α → φ. Fig. 2.6 rsinδθ ∴ tanφ= Lt tan α= Lt →δθ→ δθ QP 0 δ+rr2sin2 2
sin δθ rδθ = Lt δθ→0 δθ δδθsin r +⋅2 ⋅ r δθ sin δθ 2 2
rd⋅θ1 ==r dr +⋅⋅r 10 dr …(3) dθ 108 Engineering Mathematics through Applications
II. Length of the Perpendicular from Pole on the Tangent See the geometry of Fig. 2.6, p be the length of the perpendicular OT from the pole O to the tangent at P on the curve, then 2 2 φ 111=+ dr 1 =+2 du = 1 (i) p = rsin (ii) (iii) u , where u . prrd224θ pd2 θ r In ∆OPT, p = r sin φ result (i) …(4) 11 1 ∴ =φ=+φcosec22() 1 cot pr22 r 2 2 11dr =+1 rrd2 θ 2 11dr =+ result (ii) …(5) rrd24θ 1 Now for result (iii), given u = r du==− d11 dr therefore, ddrθθ rd2 θ
22 du1 dr implying = drdθθ4
2 du 11 ⇒ =− using result (ii) dprθ 22
2 1 du =+u2 or pd2 θ, result (iii) …(6) These three relations which involves p (length of the perpendicular from the pole upon the tangent) and r (the radius vector) both, are known as pedal equations of the curve.
III. Polar Subtangent and Polar Subnormal Let P(r, θ) be any point on the curve r = f(θ) with OP as the radius vector and let a line drawn through the pole perpendicular to the radius vector OP meets the tangent and normal in T and N respectively. Then OT is called the polar subtangent and ON is called polar subnormal. ddθθ Since the angle OPT is φ, therefore, polar subtangent OT=φ== rtan r r r2 dr dr Also in ∆ONP, polar subnormal, 1 dr ON=⋅ rtan ∠ OPN = r cot φ=⋅ r = dθ θ r d …(7) dr Geometrical Applications of Differentiation 109
Length of polar tangent N 2 dθ φ ==PT rsec φ=+ r 1 tan22 φ=+ r 1 r …(8) dr Length of polar normal 90 – φ ==PN rcosec φ=+φ r 1 cot2 Pr(, θ) φ 2 90 =+222dr =+ θ rrr 1. …(9) dθ O X 90° It is evident from the formula of subtangent that it is measured positively to the right with a supposition that observer is stationed at O and looking in the direction of P. Whereas a negative value of the subtangent shows that T is to the left as shown in the Fig. 2.8. A similar notion is applied in T case of subnormal. Fig. 2.7 T Example 10: Show that in the equiangular spiral r = aeθ cot α, the tangent is inclined at a constant angle to the radius vector. B p A Solution: The given equation r = aeθcotα O dr θαcot On differentiation gives, =⋅α=αaecot r cot Fig. 2.8 dθ dθ so that tanφ=r = tan α or φ = α dr Hence the angle φ, the angle between the tangent and the radius vector is constant. 2a Example 11: Prove that in the parabola =1–cosθ, r θ θ (i) φ=π– (ii) pa=cosec 2 2 (iii) Polar subtangent = 2acosecθ (iv) p2 = ar
2a Solution: Given =−()1cos θ r Taking log on both sides, log2a = logr + log(1 – cosθ) 1sindr θ On differentiation, 0=+ rdθ−θ1cos θ 1 dr θ θ tan or =−cot ⇒ d =− 2 rdθ 2 dr r
θ tan dθθθ ∴ tanφ=rr = −2 =− tan = tan π− dr r 22 110 Engineering Mathematics through Applications
θ implying φ=π− 2 θθθ =φ=π−= 22aa ⋅ = ⋅ Again, we know that, prsin r sin−θ sinθ sin 21cos22sin2 2 2 implying p = a cosecθ/2
22 2==aa =⋅=2 a or paarθ−θ−θ sin21cos 1cos 2 2
2a θ For polar subtangent, OT (Fig. 2.7) rtanφ= ⋅ tan π− 1cos−θ 2 θ sin 22aa =−=−=−θ22coscea . θθ θ 2sin2 cos sin 22
Example 12: Show that the angle of intersection of the following curves r = a(1 – cosθ), r = a(1 + cosθ) is π/2.
Solution: Taking logs on both sides of Ist equation, logr = loga(1 – cosθ) 1sindr θθ On differentiation, ==cot rdθ−θ1cos 2 dθθ θ ∴ tanφ=r = tan or φ= 1 dr 2 1 2 1cos+θ θ Similarly from 2nd equation, tanφ= =− cot 2 −θsin 2 πθ or φ= + 2 22 Thus the angle of intersection ‘α’ between the two curves is given by π tanα = tan(φ ~ φ )orα=() φ~ φ = . 1 2 122
Example 13: Find the pedal equation of the parabola y2 = 4a(a + x).
Solution: Equation of the tangent to the given parabola y2 = 4a(a + x) is written as
2a 2a = (Yy−=) () Xx −, where y1 in this case …(1) y y Geometrical Applications of Differentiation 111
∴ The perpendicular distance of this tangent from (0, 0), i.e. OQ (Fig. 2.3) 2a xy− xy−− y y2 ax y2 p ==1 = ++2222 14yya12a 1+ y
244ax−− ax a2 22aax()+ p ==− implying 444ax++ a22 a 42aax()+ …(2)
p2 or =+(2ax ) a Now, r2 = x2 + y2 = x2 + (4ax + 4a2) = (2a + x)2 p2 or r = (using value of 3) a Therefore, the required pedal equation, viz. a relation involving p (the length of the perpendicular from the pole to the tangent) and r (the radius vector) for this problem is p2 = ar.
x2 y2 Example 14: Find the pedal equation of the ellipse +=1. ab22
x2 y2 Solution: The equation of the tangent to given ellipse +=1 …(1) ab22 can be written as: xX yY +=1 …(2) ab22 which is comparable to AX + BY + C = 0 with its perpendicular distance from (O, O), C 1 p== AB22+2 2 x+y ab22
2 2 1=+xy or pab244 …(3)
Now, r2 = x2 + y2 or r2 – b2 = x2 + (y2 – b2)
222 rb22− xby−−() or = ab22−− ab 22 112 Engineering Mathematics through Applications
2 22−b xx2 ==a2x(using (1)) ab22− a 2 xrb222− or = …(4) aab222−
y2 ar22− Similarly, = …(5) bab222− x2 y2 On substituting the values of and into eqn.(3), we get a2 b2
−22 − 22 11=+rb 1 ar paab2222−− bab 222
ab22 ()()rb22−+− b 4 a 4 ar 22 ==+−()abr222 or pab222− …(6)
ab22 whence =+−abr222 is the desired pedal equation. p2
Example 15: Prove that the locus of the extremity of polar subnormal of the curve r = f(θ) is r = f´(θ – π/2); and hence show that the locus of the extremity of the polar subnormal of the equiangular spiral r = a eθcotα is another equiangular spiral.
Solution: Let the coordinates of N (Fig. 2.7) be (r, θ), then dr rON==polar subnormal ==θ f´() …(1) dθ π π θ=∠ = +θ θ=θ− whereas NOX or …(2) 2 2 Now eliminating θ between eqn. (1) and eqn. (2), we get a relation between r and θ as π =θ−´ rf …(3) 2 π Hence the locus of (r, θ), i.e., rf=θ−´ . 2 dr θ⋅ α Now for the curve r = a eθ·cotα, =θ=fa´() cot α⋅ ecot dθ π Hence the required locus is rf=θ−´ 2
π θ− ⋅ α cot or ra=⋅cot ae2 Geometrical Applications of Differentiation 113
π −⋅ α cot . θα =⋅aaeecot 2 cot = beθ·cot α π α where b = a cotα ·e– /2 cot is an arbitrary constant. Whence the locus of (r, θ) is another equiangular spiral.
ASSIGNMENT 2
1. Show that the tangent to the cardiod r = a(1 + cosθ) at the point θ = π/3 is parallel to the initial line. 2. Show that the angle between the tangent at any point P and the line joining P to the origin is the same as all the points of the curve log(x2 + y2) = k tan–1(y/x). 3. Show that in the curve r = aθ, the polar subnormal is constant and in the curve rθ = a, the polar subtangent is constant. 4. Find the angle of intersection of the curves r = 2 sin θ and r = 2cos θ 5. Show that the curves rn = ancosnθ and rn = bn sinnθ cut each other orthogonally. x2 y2 ab22 6. Prove that the pedal equation of the hyperbola −=1 is =−+rab222. ab22 p2 7. Show that the relation r2 = a2 – 3p2 is the pedal equation of the astroid x = acos3t, y = b sin3t. 8. Find the pedal equation of the curves: (i) r = a(1 + cosθ)(ii) r2 = a2sin2θ (iii) rm = a mcos mθ 9. Show that the pedal equation of (i) the hyperbola r2 ·cos2θ = a2 is pr = a2 (ii) the lemniscate r2 = a2cos2θ is r3 = a2p r4 (iii) of the archimedian spiral r = aθ is p2 = . ra22+ 10. Show that the length of the perpendicular from the pole on the tangent to the ellipse p 112l =+()1cose θ is given by =−+1e2 . r per22
2.4 DERIVATIVE OF ARC
2 ds dy I. Cartesian Form: =+1 . dx dx Let P(x, y) and Q(x + δx, y + δy) be two neighbouring points on the curve AB (i.e. y = f (x)) such that arc AP = s, AQ = (s + δs) then arc PQ = δs and let chord PQ = δc (Fig. 2.9). Draw PL and QM perpendicular on the X-axis and PN perpendicular on QM. 114 Engineering Mathematics through Applications
From the right angle triangle PNQ, Y (PQ)2 = (PN)2 + (NQ)2 B δ 2 δ 2 δ 2 Q ( c) = ( x) + ( y) δ c δy δ2 δ2 P c =+ y s or 1 …(1) A δ N δδxx x δδδδ222 δ2 sscs=⋅ = +y Now 1 …(2) δδδδδxcxcx ψ In the limiting case as Q → P, i.e. δc → 0, we have X O T L M 222 δδss δyδ Fig. 2.9 llt=+t1 , s = δ→δδδ δ→ lt 1 xc00xcxδ→c 0δc
2dy 2 ds =+ =+()2 or 11 y1 …(3) dx dx ds If s is measured in such a way that is positive, i.e. if x and s increases together, dx 2 ds dy then =+1 …(4) dx dx
Note: Strictly speaking ds=+ dx22 dy holds only for the case when dx > 0 otherwise ds=− dx22 + dy . For this reason, in the general case, this formula is more correctly written as ds=+ dx22 dy Again, if the equation of the curve is x = f (y), then
2 2 ds ds dxdy dx dx ==+⋅=+11 …(5) dy dx dy dx dy dy dy Further, we know that =ψtan , where ψ is the angle which the tangent makes with dx the initial axis. 2 ds dy Hence =+11tansec =+2 ψ=ψ dx dx dx or =ψcos …(6) ds dy dy dx Likewise, ==ψ⋅ψ=ψtan cos sin …(7) ds dx ds II. Parametric Curves If the equation of the curve is in parametric form, viz. x = f (t), y = φ(t), then
222 ds ds dxdy dx dx dy ==+⋅=+1 …(8) dt dx dt dx dt dt dt Geometrical Applications of Differentiation 115
Qr( + δθr , + δθ ) 2 B ds dr III. Polar Formula: =+r2 δs ddθθ N δc θ δ θ δθ Let P(r, ) and Q(r + r, + ) be any two Pr(, θ) neighbouring points on the curve AB, r = f(θ) in such a way that AP = s, arc PQ = δs and chord PQ = δc. r A Draw perpendicular PN on OQ, then in ∆OPN, δθ ON = r cosδθ, NP = r sinδθ, Ψ θ and OQ – QN = NQ = (r + δr) – r cosδθ O(0, 0) T inital aixs θ = 0 δθ =δ + 2 rr2sin Fig. 2.10 2 Now in right angle triangle PNQ, (PQ)2 = (PN)2 + (NQ)2 δθ 2 or δ=cr22(sin δθ+δ+ ) rr 2sin 2 2 δθ 2 2 δδδδ2222 δθ2 δ 2sinr ∴ sscs== rrsin ++ 2 δθ δcc δθ δ δθ δθ δθ
δθ 2 δδθδδθ22 rsin sr2sin 2 =++⋅r sin δδθδθc δθ 2 …(9) 2 In the limiting case when Q → P, i.e. δs → 0 (or δθ → 0), then 222 ds=⋅22 ⋅+ dr +⋅⋅ = + dr 11rrr 10 …(10) ddθθ d θ
ds If s is measured in such a way that is positive, i.e. s and θ increases together, then dθ 2 ds dr =+r2 …(11) ddθθ If the equation of the curve is θ = f (r), then
ds ds dθθθ dr22 d d ==+⋅=+rr221 …(12) dr dθθ dr d dr dr
ds dθ or =+1tan2 φ=φ sec,since r =φtan dr dr dr or =φcos …(13) ds 116 Engineering Mathematics through Applications
θθ θ dddrd==11 φ=φ⋅φ Likewise r cos tan cos ds dr ds r dr r dθ sinφ or = …(14) ds r θθ =φ=⋅=dd2 Further, prsin rr r …(15) ds ds 2 dr p rp22− and =φ=−φ=−cos 1 sin2 1 = …(16) ds r r
ds ds ds Example 16: For the cycloid x = a(1 – cosθ), y = a(θ + sinθ), find , and . dθ dx dy
dx dy Solution: Here =θ=+θaasin , (1 cos ), ddθθ
2 ds dx 2 dy therefore =+=θ++θ+θ a sin22 1 2cos cos ddθθ d θ
1/2 =+θa{}21() cos θθ1/2 ==aa4cos2 2 cos 22 ds ds dθθ1 θ ==2cosa ⋅= cosec dx dθθ dx2sin a 2 ds ds dθθ1 θ and == ⋅ = θ2cosa θ sec. dy d dy 222cosa 2 2 n–1 ds n n θ ds θ Example 17: Show that in the curve r = a cosn , =secan()n. Further, θ varies as dθ d dr2 (n – 1)th power of ‘r’ and anr2n ⋅ +=02–1n . ds2 Solution: Given rn = an cosnθ implies n logr = n loga + logcosnθ …(1) ndrsin nθ dr On differentiation, =−n or =−rntan θ …(2) rdθθcos n dθ
2 ds dr Now =+rrrnrn2222 =+tan θ=θ sec ddθθ
1 1 1−−11/()nn =θ⋅=⋅θ=θ()anncosn a() cos nn an() sec …(3) cosnθ hence the first result. Geometrical Applications of Differentiation 117
−−ds Again, rrrnrnannann11= ⋅sec θ= n sec θ= n cos θ⋅ sec θ= n (a constant) dθ
n ds= a ds or − means varies as (n – 1)th power of ‘r’ drθ n 1 dθ dr dr2 ddφφ dθ We know =φcos which implies =−sin φ =− sin φ⋅ ⋅ …(4) ds ds2 ds dθ ds dθπ1 π φ= =− =− θ= θ+ φ= θ+ Also tanrnnθ cot tan, i.e. n …(5) drtan n 2 2 dφ From the above relation, = n , sinφ = sin(nθ + π/2) = cosnθ. …(6) dθ
21nn− n−1 2 dr r r r dr − ∴ From (4), =−cosnn θ⋅ ⋅ =− ⋅ n ⋅ or anr2n +=21n 0 ds2 annn a a ds2
ASSIGNMENT 3 dy yc22− 1. If y2 = c2 + s2 for a curve then show that = . Also show that the perpendicular dx c from the foot of the ordinate upon the tangent is of constant length. ds 2. Find for the curves (i) ay2 = x3 (ii) y = c cosh x/c dx ds 3. Find for the curves dt xtytftsin+= cos '( ) t t (i) x = e sint, y = e cos t,(ii) xtytftcos−= sin "( ) ds 4. Find for the curve dθ a (i) r2 = a2cos2θ (ii) r = a(1+ cosθ)(iii) r = a(θ2 – 1) (iv) r = θ−2 1 (v)r = aθ (vi) rn = an sinnθ ds r22+ a 5. Prove that in the hyperbolic spiral rθ = a, = . dr r
θ α ds 6. For the curve r = ae cot , prove that (i) =αcot (ii) s = cr where c is constant, s being dr measured from the origin. ds 7. Prove that for any curve =rp2 . dθ 8. With usual meanings for r, θ, s and φ for the polar curve r = f(θ), show that dφdr2 +φ=rcosec2 0 .[Hint: Eqn (4) of Example 17] ddsθ 2 118 Engineering Mathematics through Applications
2.5 CURVATURE AND RADIUS OF CURVATURE A P In Figure 2.11, curve PQ bends more sharply than the curve AB “The measure of sharpness of bending of a curve at a particular point I is called the curvature of the curve at that point.” II About Terms: Absolute Curvature, Average Curvature;/Radius of curvature B Q Let AB be any curve with A as a fixed point on it. Fig. 2.11 Let P and Q be two neighbouring points on the curve AB. Let the arc AP = s, AQ = (s + δs) so that arc PQ = δs (A being the fixed point on the curve, arcs are measured from A.) Y Let the tangents to the curve at P and Q, makes angle ψ B and (ψ + δψ) with the initial axis respectively. In moving from P to Q, through a distance δs the tangent Qx( + δδ x , y + y ) has turned through an angle of δΨ. This angle δΨ is called Px(, y ) the total bending or total curvature of the arc PQ. As (i) Thus the angle δψ through which the tangent turns as its point of contact moves along the arc PQ is called Ψ ΨδΨ + the total bending/total curvature. X O δψ Fig. 2.12 (ii) The ratio is called the mean curvature/average δs curvature of the arc PQ. (iii) The limiting value of the mean curvature when Q → P is called the curvature of the curve at the point P. δψd ψ Thus the curvature, K of the curve at point P==lt . QP→ δsds (iv) The reciprocal of the curvature of the curve at P, provided this curvature of the curve is not zero, is called the radius of curvature at that point P(say). This is usually denoted by ρ. 1 ds Thus, ρ= = Kdψ The angle δψ is also called the angle of contingence of the arc PQ and is measured in radians. From definition it is clear that the curvature at point P depends only on the position of the curve and does not depend in any way on the system of co-ordinates. The relation between the arc length s of a curve (say AB) measured from a given fixed point (say A) on the curve and the angle ψ between the tangents at its extremities is called the intrinsic equation of the curve. ds The expression for radius of curvature is suitable only for those curves whose intrinsic dψ equations are given. Formula for radius of curvature for curves given in other different forms are discussed in succeeding section. Example 18: Find the radius of curvature at the point (s, ψ ) on the curve s = a log(tanψ + secψ) + a tanψ secψ. Geometrical Applications of Differentiation 119
