US01CMTH02 UNIT Curvature

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1 Stu mteril of BSc(Semester - I) US1CMTH (Rdius of Curvture nd Rectifiction) Prepred by Nilesh Y Ptel Hed,Mthemtics Deprtment,VPnd RPTPScience College US1CMTH UNIT- 1 Curvture Let f : I R be sufficiently mny times differentible function on n intervl I Then the points on the grph of y = f(x) is curve However, not ll curves could be represented s grph of such rel vlued function on intervls viz, the figure of circle with centre (,) nd rdius 1 in the XY -plne R is one such exmple of curve In this sitution, we hve to represent the eqution of the circle s x = cos t; y = sin t, t [, π] These re clled the prmetric equtions of the circle Also, let us think of spring put in R 3 Then the points of this spring is curve Thus formlly we hve the following definition of curve 11 Definition Let I be closed intervl nd x = x(t), y = y(t) nd z = z(t) be relvlued differentible functions defined on I Then the points (x(t), y(t), z(t)) in the spce is clled locus of the curve represented by the prmetric equtions x = x(t), y = y(t), z = z(t), t I Throughout this chpter we shll be concerned only with curves lying in the XY - plne For such curves we hve z = Hence they re described by x = x(t) nd y = y(t) A curve lying only in one plne is clled plner curve 1 Definition Let x = x(t), y = y(t) be curve If we eliminte t nd obtin reltion g(x, y) =, then this form is clled the crtesin representtion of the curve Further, if g(x, y) = cn be written in the form y = f(x) (respectively, x = f(y)), then y = f(x) (respectively, x = f(y)) is clled the crtesin eqution of the curve 13 Exmple Let I = [, 1] nd x = t, y = t Then this is curve tht cn lso be represented by the crtesin eqution y = x 14 Definition Let y = f(x) be curve Fix point A on this curve For point P on the curve, let s = rc AP be the rc length from A to P For point Q on the curve, 1

2 let s + s = rc AQ so tht s = rc P Q Let l 1, l be the tngents to the curve t the points P nd Q mking ngles ψ nd ψ + ψ respectively, with fixed line in the plne Y Clerly, the ngle between these two tngents is ψ, clled the totl bending or l totl curvture of the rc be- tween P nd Q Hence the verge bending or the verge curvture of the curve between these two points reltive to the rc length is given by ψ The bending or s the curvture of the curve t P is defined to be dψ = lim ψ = lim ψ QP s s s Derivtive of n rc O ψ A s Figure 14 P Q s ψ + ψ 1 Proposition Fix point A(x, y ) on curve given by y = f(x) For point P (x, f(x)) on the curve, let s be the rc length of rc AP (Clerly, s is function of x) Then prove tht = 1 + ( ) ψ l 1 X

3 Derivtive of n rc 3 Proof Let y = f(x) represent the given curve nd A be fixed point on it Let P (x, y) be generic point on the curve Let the rc AP = s Tke point Q(x + x, y + y) on the curve ner to P Let rc AQ = s + s From the right ngled tringle P NQ, we hve, Y P Q = P N + NQ = ( x) + ( y) ( ) ( ) P Q y = 1 + x x ( ) ( ) ( ) chord P Q s y = 1 + rc P Q x x Tking Q P, we get chord P Q rc P Q Hence, ( ) = 1 + ( ) = A s P x s y O L M X 1 + ( ) Figure 1 Q N The proof of the following corollry is left to the reder Corollry Let x = x(t) nd y = y(t) be the prmetric equtions of curve Then ( ) ( ) dt = + dt dt 3 Exercise In the Proposition 1, suppose tht the curve is represented by x = f(y) ( ) Then deduce tht the derivtive of the rc length 1 = + 4 Definition Let h(x, y) = be crtesin representtion of curve By substituting x = r cos θ nd y = r sin θ, in this form we get representtion g(r, θ) = of the curve clled polr representtion of the curve We shll be minly deling with the form r = f(θ) of the curve 5 Theorem For polr eqution r = f(θ) of curve, ( ) dr = r +

