Radius of Curvature - Differential Calculus

UNIT

1

Differential Calculus—I 1.1 INTRODUCTION In many practical situations engineers and scientists come across problems which involve quantities of varying nature. Calculus in general, and differential calculus in particular, provide the analyst with several mathematical tools and techniques in studying how the functions involved in the problem behave. The student may recall at this stage that the derivative, obtained through the basic operation of calculus, called differentiation, measures the rate of change of the functions (dependent variable) with respect to the independent variable. In this chapter we examine how the concept of the derivative can be adopted in the study of curvedness or bending of curves.

1.2 RADIUS OF CURVATURE Let P be any point on the curve C. Draw the tangent at P to the circle. The circle having the same curvature as the curve at P touching the curve at P, is called the circle of curvature. It is also called the osculating circle. The centre of the circle of the curvature is called the centre of curvature. The radius of the circle of curvature is called the radius of curvature and is denoted by ‘ρ’.

Y

C O P

Note : 1. If k (> 0) is the curvature of a curve at P, then the radius

1 . This follows from the definition k of radius of curvature and the result that the curvature of a circle is the reciprocal of its radius.

O

of curvature of the curve of ρ is

Note : 2. If for an arc of a curve, ψ decreases as s increases, then But the radius of a circle is non-negative. So to take ρ = i.e., k =

dψ . ds 1

X

Fig. 1.1

dψ is negative, i.e., k is negative. ds

1 ds = some authors regard k also as non-negative dψ k

2

ENGINEERING MATHEMATICS—II

dψ indicates the convexity and concavity of the curve in the neighbourhood of ds ds the point. Many authors take ρ = and discard negative sign if computed value is negative. dψ The sign of

∴ Radius of curvature ρ =

1 · k

1.2.1 Radius of Curvature in Cartesian Form Suppose the Cartesian equation of the curve C is given by y = f (x) and A be a fixed point on it. Let P(x, y) be a given point on C such that arc AP = s. Then we know that

dy = tan ψ dx where ψ is the angle made by the tangent to the curve C at P with the x-axis and ds = dx

R|1 + FG dy IJ S| H dx K T

2

Differentiating (1) w.r.t x, we get

U| V| W

...(1)

1 2

...(2)

dψ d2y sec 2 ψ ⋅ 2 = dx dx

d

2 = 1 + tan ψ

i ddsψ ⋅ dxds

LM1 + FG dy IJ OP 1 LM1 + FG dy IJ OP MN H dx K PQ ρ MN H dx K PQ 1 R| F dy I U| S1 + G J V ρ |T H dx K |W R|1 + FG dy IJ U| S| H dx K V| W T

1 2 2

2

=

2

=

2

Therefore,

where y1 =

ρ=

dy d2y and y2 = . dx dx 2

d2y dx 2

[By using the (1) and (2)]

3 2

3 2

...(3)

3

DIFFERENTIAL CALCULUS—I

Equation (3) becomes, ρ=

3 2 2 1

o1 + y t y2

This is the Cartesian form of the radius of curvature of the curve y = f (x) at P (x, y) on it.

1.2.2 Radius of Curvature in Parametric Form Let x = f (t) and y = g (t) be the Parametric equations of a curve C and P (x, y) be a given point on it. Then

dy dt dy = d x dt dx

...(4)

RS T

UV W

d2y d dy / dt dt ⋅ 2 = dt dx / dt dx dx

and

dx d 2 y dy d 2 x ⋅ − ⋅ dt dt 2 dt dt 2 ⋅ 1 = 2 dx dx dt dt

FG IJ H K

dx d 2 y dy d 2 x – ⋅ ⋅ d2y dt dt 2 dt dt 2 = 3 dx 2 dx dt

FG IJ H K

Substituting the values of

...(5)

dy d2y and in the Cartesian form of the radius of curvature of the dx dx 2

curve y = f (x) [Eqn. (3)]

∴

ρ=

=

R|1 + FG dy IJ o1 + y t = S|T H dx K 3 2 2 1

d2y dx 2

y2

|RS1 + FG dy / dt IJ |T H dx / dt K 2

2

2

3 2

3 2

3

2

R|FG dx IJ + FG dy IJ U| S|H dt K H dt K V| W T 2

ρ=

U| V| W

|UV |W RS dx ⋅ d y – dy ⋅ d x UV / FG dx IJ T dt dt dt dt W H dt K 2

∴

2

2

3 2

dx d 2 y dy d 2 x ⋅ − ⋅ dt dt 2 dt dt 2

...(6)

4

ENGINEERING MATHEMATICS—II 2 2 dx dy , y′ = , x″ = d x , y″ = d y 2 dt dt dt dt 2

where x′ =

ρ=

o

x′ 2 + y′ 2

t

3 2

x ′ y ″ – y′ x ″

This is the cartesian form of the radius of curvature in parametric form.