Solution: ρ= ds dψ
12 +ψψ+ψψ⋅ψ2 =×ψ+ψ⋅ψa ()sec sec tan a()sec sec tan sec tan tanψ+ sec ψ = a secψ + a secψ(sec2ψ + tan2ψ) = asecψ[1 + tan2ψ + sec2ψ] = 2asec3ψ. Radius of Curvature (Cartesian, Parametric and Polar forms) I. Cartesian Form: Let y = f(x) be the equation of the curve in implicit form, then the slope of the tangent at any point is given by =ψ==dy mytan 1 …(1) dx ψ –1 i.e. = tan y1 ψ d ==1dy12 y ++22() …(2) dx11 y11 dx y
2 ds =+dy =+2 Further, 11 y1 …(3) dx dx
3 ()1+y22 ∴ ρ=ds = ds dx = 1 ψψ …(4) ddxdy2 which is the expression for radius of curvature in cartesian form. Convention of Signs: The positive root is taken in numerator of eqn (4), therefore, radius ρ of curvature, , will be positive when y2 is positive (i.e., when the curve is concave upwards) and negative, when y2 is negative (i.e., when the curve is concave downwards). In practice, ρ numerical value of is taken. At a point of inflexion y2 is zero, therefore, curvature of the curve at the point of inflexion is zero. II Parametric Form: Let x = φ(t) and y = ψ(t) dx dy so that = x´ and = y´ …(5) dt dt
dy dydt y´ Now y=1 = = …(6) dx dt dx x´
− ==dddtdy' =1 y' = x'y" y'x" and yy21 …(7) dx dt dx x' x´ dt x' x'3 120 Engineering Mathematics through Applications
3 y' 22 31+ 3 2 ()1+y22x' ()x'+ y' 22 ρ=1 = = ∴ − …(8) y2 x'y"− y'x" x'y" x"y' x' 3 is desired expression of radius of curvature in the parametric form. Observations (i) The curvature (or radius of curvature) of the curve at any point is independent of the choice of X and Y axes since the curvature or the radius of curvature is an intrinsic property of the curve. Hence, x and y can be interchanged in the formula of curvature (κ) and radius of curvature (ρ). (ii) When the tangent at a point (where radius of curvature is desired) on the curve is parallel to Y-axis, i.e. 3 3 2 2 2 2 dx +dy 1+ 1 dy dx dy =∞ ρ= ρ= , then the formula 2 becomes inapplicable and is given by 2 dx dy dx dx2 dy2 ρ= 1 dy dx which reduces to 2 , since =∞ means = 0 (a defined quantity). dx dx dy dy2
III. Radius of Curvature at the Origin (Newtons Formula) (i) If axis of x is tangent to the curve at the origin, 2 ρ= x then ()0, 0 Lt …(9) y→0 2y dy Since axis of x as tangent at (0, 0) means = 0 at (0, 0) dx x2 2x Lt Lt 1 →==→dy Lt x 0 2y x 02 → 2 →x 0 dy Also y→0 y 0 dx y→0 (a defined quantity) dx2 0 0 form form 0 0
3 2 2 +dy 1 dx 2 ρ=(0,0) =1 = x ()0, 0 Lt ∴ dy22 dyx→02 y …(10) y→0 22 dx(0,0) dx () 0, 0
(ii) If axis of y is tangent to the curve at the origin, then 2 ρ= y ()0, 0 Lt x→0 2x (Its proof is similar to part (i)) y→0 Geometrical Applications of Differentiation 121
(iii) In case neither of the axes is tangent to the curve at the origin: Write the equation of the curve as x2 yfxf==+( ) (0) xf' (0) + f" (0) +……∞ [By Maclaurin’s series] 2! x2 =+px q +……∞ (³ f(0) = 0 in this case) 2
==d where pf'(0) dx atx = 0
d2 qf"==(0) and 2 dx atx = 0
3 2 dy 2 + 3 1 dx ()+22 ρ=0, 0 =1 p ∴ ()0, 0 dy2 q …(11) 2 dx ()0, 0
Observations (i) For ascertaining the tangents at the origin, equate to zero the lowest degree terms occurring in f(x, y) = 0 since the equation of the tangent is a linear one, i.e. of the form ax + by + c = 0 and if passes through the origin then c = 0. dy (ii) When axis of x is tangent to the curve at the origin then (y = 0), = 0 dx dx dy (iii) When axis of y is tangent to the curve at the origin then (x = 0), = 0 or =∞. dy dx
IV Some other Forms of ρ: (i) When x and y are given functions of the length of arc s. dx cosψ= ds
dψ dx2 dy 1 dx2 Differentiating with respect to s, −ψsin = or −⋅= …(12) ds ds2 dsρ ds2
dy − ds Therefore ρ= dx2 …(13) ds2 dy dx 1 dy2 (ii) Similarly, starting with sinψ= , we have = …(14) ds dsρ dx2 122 Engineering Mathematics through Applications
dx ρ= ds ∴ dy2 …(15) ds2 1 dx 1 dy (iii) Squaring and adding the values and , ρds ρds
2 2 1dx2 dy2 we get =+ …(16) ρ22ds ds 2
Some Standard Parametric Curves: x = φ(t), y = ψ(t)
S. No. Name of the curve Cartisian Form Parametric Form
x2 y2 xa=cos t ' 1. Ellipse +=1 , ab22 yb=sin t xa=cos t , 2. Circle x2 + y2 = a2 , ya=sin t} 1 −t2 x = , +2 2 2 1 t x + y = 1 2t , y = 1 +t2 =2 3. Parabola y2 = 4ax xat,, yat=2
x2 y2 xa=sec t , 4. Hyperbola −=1 ab22 yb=tan t
xct= 2 5. Rectangular xy = c =c, y t or xa= cosh t Hyperbola x2 – y2 = a2 ya= sinh t}
2 2 =3 x 2 +=y 3 xacos t , 6. Hypocycloid 1 ab yb=sin3 t
3 222 xa=cos t , 7. Astroid 333+= xya ya=sin3 t Geometrical Applications of Differentiation 123
2 xt=2 2 =−x 8. Cissoid yx1 t3 3 yt=− 3
xa=sin2 t 2 sin t 3 ya23( – x) = x r= a sin t Cissoid ya= cost cost
3at x = , + 3 9. Folium of Descrate’s x3 + y3 = 3axy 1 t , 3at2 y = 1+ t3 ax Example 19: For the curve y= , if ρ is the radius of curvature at any point (x, y ) ()a+x 2 2 2 2ρ3 y x show that =+ . axy
ax Solution: Given y = …(1) ax+ dy()axaax+−⋅12 y 2 == =a = ∴ y1 2 …(2) dx()ax+ a+ x x
dy2 3 y 3 = =−2 ⋅12 =−a =− 2 and ya2 2 3 dx2 ()ax+ a a+ x a x
3 y42 31+ ()+22x ρ=1y1 = Thus y y3 2 −2 ax 3 y 4 2 1+ ρx −=2 i.e. a y 3 x
2 22 4 2 2ρ3yyxx =+1 = + axyyx4 Hence the result.
Example 20: Apply Newton’s formula to find the radius of curvature at the origin for the cycloid x = a(θ + sinθ), y = a(1 – cosθ). 124 Engineering Mathematics through Applications
Solution: Here corresponding to θ = 0, x = 0 and y = 0, and hence the curve passes through the origin dy θ dy ==dθ asin Further dxdx a()1cos+θ dθ dy a⋅ 0 so that Lt ==0 θ = 0 dx a()11+ Means axis of x is tangent to the curve at the origin.
2 ρ= x ∴ ()0, 0 Lt x→0 2y y→0
2()θ+ θ2 = a sin Lt 0 θ→021cosa()−θ form θ→0 0
2sin1cosa()()θ+ θ + θ 0 = Lt form θ→0 2sinθ 0
a()()()()θ+θ−θ++θ+θsin sin 1cos1cos =Lt θ→0 cosθ
a00⋅+()() 1 + 11 + 1 ==4a. 1 Example 21: Show that the radii of curvature of the curve y2 = x2(a + x)/(a – x) at the origin is a 2 .
Solution: The equation of the curve, (a – x)y2 = (a + x)x2 passes through the origin. To see the nature of the tangent at the origin, equate to zero the lowest degree terms in x and y, i.e. ay2 = ax2 or y = ±x i.e., at the origin, neither of the axis are tangent to the given curve. 2 =+x +… ∴ Putting ypxq in the given equation, we get 2 2 x2 ()axpxq−++…=+() axx2 2 3 24 22x qx 2 3 or ()axpx−+++…=+2 pq axx 24 On comparing the coefficients of x2 and x3, we get ap2 = a ⇒ p = ±1 Geometrical Applications of Differentiation 125
2 and apq – p2 = 1 ⇒ q=± a 3 ()1 + p22 ∴ ρ=() where p = y | = f'(0) and q = y | = f"(0) 0, 0 q 1 x=0 2 x=0 3 ()11+ 2 ==±a2 ±2 a Hence ρ(0, 0) is numerically a 2. Alternately: Equation of the curve is ()ax+ yx22= ()ax−
1 ()+ =± ax2 ∴ yx 1 ()ax− 2
11− xx22 yx=±11 + ⋅ − aa
=± +xx +… + +… x11, [using (1 + x)n = 1 + nx + …] 22aa
xxx2 =±x1 + + + +… 224aaa2
xx23 i.e. yx=± + + +… aa42
2 =± +23xx + +… So that y1 1 aa42
=±26 +x +… and y 2 aa42
2 At (0, 0) y = ±1, y =± 1 2 a
3 3 ()+ 22 + 1 y1 ()112 a ρ= = =±⋅=±22a 2 ∴ ()0, 0 2 y2 ± 2 a Hence ρ(0, 0) is numerically a 2. 126 Engineering Mathematics through Applications
x2 y2 ab22 Example 22: Prove that for the ellipse +=1, ρ= , p being the perpendicular ab22 p3 distance from the centre on the tangent at (x, y). [MDU, 2005; NIT Kurukshetra, 2007]
x2 y2 Solution: For the given ellipse +=1, on differentiating with respect to x, we get ab22
2 dy ==−bx y1 …(1) dx a2 y
−bx2 yx− 2 4 2222bx42y b dy==dbxb − − a y =− + =− Further, y2 23 2 2 23 …(2) dx2222 dx a y a y ay a b ay Therefore, at any general point P(x, y), the radius of curvature
3 22 2 +−bx 3 1 2 2 ()1+2 ay ρ=y1 = 4 y2 − b ay23
3/2 3 bx42y2 2 + 244 =yba −b4 ay23
3 x2 y2 2 =−ab22 + …(3) ab44 x2 y2 The equation of the tangent at (x , y ) to the ellipse +=1 is given by 1 1 ab22 x+=y xy111 …(4) ab22 (comparable the general form of tangent ax + by + c = 0) Now perpendicular distance of the tangent (4) from (0, 0) is
xy 1 ⋅+0011 ⋅− 22 −1 p==ab 2222 xy x1y1 1+1+ …(5) ab22 ab44 Geometrical Applications of Differentiation 127
c Which comparable to p= ab22+ On using eqn (5), radius of curvature at P(x, y) becomes ab22 ρ= . p3 xa=φcos Alternately we can use parametric coordinates with yb=φsin } ρ ρ Example 23: If 1, 2 be the radii of curvature at the extremities of two conjugate semi- ρ 2/3 ρ 2/3 2/3 2 2 diameters of an ellipse, prove that ( 1 + 2 ) (ab) = (a + b ) [NIT Kurukshetra, 2006] Solution: Clearly in the geometry, CP and CD are π two conjugate semi-diameters to each other to the + 2Y φ sin ellipse with a and b as the semi-major and semi- π,b ( (0, b) a c + o minor axis respectively. φ 2 s osD (90 P φφ c – φ) , a b s 90° i Thus for P(x, y), xa=φcos n =φ} …(1) φ ) ybsin X C (,a 0) dx (0, 0) We get x'==−φ asin and y' = b cosφ dφ …(2) P´ D´ dx2 and x"==−φ acos and y" = –bsinφ Fig. 2.13 dφ2 ∴ Radius of curvature at P(x, y),
33 ()xy22+φ+φ22() a 2222sin b cos ρ='' = 1 (xy"''−−φ−φ−−φφ yx")()()()() asin b sin a cos b cos
3 ()ab22sinφ+ 22 cos φ 2 = …(3) ab Now for position D radius vector CD encloses an angle (90 + φ) with the initial axis (instead φ as in case of P) Therefore, radius of curvature at D
3 ππ 2 22 2 2 3 absin+φ + cos +φ 2 ()22cosφ+ 22 sin φ ρ=22 =ab …(4) 2 ab ab
22φ+ 2 2 φ 22 φ+ 22 φ 22()absin cos() ab cos sin Whence ρ+ρ=33 + 12 22 ()ab33() ab
ρ2/3 ρ 2/3 2/3 2 2 i.e. ( 1 + 2 ) (ab) = (a + b ). 128 Engineering Mathematics through Applications
x2 y2 Example 24: Show that the radius of curvature at P(x, y ) on the ellipse +=1 is ab22 CD3 given by ρ = where CD is the semi-diameter conjugate to CP. ab Solution: In the previous problem ρ at P(x, y) is 3 ()ab22sinφ+ 2 cos 2 φ 2 ρ= ab Now the distance CD with C(0, 0) and D (acos(90 + φ), b sin(90 + φ)), i.e. D(–a sinφ, b cosφ) is given by CD=φ+φ a22sin b 2 cos 2 CD3 Now clearly ρ= . ab ρ ρ Example 25: If 1 and 2 be the radii of curvature at the ends of a focal chord of the 2 ρ–2/3 ρ –2/3 –2/3 parabola y = 4ax, then show that 1 + 2 = (2a) . [MDU, 2006; KUK 2008] Solution: The given parabola y2 = 4ax passing through any general point P(x, y) in its parametric form is given as follows: x = at2, y = 2at xatya´2,´2== So that xay´´== 2 , ´´ 0 } …(1)
3 3 222 22 2 2 ()xy´´+ ()+ 3 ∴ ρ 44at a 2 at P(x, y) = ==−+21at()2 …(2) x'y´´ − x´´y´ 2at⋅− 022 a ⋅ a ρ ρ Y If at P(x, y) is denoted by 1, then −2/3 Px(, y) −2 −− ρ=3 2/3 + 2 1 =(2a ) 1 (2at ) (1 ) …(3) 1 +t2 Further, the parametric coordinates of point Q at the 2nd end of the focal chord would be θ O X a 2a (0, 0) Sa(, 0) x = and y = …(4) t2 −t Q The general equation of the line passing though P(t1) and Q(t2) with parametric variables t1 and t2, Fig. 2.14 (t1 + t2)y = 2x + 2at1t2 =−1 But if it pass through S(a, 0) where x = a, y = 0, we get (t1 + t2) · 0 = 2a + 2at1t2, i.e. t2 t1 ρ ρ With above arguments, at Q if denoted by 2, then −2 3 2 −2 ()2a −2 t ρ=3 =()2a3 2 1 2 + +1 t 1 …(5) t2 Geometrical Applications of Differentiation 129
Adding (4) and (5), we get
−−22 −−222 ρ+ρ=33()331 +t =() 1222aa tt22++11 Hence the result.
2 2 1dx2 dy2 Example 26: If x, y are given as functions of the arc s, show that =+. ρ22ddss 2 22 22 s+c Hence show that for catenary xc=log+ s s + c and y= s22 +c , ρ is . c dy Solution: We know that tanψ= (standard result) dx 2 ds dy which implies =+11tansec =+2 ψ=ψ dx dx dx =ψcos ds …(1) dy dy dx and =⋅=tan ψ⋅ψ=ψ cos sin ds dx ds dx2 dψ Further =−sin ψ ds2 ds …(2) dy2 dψ and =ψcos ds2 ds Whence on squaring and adding expressions under (2), we get 2 2 2 dx2 dy2 dψ 1 +== …(3) ds22 ds ds ρ 2
22 Now xc=++log s s c
dx=⋅++1 d 22 ⇒ cssc dsssc++22 ds =⋅+c 1 12s ssc++222 sc 22 + = c
sc22+ 1 c− dx2 d c2 cs ==⋅=− 2s Further, ds2 ds sc22+ 33 …(4) ()sc22++22() sc 22 and ysc=+22 130 Engineering Mathematics through Applications
dy = s ⇒ ds sc22+
dy2 ds c2 == Further ds2 ds sc22+ 3 …(5) ()sc22+2
Now using (4) and (5), in (3), we get
2 2 1dx2 dy2 =+ ρ22ds ds 2
1 or ρ= 2 2 dx2 dy2 + ds ds2 = 1 − 222 cs +c 33 ()sc22++22() sc 22
1 +sc22 == cs22+ c 4 c 3 . ()sc22+
Example 27: Show that for a cycloid x = a(θ – sinθ), y = a(1 – cosθ), radius of curvature at any point is twice the portion of the normal intercepted between the curve and the x-axis.
Solution: Here x = a(θ – sin θ), y = a(1 – cosθ) …(1) ∴ x' = a(1 – cosθ), y' = a sinθ x" = a sinθ, y" = acosθ …(2)
3 ()x'2+ y' 22 Thus ρ= x'y"− y'x"
3 2 22 {}aa()1cos−θ+() sin θ = aaaa()1coscos−θ θ−θ⋅θ sinsin
3 −θ21cosa2()2 = aa2222cosθ−() cos θ+ sin θ
3 3 ()21cosa22()−θ2 = −−θa2()1cos Geometrical Applications of Differentiation 131
θ 1Q−θ= 2 =− ⋅() − θ 2 (1 cos ) 2 sin 22a 1 cos 2 θ =−4sina 2 Now the length of the normal intercepted between the curve and the x-axis, =+2 py1 y1 dy 2 =+dθ =dy yy1as 1 dx dx dθ
2 θ =−θ+asin a()1cos1 a()1cos−θ θθ =⋅22 + a2sin 1 cot 22 θθθ ==22 i.e., pa2 sin cosce 2 a sin …(4) 222 From eqn (3) and (4), the radius of curvature ρ at any general point is twice the length of the normal (p). x Example 28: Prove that the radius of curvature for the catenary y=ccosh is equal to the c portion of the normal intercepted between the curve and the X-axis and that it varies as the square of the ordinate. Solution: Equation of the curve is x yc=cosh …(1) c dx== ∴ yy1 sinh …(2) dx c dx==1 and yy12 cosh …(3) dx c c
33 22 3+22xx ()+221sinh cosh 1y1 x ρ= =cc = =ccosh2 Thus 11xx …(4) yc2cosh cosh cc cc Now portion of the normal intercepted between the curve and the X-axis is =+2 ny1 y1
=+xx2 ccosh 1 sinh using (2) cc 132 Engineering Mathematics through Applications
xx x =⋅=cccosh cosh cosh2 …(5) cc c Clearly from eqn (4) and (5) we see that ρ (radius of curvature) = n (length of the normal) x y2 ==ccosh2 , using (1) cc ∴ρ varies as square of the ordinate.