4 4 Proof Let r = f(θ) represent the given curve nd A be fixed point on it Let P (r, θ) be generic point on the curve Let rc AP = s Tke point Q(r + r, θ + ) on the curve ner to P Let rc AQ = s + s From the right ngled tringle ONP s shown in figure, we hve, nd Also, from the figure, sin = P N OP = P N r cos = ON OP = ON r Y Q(r + r, θ + ) r + r N θ O X P N = r sin ON = r cos r Figure 5 s P (r, θ) s A NQ = OQ ON = r + r r cos = r(1 cos ) + r = r sin + r Now from the right ngled tringle P NQ, we hve, P Q = P N + NQ P Q = r sin + (r sin + r) ( ) ( ) [ ( ) ( P Q sin = r sin ( )) + r sin ( ) ( ) ( ) chord P Q rc P Q sin = r rc P Q [ ( ) ( sin ( )) + r sin Tking Q P, we get chord P Q rc P Q Hence, ( ) = r + ( ) dr = r + ( ) dr ] + r ] + r

5 3 Rdius of curvture 5 6 Exmple For the curve r n = n cos nθ, prove tht = (sec nθ) n 1 n Solution Here r n = n cos nθ Tking log on both the sides, By differentiting this we get, n dr r 7 Exercise n log r = n log + log(cos nθ) sin nθ = n cos nθ r 1 = dr = r tn nθ r + r1 = r (1 + tn nθ) = r sec nθ = r + ( dr ) = r sec nθ = (cos nθ) 1 n sec nθ = (sec nθ) n 1 n 1 Show tht curvture of circle is constnt nd is equl to the reciprocl of its rdius Show tht curvture of stright line is zero 3 Find for the following curves (i) y = cosh x ( (ii) y = log 4 Find for the following curves dt (i) x = (t sin t); y = (1 cos t) (ii) x = (cos t + t sin t); y = (sin t t cos t) (iii) x = e t sin t; y = e t cos t 5 Find for the following curves (i) r = (1 cos θ) (ii) r = cos θ 3 Rdius of curvture x ) 31 Definition Let P be point on curve such tht the curvture of the curve t P is nonzero Then the rdius of the curvture t P is defined to be the reciprocl of the curvture t P nd is denoted by ρ Tht is, ρ = dψ 3 Theorem Let y = f(x) be curve nd P be point on it Then prove tht the rdius of curvture t P is given by where y 1 = nd y = d y ρ = (1 + y 1) 3 y,

6 6 Proof Let y = f(x) be the given curve Then tn ψ = Differentiting with respect to s, we get, sec ψ dψ = d (1 + tn ψ) dψ = y (1 + y 1) dψ = y ρ = dψ = 1 + y 1 ρ = 1 + y 1 y y ( ) = d 1 + y 1 ρ = (1 + y 1) 3/ y ( ) = y 33 Theorem Let r = f(θ) be polr form of curve with point P on it Then prove tht the rdius of curvture t P is given by where r 1 = f (θ) nd r = f (θ) ρ = (r + r 1) 3/ r + r 1 rr, Proof From the figure it is cler tht ψ = θ + φ Hence, dψ = + dφ = + dφ = ( 1 + dφ ) (331) We know tht tn φ = r r 1 Differentiting this with respect to θ, we get, sec φ dφ = r 1 rr r 1 dφ = r 1 rr r tn φ = r 1 rr r 1 O 1 θ 1 + r r 1 r Figure33 = r 1 rr r 1 + r We lso know tht = r + r1 Hence by (331), we get, ( ) dψ = r 1 rr = r + r1 rr r + r1 r1 + r (r + r1) 3/ ψ φ P (r, θ)