WORKED OUT EXAMPLES 1. Find the radius of curvature at any point on the curve y = a log sec Solution Radius of curvature ρ = Here,

o

t

1 + y12

3 2

y2

y = a log sec

FG x IJ H aK

1

y1 = a ×

sec

FG x IJ H aK

⋅ sec

FG x IJ tan FG x IJ ⋅ 1 H aK H aK a

FG x IJ H aK F xI 1 = sec G J ⋅ H aK a

y 1 = tan y2

2

RS1 + tan FG x IJ UV H aKW T ρ = 1 F xI sec G J H aK a RSsec FG x IJ UV T H aKW = = 1 F xI sec G J H aK a 3 2

2

Hence

2

32

2

2

∴

Radius of curvature = a sec

FG x IJ H aK

FG x IJ H aK F xI sec G J H aK F xI = a sec G J H aK a sec 3 2

FG x IJ . H aK

5

DIFFERENTIAL CALCULUS—I

FG x IJ , show that ρ = H cK d1 + y i ρ =

2. For the curve y = c cos h

y2 · c

3 2 2 1

Solution

y2

FG x IJ H cK F xI 1 F xI y = c sin h G J × = sin h G J H cK c H cK F xI 1 y = cos h G J × H cK c RS1 + sin h FG x IJ UV c FG cos h x IJ H cKW = H cK T ρ = x 1 F xI cos h cos h G J c H cK c F xII F xI 1 F = c cos h G J = G c cos h G J J H cKK H cK c H y = c cos h

Here,

1

and

2

3 2

2

2

3 2

2

2

=

1 2 ⋅y c

y2 ρ = · c

∴

Hence proved.

3. Find the radius of curvature at (1, –1) on the curve y = x2 – 3x + 1.

d1 + y i ρ =

3 2 2 1

Solution. Where

y2

Here, Now,

( y1)(1, ( y2)(1,

∴

ρ(1,

at (1, – 1)

y = x2 – 3x + 1 y 1 = 2x – 3, y2 = 2 –1) = – 1 –1) = 2

–1)

b1 + 1g =

3 2

2

=

=

2 2 2

2

4. Find the radius of curvature at (a, 0) on y = x3 (x – a).

Solution. We have

ρ =

d

1 + y12 y2

i

3 2

at (a, 0)

6

ENGINEERING MATHEMATICS—II

y = x3(x – a) = x4 – x3a y 1 = 4x3 – 3ax2 y 2 = 12x2 – 6ax ( y1)(a, 0) = 4a3 – 3a3 = a3 ( y2)(a, 0) = 12a2 – 6a2 = 6a2

Here, and Now

RS1 + da i UV T W = 2

3

∴

ρ(a,

0)

6a 2

o

1 + a6

=

6a

2

t

3 2

d

1 + y12

Solution. We have

ρ =

Here

y = a sec y1 y1

and

y2

i

· πa on y = a sec 4

5. Find the radius of curvature at x =

∴

3 2

3 2

at x =

y2

πa 4

FG x IJ H aK F xI F xI 1 = a sec G J ⋅ tan G J × H aK H aK a F xI F xI = sec G J tan G J H aK H aK x 1 F xI F xI 1 × + sec G J ⋅ tan G J ⋅ = sec H aK H aK a a a 1L F xI F xI F xIO = Msec G J + sec G J tan G J P H aK H aK H aKQ aN 3

2

3

and y2 =

2

π π πa , y = sec ⋅ tan = 2 1 4 4 4

At x =

e

j

1 3 2 2 2+ 2 = a a

RS1 + e 2 j UV W T 2

∴

ρ

x=

πa 4

FG x IJ . H aK

=

=

3 2 a

3 a. 2

3 2

=

3 3 3 2

⋅a

7

DIFFERENTIAL CALCULUS—I

FG x IJ . H 2K

π on y = 2 log sin 3

6. Find ρ at x =

Solution. We have ρ =

d1 + y i

3 2 2 1

y2

y = 2 log sin

The curve is

at x =

FG x IJ H 2K

1

× cos

sin

y2

At

x=

π ,y 3 1

and

2

y2 =

–1 π cosec 2 = – 2 2 6

=

RS1 + e 3j UV T W

=

b1 + 3g

2

∴

ρ

x=

π 3

Solution. We have ρ =

3 2

–2 3 2

–2

=

o

1 + y12

t

y2

4×2 = – 4. –2

FG 3a , 3a IJ H 2 2K

7. Find the radius of curvature at

x3

FG x IJ × 1 H 2K 2

FG x IJ H 2K F xI = cot G J H 2K F xI 1 = – cosec G J × H 2K 2 F πI = cot G J = 3 H 6K

y1 = 2 ⋅

and

π 3

3 2

at

on x3 + y3 = 3axy.