ASSIGNMENT 4
1. Find the radius of curvature at the point (s, ψ) on the following curves: 1 (i) sa=ψ8sin2 (Cardioid) (ii) s = 4a sinψ (Cycloid) 6 (iii) s = c log sec ψ (Tractrix) (iv) s = a(emψ – 1)
3 ()xy22+ 2 2. Show that for the rectangular hyperbola xy = c 2, ρ= . [MDU, 2004] 2c2 θ 3. In the cycloid x = a(θ + sinθ), y = a(1 – cosθ), prove that ρ=4cosa . 2 4. Show that the radius of curvature at a point (a cos3θ, a sin3θ) on the curve x2/3 + y2/3 = a2/3 is 3a sinθ cosθ. 5. Find the radius of curvature at the point x (i)(at2, 2at) of the parabola y2 = 4ax,(ii) (0, c) of the catenary yc=cosh . c a 6. Show that the radius of curvature at (a/4, a/4) on the curve xya+= is . 2
− 2 ax x y 7. Prove that the radius of curvature of the catenary yee=+()aa is and that of the 2 a catenary of uniform strength y = c log sec(x/c) is csec(x/c). 8. Find the radius of curvature at the origin for (i) x3 + y3 – 2x2 + 6y = 0 (ii)2x4 + 3y4 + 4x2y + xy – y2 + 2x = 0. 9. Show that the radius of curvature of the lemniscate (x2 + y2)2 = a2(x2 – y2) at the point 2 where tangent is parallel to X-axis is a . 3 at 10. The coordinates of a point on a curve are given by xa=−sin tb sin , b at ya=−cos tb cos . Show that the radius of curvature at the point with parameter t is b ab a− b 4sint . ab+ 2 b 11. Prove that the radius of curvature at any point of the astroid x2/3 + y2/3 = a2/3 is three times the length of the perpendicular from the origin to the tangent at that point. [Hint: p = x1/3 y1/3(x2/3 + y2/3)1/2 = (axy)1/3 ] Geometrical Applications of Differentiation 133
x2 y2 12. Show that for an ellipse +=1, the radius of curvature at the end of the major axis ab22 is equal to the semi-latus-rectum of the ellipse. b2 [Hint: ρ at ()a,0 = (the semi-latus-rectum)] a dy2 13. Prove that the curvature at a point on the curve y = f (x) is given by cos3 ψ . dx2
V Radius of Curvature for Polar Curve r = f(θ). With usual notations, from the geometry ψ = θ + φ …(1) On differentiation with respect to s, ψφθ 1 ==+ddd ρds ds ds θθθφφ =+ddddd = + 1 …(2) ds dθθ ds ds d Also we had derived the result θ Y φ= ⋅dr = tan r dr r1 P φ −1 r φ=tan r i.e., r1 ψ θ T φ X d = 1 dr O so that θθ2 p ddrr1 1 + N r1 Fig. 2.15 2 − =⋅rrrrr1112 ()22+ 2 rr11 r
2− =rrr12 ()22+ …(3) rr1
ds =+22 Also rr1 …(4) dθ
222−+− 11=+=rrrr122 rrr 12 Using (3) and (4), 131 ρ+22+2rr22 22+2 ()rr1 1 ()rr1
3 ()rr22+2 ρ= 1 Hence 22+− rrrr212 134 Engineering Mathematics through Applications
Corollary 1 If equation of a curve is given in the form u = f(θ), where u = or even in case of vice versa r 2 2 =1112 =−1 =−du =− d u + du i.e., ru,then r12 , r uududud2223θθθ
3 ()rr22+2 ρ= 1 then 22+− reduces to rrrr212
3 22 11du + 3 uud24θ ()22+ 2 ρ= = uu1 2223()+ 11 11 2 uu u2…(5) +−−⋅+2du d u du uuduudud24θθθ 223
2 ==du d u where uu12,. ddθθ2
VI Radius of Curvature for Pedal Curve p = f (r) From the (Fig. 2.15), we see that 1 ddψφdθ ψ = θ + φ and ==+ …(6) ρds ds ds Also in triangle OPN, p = r sinφ dp dφ so that =φ+φsinr cos dr dr
dp ddrθ dφ =+rr dr ds ds dr
dp drθ dφ =+=r dr ds ds ρ …(7)
dr Hence ρ=r . …(8) dp
Example 29: Find the radius of curvature for the following curves (i) rn = ancosnθ (ii) r = aeθ ·cotα (iii) r2 = a2cos2θ
Solution (i) Given rn = ancosnθ …(1) Taking logs on both sides nlogr = n loga + log cos nθ Geometrical Applications of Differentiation 135
ndr 1 Differentiating, ⋅=+0sin ×−θnn, i.e. r = –rtannθ …(2) rndθ cos θ 1 Again differentiating, 2 θ θ 2 θ 2 θ r2 = –rnsec n – r1tann = – rnsec n + rtan n …(3)
33 ()rr22++θ22() rr 22tan 2 n ∴ ρ=1 = 22+− 222 + θ+ 2 2 θ− 22 θ rrrrrr22tansectan12 nrnnr n
33θθ ===⋅rnsec rnr sec 1 ()nr+θ++θ1sec22 n() n 1 n 1cos n
nn ρ=ra ⋅ = a or ()nrnr++11nn()−1 In other words, here ρ varies as the (n – 1)th power of the radius vector.
dr Alternately: Change the given equation into its pedal form and then find r ⋅ dp
dr Here in this problem, ==−rrntan θ dθ 1 θ− π π φ=dr = =− θ= + θ φ= + θ ∴ tanrnn cot tanor n dr rtan nθ 2 2
π n n+1 =φ=+θ=θ=⋅= rr Now prsin r sin n r cos n r 2 aann ()nr+ 1n Differentiating it, dp = dr an dr ann a ∴ ρ=rr = ⋅ = dpnr()++11nn() nr−1 (ii) so that r = aeθcotα, 2 dr θα dr θα =αaecot cot , =αaecot2 cot , dθ dθ2 2 dr ⋅θ α rae222cot2+= ()1cot +α = a2e2·θcotα cosec2α, dθ
3 ()ae22⋅θ cot α cosec 2α 2 ∴ ρ= =aeθαcot cosec α= r cosec α. ae22⋅θ cot α cosec 2α Extension: Show that in equiangular spiral r = aeθcotα, radius of curvature subtends a right angle at the pole. drθ Here tanφ=r = = tan α, φ = α (Fig. 2.16) dr rcotα 136 Engineering Mathematics through Applications
QP Y Now =α=−α=∠cosec sec() 90 sec OPQ and OP Q whence ∠POQ = 90° (iii) Given r2 = a2cos2θ …(1) Differentiating (1) with respect to θ, we get 90 – φ P dr 90° ra=−2 sin 2 θ θ r Ψ θ …(2) X d O T Further differentiating with respect to θ, we get Fig. 2.16 2 dr2 dr ra+=−θ 2cos22 ddθθ2
2 dr2 dr or rr=−2 2 − …(3) ddθθ2
3 2 2 2 +dr r θ ρ= d Now 2 dr d2 r rr2 +−2 ddθθ2
3 2 2 2 +dr r θ =d 22 dr dr rr22+++22 ddθθ
1 2 2 =+1 2 dr r …(4) 3 dθ On squaring and adding (1) and (2), we get
2 dr rr42+= a 4 dθ
2 22+=dr 4 or rr a dθ
1 2 2 2 2 +=dr a or r …(5) drθ
a2 Using (5), we get ρ= 3r Geometrical Applications of Differentiation 137
Corollary: Radius of curvature of the lemniscate r2 = a2cos2θ at the point where tangent is 2a2 parallel to X-axis is . Here in this curve tangent and the radius vector coincides at 3 πaa22 θ = ±π/4. Therefore ρθ= = = 2. 433r ρ ρ Example 30: If 1, 2 are the radii of curvature at the extremities of focal chord of the conic l =1+cos()eθ, prove that when e = 1, ρ –2/3 + ρ–2/3 = l–2/3. r 1 2 1 Solution: Let r = …(1) u so that for l/r = (1 + e cosθ) we get =+ θ ue(1 cos )/ l =− θ ue1 sin / l and =− θ …(2) ue2 cos / l
1 ==dr d −1 =−u1 Now r = implies ru1 () u ddθθ u2
=−12 ⋅ + 2 and ruu221 …(3) uu23
3 ()rr22+2 ∴ ρ= 1 22+− rrrr212
3 2 2 1u +−1 uu22 = 2 11uuu22 +−2121 − −+ uuuuu22 23
3/2 ()uu22+ = 1 3()+ …(4) uu u2
3 +θ2 22 θ2 1cosee+ sin ll2 Now ρ= +θ+θ3 θ 1cos1coscoseee− lll
3 +12cosee2 + θ2 2 ρ= l 3 1cos1+θe …(5) ll 138 Engineering Mathematics through Applications
l θπ = /2 The general equation of the conic =+()1cos,e θrepresents r θ a parabola for e = 1 P() an ellipse for e < 1 θπ = θ θ = 0 a hyperbola for e > 1 O
3 3 2 3 ⋅+θ2() 2 ρ=l21cos = ⋅ 2 = l Thus for e = 1, 3 l 3 ()1cos+θ ()1cos+θ2cos3θ 2 General Geometry θπ …(6) Q( + ) of the conic Now if ρ at P is termed as ρ and ρ at Q is termed as ρ , then 1 2 Fig. 2.17 ρ–2/3 –2/3 2θ 1 = l cos /2
− 2 −−22θ+π θ ρ=33322 = and 2llcos sin 22 ρ –2/3 ρ –2/3 –2/3 Add the two, 1 + 2 = l Example 31: If φ be the angle which the radius vector of the curve r = f(θ) makes with the r dφ tangent prove that =sinφ 1+ , where ρ is the radius of curvature of the curve. Also ρ dθ a apply the result to show that ρ = for the circle r = a cosθ. 2 Solution: See the geometry given in Fig. 2.18. In the ∆OPT, ψ = θ + φ …(1) ψ is the angle which the tangent to the curve makes at with the initial axis θ is the angle which the radius vector OP makes with the initial axis; φ is the angle which the radius vector encloses with the tangent at P(r, θ).
ddψφdθ Y Equation (1) implies =+ B ds ds ds α Qr( + δθδθ r , + ) 1 ddθθdφ δs =+ N δc i.e. ρθ ds d ds Pr(, θ) θ φ 1 =+d d s or 1 …(2) A φ ρθds d δθ ∆ r In the PNQ, for the limiting arc when Q approaches to ψ P(back), i.e. when δθ → 0 θ X O T δ 2 (0, 0) c = ds=+2 dr Lt 1 and r Fig. 2.18 δθ→0 δs ddθθ dθ Lt tanα= tan θ, tan θ=r δθ→0 dr ds ds dθ Thus =⋅ dr dθ dr Geometrical Applications of Differentiation 139
2 dr dθ =+r2 ⋅ ddrθ
dθ 2 =+r 1 dr =φ+tan2 1 ds dr i.e. =φsec or =φcos …(3) dr ds dθ sinφ dθ Now tanφ=r or =r dr cosφdr θ φ= φd dr dθ i.e. sin cos r =⋅r Using (3) dr ds dr sin φdθ or = …(4) rds Now on using (4), (2) becomes φ r =φ+d sin 1 …(5) ρθd θ θ Further, the circle r = a cos implies r1 = –a sin θπ φ=ra =cos =− θ= +θ Thus, tan−θ cot tan ra1 sin 2 π φ= +θ dφ i.e. and = 1 …(6) 2 dθ θ ∴ ρ=ra = cos φπ +φd++θ 1sin(1 1) sin dθ 2 aacosθ == 2cosθ 2 VII Radius of Curvature for Tangential Polar Equation p = f(ψ). A relation between perpendicular p from the origin on any tangent to a curve and ψ which this tangent makes with X-axis is called the tangential polar Y equation of the curve. Let p be the length of perpendicular OL drawn from the Px(, y) origin on the tangent to the curve at the point P(x, y), then OL L makes an angle (ψ – π/2) with the positive direction of X-axis. π/2 ∴ The equation of the tangent PT p ψ π ψ π p = X cos( – /2) + Y sin( – /2) ψ ψπ – /2 X = Xsinψ – Ycosψ …(1) – O T where X, Y are the coordinates of any point on the tangent. Fig. 2.19 140 Engineering Mathematics through Applications
As P(x, y) lies on (1), therefore p = x sinψ – y cosψ …(2) Differentiating both sides of (2) with respect to ψ, dpdx dy =ψ+ψ⋅+ψ−ψxycos sin sin cos dddψψψ dx dy =ψ+ψ+ψ⋅−ψ()xycos sin sin cos ddψψ dx dsdy ds =ψ+ψ+ψ⋅⋅−ψ()xycos sin sin cos ds dψψ ds d = (xcosψ + y sinψ) + sinψ·ρ·cosψ – cosψ·ρ·sinψ = (xcosψ + y sinψ) + 0 …(3) Differentiating (3) again with respect to ψ, 2 dp =− ψ+ ψ⋅dx + ψ+ ψ⋅dy xysin cos cos sin dddψψψ2 =− ψ+ ψ+ ψ⋅dx + ψdy ( xysin cos) cos sin ddψψ 2 dp =− ψ+ ψ+ρ22 ψ+ ψ dx dy ()xysin cos (cos sin ) Q=ρcos Ψ , =ρ sin Ψ dψ2 ddΨΨ 2 dp =− +ρ p dψ2 2 ρ= + dp or p …(4) dψ2 Alternately: Using the relation between p and ψ. dp=⋅⋅ dp dr ds ddrdsdψψ dp =⋅φ⋅ρcos dr dp dr dr =φ⋅cosr , As ρ=r dr dp dp = rcosφ 2 dp ∴ prrr222222+=φ+φ= sin cos , (As p = r sinφ) …(5) dψ On differentiating (5) with respect to p, 2 2 dp d p dψ dr +=ρdp 22pr+= 2 or p …(6) dψψ d2 dp dp dψ2 Hence the result. Geometrical Applications of Differentiation 141
Example 32: Find the radius of curvature for the hyperbola p2 = a2 cos2ψ – b2sin2ψ
Solution: The curve is p2 = a2 cos2ψ – b2 sin2ψ …(1) ab22()1cos2+ψ() 1cos2 −ψ or p2=− 22
222221 or pabab=()() −++cos2 ψ …(2) 2 ∴ On differentiating both sides with respect to ψ,
dp 122 22sin2pab=−() + ψ dψ2 dp 1 ∴pab=−()22 +sin 2 ψ dψ 2 …(3) 11 Also from (2), pabab22222−−=+()()cos2 ψ …(4) 22 On squaring and adding (3) and (4), 2 242222222dp 1122 ppabpabab++()() − − − =() + dψ 44 2 dp 122 or pppababab242222222+−() − −()() + − − =0 …(5) dψ 4 On dividing both sides by p2, we get 2 dp ab22 +−pab222() − = dpψ 2 Differentiating both sides with respect to ψ, dp d2 p dp2ab22 dp 22⋅+p =− ddψψ23 d ψ pd ψ dp2 ab22 dp or +=−p (on cancelling 2 throughout) dpψ23 dψ ab22 ∴ ρ=p + p2 numerically.
ASSIGNMENT 5
1. Find the radius of curvature at point (r, θ) on the following curves: (i) r(1 + cosθ) = a (sine spiral) (ii) rm = am sinmθ (iii) r = a(2cosθ – 1) (iv) r2cos2θ = a2 (v) θ = a–1(r2 – a2)1/2 – cos–1(a/r) 2. Find the radius of curvature to the curve r = a(1 + cosθ) at the points where tangent is parallel to the initial line. 142 Engineering Mathematics through Applications
3. Find the radius of the curve r2 = a2sin2θ at the point where the radii vector is perpendicular to the tangent. [Hint: such points are (a, π/4). First find general values of ρ(r, θ) and then take θ = π/4] 4. Find the radius of curvature at (r, θ) for the polar curve r θ2 = a. 33 ()22+θ+22() 2 1 θ2 uu1 a 4 Hint: u == , ρ= = 332()+θθ+ ra uu u2()2 cos22θθ sin 5. Find the radius of curvature at the point (r, θ) of the curve u2=+. ab22 6. Find the radius of curvature at the point (p, r) on the following curves: (i) r3 = 2ap2 (Cardiod); (ii) r3 = a2p (Lemniscate) (iii) pr = a2 (Hyperbola); (iv) pan = rn+1 (Sine spiral) r4 111 r2 (v) p2= (Archimedian spiral) (iv) =+− (Ellipse) ()ra22+ pabab22222 7. Find the radius of curvature for the curves (i) p = a(1 + sinψ)(ii) p2 = a2cos2ψ + b2sin2ψ 8. Show that for the Epicycloid p = a sinbψ, ρ varies as p.
2.6 CENTRE OF CURVATURE, CIRCLE OF CURVATURE, EVOLUTE AND CHORD OF CURVATURE I Centre of Curvature Centre of curvature Cxy(), for any point P(x, y) of a curve is the point on the positive direction of the normal to the tangent at P at a distance ρ from it (Figure 2.20). Angle NCP = 90° – Angle NPC = Angle NPT = ψ ∴ xOMOLMLOLNP==−=− = x – ρsinψ …(1) 3 ()+22 1y1 y y 1 =−x ⋅ 1 QQtanψ=y , sin ψ=1 , cos ψ= +2 1 ++22 y2 1y111yy11 yy()1+2 =−11 x …(2) Y y2 ==+=+ (,xy ) and yMCMNNCLPNC CP = y + ρ cosψ …(3) NΨ 3 ρ(,xy ) ()1+y22 =+1 ⋅ 1 y y +2 y2 1y 1 Ψ ()1+y2 X =+ 1 O T ML y …(4) x y2 x II Circle of Curvature Fig. 2.20 The circle with its centre at the centre of curvature C and radius equal to ρ is called the ‘Circle 2 2 of Curvature’ of the curve at the point P and its equation at P is ()xx−+−=ρ() yy 2. Geometrical Applications of Differentiation 143
III Evolute P6 The locus of the centre of curvature of the given curve is C6 C5 P5 called its ‘evolute’ and in turn the curve is termed as an C4 C ‘involute’ of its evolute. Y 3 P4 C2 If C0, C1, C2, …, etc. are the centre of curvature of the C1 P3 C0 curve y = f (x) corresponding to the points P0, P1, P2, …, etc. respectively, we see that in moving from P0 to P1, P P2, …, etc. Center of curvature moves along the curve 2 C C C … in turn which is called the evolute of y = f (x). P0 0 1 2 P As x, y, ρ and ψ depends upon s, therefore the equation 1 X (1) and (3) may be treated as parametric equation of the O evolute. Fig. 2.21 Two Important Properties of Evolutes (a) The normal at any point of a curve is the tangent to its evolute at the corresponding centre of curvature. (figure 2.22) (b) The length of the arc of the evolute between any two points is equal to the difference between the radii of curvature at the corresponding points of original curve ρ ρ i.e. S 2 – S1 = 2 – 1, ρ ρ where S2 – S1 = length of the arc C1C2 and 2 – 1 = difference between radii of curvature at point P1 and P2. (Fig. 2.23).
Y te C2 Y lu vo E C1 Cx y ( C , 0 )
Px(, y)
P2 P 0 P1 O X X O Fig. 2.22 Fig. 2.23
IV Chord of Curvature
The length intercepted by the circle of curvature of the curve at P, D Q on a straight line drawn through P in any given direction is called φ C Chord of curvature through P in that direction. P Thus, if the chord of curvature PQ (Fig. 2.24) makes an angle φ, with the normal PCD, then its length PQ is given by PQ = PDcosφ = 2ρcos φ …(1) T [³ ∠DQP being in a semi-circle is a right angle] Fig. 2.24 Geometrical Applications of Differentiation 173
122 224 =−()28xa + ± axa + 2 On neglecting the term with negative sign before the radical on the right side, because if y2 is negative y will become imaginary. Y ∴ 2222241 yxaaxa=−()28 + + + 2 On differentiating the above equation, we get x = –a dy 116ax2 (,a 0) 24yx=−+ X 22 4 dx 228ax+ a O xa = 90° 2 dy = 8ax Now = 0 implies 4x dx 8ax22+ a 4 3 or xa=± Fig. 2.34 8 3 Hence the tangent to the curve are parallel to the axis of X at xa=± and, are thus 8 the extreme values of y (i.e. point of maxima-minima). Further the tangents are parallel to Y-axis at x = –a and x = a which can be verified by shifting the origin from (0, 0) to (x – a, 0) and equating to zero the lowest degree in x and y for tangent at x = –a. Similarly, for x = a by replacing x by (x + a). (5) Regions: For –∞ < x < –a and a < x < ∞, y 2 is negative, whereas for –a < x< 0 and 0 < x < a, y first increases and then diminishes to zero.
Example 62: Trace the curve y2(a2 – x2) = a3x.