7 3 Rdius of curvture 7 Hence, ρ = dψ = (r + r 1) 3/ r + r 1 rr 34 Exmple Prove tht if ρ is the rdius of curvture t ny point P on the prbol y = 4x nd S is its focus, then prove tht ρ SP 3 Solution Let P (x, y) be ny point on the give prbol If the coordintes of the focus S is given by (, ), then SP = (x ) + y = x x + + 4x = x + Now we find ρ for the given prbol y = 4x Here yy 1 = 4 Tht is, y 1 = Also, y y = y y 1 = 4 Hence, y 3 ρ = (1 + y 1) 3/ y This proves tht ρ SP 3 = (1 + 4 y ) 3/ = (y + 4 y 3 4 ρ = (4x + 4 ) = 643 (x + ) = 4 )3/ 4(x + )3 = 4 SP 3 35 Exmple Show tht the rdius of curvture t ny point of the curve x = e θ (cos θ sin θ), y = e θ (sin θ + cos θ) is twice the perpendiculr distnce of the tngent t the point form the origin Solution Here = eθ (cos θ sin θ) + e θ ( sin θ cos θ) = e θ sin θ Similrly, = eθ cos θ Hence y 1 = = cot θ nd y = cosec θ = cosec3 θ e θ Thus, ρ = (1 + y 1) 3/ y = (1 + cot θ) 3/ ( cosec 3 θ e θ ) = e θ Now the eqution of the tngent t point is y e θ (sin θ + cos θ) = (x eθ (cos θ sin θ)) y e θ (sin θ + cos θ) = cot θ(x e θ (cos θ sin θ)) y sin θ + x cos θ e θ = Hence the length of the perpendiculr distnce of the tngent from the origin is e p = θ = e θ Hence ρ = p cos θ+sin θ

8 8 36 Exmple For the cycloid x = (θ + sin θ), y = (1 cos θ) prove tht ρ = 4 cos( θ ) Also show tht ρ 1 + ρ = 16, where ρ 1, ρ re the rdii of curvture t the points where the tngents re perpendiculr Solution = (1 + cos θ), = sin θ Therefore, sin θ y 1 = (1 + cos θ) = sin ( θ ) ( cos θ ) = tn θ cos θ ( ) ( ) y = 1 sec θ = 1 1 = cos θ cos θ Hence, 1 4 cos 4 θ ρ = (1 + ( y 1) 3/ = 1 + tn θ ) 3/ 4 cos 4 θ y = 4 θ θ sec3 cos4 = 4 cos( θ) If P (θ 1 ) nd Q(θ ) re the points t which the Y tngents re perpendiculr, then ρ 1 = 4 cos( θ 1 ) nd ρ = 4 cos( θ ) If the tngents t these points mke the ngles ψ 1 nd ψ with the X-xis respectively, then P Q tn ψ 1 = = tn θ 1 ψ Therefore, ψ 1 = θ 1 ψ 1 O But X ψ 1 ψ = π Therefore, θ 1 + θ = π Hence, Figure 36 [ ( ρ 1 + ρ = 16 cos θ 1 π + cos θ )] 1 = 16 [ ] cos θ 1 + sin θ 1 = Exmple For the curve r = (1 cos θ), prove tht ρ r Also prove tht if ρ 1 nd ρ re rdii of the curvture t the en of chord through the pole, ρ 1 + ρ = 16 9 Solution Here r 1 = sin θ nd r = cos θ Hence, ρ = (r + r1) 3/ r + r1 rr ( (1 cos θ) + sin θ) 3/ = (1 cos θ) + sin θ (1 cos θ) cos θ = ( (1 cos θ)) 3/ 3 (1 cos θ) = (4 sin θ )3/ 6 sin θ

9 5 Length of n rc 9 Thus, = 4 3 sin θ ρ = 16 9 sin θ = 8 9 (1 cos θ) = 8r 9 ρ r Let P (r 1, θ 1 ) nd P (r, θ ) be the en of the chord through the pole Then θ θ 1 = π Then ρ i = 16 9 sin θ i, (i = 1, ) Hence ρ 1 + ρ = 16 ( 9 sin θ 1 + θ ) sin = 16 [ 9 sin θ ( )] 1 π + + θ1 sin = 16 ( 9 sin θ 1 + θ ) cos 1 = 16 9 Rectifiction 4 Derivtive of n rc: Revisited Rectifiction is the process of computing the length of n rc of curve The curves my hve different representtions like crtesin, polr nd prmetric So, we shll be deling with ll the three forms Besides, the curve could be expressed s combintion of rcs of two different curves yielding new closed curve In this cse, the length of rc will be its perimeter The ide of finding the length of rc is simple We hve obtined the derivtive of n rc in Section erlier It is the derivtive of the length of rc s with respect to the independent vrible If we integrte the sme, we shll get the length of rc A curve is sid to be rectifible if it is possible to find its length 5 Length of n rc of curve 51 Theorem Let y = f(x) be crtesin representtion of curve C Then the length of rc of C between two points A nd B corresponding to the x-coordintes nd b respectively, is given by b ( ) rc AB = 1 + Proof Let s(x) be the length of rc of curve between fixed point A on the curve nd the generic point P (x, f(x)) Then integrting (??) from to b, we hve, b ( ) b 1 + =