Solution: 1. Symmetry: The given curve is symmetrical about X–axis only. 2. Origin: It passes through the origin and at the origin, the tangent to the curve is a cusp, viz. x = 0 (i.e. Y-axis). 3. Asymptotes Y (i) On equating to zero the coefficient of highest powers of x, we get y2 = 0, hence X-axis itself xa = – is the asymptote to the curve. (ii) Again, on equating to zero the coefficient of xa = X highest power of y, we get a2 – x2 = 0, i.e., O x = a and x = –a are the two asymptotes parallel (0, 0) to Y-axis. (iii) It has no oblique asymptote. 4. Points: The curve does not meet the axes except at the origin and also there are no such points Fig. 2.35 where tangents are parallel to either of the axes except at the origin. 172 Engineering Mathematics through Applications
1. Symmetry: The curve does not have any symmetry. 2. Origin: It does not pass through the origin as equation of the curve has a constant term. 3. Asymptote: The curve does not possess any asymptote. 4. Points: The curve intersects with the Y-axis (i.e. x = 0) at y = 1, y = 2, y = 3. It intersects the X-axis (i.e. y = 0) at x = – 6. Y On simplification, x = y3 – 6y2 + 11y – 6 = 0 (0, 3) (–.384, 2.6) dx (0, 2) and = 0 implies for 3y2 – 12y + 11 = 0 dy (.384, 1.4) (0, 1) X 12±−×× 144 4 3 11 (–6, 0) O i.e. y = 23× 1 or y=±22.6,1.4 =() Fig. 2.33 3 for y = 2.6, x = –0.384 y = 1·4, x = 0.384 Hence the tangent parallel to the axis of y are at points (–.384, 2.6) and (.384, 1.4) respectively.
Example 61: Trace the curve (x2 + y2)2 = a2(x2 – y2) (i.e. r2 = a2cos2θ in polar form)
Solution: 1. Symmetry: The given curve is symmetrical about both X and Y-axes. 2. Origin: The curve passes through the origin and at the origin the tangent to the curve are y = ±x (node). 3. Asymptotes: The given curve has no asymptotes. 4. Points: (a) Intersection with axes: (i) Intersection with X-axis (i.e., y = 0) is at the points x = 0, ±a (ii) Intersection with Y-axis (i.e., x = 0) is at the point y = 0. Hence the given curve intersects the axis at (0, 0), (a, 0), (–a, 0). dy dy (b) Points where = 0 or =∞. dx dx Rewrite the given equation after simplification as: y4 + (2x2 + a2)y2 + (x4 – a2x2) = 0
2 −+±()()()224xa22 xa 22 +− xax 422 − or y2= 2 −+±++−+()24444xa22 xa 24 axx 224 ax 22 = 2 Geometrical Applications of Differentiation 171
111−−−−3 =± −22 ⋅11 + + − 2 ⋅− ⋅ + 2 xax{}()331() ax() ax() ax 22 +⋅1(3ax − )1/2 ( ax + )− 1/2
11 1 ()33ax−−++−22() ax() ax() 3 ax 2 =± −x + 311 2()ax+−22()3 ax ()ax+2
1 ()()33ax−++ ax() ax − 2 =± −x + 311 2()ax+−22()3 ax ()ax+2 −+−()() + =±23axaxax 3 1 ()ax+−22()3 ax −+23ax() a22 −+− ax 3 ax x =± 31 ()ax+−22()3 ax ()322− dy =± ax 31 dx ()ax+−22()3 ax dy Now = 0 implies 3a2 – x2 = 0, i.e. xa=± 3 . dx (Leaving the value xa=− 3, since for –∞ < x < –a, y is imaginary). Thus, xa=3 is the point of maxima and at this point, the tangent is parallel to the axis of X.
Similarly dy =∞ implies a + x = 0 and dx Y 3a – x = 0, i.e.x = 3a Hence the tangent to the curve is parallel to Y-axis at (3a, 0) and also at (–a, 0) which is behaving as an asymptote. x 3
= ()− y xa = 3 =± 3ax Regions: For yx , (3a, 0) ax+ X xa = – O (,a 0)(2a , 0) (i) If x < –a, y is imaginary y ∞ = (ii) If –a < x < 0, y decreases from – to 0 (i.e., a – 3 x defined entity) (iii) If 0 < x < 3a, y is positive and it first increases and then decreases to zero. (iv) If x > 3a, y is imaginary.
Example 60: Trace the curve x = (y – 1)(y – 2)(y – 3). Fig. 2.32
Solution: 170 Engineering Mathematics through Applications
ax()− ax y2= or ()2ax− , i.e. y2(2a – x) + ax(a – x) = 0 On equating to zero the lowest degree term; we get a2x = 0, i.e. x = 0 or in other words the tangent is parallel to Y-axis at (–a, 0). ax() a+ x 6. Regions: From the given equation, we have y =± ax− Now + ()is–vexa (i) When –∞ < x < – a on the left hand side of above expression, ax is –ve and, ()is+veax− therefore, y is +ve and goes on increasing with the decrease in x. ax is –ve (ii) When –a < x < 0, then ()+ and hence the square root expression is ax is +ve , ()ax− is +ve negative, i.e. y is imaginary. ax is +ve (iii) When 0 < x < a, then ()+ and hence the square root expression is positive axis +ve , ()ax−is +ve and y is increasing with the increase in x. ax is +ve (iv) When a < x < ∞, then ()+ and hence the square root expression is axis +ve , ()ax−is –ve negative and y is imaginary.
Example 59: Trace the curve y2(a + x) = x2(3a – x). Solution 1. Symmetry: The curve is symmetrical about X-axis only. 2. Origin: The equation of the curve does not have any constant term, so it passes through the origin and the tangent at the origin are y2 – 3x2 = 0 or yx=± 3 . The origin is a node. 3. Asymptote: The curve possesses asymptote parallel to Y-axis, viz. a + x = 0. 4. Points: Intersection with the X-axis y = 0 ⇒ x = 0, 0, 3a Thus the curve passes through (0, 0) and (3a, 0). 1 ()− 2 =±xax3 Also from given equation y 1 ()ax+ 2 so that
11−− 11 dy =±d −22 + + ⋅ − 22 + xaxax{}()313() () axax() dx dx Geometrical Applications of Differentiation 169
2. Origin: As there is no constant term in the equation of the curve, it passes through the origin. The tangents at the origin are obtained by equating to zero the lowest degree term in the equation of curve. a(y2 – x2) = 0, i.e.y = ±x (The tangent is a node) 3. Asymptotes: From (1), a + x = 0, i.e.x = – a is an asymptote parallel to Y-axis. 4. Points: The curve intersects X-axis at x2(a – x) = 0, i.e.x = 0 and x = a (whereas it does not intersects Y-axis)
1 ()()axax+−−2 2 ax dy = 1/2 3/2 Further 33 becomes infinite for (a – x) (a + x) = 0. dx ()−+22() ax ax Y i.e. at x = a, Y-axis is tangent at the point (a, 0).
x − y = =± ax = y xa = 5. Regions: yx – ax+ x A X y has imaginanry values for x > a and from x = 0 to x = a x = – a (,a 0) O it first increases from O onwards than becomes zero at x = a Again for –∞ < x < –a , y is imaginary, see Fig. 2.30. () ax a + x Fig. 2.30 Example 58: Trace the curve y=2 . ()a–x ax() a+ x Solution: Given curve y2= can be rewritten as ()ax− (a – x)y2 – ax(a + x) = 0 1. Symmetry: The given curve is symmetrical about the X-axis as it contains y in even powers only. Rest of the symmetries are missing. 2. Origin: It passes through the origin (as there is no constant term in the equation of the given curve). On equating to zero the lowest degree terms in x and y, i.e. a2x = 0 or x = 0, i.e. Y-axis is tangent to the given curve at the origin. 3. Asymptotes: The curve has one asymptote parallel to Y-axis only and is given by a –x = 0, i.e. x = a. Y 4. Points of Intersection: Intersection with X-axis, i.e. y = 0 implies ax(a + x) = 0 or x = 0, – a, i.e. the curve meets the X-axis at (– a , 0) (,a 0) –X X (0, 0), (–a, 0), whereas it does not meet the xa = – O (0, 0) xa = Y-axis. To see the tangents at (–a, 0), shift the origin (0, 0) to (x–a, 0), i.e. ax()−+− a() a x a y2= ()axa−− Fig. 2.31 168 Engineering Mathematics through Applications
(iv) Symmetry about the line y = –x: If the given equation of the curve remains unchanged by interchanging y by –x, the graph of the curve is said to be symmetrical about the line y = –x. e.g. x4 + y4 = 4a2x2y2. (v) Symmetry about opposite Quadrants: If the given equation of the curve remains unchanged on replacing x by –x and y by –y, the graph of the curve is said to be symmetrical about the quadrants. e.g. (i) x5 + y5 = 5ax2y (ii) xy = c2 (iii) y = sinx (iv) x = y3 (vi) Symmetry about the Origin: The graph of a curve is said be symmetric with respect to origin if whenever (x, y) is a point on the graph, (–x, –y) is also a point on the graph, also if (–x, y) is a point on the graph then (x, –y) is also a point on it. or In other words, if the rotation of the graph through 180° leaves it unchanged, the curve is termed as symmetrical about the origin. 2. Origin (i) See if the curve passes though the origin. (A curve passes through the origin if there is no constant term in the equation of the curve). (ii) Say it passes through the origin, then find the equation of the tangent at the origin by equating to zero the lowest degree terms in the equation of the curve. (iii) If the origin is a double point, see is it a node, cusp or a conjugate point. 3. Asymptotes (i) Check whether the curve possesses asymptotes parallel to X-axis or Y-axis. (ii) Find oblique asymptotes if any. 4. Points (i) Find the points where the curve intersects X-axis and Y-axis, if any. (ii) Find the real points where it intersects the asymptote, if exist. (iii) Find the points where the asymptotes intersects the axes, if any. (iv) Locate the points where tangents to the curve are parallel to X-axis and Y-axis by dy dy putting ==∞0and , respectively. dx dx dy23 dy (v) Find points of inflexion, i.e. =≠0and 0 dx23 dx (vi) Find the region/regions in which no portion of the curve exists.
Example 57: Trace the curve y2(a + x) = x2(a – x).
Solution: The given curve y2(a + x) = x2(a – x) …(1) can also be written as x3 + xy2 + a(y2 – x2) = 0 …(2) 1. Symmetry: The curve is symmetrical about X-axis as y appears in the equation in even powers only. Rest of the symmetries are missing. Geometrical Applications of Differentiation 167
6. Find the equation of the conic on which lie the eight points of intersection of quartic curve xy (x2 – y2) + a2y2 + b2x2 – a2b2 = 0 with its asymptotes.
[Hint: Joint equation of asymptotes Fn = 0 is xy(x + y)(x – y) = 0] 7. Find the asymptotes of the following curves (i) r sinθ = 2cosθ (ii) rθ = a (iii) r cosθ = a sin2θ (iv) r cos2θ = a sin3θ (v) r cosθ = a cos2θ (vi) r(1 – eθ) = a (vii) r = a logθ (viii) r logθ = a (ix) r(θ + π) = aeθ (x) rθcosθ = a cos2θ
2.8 TRACING OF CURVES By tracing of curve, we mean simply to sketch a rough graph of the curve from its cartesian, polar or parametric equation without having to plot a large number of points on it. For that purpose, we collect certain inferences and facts from the given equation of the curve itself and then with the help of these facts we sketch the approximate graph. I. Tracing in Cartesian Co-ordinate System For this, we see the following behavioural characteristics of the given curve: 1. Symmetry (i) Symmetry about X-axis (ii) Symmetry about Y-axis (iii) Symmetry about the line y = x (iv) Symmetry about the y = –x (v) Symmetry in opposite quadrants (vi) Symmetry about the origin (i) Symmetry about X-axis: If in the given equation of the curve, y occurs in even powers only, the graph of the curve is said to be symmetrical about X-axis. or In other words, if to every point (x, y) on the curve, there is a point (x, –y) on it, i.e. the portion of the curve below X-axis, if folded coincides completely with the portion above X-axis. e.g. y2 = 4ax;(ii) y2(4 – x) = x2(x + 4). (ii) Symmetry about Y-axis: If in the given equation of the curve, x-occurs in even powers only, the graph of the curve is said to be symmetrical about Y-axis. or In other words, if to every point (x, y) on the curve, there is a point (–x, y) on it i.e. the portion of the curve left to Y-axis if folded, coincides completely with the portion to the right side of Y-axis. ()x2+1 2 y= e.g. (i) x = 4ay,(ii) ()x2−1 (iii) Symmetry about the line y = x: If the given equation of the curve remains unaltered by interchange of x and y, the graph is said to be symmetrical about the line y = x. e.g. x3 + y 3 = 3axy 166 Engineering Mathematics through Applications
Example 56: Prove that the curve r(1 – cosθ) = a has no asymptotes. a Solution: The given equation of the curve is r = 1cos−θ 11cos−θ So that u== ra 1cos−θ and u→→0implies 0 or cosθ → 1orcosθ → cos0 = cos2nπ a θ π i.e. 1 = 2n 1cos−θdu sin θ Now u ==so that aadθ
daaθ− dθ and Lt−=−→→+∞ Lt . Means Lt − does not exist. θ→θ θ→2n π θπ θ→θ 1 dusin sin2 n 1 du a ∴ The curves r = does not have any asymptote. 1cos−θ
ASSIGNMENT 7
1. Find the asymptotes, parallel to the axes, of the following curves (i)(a/x)2 + (b/y)2 = 1 (ii) x2y3 + x3y2 = x3 + y3 (iii) x2y2 – y2 – 2 = 0 (iv) y = x(x – 2)(x – 3) (v) y3 + x2y + 2xy2 – y + 1 = 0 (vi) y4 + x2y2 + 2xy2 – 4x2 – y + 1 = 0 2. Find the asymptotes of the following curves −2 (i) ye=x (ii) y = logx (iii) y = tanx (iv) y = cosecx 3. Find the asymptotes of the following curves (i) y2(x – 2a) = x3 – a3 [MDU, 2004, 2005] (ii) x2y + xy2 + xy + y2 + 3x = 0 (iii) y3 – xy2 – x2y + x3 + x2 – y2 = 0 [MDU 2006; KUK, 2007] (iv)(y – a)2 (x2 – a2) = x4 + a4 4. Show that the points of intersection of the curve 2y3 – 2x2y – 4xy2 + 4x3 – 14xy + 6y2 + 4x2 + 6y + 1 = 0 and its asymptotes lie on the straight line 8x + 2y + 1 = 0. [Hint: Inclined asymptotes are y = 2x, y = x – 1, y = –x – 2, and their joint equation, 3 2 2 3 2 2 i.e. Fn = 0 is 2(y – x y – 2xy + 2x – 7xy + 3y + 2x + 2y – 4x = 0)] 5. Find the asymptotes of the curve 4(x4 + y4) – 17x2y2 – 4x(4y2 – x2) + 2(x2 – 2) = 0 and show that they pass through the points of intersection of the curve with the ellipse x2 + 4y2 = 4.
[Hint: Joint equation of the asymptotes viz. Fn = 0 is {(x4 + y4 – 17x2y2) – 4x(4y2 – x2} + (x2 – 4y2) = 0] Geometrical Applications of Differentiation 165
du ()ab+θ−θ−θθsin() sin cos() b cos or = dθ ()ab+θsin 2 du ()absinθ+ or =− dθ (ab+θsin )2 2 dθ ()ab+θsin p=−=Lt Lt Now θ→θ π θ+ 1duθ→()21n + ( asin b) 2 π2 ab++sin() 2 n 1 n2 ab+−()1 =2 = π()−+n …(4) ansin() 2++ 1 bab1 2 (³ sin(2n + 1)π/2 = sin(nπ + π/2) = (–1)n sin π/2 = (–1)n) By definition, θ θ p = r sin( 1 – ) …(5) 2 +−n ab()1 π ∴ =+−θrnsin() 2 1 , using (4) ab()−+1 n 2
π =π+−θrnsin 2 π =−n −θ r(1)sin 2 = r(–1)n cosθ …(6)
()ab+2 Now if n is even, (–1)n = 1 and we get =θrcos , ()ab+ i.e. rcosθ = (a + b) …(7) ()ab−2 If n is odd, (–1)n = –1 and we get =−rcos θ ()−+ab i.e. rcosθ = (a – b) …(8) Hence r cos θ = (a + b) and r cos θ = (a – b) are the two asymptotes of the given curve. cosθ Corollary. If a and b are equal, in that case, u → 0 will imply →0 means a()1sin+θ π π cosθ θ→θ = π+ ()+ 0 1 2n instead 21n , otherwise +θ will become . Since for odd 2 2 ()1sin 0 π θ=() + θ 35ππ θ values of n in 121n , sin will be negative for , , …, etc. leading to (1 + sin ) as 2 22 θ zero for these values of 1. 164 Engineering Mathematics through Applications i.e. y3 + (y – 2)x2 + 2y2 = 0 Now asymptotes parallel to X-axis is y – 2 = 0, i.e.y = 2 or r sinθ = 2 (On equating the co-efficient of highest powers of x equal to zero) However, there is no asymptote parallel to Y-axis as the highest coefficient of y is merely a constant.
Example 54: Find the asymptote of a polar curve r = a tanθ.
1cosθ Solution: Given r = a tanθ, i.e. = …(1) rasinθ
= 1 cosθ π Let u then u → 0 implies → 0, i.e. cosθ → 0 impling θ→()21n + r asinθ 2 du 11 Now =−()cosec2 θ=− daθθ asin2
dθ ∴paa=−=−−θ=θLt Lt() sin22 (sin ) θ→θ θ→θ 1 11du = a[sin(2n + 1)π/2]2 = a(–1)2n = a …(3) By definition, θ θ p = r sin( 1 – ) …(4) From (3) and (4) a = r sin((2nπ + π)/2 – θ)) a = r sin(nπ + π/2 – θ) = r(–1)nsin(π)/2 – θ) a = ±r cos θ or r cos θ = ±a …(5)
Example 55: Find the asymptote of the curve r = a secθ + b tanθ.
Solution: The equation of the curve is r = a secθ + b tanθ …(1) 11ab sinsinθ+ ab θ Let r==+=so that uucosθθ cos cos θ cosθ ∴ u = ()ab+θsin …(2)
cosθ Now u → 0 implies →0 i.e. cos θ → 0 ()ab+θsin ∴ θ → π θ (2n + 1) /2 = 1 …(3) From (2),
du d cosθ = ddabθθ+( sin θ) Geometrical Applications of Differentiation 163
→ θ → θ → π θ Now u 0 implies sin 0, i.e. n (= 1) …(2)
du 1 cos2θθ+θθ⋅() cos sin sin 2 2 Also = dθθ2cos22
1 ()cos2θθ+θ cos sin sin 2 θ+θ sin sin 2 θ = 2cos22θ
1 cos() 2θ−θ +{} 2sin θ sin 2 θ = 1 2 2cos22θ
1 cosθ+{} cos θ− cos3 θ du = 1 2 dθθ2cos22
dθ ∴p =−Lt θ→θ 1 du
2cos22θ =−Lt θ→θ 1 1cosθ+{} cos θ− cos3 θ 2
−π2 =− 2cos 2n 1 cosnnnπ+{} cos π− cos3 π 2 − =2 1 (1)−+nnn{} (1) −−− (1)3 2 −22 p==+ i.e. ()−−11nn()1 …(3) Thus by definition, θ θ π θ n + 1 θ p = r sin( 1 – ) = r sin(n – ) = r(– 1) sin …(4) Therefore by (3) and (4), we get
2 n+1 θ +=−()1sinr θ or r sin = 2 ()−1n1
Alternately: rsinθ = 2 cos2θ = 2(cos2θ – sin2θ) or r2·rsinθ = 2(r2cos2θ – r2sin2θ) i.e. (x2 + y2)y = 2(x2 – y2) x2y + y3 – 2x2 + 2y2 = 0 162 Engineering Mathematics through Applications but here Y-axis, i.e. x = 0 is the tangent at the origin, means b = 0 and equation of the curve reduces to (x – y)(x – 2y)(x – 3y) + ax = 0. As it passes through (3, 2) implies, a = –1. ∴ The desired curve (x – y)(x – 2y)(x – 3y) – x = 0 or x3 – 6x2y + 11xy2 – 6y3 – x = 0.
VI. Asymptotes of Polar Curves θ θ θ Asymptotes of the curve r = f ( ) is p = r sin( 1 – ) dθ 1 where p =−Lt with u = and θ as the root of the equation u = 0. θ→θ 1 1 du r Note: Results to be remembered: (i)sinθ = 0 ⇒ θ = nπ, n is any interger; (ii)cosθ = 0 ⇒ θ = (2n + 1)π/2 (iii) sin(nπ + θ) = (–1)n sinθ;(iv) cos(nπ + θ) = (–1)ncosθ (v)cosθ = cosα ⇒ θ = 2nπ ± α;(vi)sinθ = sinα ⇒ θ = nπ + (–1)nα 1 (vii)tanθ = tan α ⇒ θ = nπ + α;(viii) = (–1)n. (–1)n
Example 52: Find the asymptote of the curve r sinnθ = a. a Solution: Given r = sinnθ 1sinnθ Let u== …(1) ra sin nθ mπ ∴ u → 0 ⇒ → 0 , i.e. nθ → mπ or θ= …(2) a 1 n du n Now =θcosn daθ θ =−d a 1 a 1 ∴ p Lt =−Lt ⋅ =− ⋅ …(3) θ→θ θ→θ θ 1 du nn1cos nmcos π By definition, mπ p = r sin(θ – θ) =−θrsin …(4) 1 n From (3) and (4), ππ ma−θ =−11 θ− ma = rrsin or sin . nnmcosππ nnm cos
Example 53: Obtain the Asymptotes of the curve r sinθ = 2cos2θ. [KUK, 2009]
2cos2θ Solution: Given r = …(1) sinθ 11sinθ Let ru===so that ur2cos2θ Geometrical Applications of Differentiation 161
Thus this asymptotes cuts the curve in n(n – 2) = 3(3 – 2) = 3 points and these three points of intersection lie on the curve of intersection Fn – 2 = 0, i.e. on the line x – y = 0. Example 50: Find the equation of the quartic curve which has x = 0, y = 0, y = x, and y = –x four asymptotes which pass through (a, b) and which cuts the curve in eight points that lie on the circle x2 + y2 = a2.
Solution: The joint equation of the asymptotes, i.e. Fn = 0 is xy(y – x)(y + x) = 0 or xy(x2 – y2) = 0 …(1) The given equation of common points of intersection is x2 + y2 – a2 = 0 …(2) Here the equation of the quartic whose asymptotes are given by (1) and whose intersection with the asymptotes lie on (2), is given by 2 2 λ 2 2 2 Fn + Fn – 2 = 0, i.e.xy(x – y ) + (x + y – a ) = 0 …(3) whence λ is a constant. Now this curve pass through the point (a, b) means aa()22− b ab(a2 – b2) + λ(a2 + b2 – a2) = 0 or λ= …(4) b Whence with above value of λ, the equation of the quartic becomes bxy(x2 – y2) + a(a2 – b2)(x2 + y2 – a2) = 0. Example 51: Find the equation of the cubic which has the same as asymptotes the curve x3 – 6x2y + 11xy2 – 6y3 + x + y – 1 = 0 and which touches the axis of y at the origin and pass through the point (3, 2). Solution: Rewrite the given equation as (x3 – 6x2y + 11xy2 – 6y3) + (x + y) – 1 = 0 φ 2 3 φ so that 3(m) = 1 – 6m + 11m – 6m and 2(m) = 0 φ 2 3 Further 3(m) = 0 implies 1 – 6m + 11m – 6m = 0 or (1 – m)(1 – 2m)(1 –3m) = 0 φ 2 i.e., m = 1, 1/2, 1/3. 31/m = –6 + 22m – 18m φ()m Now, c=−2 =0, since φ (m) = 0 and φ' (m) is defined for all m. φ 2 3 '3()m ==11 Whence asymptotes are y = x, yxyx,, and their combined equation, i.e. Fn = 0 is, 23
F3 = (x – y)(x – 2y)(x – 3y) = 0. Therefore, the equation of the curve having Fn = 0 as its asymptotes can be written as Fn + F n – 2 = 0, where Fn – 2 is an expression of degree n – 2, i.e. of degree 3 – 2 = 1. Let this equation be F1 = ax + by + c = 0. Whence, the equation of the curve becomes F3 + F1 = 0, i.e. (x – y)(x – 2y)(x – 3y) + (ax + by + c) = 0. Using the condition it passes through the origin, i.e. c = 0 results in the equation as (x – y)(x – 2y)(x – 3y) + (ax + by) = 0. The equation of the tangent to the curve at the origin is ax + by = 0 (equating the lowest degree term to zero) 160 Engineering Mathematics through Applications
φ φ φ Whence n(m) = 0 and c 'n(m) + n – 1(m) = 0 …(4) giving values ‘m’ and corresponding values of ‘c’. Now clearly with (4), equation (3) simply reduces to 2 nn−−23cφ+φ+φ+() () [] …+…= xmcmmx"nn'−−12() n 0 …(5) 2! which is an equation of degree (n – 2) and determines (n – 2) values of x. Hence (1) cuts (2) in (n – 2) points.
Corollary 1. If a curve of nth degree has n asymptotes, then they cuts the curve in n(n – 2) points. 2. If the equation of the curve of nth degree can be put in the form Fn + Fn – 2 = 0, where Fn – 2 is of degree (n – 2) at the most and Fn consists of n non-repeated linear factors, then the n(n – 2), points of intersections of the curve and its asymptotes lie on the curve Fn – 2 = 0. If the joint equation of the asymptote is Fn = 0 and the equation of the curve be Fn + Fn – 2 = 0, then the n(n – 2) points of intersection of the asymptote with the curve must separately satisfy Fn + Fn – 2 = 0 and Fn = 0 which precisely means they lie on Fn – 2 = 0, e.g. (i) The asymptotes of a cubic curve, cuts the curves in 3(3 – 2) = 3 points which lie on the curve of degree 3 – 2 = 1, i.e. on a straight line. (ii) The asymptotes of a biquadratic (or quartic curve), cuts the curve in 4(4 – 2) = 8 points which lie on a curve of degree 4 – 2 = 2, i.e. on a conic.
Example 49: Find the asymptotes of the curve x2y – xy2 – xy + y2 + x – y = 0 and show that they cut the curve again in three points which lie on the line x + y = 0. [NIT Kurukshetra, 2008]
Solution: Rewrite the given equation x2y – xy2 – xy + y2 + x – y = 0 as yx2 + (1 – x)y2 – xy + x – y = 0 Clearly asymptotes parallel to X-axis and Y-axis are y = 0 and x = 1 respectively. For oblique asymptotes: φ 2 φ 2 φ 3(m) = m – m and 2(m) = –m + m implying '3(m) = 1 – 2m φ()m−+mm2 For mc== 0, 0 so that c=−2 =− ⇒ φ − For mc== 1, 0 '3()m 12m Therefore, equation of asymptotes y = mx + c becomes y = 0 , y = x. Now the joint equation of the asymptotes, y(x – 1)(y – x) = 0 or x2y – xy2 – xy + y2 = 0 Clearly given equation of the curve, i.e. (x2y – xy2 – xy + y2) + (x – y) = 0 is expressible like 2 2 2 Fn + Fn – 2 = 0 wherein we have obtained, Fn = 0, i.e. x y – xy – xy + y = 0 as the joint equation of the asymptotes. Geometrical Applications of Differentiation 159
φ()m −2am2 cama=−2 =− = =± and φ'() . 3mm2 Therefore the two oblique asymptotes are y = x + a and y = –x – a.
Example 47: Show that the parabola y2 – 4ax = 0 has no asymptotes.
Solution: Since in the equation of the parabola the coefficient of y2, the highest degree term in y, is merely a constant, therefore, there is no asymptote parallel to Y-axis. Again, since the coefficient of x, the highest degree term in x is also a constant, therefore, there is no asymptote parallel to X-axis. For oblique asymptote: φ 2 φ 2(m) = m and 1(m) = –4a. φ Now 2(m) = 0 implies m = 0, 0 But asymptotes corresponding to m = 0 (if any) are parallel to X-axis. But it has already been ruled out. Hence the curve has no oblique asymptotes also.
Example 48: Find the asymptotes of y3 – x2y – 2xy2 + 2x3 – 7xy + 3y2 + 2x2 + 2x + 2y + 1 = 0.
Solution: As per article 2.6 (IV), substitute y = mx + c in the equation of the curve and equate to zero coefficients of the two highest powers of x. Determine m and c. Therefore, the given equation becomes, (mx + c)3 – (mx + c)x2 – 2(mx + c)2x + 2x3 – 7(mx + c)x + 3(mx + c)2 + 2x2 + 2x + 2(mx + c) + 1 = 0 or (m3 – 2m2 – m + 2)x3 + (3m2c – 4mc – c + 3m2 – 7m + 2) + … = 0 Therefore on equating the coefficient to zero, we get m3 – 2m2 – m + 2 = 0 …(i) and (3m2 – 4m – 1)c + (3m2 – 7m + 2) = 0 …(ii) The first equation gives, m = 1, –1, 2 From (ii) equation for m = 1, c = –1; m = –1, c = –2; m = 2, c = 0. Hence the asymptotes are y = x – 1, y = –x – 2, y = 2x. V. Intersection of a Curve with its Asymptotes An asymptote of a curve of degree n cuts the curve in at the most (n – 2) points. From Article. 2.6 (IV), we see that if y = mx + c …(1) is an asymptote to the algebraic curve nφ n – 1φ n – 2φ x (y/x) + x n – 1(y/x) + x n – 2(y/x) + …… = 0 …(2) Means the two curves intersects such that two of the roots of equation 2 nnφ+−−1' φ+φ+ n 2c φ+φ+φ+…= " ' xmxcmnnnnnn() ()−−−112 () m x () mc () m () m 0 …(3) 2! are at infinity. 158 Engineering Mathematics through Applications
0 Sometimes it is form, then find c from 0 2 cφ+φ+φ="'() () () nnmc−−12 m n m0 2! φ Note: Asymptotes corresponding to m = 0 (as a root of n(m) = 0) is parallel to X-axis and are obtained directly. [Using Article 2.6 II]
Example 45: Find the asymptotes of the curve y3 + x2y + 2xy2 – y + 1 = 0.
Solution: Here the line y = 0 is the asymptote parallel to X-axis whereas there is no asymptote parallel to Y-axis. For Oblique Asymptotes: In the given equation of curve, expression containing the third degree terms is y3 + x2y + 2xy2 φ 3 2 Thus, 3(m) = m + 2m + m (by taking y = m, x = 1) φ 2 φ so that '3(m) = 3m + 4m + 1 and "3(m) = 6m + 4 φ φ Likewise, 2(m) = 0, 1(m) = –m φ ⇒ 3 2 3 = 0 m + 2m + m = 0 or m = –1, –1, 0 φ Now for equal values of m in n(m), corresponding values of ‘c’ are obtained from 2 cφ+φ+φ="'() () () 321mcm m0 2! c2 m ⇒ ()64mcom++⋅−= 0or c2 = 2 32m + m −1 For m = –1, c2 ===1 implying c = ±1 3232m +−+ and for m = 0, already we had obtained the parallel asymptote. Therefore, the asymptotes are y = 0, y = –x + 1, y = –x – 1.
Example 46: Find the asymptotes of the curve y2(x – 2a) = x3 – a3.
Solution: Asymptote parallel to Y-axis: Equating to zero the coefficients of highest powers of y ∴ x – 2a = 0 Asymptote parallel to X-axis: There is no asymptote parallel to X-axis (as coefficient of highest power of x is merely a constant). For oblique asymptote: The given equation is y2(x – 2a) – x3 + a3 = 0 ∴ φ 2 φ 3(m) = m – 1 and 3' (m) = 2m φ 2 and 2(m) = – 2am φ ⇒ 2 Now 3(m) = 0 m – 1 = 0, i.e. m = ±1. Geometrical Applications of Differentiation 157
==−y() mcymxLt , Lt …(3) xx→∞x →∞
y c Thus on putting =+m into (1), we get xx
nφ++φccn−1 ++……+φ++φ+ cc xmnn x−110 m xm m = 0 …(4) xx xx On expanding each term by Taylor’s series, we get
2 n φ+φ+φ+…+φ() cc'"() 1()n−1 () +φ+… c '() xmnn m n m x n−−11 m n m xx2! 2 x +n−2 φ() +c φ' () +… +…+= xmnn−−22 m 0 x nφ n – 1 φ φ or x n(m) + x {c · 'n(m) + n – 1(m)} 2 +φ+φ+φ+……=n−2"c() ' () () xmcmmnn−−12 n 0…(5) 2! As the line (2) is an asymptote to the curve (1), means it cuts the curve in two points at infinity, precisely means the equation (5) has two of its roots at infinity for which the coefficients of two highest powers of x should be zero. φ φ φ i.e. n(m) = 0 and c· n´(m) + n – 1(m) = 0 …(6)
If from above m1, m 2, m 3, …, mn be the n values of ‘m’ and c1, c 2, c3, …,cn be the corresponding values of ‘c’, then the equation of the asymptotes will be
y = m1x + c1, y = m2x + c2, y = m3x + c3, …, y = mnx + cn Observations φ φ ≠ (i) when n' (m) = 0 but n – 1(m) 0, the finite value of ‘c’ can not be determined from (6) and there is no asymptote in this case. φ (ii) If n(m) = 0 gives two equal values of ‘m’, then the corresponding value of ‘c’ can't be obtained from (6). In this case, ‘c’ will be obtained by equating to zero the coefficient of xn – 2, i.e. 2 cφ+φ+φ="'() () () nnmc−−12 m n m0 2! Working Rule φ (i) Find the polynomial n(m) by putting x = 1, y = m in the highest degree term of the φ given equation. Put n(m) = 0 and solve it for various values of m say m1, m2 , m3, … φ (ii) Likewise, find n – 1(m) from the next lower degree terms of the equation and so on. (iii) Find the values of c1, c 2, c 3, … corresponding to m1, m 2, m 3; … from the relation,
φ−()m c=− n1 φ ≠ φ’(), provided n(m) 0. n m 156 Engineering Mathematics through Applications
1 Example 44: Find the asymptotes of the curve yxe+1= x .
1 Solution: The equation of the curve is yxe=−x 1 For Asymptotes parallel to X-axis, let x → ±∞ 1 As x → ∞, y → ∞ QLtex = 1 x→∞ also x → ∞, y → –∞ In both the cases, y does not tend to any value and hence there is no asymptotes parallel to X-axis. For asymptotes parallel to Y-axis, let y → ±∞ 1 11111 Then yxe=−=+++x 11 x +……− 1 xx2!23 3! x 11 1 1 =+x + +…… 2!xx 3! 2 y → +∞ as x → 0+ 1 – Q=−→−()x → ∞ →0 yxeLt− 1 1 Also y –1 (not )asx x→0 ∴x = 0 is an asymptote (y → ∞).
1 Oblique asymptote: Given yxe=−x1 1 ==−=−=yx 1 Implying, meLt Lt 1 0 1 xx→∞xx →∞ 1 =−=−=−−x Also cymxyxxexLt()() Lt Lt 1 xxx→∞ →∞ →∞ =+++……−−111 Ltxx 1 1 x→∞ xx2! 2
=+……=11 Lt 0 x→∞ 2! x ∴ y = mx + c = x is an oblique asymptote
IV. Oblique (Inclined) Asymptotes of the General Rational Algebraic Equation Let the general rational algebraic equation be of the form nφ n – 1φ n – 2φ x n(y/x) + x n – 1(y/x) + x n – 2 (y/x) + …… = 0 …(1) φ where r(y/x) is a polynomial of degree r in (y/x). Let the straight line y = mx + c …(2) be an asymptote of the curve (1), where m and c are finite and Geometrical Applications of Differentiation 155
Now asymptotes parallel to X-axis are obtained by equating Y to zero the coefficient of highest powers of x in the equation, i.e. y2 = a2 implying y = ±a. Baa(– , ) Aa(, a) ya = Likewise, asymptotes parallel to Y-axis are x2 = a2 implies x = ±a X Thus clearly ABCD is a square of sides 2a. O(0, 0)
Example 42: Find the asymptotes of the curve y = ex. ya = – Da(, –) a C (–aa , – ) Solution: The given equation of the curve is y = ex. xa = – xa = (a) For Asymptotes Parallel to X-axis let x → ±∞, then (i) x → ∞ implies y → ∞ Fig. 2.29 (ii) x → –∞ implies y → 0(= e – ∞) Hence as per definition, y = 0 (for x → –∞) is an asymptote parallel to X-axis (b) For Asymptote Parallel to Y-axis Rewrite the given equation as x = logy and let y → ±∞. Now in the case for y → ±∞, x does not tend to any finite value. Whence there is no asymptote parallel to Y-axis. yeexx (c) Oblique Asymptote: Lt===∞ Lt Lt (No finite value), hence no oblique xx→∞xx →∞ x →∞ 1 asymptote (L' Hospital Rule). Example 43: Find the asymptote of curve y = secx.
Solution: The equation of the curve is y = sec x For asymptotes parallel to X-axis: As x → ±∞, y does not tend to any unique finite value. ∴ There is no asymptote parallel to X-axis. For asymptote parallel to Y-axis: 1 π As y → ±∞, x → (2n + 1)π/2 Qyx==sec →±∞→+ as xn()2 1 cosx 2 ∴ x = (2n + 1)π/2 gives asymptotes parallel to Y-axis. Oblique asymptotes: 1 yx==sec cosx y 1 ∴ = xxxcos y==1 Lt Lt 0 xx→∞xxx →∞ cos i.e., m = 0 which implies asymptotes parallel to X-axis, however we had proved above that there is no asymptote parallel to X-axis. Hence the curve has no oblique asymptotes. 154 Engineering Mathematics through Applications
If in equation (5), y happen to be such that a0 = 0 and a1y + b1 = 0 or in other words if two of the coefficients of highest powers of x vanishes meaning there by two of its roots are at infinity. Hence by definition,
a1y + b1 = 0 …(6) will be asymptote parallel to X-axis. 2 Again if in equation (5), y happen to be such that a0, a1, b1 are all zero and a2y + b2y + c2 = 0 or in other words coefficients of next two highest powers of x vanishes meaning there by three of its roots are at infinity. Hence by definition 2 a2y + b2y + c2 = 0 …(7) will be asymptotes parallel to X-axis and so on. Working Rule (i) For finding asymptotes parallel to axis of X, equate to zero the coefficients of highest powers of x in the equation, provided it is not merely a constant. (ii) For finding asymptote parallel to Y-axis, equate to zero the coefficients of highest powers of y in the equation, provided it is not mearly a constant.
Alternate Method for Finding Asymptotes Parallel to Axes Let the Lt.x = α …(8) be an asymptote parallel to Y-axis to the curve Y y = f (x) then we are to find the value of α. Let PM be the distance of the point P(x, y) on the curve Px(, y) from the line (8), then M PM = (x – α) Now by definition of an asymptote, if the line (8) is an T asymptote to the curve, then PM → 0 as P → ∞. ∴ As P → ∞, PM = (x – α) → 0orx → α α or in words if P → ∞ as x → α, only y coordinates tends to O X infinity (i.e. +∞ or –∞), and thus L → α → ∞ Lt x =α or x as y Fig. 2.28 y→∞ (i) Hence to find the asymptotes parallel to Y-axis, we find from the given equation, the α α → ∞ ∞ α α definite values 1, 2, … to which x tends as y , – . Then x = 1, 2, etc. are called the asymptotes parallel to X-axis. β β (ii) Asymptotes parallel to X-axis: From the given equation, find the definite values 1, 2, β → ∞ ∞ β β β 3, … to which y tends as x or – ; then y = 1, y = 2, y = 3, … are asymptotes parallel to Y-axis.