10 1 = b = [ s ] b = s(b) s() = rc AB rc AA = rc AB 5 Theorem Let r = f(θ) be polr representtion of curve C Then the length of rc of C between two points A nd B corresponding to the ngles θ = θ nd θ = θ 1 respectively, is given by θ 1 ( ) dr rc AB = r + θ 53 Exmple Find the length of rc of the prbol y = 4x, ( > ), mesured from the vertex to one extremity of its ltus rectum Solution We cn write the given eqution s x = y ( ) = 1 + y 4 = 1 y + 4 Then = y Therefore, From the figure, we see tht coordintes of the vertex O nd top end of the ltus rectum L re (, ) nd (, ) respectively Hence the required length of rc is O L(, ) (, ) L (, ) X rc OL = = 1 = 1 = ( ) y + 4 Y Figure 53 [ y y log(y + ] y + 4 ) [ + log( + ) log ]

11 5 Length of n rc Exmple [ ( (1 + )] ) = + log ( ) = + log(1 + ) () Find the entire length of the stroid x /3 + y /3 = /3 (b) Prove tht the length of the curve x /3 +y /3 = /3 mesured from (, ) to the point (x, y) is given by 3 (x ) 1/3 Solution () B(, ) P (x, y) O A(, ) Here, Figure 54 x /3 + y /3 = /3 3 x 1/3 + y 1/3 3 = x 1/3 = = y1/3 y 1/3 x 1/3 1 + ( ) = 1 + y/3 From the figure, the entire length of the stroid is 4 rc AB = = 4 /3 = x/3 x = /3 /3 x /3 ( ) 1/3 x 1/3 = 4 1/3 [ x /3 /3 ]

12 1 ( ) = 4 1/3 /3 /3 = 6 Since the length of n rc is lwys positive, we infer tht the entire length of the stroid is 6 (b) The required rc length = rc BP = = x x 1 + ( ) 1/3 x 1/3 = 1/3 [ x /3 /3 1/3 x/3 = /3 = 3 1/3 x /3 = 3 (x ) 1/3 55 Exmple Show tht the entire length of the curve x ( x ) = 8 y is π ] x Solution The given curve is symmetric bout ll X-xis, Y - xis nd the origin Putting y =, we get x {, ±} Also since y = x ( x ) 8, we hve y = x x Hence x is the only possibility for getting y rel The shpe of the given curve is s shown in the figure It contins two equl loops Now, (, )B Y X O A(, ) Figure 55 8 y = x ( x ) 16 y = x( x ) + x ( x)

13 5 Length of n rc 13 From the figure (55), we sy tht = x( x ) 8 y ( ) ( x( x ) 1 + = y the entire length = 4 rc OA = ) = 1 + x ( x ) 8 x ( x ) = 84 8 x x + 4x 4 8 ( x ) = (3 x ) 8 ( x ) ( ) = 4 = 4 3 x x [ ] x + x x [ ] = 4 x + x [ ( x ) = 4 sin ( x ( x + x ) )] sin 1 [ ( x ) = 4 sin 1 + x ] x = 4 π = π 56 Exmple Find the length of the crdioid r = (1 + cos θ) lying outside the circle r = cos θ