Example 41: Find the asymptotes parallel to the axis for the curve x2y2 = a2(x2 + y2) and show that they form a square of sides 2a.
Solution: The given equation of the curve can be written as x2(y2 – a2) – a2y2 = 0 or y2(x2 – a2) – a2x2 = 0 Geometrical Applications of Differentiation 153
See figure 2.27 (ii), clearly, the straight line LM is fixed and is at a finite distance from the origin while if the tangent to the curve cuts this line then the tangent to the curve tends as its point of contact receeds. Y Y M B M α P
X O P P
T L (0, 0) A X O (i) (ii) Fig. 2.27 However, in layman’s language, asymptotes are the tangents at infinity.
II. The General Equation of an Asymptote Let the equation of the given curve be y = f(x) …(1) Then the equation of the tangent to the curve at the point (x, y) will be dy dy dy Yy−=() Xx − or YXyx=+− …(2) dx dx dx then as per definition, the straight line at a finite distance which meets the curve at two points both of which are situated at infinite distance from the origin, precisely means for x → ∞, if dy dy Lt = m and −= →∞ Lt yx c x dx x→∞ dx The equation of the tangent will take the form Y = mX + c …(3) and is called the asymptotes to the curve. ydyc Note: m of an asymptote is also equal to the limit of (y/x) as x tends to infinity as lim−−= 0 or x→∞ xdxx y lim = m x→∞ x III. Asymptotes Parallel to Axes of Coordinates The most general algebraic equation of the curve will be n n – 1 n n – 1 n – 2 n – 1 (a0x + a1x y + … + any ) + (b1x + b2x y + … + bny ) + … + (ln – 1x + lny) + kn = 0 …(4) Which on rearrangement may be written as n n – 1 2 n – 2 a0x + (a1y + b1)x + (a2y + b2y + c2)x + … + kn = 0 …(5) 152 Engineering Mathematics through Applications
θ θ Hint: Take parametric equation as x = a sec , y = b tan ; Prove that center of curvature +ab22 ab 22 + ()xy,sec,tan=θ−θ33 and then use sec2θ – tan2θ = 1 . ab t 3. Show that the evolute of the tractrix xc=+cos t log tan , yc = sin t 2
x is the catenary y = c cosh . c 4. Show that the evolute of the rectangular hyperbola xy = c2, (i.e. x = ct, y = c/t) is the curve (x + y)2/3 – (x – y)2/3 = (4c)2/3. xa=θ+θ()sin , θ θ 5. Show that the evolute of the cycloid ya=−θ()1cos is the curve x = a( – sin ), (y – 2a) = a(θ + cosθ). xa=+()cos tt sin t , 6. Find the evolute of the curve ya=−()sin tt cos t. x 7. Prove that the chord of curvature parallel to Y-axis for the curve ya=log sec is of a constant length.
8. If Cx, C y be the chord of curvature parallel to the axis of X and Y respectively at any = x 2 2 2 4 point of the curve yccosh ; prove that 4c (Cx + Cy ) = Cy . c θ 9. If Co and Cp denotes the length of the chord of curvature of the cardiod r = a(1 + cos ) through the pole along the radius vector and perpendicular to the radius vector at any 2 2 point respectively then show that 3(Co + Cp ) = 8aCo.
2.7 ASYMPTOTES I. Introduction and Definition Several times we come across examples where in the curve or its branch tends to infinity, like in the case of parabola or hyperbola. So, in such cases, it becomes interesting to know what happens to the tangent to the curve, when the point at which the tangent drawnn to the curve moves away and away from the origin. There are three possibilities that the tangent may go further away and away from the origin, or it may keep oscillating, or it may tend to a definite straight line. In the last case, the straight line to which the tangent tends is called the asymptote to the curve. The formal definition is as follows: Definition 1: A straight line, at a finite distance from the origin is said to be an asymptote to an infinite branch of a curve, if the perpendicular distance of a point P on that branch from the straight line tends to zero, as P → ∞ along the branch (see figure 2.27 (i)). Definition 2: An asymptote of a curve is a straight line at a finite distance from the origin, to which the tangent to the curve tends as the point of contact recedes to infinity. Geometrical Applications of Differentiation 151
θ Differenting, −=−1sindr dθ+θπθ1cos θ+θ()or r =φ=tan = tan − rd 1cos dr sinθ 2 2 πθ which implies φ= − …(2) 22 θπ θ == φ= − = Now, say fr() p r sin r sin r cos 22 2 2a θ or fr()== p ar, Using =+1 cos θ= 2cos2 …(3) r 2
dp 1 a implying f'() r == …(4) dr2 r
fr() ar Now, C , the length of chord of curvature through the pole = 2 =⋅24 =r 0 f'() r 1a 2r which is clearly 4 times the focal distance of the point taking pole as the focus. dr2 r Again ρ=rr = ⋅ dp a Also we know that here, ψ = θ + φ = θ + (π/2 – θ/2) = (π/2 + θ/2)
Thus, the length of the chord of curvature parallel to X-axis (C0),
dr πθ 2ρ sinψ =+2sinr dp 22 θ = r 22r cos a 2
θ =⋅=ra 2 2a 44rr, Using 2cos=+() 1 cos θ= ar 2 r Hence the length of the chord parallel to X-axis has the same length viz. 4 times the focal distance of the point.
ASSIGNMENTS 6
33aa 1. Find the centre of curvature of the point , of the folium x3 + y3 = 3axy. 22 [KUK, 2009] x2 y2 2. Find the evolute of the hyperbola −=1 ab22 150 Engineering Mathematics through Applications
=1 2 x and y2 sec aa
3 2 3 + 2x ()+221tan 1y1 a x ∴ ρ= = asec 1 x ya2 sec2 aa
===ψψ=xxdy Also y1 tan tan implies adx a Hence, the chord of curvature parallel to Y-axis xx 2ρcos ψ =2seccosa = 2a. aa
Example 39: Show that the length of the chord of curvature through the pole of the equiangular spiral r = aeθ·cotα is 2r.
Solution: Given, equation of the curve r = aeθ·cotα ∴ α θ·cotα α r1 = a cot e = r cot α 2α and r2 = r1cot = r cot drθ Further tanφ=r = = tan α, ∴ φ = α dr r1
33 ()rr22++α22() rr 222cot ∴ ρ=1 = 22+− 22222 + α−α rrrrrr22cotcot12 r
3 r32()1cot+α2 ==αrcosec r22()1cot+α Now, the length of the chord of curvature through the pole (Co) is equal to 2ρsinφ = 2rcosecα·sinα = 2r (as φ = α).
Example 40: Show that the chord of curvature, through the focus of a parabola is 4 times the focal distance of the point. Further, show that focal chord parallel to the axis has the same length. 2a =+()1cos θ Solution: Let the equation of the parabola be r …(1) First, we wish to find the pedal equation of the parabola with the pole as the focus and initial line as the X-axis So, take logs on both sides, log2a – log r = log(1 + cos θ) Geometrical Applications of Differentiation 149
Example 37: Show that the circle of curvature at the origin for the curve x + y = ax2 + by2 + cx3 is (a + b)(x2 + y2) = 2(x + y).
Solution: Equation of the curve is (x + y) = ax2 + by2 + cx3 …(1) x2 Put ypxq=+ +… …(2) 2
2 xx22 We get xpxq+++…=+ axbpxq23 + +…+ cx …(3) 22 On equating the coefficients of x and x2 on both sides p + 1 = 0, i.e.p = –1 q and =+abp2, i.e.q = 2(a + b) …(4) 2 Therefore the radius of curvature at the origin is given by
3 3 +22 2 ()1 p ()11+ 2 ρ=() = = …(5) 0,0 qabab2()++()
Further y1 = p = –1,
y2 = q = 2(a + b), ()+2 −+ yy111 11() 1 1 xx()=− =−0 = Thus 0,0 yabab2(++)( ) …(6) 2 ()0,0
()+2 + 1 y1 ()11 1 yy()=+ =+0 = and 0,0 yabab2(++)( ) …(7) 2 ()0,0 Hence the equation of the circle of curvature at the origin, 2 ()xx−+−=ρ2() yy 2 becomes
22 −+−=112 xy 2 ab++ ab ()ab+ i.e. [(a + b)x – 1]2 + [(a + b) y – 1]2 = 2 or (a + b)2 (x2 + y2) – 2(a + b)(x + y) + 2 = 2 i.e. (a + b)(x2 + y2) = 2(x + y).
Example 38: Show that the chord of curvature parallel to Y-axis for the curve y = a log sec(x/a) is 2a.
Solution: Given, y = a log sec(x/a) axxx 1 y=⋅sec tan ⋅= tan Thus 1 xaaaa sec ()a 148 Engineering Mathematics through Applications
a2 =θ−θ−θbsin() 1 cos23 sin b ()ba22− =θsin3 …(5) b
−22 − 22 ab33θθ ba Whence the centre of curvature Cxy(), is given by cos , sin . ab From (4) and (5), we get ax=−() a22 b cos 3 θ and by=−() a22 − b sin 3 θ
2222 which implies ()ax3+=−() by3() a22 b33()()cos 2 θ+θ=− sin 2 a 22 b Hence the equation of evolute is (ax)2/3 + (by)2/3 = (a2 – b2)2/3
Example 36: Find the circle of curvature for the curve x1/2 + y1/2 = a1/2 at the point (a/4, a/4).
Solution: Given x1/2 + y1/2 = a1/2 …(1)
1 dy y 2 dy Thus =− implying =−1 …(2) aa dx x dx , 44 11dy − xy 2 2 22ydx x dy =4 and dy=− =1 + y implying …(3) 1 2 aa dx2 x2 x x dx, a 44
3 3 2 2 ()+ 22 11+−() 1y1 3 ρ= = =2a Now aa 2; …(4) , 4 44 y2 4 a 2 ()+2 (1)1−+−() 1 yy111 aaaa 3 xxaa =− =− =+= ; , 4 44 y2 4424…(5) a 2 ()+ 2 11+−() 1y1 aaaa23 yyaa =+ =+ =+ = , 4 and 44 y2 4444 …(6) a Therefore the equation of circle of curvature 2 ()xx−+−=ρ2() yy 2 becomes 22 33aaa2 xy−+−=23 4416 22 33aaa2 or xy−+−=. 442 Geometrical Applications of Differentiation 147
x2 y2 Example 35: Find the coordinates of the center of curvature of the ellipse +=1 ab22 or x = a cosθ, y = bsinθ.
Hence show that the equation of its evolute is (ax)2/3 + (by)2/3 = (a2 – b2)2/3.
Solution: Given x = a cosθ, y = b sin θ …(1) dy θ ==dy dθ =bbcos =−θ y1 cot …(2) dxdx −θ asin a dθ
ddbdθ bb1 and yy==−θcot =θ=−θcosec23 cosce …(3) 21dx dθ a dx aaa−θsin 2
yy()1+2 ∴ xx=−11 y2
−bb2 cotθ+ 1 cot2 θ aa2 =θ−acos −b cosec3θ a2
b2 =θ−θθ+θaacos sin22 cos 1 cot a2 b2 =θ−θ−θacos() 1 sin23 cos a ab22− =θcos3 …(4) a ()+2 1y1 and yy=+ y2
b2 1cot+θ2 a2 =θ+bsin b −θcosec3 a2
222θ =θ−θ+ab3cos bsin sin 1 ba22sin θ
a2 =θ−θ−θθbbsin sin32 sin cos b 146 Engineering Mathematics through Applications
y and t3 =− …(7) 2a On taking cub of (6), squaring (7) and equating the two −3 2 xa2 =−y 3 i.e. 27ay2=− 4() x 2 a 32aa Hence the locus of ()xy,, i.e. the equation of evolute is 27ay2 = 4(x – 2a)3.
Example 34: Find the center of curvature and evolute of the hypocycloid x2/3 + y 2/3 = a2/3 (or the astroid x = a cos3θ, y = a sin3θ).
Solution: The given equation in its parametric form is x = a cos3θ, y = asin3θ dy 2 θ⋅ θ ==dθ 3sina cos =−θ ∴ y1 tan dx −θθ3cosa 2 sin dθ θ =−θ=−θddd() () and y2 tan tan dx dθ dx 1 1 =⋅−θ(sec)2 = −θθ3cosa 2 sin 3cosa 4θθ sin
Hence the centre of curvature ()xy, at the point ‘θ’, is given by
yy()1tan1tan+θ+θ22() xx=−11 = acos3 θ+ = a (cos3θ + 3sin2θ cosθ) 1 y2 3cosa 4θθ sin and ()11tan++θy22() yy=+1 = asin3 θ+ 3θ 2θ θ 1 = a(sin + 3cos ·sin ) y2 3cosa 4θθ sin
Further xya+=()cos θ+θ sin 3 and xya−=()cos θ−θ sin 3
22 2 22 2 Implying ()xy+=3 a3()cos θ+θ sin and ()xy−=3 a3()cos θ−θ sin
22 332 So that ()xy++−=() xy2 a3
Hence the locus of ()xy, which is called evolute of the curve, given by (x + y)2/3 + (x – y)2/3 = 2a2/3 Geometrical Applications of Differentiation 145
drp dr =ρ2sin φ=⋅ 2rp = 2 dp r dp …(6)
dr Since pr=φsin and ρ= r dp dp pfr() Also, p = f(r) ⇒ = f'() r ; and p = r sinφ⇒sinφ= = dr rr dr 1 fr() fr () ∴ Crr=ρ2 sin φ=⋅ 2 ⋅ sin φ= 2 ⋅ ⋅ = 2 …(7) 0 dp f'() r r f' () r
22− ρ dr dr rp 22dr 22 and Cr=ρ2cos φ= 2 1 − sin2 φ= 2 r =−⋅=−⋅22rp rp …(8) p dp dp r2 dp r
Example 33: Find the coordinates of the center of curvature of the parabola y2 =4ax. Also find the equation of the evolute of the parabola.
Solution: Equation of the parabola is y2 = 4ax i.e. x = at2 and y = 2at (in porometric form) …(1) dy ===dt 21a So that y1 …(2) dx 2at t dt
==dddt1111 =−⋅=− and yy21 …(3) dx dt t dx t2322 at at
+1 ()+2 1 yy111 12 xx=− = at2 −⋅ t ∴ 1 yt2 − 2at3
3 =+2221at + = + at132 at a …(4) tt2
1 1+ ()1+y2 2 y=+ y1 =22 at +t =− at3 and 1 …(5) y2 − 2at3 Hence the coordinate of centre of curvature at any point (x, y) of the parabola are (3at2 + 2a, –2at3). Now, from eqn (4) and (5), we have ()xa−2 t2= …(6) 3a 144 Engineering Mathematics through Applications
(i) Length of Chord of Curvature in Cartisian Co-ordinates: Let the tangent at P makes an angle ψ with X-axis, then the chord of curvature PR(parallel to X-axis) makes an angle of (90° – ψ) with the normal PCD. Likewise, chord of curvature PQ (parallel to Y-axis) makes a angle ψ with the normal PCD. ∴ Cx = length of the chord of curvature paraller to X-axis (PR) = PD cos(90° – ψ) = 2ρsinψ Y D Q 3 ()+22 C 1y1 y R P =⋅2 ⋅ 1 +2 y2 1y1
21yy()+2 T Ψ =11 X …(2) O y2 Fig. 2.25
and Cy = Length of the chord of curvature parallel to Y-axis (PQ) 3 ()++222() 121yy111 =ψ=ρψ=⋅PDcos2cos2 = …(3) +2 yy221y1 (ii) Length of Chord of Curvature in Polar Coordinates: If φ is the angle between the tangent and the radius vector, then certainly, PL the chord of curvature through pole O, makes an angle of (90° – φ) with PCD, the normal to the curve at P. And whence, PM the chord of curvature perpendicular to the radius vector OP, makes an angle φ with
the normal PCD (Fig 2.26). M ∴ C0 = Length of the chord of φ curvature D C φ PL through the pole (along φ 90° P the φ radius vector) L = PD cos(90 – φ) = 2ρ sinφ r 3 ()22+2 rr1 r =⋅2 ⋅ 22+− 22+ θ Ψ rrrr212rr1 OT 22+ 2rr() r1 = Fig. 2.26 22+− …(4) rrrr212
and Cp = Length of the chord of curvature PM perpendicular to radius vector.
3 22++2 22 ()rr111r 2 rrr() =φ=ρφ=⋅⋅PDcos 2 cos 2 1 = …(5) 22+−22+ 22 +− rrrr2212rr1 rrrr 12 (iii) Length of Chord of Curvature using Pedal Equations (p – f(r)): Co = Length of chord of curvature PL through the pole along radius vector Geometrical Applications of Differentiation 199
Y
(0, a) A Y xa = – xa =
X' X (v) (vi) O 0.31a X´ (,a 0) O X
Y´ A'(0, – a )
Y'
Fig. 2.79 Fig. 2.80
Y
A C
(vii) X' X O
B Y'
Fig. 2.81 198 Engineering Mathematics through Applications
O (ix) X (x) X O
Fig. 2.72 Fig. 2.73 [Hint: r = a is a circular asymptote. [For r = a eθcotα, tanφ = tanα i.e. aθ2 ∴ Lt= a. φ = α, meaning thereby that the θ→∞ 1 +θ2 The part of the curve for negative value tangent to the the curve at any is its reflection in the initial line] point of it always makes a constant angle α with the radius vector of that point]
Assignment 10
Y Y θ = – π θ = π θ = 0 θ = 2π A B A B
θ θ /2 = /2 = π 2 – π 2a π 3 a π = 2a /2 = 2a (i) /2 θ (ii) θ π θ π xa = – O = 0 xa = X O θπ = X
Fig. 2.75 Fig. 2.76
Y
Y
θπ = B 3 (iii) (iv) t = ± 2a X' X O (3a, A 0) A C X θ = 0 O θπ = 2
Y' Fig. 2.77 Fig. 2.78 Geometrical Applications of Differentiation 197
θπ = /2 Y
xa = θπ = B (,a 0)θ = 0 (iii) (,a π) O A (iv) (0, π/2) O X
(2a, 3π/2) C θ = – π /2
Fig. 2.67 Fig. 2.68
Y (4a , 2π /3) D C π (2a , π/2) θ = 2
B (4a /3, π/3) θ = π θ = 0 O (,a 0) θ = 0 X' X (v) (vi) O X´ A X
Y'
Fig. 2.69 Fig. 2.70
θ = π/2
π a, bπ H2K a+, b2π H23K a− , H 2 3K C D θ θ = π/2 B = /4 E 3π =π /4 θ F π (vii) θ = π (ab – , ) θ = 0 (viii) O A X O θ = – 3π θπ = 3 /2 /4 (ab – ) θ = – π /2 Fig. 2.71 Fig. 2.72 [Since, |cosθ| ≤ 1 in all cases, therefore, (a – b) is always positive for all values of θ] 196 Engineering Mathematics through Applications
Y
) 2 a/ Y 3 , /2 a
a 2
3
( =
x
X' O X (xiii) (xiv) X' X O
xya + [Cissoid] + = 0 Y' Y' Fig. 2.61 Fig. 2.62
Y Y
O A(1, 0) X' X (,OC ) (xv) (xvi) X O
Y' Fig. 2.63 Fig. 2.64 x [Hint: As cosh≥ 1, therefore there is no part of the curve is below the line y = c] c
Assignemnt 9
θ θπ = /2 = /3 π 2 π Y = /3 θ θπ = /2 θπ θπ = 3 /4 = /4 θ /6 = 5 2 π 5π C θ = /6 O A 6 θπ = 1 θ = 0 O (i) (ii) θπ = θ = 0 X X B A 3 4 B
θπππ: 0 /6 /4 /2 … θ = 0 πππ /6 /3 /2 … ra: a/ 2 0 imaginary … ra– = 0 0 a … Fig. 2.65 Fig. 2.66 Geometrical Applications of Differentiation 195
(0, a) ya = Y
(vii) X(viii) O ABC
ya = –
Fig. 2.55 Fig. 2.56
Y a
=
x y = 1
(2a , 0) X´ X (xi) O (x) O
y = – 1
Y´
Fig. 2.57 Fig. 2.58
Y Y a
=
x
1
=
– 1 –
x = y x = – x x = (xi) X' X (xii) (– a, 0) y O X O y = – 1
Y´
Fig. 2.59 Fig. 2.60 194 Engineering Mathematics through Applications
7. (i) r sin(θ – nπ) = 2 (ii) r sinθ = a (iii) r cosθ = a nππ3 π arnsin+= sin (2 +−θ 1) θ θ (iv) 24 4 ,orr sin = ± a/2, r cos = ± a/2 (v) rcosθ + a = 0, (vi) r sinθ + a = 0 (vii) θ = 0 (viii) a = r sin(θ – 1) (ix) r sinθ · eπ + a = 0 (x)(2n + 1)π r cosθ + 2a = 0
Assignment 8 Y Y (0, 2a) ya = 2 − 33aa3 33aa3 G , J G , J H 4 2K H4 2K X´ (0, 0) 1. (i) (ii) X O (,a 0)
O X xa =
Fig. 2.49 Fig. 2.50 Y
Y x–a = xa =
(/3,aa 2/3 y = – x x y = X' X A X O O (,a 0) (iii) (a /3, 0) (iv) (aa /3, – 2 /3)
x = 0 Y' Fig. 2.51 Fig. 2.52
a
–
= Y
x
X O (4a , 0) (v) (vi) X O (,a 0)
Y´ Fig. 2.53 Fig. 2.54 Geometrical Applications of Differentiation 193
Assignment 4
1 ψ 1. (i) 4sina ψ (ii)4acosψ (iii) c tanψ (iv) maem 3 8. (i) 3/2 (ii) 1
Assignment 5
arn −+n 1 am 1. (i) (ii) m−1 n+1 ()mr+1 3 1 aar2()32− 2 ab22 (iii) (iv) (v) (r2 – a2)1/2 32()ar− p3
3 2a ()ba42cosθ+ 42 sin θ 2 2. 5. 3 uab344 2 a2 6. (i) 2ar (ii) 3 3r r3 arnn−+1 (iii) (iv) a2 n+1 3 ()ra22+2 ab22 (v) (vi) 3 ()ra22+2 p ab22 7. (i) a (ii) p3
Assignment 6
21 21 1. aa, 2. (ax)2/3 – (by)2/3 = (a2 + b2)2/3 6. x2 + y2 = a2 16 16
Assignment 7
1. (i) x = ± a, y = ± a (ii) y = ± 1, x = ± 1 (iii) y = 0, x = ± 1 (iv)NO (v)y = 0 (vi) y = ± 2 2. (i) y = 0 (ii) x = 0 (y → – ∞) π (iii) xn=+()21, n is an integer (iv) x = nπ, n is any integer 2 3. (i) x = 2a, x + y + a = 0, x – y + a = 0 (ii) y = 0, x + 1 = 0, x + y = 0 (iii) y = ± x, y = x + 1 (iv) x + a = 0, x – a = 0, x – y + a = 0 , x + y – a = 0 5. x – 2y = 0, x + 2y = 0, 2x – y + 1 = 0, 2x + y + 1 = 0 x2 y2 6. +=1 ab22 192 Engineering Mathematics through Applications
t sin0 Note: For t = 0, log tan becomes log==−∞ log0 2 cos0
ASSIGNMENT 10 Trace the following curves: (i) x = a(θ + sinθ), y = a(1 – cosθ), (ii) x = a(θ – sinθ), y = a(1 + cosθ) t3 (iii) x = a(θ – sinθ), y = a(1 – cosθ), (iv) x = t2, yt=− 3 at()11−−22 att() (v)xy==, (vi) x = a[cosθ – log(1 + cosθ)], y = a sinθ 11++tt22() (vii) x = a sin2θ(1 + cos2θ), y = a cos2θ(1 – cos2θ)
ANSWERS
Assignment 1
1. (0, ∞) 2. (a, b) 6. t t sin2 t t Na=2tan SN= 2 a 2 13. 2sina ; ;ST = a sint; t 2 2 2tan 2 −a 14. a sin3·tanθ 15. ST== SN 17. π/4 2
Assignment 2
π 1. 8. (i) r3 = 2ap2` (ii) r3 = a2p (iii) rm + 1 = pam 2
Assignment 3
+ 9 x x 2. (i) 1 (ii) cosh 3. (i) t (ii) f’(t) + f”(t) 4 a c 2 e
θ a()θ+21 θ2 4. (i) a sec2θ (ii) 2cosa (iii) a( + 1) (iv) 2 2 ()θ−21 n−1 (v) ra22+ (vi) an()cosec θn Geometrical Applications of Differentiation 191
5. Region: As |sint| ≤ 1 ∴ –a ≤ y (= sint) ≤ a and hence the curve entirely lies between the line y = ± a. 6. Special Points: dx11 t t 1 =−atsin + a ⋅ 2 ⋅ tan ⋅ sec2 ⋅ Here t dt 2222tan2 2 1 =−atsin + tt 2sin cos 22 11sincos−22tt =−atsin + = a = a sinttt sin sin dy dy dt atcos and thus == =tan t dxdx cos2 t a dt sin t dy π (0, a) =∞ when t =± . Hence at the dx 2 t = π /2 point.(0, ± a), the tangent to the curve is ∞ 0 t = π → parallel to Y-axis. (x → – )t = (x ∞) dy → ∞ O = 0 means t = 0 (x ), i.e. X-axis as dx asymptote already proved. (0, –a ) dy2 dddt dy t = – π /2 Further, == ()tan t dx2 dx dx dt dx Fig. 2.48 ddttsin ==⋅()tantt sec2 dt dx acos2 t sint = atcos4 which is a positive value for t in [ 0, π]. Hence, the curve is concave upward for values of t in [0, π]. dy See the table for values of x, y and corresponding to various round values of t. dx dy txy dx π − 0–a–∞ 2 0–∞00 π 0a∞ 2 π∞00 190 Engineering Mathematics through Applications
dy2 Clearly, is negative for all values θ. dx2 Therefore the curve is concave downwards. The curve consists of ‘Congruent arches’ on both sides of Y-axis which extends to ∞. Note: (i) All the four cycloid, viz. xa=θ+θ( sin) xa =θ( – sin θ) , ya=+θ(1cos) ya =( 1–cos θ) xa=θ+θ( sin) xa =θ( – sin θ) , and ya=θ=+θ(1 – cos) ya( 1 cos ) are symmetrical about Y-axis and lie within the range y = 0 to y = 2a. xa=θ( –sin θ) , (ii) Further, the cycloid, for one full branch in the range 0 ≤ θ ≤ 2π is the replica of ya=−θ(1cos) xa=θ+θ()sin , for one full branch in the range –π ≤ θ ≤ π. ya=+θ()1 cos xa=θ−θ( sin) , Likewise, the inverted cycloid, for one full arc in the interval 0 ≤ θ ≤ 2π is the replica ya=+θ(1cos) xa=θ+θ()sin , of the inverted cycloid for one full arc in the interval –π ≤ θ ≤ π ya=−θ(1cos) 1t Example 75: Trace the curve tractrix xta=a cos + log tan2, y = a sint 22 Solution: 1. Symmetry: (i) On putting t as –t, x remains unchanged where as y changes to –y. Hence the given curve is symmetrical about X-axis. (ii) On changing t to (π – t), y remains unchanged whereas x becomes –x. Hence the curve is symmetrical about Y-axis (also). a 2. Origin: y = 0 implies t = 0. For t = 0, xa=+log0 →−∞. 2 Hence the curve does not pass through the origin. 3. Asymptotes: For y = 0 (at t = 0) x → –∞ Means y = 0, i.e., axis of x is an asymptote to the curve. 4. Intersection with axis: X-axis has already been proved as an asymptote, the curve meets the Y-axis, i.e., x = 0 which implies
1 t π atcos+= a logtan2 0 or t =± 22 2 and then y = a sin(±π/2) = ±a Therefore, the curve meets the Y-axis in the point (0, ±a). Geometrical Applications of Differentiation 189
Therefore, the curve lies entirely between the lines y = 0 and y = 2a. 5. Special Points: (i) Intersection with the X-axis: Putting y = 0, we get θ = π which gives x = aπ ∴ Intersection with X-axis is at (aπ, 0) (ii) Intersection with Y-axis: Putting x = 0, we get (θ + sinθ) = 0, i.e. θ = 0 and for θ = 0, y = a(1 + cosθ) = 2a ∴ Intersection with the Y-axis is at (0, 2a) dy dy (iii) Points where ==∞0and . dx dx dy θθ 2sin cos dy θasinθθ ==−d =−22 =−tan dx ()+θ θ dx a 1cos2cos2 2 dθ 2 dy For various values of θ, the corresponding values of x, y and and point on the dx curve are as follows: dy θ –π y dx –π – aπ 0 ∞ π π − −+ a1 a 1 2 2 00 2a0 π π+ a1a –1 22 π aπ 0– ∞ dy Chang in θ Change in x Change in y Portion of the curve traced dx –π to 0 increases increases ∞ to 0 A to B from –aπ to 0 from 0 to 2a 0 to π increases decreases 0 to – ∞ B to C from 0 to aπ from 2a to 0 (iv) Points of inflexion:
dy θ θπ As =−tan = /2 dx 2 Ba(0, 2 ) θ = 0 dy2 θθ1 d ⇒ =−sec2 dx2 22 dx θθ 2a =−1122 sec sec θ π 2222a A´ θπ = – C = (–aπθ , ) (,aπθ ) 1 θ =− sec4 42a Fig. 2.47 188 Engineering Mathematics through Applications
(iii) If on changing t to –t; x becomes –x and y becomes –y, the curve is symmetrical in opposite quadrants. = xct e.g. The rectangular hyperbola, =cis symmetrical in opposite quadrants. y t 2. Origin: If on putting x = 0, a real value of t can be found, which makes y = 0 or vice versa, then the curve passes through the origin. Alternately, put both x and y equal to zero and find the value of t. If there is any common value of ‘t’, then the curve passes through the origin. 3. Asymptotes: Find asymptotes if any. (If the curve x = φ(t) and y = ψ(t) are purely in terms of cos t and sint and no terms of tan t, cott, cosect sect, then there will be no asymptotes as sint and cost both has finite values for all finite t.) 4. Points: (i) Points of intersection with X-axis (i.e. put y = 0 and find value of ‘x’) (ii) Points of intersection with Y-axis (i.e., put x = 0 and find value of y) dy dy (iii) Points where = 0 and =∞. (i.e. find those values of ‘t’ for which tangents dx dx are parallel to X-axis and Y-axis.) dy2 dy2 (iv) Points of inflexion, i.e. point where < 0 or > 0 (viz. see positions, concave dx2 dx upwards and downwards respectively) dy dy 5. Region: See the points where = 0 and =∞. If easily possible, find the greatest dx dx dy and least values of x and y, with a table giving values of ‘x’, ‘y’ and at four-five dx broad values of ‘t’. Example 74: Trace the cycloid x = a(θ + sinθ), y = a(1 + cosθ). Solution: The given equation of the curve (in parametric form) is x = a(θ + sinθ), y = a(1 + cosθ) …(1) The cycloid is the curve described by a point on the circumference of a circle which rolls without sliding on a fixed straight line. The fixed line, viz. axis of x is called the ‘base’ and the farthest point on the axis perpendicular to this fixed line is called ‘vertex’ of the cycloid. Therefore, first we shall trace the curve for values of θ from –π to π. 1. Symmetry: On changing θ to –θ in equation (1), x changes to –x and y remains unaltered. ∴ The curve is symmetrical about Y-axis. 2. Origin: Taking y = 0, a(1 + cosθ) = 0 implies θ = ±π. For θ = ±π, x = a(θ + sinθ) = ±aπ ≠ 0 ∴ The curve does not pass through the origin. 3. Asymptotes: The curve has no asymptotes. 4. Region: As |cosθ|≤ 1, thus the greatest value of y is 2a and least value is zero. Geometrical Applications of Differentiation 187
Note: Find the tangent at x = 2a. For finding tangent at the origin, replace x by (x – 2a), the equation of the curve becomes y2(x – 3a) + (x – 2a)2x = 0 or y2(x – 3a) + x(x2 + 4a2 – 4ax) = 0 Clearly the lowest-degree term equated to zero implies 4a2x = 0 mean at x = 2a, y-axis is tangent to the curve. − = 2ax 5. Regions: From the curve, yx , Y xa− (i) For x < a, (x – a) is negative, (2a – x) is positive xa = means radical part is negative and thus y is imaginary. (2a , 0) X´ X (ii) For a < x < 2a, both the factor (2a – x) and (x – a) O are positive, y is defined for all values x in the ray. x = 2a Further, y decreases from ∞ to O as x increases from a to 2a. (iii) For x > 2a, factor (2a – x) is negative whence y is Y´ imaginary. Fig. 2.46
ASSIGNMENT 9 Find the asymptotes of the following curves: (i) r = a sin3θ; [NIT Kurukshetra, 2008] (ii) r2 = a2 cos2θ; sin2 θ (iii) r = a(1 – sinθ); (iv) ra= (cissiod); cosθ 2a (v) =+()1cos θ, parabola; (vi) r2 cosθ = a2 sin3θ; r (vii) r = a + b cosθ, a > b; (viii) r2 cos2θ = a2 [cube of rectangular hyperbola, i.e. x2 – y2 = a2] aθ2 (ix) r = ;(x)r = aemθ (or r = aeθcotα or r = aeθ) 1 +θ2 (xi) r2θ = a2
III. Procedure for Tracing Curves in Parametric Form: x = φ(t), y = ψ(t)
1. Symmetry (i) If on changing t to –t or t to (π – t); x = φ(t) remains unchanged and y changes to –y, then the curve is symmetrical about X-axis. e.g. The parabola x = at2, y = 2at is symmetrical about X-axis. (ii) Similarly, if on changing t to –t or t to (π – t); x changes to –x and y remains unchanged, then the curve is symmetrical about Y-axis. xat=+()sin t , e.g. The cycloid i.e. symmetrical about Y-axis, as by changing t to ya=+()1cos t, –t, the equation of the curve remains the same. xa=cos t , The ellipse is symmetrical about Y-axis as by changing t to π – t, the yb=sin t equation of the curve remains the same. 186 Engineering Mathematics through Applications
By definition, θ θ p = r sin ( 1 – ) ⇒ –a = r sin(0 – θ)orrsinθ = a, i.e. y = a
a sinθ sinθ r = ⇒ rasinθ= < a , Q<1 4. Regions: θ θ θ Therefore the given curve entirely lies below the asymptote r sinθ (=y) = a. 5. Points: a r θ dr (i) tanφ= = =−θ (Here r = is –ve, means r decreases as θ increases) a 1 θ r1 − d θ2 Y ya = (ii) Various values of r for various θ’s π π π 2π 3π θ:0 π 2π …∞ X 6 3 2 3 2 6a 3a 2a 3a a 2a a r: ∞ …0 Fig. 2.45 π π π 2π π 3π 2π Clearly as θ increases from 0 to ∞, r is +ve and goes on decreasing from ∞ to 0. For negative values of θ where r is also negative is oblained by turning the curve for positive values of θ through two right angles and this branch is shown by doted lines.
Example 73: Trace the curve r = a(secθ + cosθ). [NIT Kurukshetra, 2004]
Solution: Sometime it is helpful to change from polar to cartesian coordinates and vice versa. In this case, changing to cartesian to polar coordinate system by putting x = rcosθ, y = rsinθ so that r = a(secθ + cos θ) becomes
rx 2 2 2 2 2 2 2 ra=+or r x = a(r + x ), i.e.x(x + y ) = a(2x + y ). xr Rewriting as y2(x – a) = x2(2a – x). Now it is easily discussionable under cartesian coordinate system. 1. Symmetry: The curve does not possess any kind of symmetry. 2. Origin: The curve pass through origin as there is no constant term in its equation and the tangent at the origin are given by 2x2 + y2 = 0 means the tangents are imaginary and the origin is an isolated point. 3. Asymptotes: The line x – a = 0 is the only asymptote to curve. Even in case of polar, r → ∞ as 1cosθππ 3 =→θ→0as or . ra()1cos+θ2 2 2 4. Points: The X-axis (i.e. y = 0) cuts the curve in points (0, 0) and (2a, 0). Geometrical Applications of Differentiation 185
π Since given a < b, i.e. a – b is always negative. θ =
2 π 3
π α
∴ π θ 3
= =
The point (a – b, ) lies on the half ray = 0 =
θ
θ θ π 4 b D C = Without much loss of generality, let a > . θ 2 B ∴ r is positive for all values of θ from 0 to E θπ = (ab + , 0) π O θ = 0 2 3π A and negative when θ= or π (see F G 3 4 the table in r, θ). ∴ r must vanish somewhere between 2π 3π θ= and θ= or in other words, 3 4 Fig. 2.44 − a 2π 3π α=cos 1 − must lie between and . b 3 4 dr 6. ‘Value of φ’: We have =−bsin θ dθ dr Thus is negative for all values of θ between 0 and π dθ ∴ r decreases as θ increases from 0 to π. dabθ+cos θ Also tanφ=r ⋅ =⋅ dr−θ bsin tanφ → ∞ when sinθ = 0, i.e. θ = 0 or π, i.e the points (a + b, 0) and (a – b, π) and the tangents are perpendicular to the initial line in these positions. Note: It can be discussed with a = 2 and b = 3, i.e. r = 2 + 3 cosθ can be discussed simply by replacing a = 2 and b = 3 in the above problem as a particular case. Example 72: Trace the curve (reciprocal spiral), rθ = a.
Solution: 1. Symmetry: The curve is symmetrical about the axis, θ = π/2, i.e. perpendicular to the initial axis, since by changing θ to –θ and r to –r, the equation of the curve remains unchanged. 2. Origin: The curve does not passes through the pole, since r is not zero for any of the finite value of θ.