14 14 Solution First we find the ngle between two curves t the point of their intersection By compring them, we get (1 + cos θ) = cos θ cos θ = 1 θ = π ± π 3 θ = π, 4π From the figure, 3 3 we see tht the given curve is symmetric bout the polr xis nd the required rc length is rc ABC = rc BA Now for the curve r = (1 + cos θ), r + Hence the required rc length θ = π θ = π 3 θ = 4π 3 Figure 56 ( ) dr = (1 + cos θ) + sin θ = (1 + cos θ) = 4 cos θ rc BA = = π/3 π/3 r + cos( θ ) A C ( ) dr O B θ = = 8 [ sin θ ] π/3 = 8 ( sin π 3 ) = Intrinsic eqution 61 Definition Let A be fixed point on curve C nd P be generic point on the curve Let ψ(p ) denote the ngle between the tngents to the curve t points A nd P Also, let s = rc AP Then the reltion between s nd ψ is clled the intrinsic eqution of the curve It is customry to fix origin (or pole) s the fixed point A if it lies on the curve Otherwise we mention the fixed point explicitly We follow this convention throughout this section including exercise Now we obtin the intrinsic equtions of the curve represented in different forms

15 6 Intrinsic eqution 15 Let A(, b) be fixed point nd P (x, y) be generic point on the curve y = f(x) We develop the eqution in prticulr cse when the tngent to the curve t A is prllel to the X-xis Then nd s = I Crtesin form x 1 + Y O ( A ψ ψ P X Figure 61 ) (611) tn ψ = (61) Eliminting x from (611) nd (61) we get reltion F (s, ψ) =, which is the intrinsic eqution of the curve in crtesin form If the curve is represented in the form x = f(y) or in prmetric form, then the intrinsic eqution cn be obtined similrly by eliminting y or the prmeter t respectively However, in the polr form, the coordintes re chnged, so we give intrinsic eqution in this cse seprtely Let A(r 1, θ 1 ) be fixed point nd P (r, θ) be generic point on the curve r = f(θ) We develop the eqution in prticulr cse when the tngent to the curve t A is prllel to the polr xis Then Now from the figure, s = II Polr form θ θ 1 r + O θ A Figure 61 ( ) dr (613) ψ = θ + φ, (614) where φ is the ngle between the rdius vector nd the tngent t point P We lso know tht tn φ = r dr (615) Eliminting φ nd θ from (613), (614) nd (615) we get which is the intrinsic eqution of the curve in polr form F (s, ψ) =, (616) ψ ψ P φ

16 16 6 Exmple Find the intrinsic eqution of the Crdioid r = (1 + cos θ) Hence prove tht s + 9ρ = 16, where ρ is the rdius of curvture t ny point of the curve Solution Here r = (1 + cos θ) Therefore, dr tn φ = r = 1+cos θ dr sin θ = cos θ sin θ cos θ = sin θ Hence = cot θ = tn ( π + θ ) Hence, φ = π + θ Also, Now, ψ = θ + φ = θ + π + θ = 3θ + π (61) s = = θ θ 1 θ r + ( ) dr (1 + cos θ) + sin θ = θ θ = θ = (1 + cos θ) 4 cos θ cos θ = [ sin θ ] θ = 4 sin θ ( ψ = 4 sin 3 π ), (by (61)) (6) 6 which is the required intrinsic eqution By differentiting (6), we hve ρ = = dψ 4 cos ( ψ ) ( π Hence 3ρ = 4 cos ψ ) π 3 6 So, s + 9ρ = Exmple Show tht the intrinsic eqution of the curve y 3 = x, is 7s = 8(sec 3 ψ 1) Solution We cn write the given eqution s x = 1 y 3/ Then = 3 y Here the tngent to the curve t the origin is Y -xis Therefore, tn ψ =, Tht is, tn ψ = 3 y (631)

17 6 Intrinsic eqution 17 Now, s = y y 1 + ( ) = 1 + 9y 4 [ ( = y ) ] 3/ y [ ( 7s = y ) 3/ 1] 4 7s = 8 [ (1 + tn ψ) 3/ 1 ] 7s = 8(sec 3 ψ 1) Y O ψ P Figure 63 X

JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 6 (First moments of an arc) A.J.Hobson

JUST THE MATHS UNIT NUMBER INTEGRATION APPLICATIONS 6 (First moments of an arc) A.J.Hobson JUST THE MATHS UNIT NUMBER 13.6 INTEGRATION APPLICATIONS 6 (First moments of n rc) by A.J.Hobson 13.6.1 Introduction 13.6. First moment of n rc bout the y-xis 13.6.3 First moment of n rc bout the x-xis