1 θ 3. Asymptotes: Given rθ = a, i.e. ==u ra → θ → θ u 0 implies 0, i.e. 1 = 0 du 1 and = daθ dθ paa=−=−=−Lt Lt () so that θ→θ θ→θ 11du 184 Engineering Mathematics through Applications
5. Points: θ = π/2 θ = 3π /4 dθθ1tan2 θπ (i) tanφ=ra = sin 2 θ⋅ = = /4 θ dr2cos2 a 2 D 7 2 A π φ= ππππ357 O ∴ (i.e., tanφ = ∞) when θ= ,,,.θπ = 8 θ = 0 2 1 44 4 4 5 4 (ii) Values of r at different points: 6 θ – Variation in Variation in r Portion of curve 3 3π BC/4 = traced θ =
7π π /4 0 to 0 to aO to A Fig. 2.43 4 π π to a to 0 A to O 4 2 π π to 3 0 to –aO to B 2 4 3π to π –a to 0 B to O 4 and rest by symmetry
Example 71: Trace the curve (Lemicon) r = a + b cosθ, when a < b. [NIT Kurukshetra, 2007]
Solution: The given equation of the curve is r = a + b cosθ, where a < b, i.e. (a – b) is negative 1. Symmetry: The curve is symmetrical about the initial line since with the change in θ to – θ, equation of the curve remains unchanged. a 2. Origin or Pole: When r = 0, cosθ=− is a real value numerically less than 1. b − b ∴ The curve passes through the pole and θ=cos 1 is the tangent to the curve at the pole. a 1 1 3. Asymptotes: Taking r = so that u = u ab+θcos Here u = 0 does not give any finite value of θ ∴ r does not tands to ∞ for any finite value of θ and thus has no asymptote. 4. Region: As –1 ≤ cosθ ≤ 1 ∴ r ≤ (a + b) and hence the entire curve lies within the circle r = (a + b). 5. Points: Values of r for various values of θ are as follows: π π π 2π 3π θ =0 π 4 3 2 3 4 +b +b −b −b r = a + b a a a a a (a – b) 2 2 2 2 Geometrical Applications of Differentiation 183
π π 3π and φ= when θ = 0, , π, . 2 2 2 (ii) Change in r and φ with the change in θ
Variation in θ Variation in r Variation in φ Curve traced π π From 0 to a to 0 to 0 A to O 4 2 π π π to 0 to –a 0 to O to B 4 2 2 π 3π π to –a to 0 to 0 B to O 2 4 2 3π π to π 0 to a 0 to O to C 4 2 Rest by symmetry θπ = /2 θ π π D [As increases from to 5 /4, portion of the θπ /4 = π π 3 = 5 /4 θ curve is traced from C to O. Increases from 7 6 4 4 1 θπ θ 3π = C O A = 0 to , portion of the curve is traced from O to 2 5 3π 7π 8 θ π/4 2 D, increases from to , portion is traced 5 3 2 4 θ = from D to O and finally O to A.]. B Example 70: Trace the curve r = a sin2θ. Fig. 2.42 Solution: 1. Symmetry: π (i) The curve is symmetrical about the line θ= as by change θ to π – θ, the equation 2 of the curve remains the same. (ii) The curve is symmetrical about pole also as by changing θ to (π + θ) the equation remains unaltered. π 3π (iii) Further, the curve is symmetrical about the line and θ= . 4 4 nπ 2. Pole: Putting r = 0 results in a finite value of ‘θ’, i.e. sin2θ = 0 or θ= . Therefore, the 2 curve passes through the pole and the tangents at the pole are θ = 0, π/2, 3π/2, 2π. 3. Region: The curve wholly lies within the circle r = a as |sin2θ| ≤ 1. 4. Asymptotes: The curve does not have any asymptote. 182 Engineering Mathematics through Applications
3. Asymptotes: Since r does not tends to infinity for any finite value of θ. Therefore, the curve has no asymptotes. 4. Regions: |cosθ| ≤ 1 ∴ The curve wholly lies within the circle r = 2a. When θ increases from 0 to π, r remains positive and increases from 0 to 2a. But when θ increases from π to 2π, r is positive and decreases from 2a to 0.
5. Special Points: For various values of θ, r is as θπ = /2 follows: A (,a π/2) θ : = 0 π/3 π/2 2π/3 π a r : = 0 a/2 a 3a/2 2a. θπ = B O θ = 0 [As the curve is symmetrical about the initial axis, (2a , π) (0, 0) we need not to trace the curve for θ from π to 2π and rest is completed by symmetry]. (,a 3/2)π dr C 6. Value of φ: As =θasin dθ Fig. 2.41 θ a 2sin2 dθ 1 θ tanφ=ra =() 1 − cos θ ==2 θ θθtan dr asin a 2sin cos 2 22 i.e. φ = θ/2 Now φ = 0 when θ = 0, φ = π/2 when θ = π ∴ At (0, 0) the tangent coincides with radius vector, however at the point (2a, π) it is perpendicular to the line θ = π (the radius vector).
Example 69: Trace the curve r = a cos2θ.
Solution: 1. Symmetry: The curve is symmetrical about the initial line as the equation remains unaltered by changing θ to – θ. 2. Pole: Putting r = 0, we get cos2θ = 0, i.e.,2θ = (2n + 1)π/2. π π Thus the curve passes through the pole and the tangents at the pole are θ= , 3 , 4 4 5π 7 π , . 4 4 3. Region (Limits): The curve wholly lies within the circle r = a, since |cos2θ| ≤ 1. 4. Points: dθ acos2θ 1 1 (i) tanφ=r ⋅ ==−θ=π+θcot2 tan() 2 dr −2θ2sina 2 2 ππππ357 ∴ φ = 0 when θ= ,,, 44 4 4 Geometrical Applications of Differentiation 181
(iii) The curve is symmetrical about the line through pole and perpendicular to the initial line if on changing θ to π – θ, the equation of the curve remains unchanged (i.e., curve involving π terms of sin or cosec only). This is also called symmetry about half ray, θ= . 2 e.g. r = a sin3θ, r = a(1 + sinθ) Also, alternately the curve is symmetrical about the line through pole and ⊥ to initial axis if on changing r to –r and θ to –θ together, the equation of the curve does not changes. e.g. rθ = a (Hyperbolic or Reciprocal Spiral), r = a sinθ (iv) The curve is symmetrical about pole (i.e. symmetrical in opposite quadrant) if on changing r to –r or θ to (π + θ), the equation of the curve remains unchanged. e.g.r2 = a2cos2θ, r = a cos2θ (r → –r)(θ → π + θ) 2. Pole (Origin): If by putting r = 0, we get real value, the curve passes through the pole otherwise it does not. Also, then that real value of θ for which r = 0, is the tangent to the curve at the pole. For example, say curve r = a(1 – cosθ). On putting r = 0 implies cosθ = 1, i.e. θ = 0, whence the initial axis is the tangent to the curve at the pole. 3. Asymptotes: Already discussed under article 2.6. 4. Limits (Regions): r and θ are confined between certain limits. For example, the curve r = a cos2θ wholly lies within the circle r = 2a since |cos2θ| ≤ 1 for all real values of θ. Find the region in which the curve does not lie. For example in case of r2 = a2cos2θ, ππ3 cos2θ is negative within interval <θ< , and therefore r is imaginary, Thus, no 44 ππ portion of the curve lies in between <θ< 3 . 44 5. Points: Find values of r for various values of a. Also, determine points where tangent coincides with the radius vector or is perpendicular to it (i.e. find points where dθ tanφ=r ⋅ = 0 or ∞). dr Example 68: Trace the curves (i) r = a (1 – cosθ)(ii) r = a (1 + cosθ).
Solution: The equation of the curve is r = a(1 – cosθ) 1. Symmetry: The curve is symmetrical about the initial line because the equation of the curve remains unaltered when θ is changed to –θ. 2. Pole or Origin: (i) When θ = 0, r = 0. Hence the curve passes through the pole and the tangent at the pole is the line θ = 0, i.e. the initial line itself. π (ii) The curve meets the initial line θ = 0 at (0, 0) and the line θ= and π at the points 2 (a, π/2) and (2a, π) respectively. 180 Engineering Mathematics through Applications
If x2 ≥ 1, i.e.x ≥ 1, or x ≤ –1 then y2 – 1 ≤ 0, i.e. y2 ≤ 1or–1 ≤ y ≤ 1. ≥ ≤ (precisely means x 1 or x –1, y lies between –1 and 1) With these all above inference, the shape of the curve becomes as follows:
Y x = y (–1, 1) (0, 1) (1, 1)
135° X´ X (–1, 0) (1, 0) O
(1, – 1) ( –1, –1) (0, –1)
Y´ Fig. 2.40
ASSIGNMENT 8 Trace the following: (i) a2x2 = y3(2a – y) [KUK, 2008] (ii) ay2 = x(a – x) 2 2 2 2 2 2 2 (iii)3ay = x(x – a) (iv) x y = a (y – x ) (v) x2 = y2(x + a)3 (vi) x2(x – a) = ay2 (vii) y2 = (x – 1)(x – 2)(x – 3) *(viii) y2(x2 + y2) + a2(x2 – y2) = 0 (ix) y2(x – a) = x2(2a – x)(x)x2y2 = x2 – a2 (xi) y = x2/(1 – x2) (xii) y2(a – x) = x2(a + x)(xiii) x3 + y3 – 3axy = 0 (xiv) y2(2a – x) = x3 (cissoid) (xv) y = logx x (xvi) yc=cosh . *[NIT Kurukshetra, 2005] c
II. Procedure for Tracing of Polar Curves: r = f (θ)
1. Symmetry (i) The curve is symmetrical about any line θ = α if on changing θ to (2α – θ), the equation of the curve remains unchanged. (ii) The curve is symmetrical about the initial line (θ = 0) on changing θ to – θ, the equation of the curve remains unchanged. In other words, curve involving terms of cosθ or secθ. e.g. r = a(1 + cosθ), r = a cos2θ Geometrical Applications of Differentiation 179
Example 67: Trace the curve x4 + y4 = x2 + y2. Solution: 1. Symmetry: (i) The curve is symmetrical about X-axis and Y-axis (as x and y occurs both in even powers only) (ii) The curve is symmetrical about the line y = x and the line y = –x (as by putting y = x and y = –x, the equation of the curve remains unaltered). (iii) Symmetrical in opposite quadrants (as by putting x to –x and y to –y, the equation of the curve remains unchanged). 2. Origin: The equation passes through the origin and for tangents at the origin, x2 + y2 = 0 (the lowest degree terms equated to zero) or y = ±ix, i.e. the tangents at the origin are isolated points. 3. Asymptotes: The curve does not possess any asymptote. 4. Points: (a) Intersection with the symmetrical lines: (i) The curve meets the X-axis in (0, 0), (1, 0) and (–1, 0). (ii) The curve intersects the Y-axis in (0, 0), (0, 1) and (0, –1). (iii) The curve cuts the symmetrical line y = x in points (1, 1) and (–1, –1). (iv) The curve cuts the symmetrical line y = –x in points (1, –1) and (–1, 1). dy dy (b) Points where = 0 and =∞ dx dx Differentiating the given curve x4 + y4 = x2 + y2 with respect to ‘x’, we get
dy dy dy xx()12− 2 44xy33+=+ 22 xy or = dx dx dx y()21 y2−
dy 1 = 0⇒ x = 0 and x =± . Here x = 0 means, tangent at (0, 1) and (0, –1) are dx 2 parallel to X-axis. dy 1 Similarly, =∞ ⇒ y = 0 and y =± . Here y = 0 means tangents at (1, 0), (–1, 0) dx 2 are parallel to Y-axis. dy Further, =−1 at (1, 1) means the tangent makes an angle of 135° with the X- dx axis or is perpendicular to the line y = x. 6 Regions: On rewriting the given equation, y4 – y2 = x2 – x4 as y2(y2 – 1) = x2(1 – x2) x2 or yx22−=11() − . y2 If x2 ≤ 1, i.e. –1 ≤ x ≤ 1, then y2 – 1 ≥ 0ory2 ≥ 1ory ≥ 1, or y ≤ –1 (Precisely mean, when x lies between –1 and 1, y ≥ 1 or y ≤ 1) 178 Engineering Mathematics through Applications
Likewise, taking negative sign in (2), the equation of the asymptote becomes y = –x + 3a …(5) −11a2 where, in, yy−= , and the curve lies below the asymptote y = –x + 3a for x ca 2x positive and above the asymptote y = –x + 3a for x negative. 4. Points: (i) Intersection with X-axis: put y = 0 in the given equation giving x = 0, a, 2a. Thus the curve meets the X-axis at O(0, 0), A(a, 0), B(2a, 0). (ii) Intersection with Y-axis: The curve meets the Y-axis at the origin where tangents are parallel to X-axis: Tangent at (a, 0) by the origin shift property, puting x to (x – a) in the given equation of the curve, then equating the lowest degree term equal to zero, i.e. see ax = 0 means Y-axis is tangent at point (a, 0). Similarly Y-axis is tangent at (2a, 0). 5. Regions: On solving for y, taking positive value of the square root. xx()()−− a x2 a y= , ()xa+3 Case: (i) When x < –3a, y is real (ii) When –3a < x < a, y is imaginary (iii) When 0 < x < 0, y is real (iv) When a < x < 2a, y is imaginary (v) When x > 2a, y is real See the shape of the curve as Fig. 3.39. Y
(0, 3a ) y = – x + 3 a
(2a , 0) B (3a , 0) X´ X O A
a 3 – x y = (0, –3a )
Y´ Fig. 3.39 Geometrical Applications of Differentiation 177
Example 66: Trace the curve y2(x + 3a) = x(x – a)(x – 2a). Solution: 1. Symmetry: The curve is symmetrical about X-axis as in the given equation of the curve, y occurs in even powers only . 2. Origin: The curve passes through the origin and the tangent at the origin is x = 0, i.e., Y-axis 3. Asymptotes: x + 3a = 0 or x = –3a is the asymptote parallel to Y-axis. Position of the curve with respect to curve, we see xx()()−− a x2 a that xa+=3 y2 − 3 = 60a y2 (if we take x = –3a) …(1) which is negative when y > 0 or y < 0. Whence the curve lies to the left of the asymptote x = –3a. Further for oblique asymptotes, xx()()−− a x2 a y2= ()xa+3 −1 aaa23 yx22=−11 − 1 + xx x 11− 1 aaa2223 2 or yx=±11 − − 1 + xx x 22 =± −111aa − +… − 12112 a − a … × x11 2xx 4 2! 2 x 4 2! x
2 −+13aa 313 … 1 242!xx 2 =± −311aa + +… x1 xx22 2 =±−++…11a or yxa3 …(2) 2x Taking positive sign, equation of one branch is 11a2 yx=−3 a + +… …(3) 2x ∴ The equation of the asymptote is y = x – 3a …(4) 11a2 As ()yy−= is positive for large values of x, if x > 0 and negative for large values of ca 2x x, if x < 0. Thus the curve lies above the asymptote y = x – 3a for x > 0 and below the asymptote for x < 0. 176 Engineering Mathematics through Applications
2. Origin: The curve passes through the origin and at the origin, x2y = 0, i.e. x = 0, x = 0, y = 0 ∴ Y-axis is a cuspidal point. 3. Asymptotes: There is no asymptote parallel to the axes. φ 5 φ For oblique asymptote, 5(m) = 1 + m , 4(m) = 0 φ 5 5(m) = 0 implies m + 1 = 0 or m = –1 (as the real root) 0 Now for c, cφ' (m) + φ (m) = 0, i.e. c ==0 5 4 5m4 Therefore, y = –x or x + y = 0 is the equation of the asymptote. 4. Position of the curve with respect to the asymptote In the second quadrant x is negative and y is positive. Therefore x2y is positive which implies that (x5 + y5) must also be positive, i.e. greater than 0, whence y must numerically be greater than x. Thus, in the 2nd quadrant, the curve lies above the asymptote y = –x. Because of symmetry, in 4th quadrant, the curve approaches the other end from below the asymptote y = –x. 5. Points: (i) Points of intersection with axes: (0, 0) is the only point of intersection. (ii) Intersection with the line y = x: With y = x, the given equation becomes 5 2x5 = 5a2x3 implying x = 0 and xa=± 2
5 ∴ y = 0 and xa=± 2 ±±55 Hence intersection with the line y = x are (0, 0) and aa, 22 (iii) Intersection with the asymptote, y = –x. (0, 0) is the point of intersection with y = –x. 6. Regions: Y x y = On transforming to polar coordinates with = y – x = r cos θ, y = r sin θ, we get the equation of curve x as: 5aa, 5 5cossina22θθ 2 2 r2= cos55θ+ sin θ X 3π 5a, – 5 a Clearly, for values of θ between and π, r2 is 2 2 4 negative and so r is imaginary. Thus, no portion of the curve lies in between θ = 3π/4 to θ = π. Fig. 2.38 However, above things have already been stated in step 4. Geometrical Applications of Differentiation 175
2. Origin: The curve passes through the origin (as there is no constant term in the equation of the curve) and the tangents at the origin are 3ax2 = 0, i.e. x = 0, 0. Thus Y-axis is the tangent (a cusp.) at the origin. φ 3 φ ⇒ 2 3. Asymptotes: Here 3(m) = 1 + m , 3(m) = 0 (m + 1)(m – m + 1) = 0, i.e.m = –1 φ()m −3a ca=−2 =− = and φ′() 2 3 m3m Hence y = –x + a, i.e. x + y – a = 0 is the oblique asymptote. This asymptote intersects the axis at (a, 0) and (0, a). 4. Special Points: The given curve is written as y3 = x2(3a – x) …(1) Thus y = 0 (i.e. intersection with X-axis) gives x2(3a – x) = 0, i.e. x = 0 and x = 3a Further, y = x2/3 (3a – x)1/3 …(2)
− dy 21−1212 i.e. =−−−xax33()3333() axx dx 33 12 ()−−33 =23ax xx 2 1 33()ax− 3 x3 ) −− , a (2a , 22/3a) =62axx (0 2 31 33()ax− x3 (,a 0) ()− O (3a , 0) =2ax (0, 0) 1 2 …(3) x 3()−3 + xax3 y – a dy = Now = 0 when x = 2a 0 dx Fig. 2.37 dy and =∞ when x = 0 and x = 3a dx From (2), x = 2a, gives y = (2a)2/3 (3a – 2a)2/3 = (2a)2/3 a1/3 = 22/3a Hence at the point (2a, 22/3a) the tangent to the curve is parallel to X-axis whereas at (0, 0) and (3a, 0), the tangent to the curve is parallel to Y-axis. 5. Regions: From the given equation y3 = x2(3a – x), it is clear that (i) When x is –ve, y is +ve as x decreases from 0 to –∞, y increases from 0 to ∞ (ii) When x is +ve and lies between 0 and 3a, y is also +ve. (iii) When x > 3a, then y is –ve. This is possible only if the curve crosses the line x = 3a which is tangent to the curve at (3a, 0). Therefore (3a, 0) is a point of inflexion, as x → ∞, y → ∞. Example 65: Trace the curve x5 + y5 = 5a2x2y. Solution: 1. Symmetry: The curve is symmetrical in opposite quadrant because the equation of the curve remains unchanged on replacing x by –x and y by –y. 174 Engineering Mathematics through Applications
5. Regions: From the given curve, ax3 y=± ()()axax+− (i) For x → –∞, y → ∞ (ii) –a < x < 0, y is imaginary (iii) 0 < x < a, y is increasing and tends to infinity as x → a (iv) x > a, y is imaginary. ()x2+1 Example 63: Trace the curve y= ()x2–1 [KUK, 2007, 2009] Solution: 1. Symmetry: The given curve y(x2 – 1) = (x2 + 1) is symmetrical about Y-axis only (as it contain even powers of x) 2. Origin: The curve does not pass through the origin. 3. Asymptotes: Asymptote parallel to Y-axis is given by x2 – 1 = 0, i.e., the straight line x = ±1 and asymptotes parallel to X-axis is given by equating to zero the coefficients of x2, viz. y – 1 = 0 or the straight line y = 1. The curve does not possess any oblique asymptote. 4. Points: The curve intersects the Y-axis at (0, –1). ()()xxxx22−−+12 12 dy = Here 2 dx ()x2−1 Y x = –1 x = 1 211xx−−−22 x − ==4x 22 ()()xx+−11()() xx +− 11 y = 1 dy y = 1 = 0 implies x = 0 for which y = –1 O (0, 0) dx –X X Hence at the point (0, –1), the tangent to the curve is parallel to the axis of X, whereas, (0, –1) dy =∞ implies x = 1, –1 for which y becomes dx infinity, i.e. these lines are asymptotes parallel to Y-axis which has already been proved. –Y Fig. 2.36 xy==−∞0 to –1, 1 to – ; 5. Regions: For , (Since for –1 < x < 1, x2 – 1 < 1) xy==−∞0to1, 1to– ;}
xy=−1to −∞ , →∞ For xy=∞→∞1to , }
Example 64: Trace the curve x3 + y3 = 3ax2. Solution: 1. Symmetry: It has none of the symmetries.