FG 3a , 3a IJ . H2 2K

y3

+ = 3axy Here, Differentiating with respect to x 3x2 + 3y2 y1 = 3a (xy1 + y) 3 ( y2 – ax) y1 = 3 (ay – x2) ⇒

y1 =

ay – x 2 y 2 − ax

...(1)

8

ENGINEERING MATHEMATICS—II

Again differentiating w.r.t x. ⇒

y2 =

Now, from (1), at

dy

2

ib

g d

− ax ⋅ ay1 − 2 x − ay − x 2

FG 3a , 3a IJ H 2 2K F 3a I F 3a I aG J −G J H 2K H 2K y = FG 3a IJ − a FG 3a IJ H 2K H 2K

dy

2

− ax

i

i b2 yy

1

−a

g

2

2

2

1

=

6a 2 − 9a 2 9a 2 − 6a 2

d

– 9a 2 − 6a 2 =

From (2), at

FG 3a , 3a IJ H 2 2K y2 =

d9 a

F 9a GH 2

2

− 6a

−

2

3a 2 2

i

i

= –1

I b– a − 3ag − F 3a JK GH 2 F 9a − 3a I GH 4 2 JK

2

2

–

3 2 3a 2 a × 4a – × 4a 4 4

F 3a I GH 4 JK

=

=

2

2

2

– 6a 3 – 32 = 3a 9a 4 16

Using these

ρ F 3a

GH 2 , 32a IJK

{1 + b– 1g } FG − 32 IJ H 3a K

3 2 2

=

= – ∴ Radius of curvature at

2 2 × 3a − 3a = 32 8 2

FG 3a , 3a IJ H 2 2K

is

3a · 8 2

2

2

−

9a 2 4

I b– 3a − ag JK

...(2)

9

DIFFERENTIAL CALCULUS—I

8. Find the radius of curvature of b2x2 + a2y2 = a2b2 at its point of intersection with the y-axis. Solution. We have ρ = Here,

o1 + t

3 2 2 y1

at x = 0

y2

b2x2 + a2y2 = a2b2

When x = 0,

a2y2 = a2b2 y2 = b 2

⇒

y =±b

i.e., the point is (0, b) or (0, – b) The curve is b2x2 + a2y2 = a2b2. Differentiating w.r. to x 2b2x + 2a2yy1 = 0

b2 x a2 y Differentiating again w.r. to x y1 = –

y2 =

F y − xy I GH y JK – b b0g =0 a bbg – b F b − 0I G J a H b K – b2 a2

1

2

2

Now at (0, b),

y1 =

2

2

and

y2 =

2

2

–b a2 i.e., Radius of curvature at (0, b) is =

∴

ρ(0,

b1 + 0g FG – b IJ Ha K

3 2

b)

=

=

– a2 b

2

∴ Radius of curvature is

a2 b

Next consider (0, – b), y1 =

– b2 0 × =0 2 –b a

y2 =

– b2 a2

FG – b – 0IJ = a H b K b

2

2

10

ENGINEERING MATHEMATICS—II

ρ(0,

– b)

=

b1 + 0g FG b IJ Ha K

3 2

=

a2 b

2

∴ Radius of curvature of (0, – b) is

a2 . b

9. Show that at any point P on the rectangular hyperbola xy = c2, ρ =

r3 where r is the 2c 2

distance of the point from the origin. Solution. The curve is xy = c2 Differentiating w.r. to x xy1 + y = 0

y x

y1 = – Again differentiating w.r.t. x

=

RS xy − y UV T x W R| – xy − y U| 2y –S x = V x |W x |T R|1 + FG y IJ U| S| H x K V| W T

=

d

=

dx

1

y2 = –

=

ρ=

where

x2

∴

+

y2

=

r2

2

2

2

d1 + i

and

2

3 2 2 y1

y2

xy =

3 2

2y x2

x2 + y2

i

3 2

i

3 2

2y x3 × 2 x 2

+ y2 2 xy

c2.

ρ =

10. Show that, for the ellipse

r3 . 2c2 x2 y2 a 2b 2 + 2 = 1, ρ = where p is the length of the perpen2 p3 a b

dicular from the centre upon the tangent at (x, y) to the ellipse.

11

DIFFERENTIAL CALCULUS—I

Solution. The ellipse is

x2 y2 + = 1 a 2 b2

Differentiating w.r.t. x

2 x 2 yy1 + 2 = 0 a2 b ⇒

y1 =

Again Differentiating w.r. to x y2 =

b2 x a2 y

–

LM y − xy OP a N y Q LM y + b ⋅ x OP –b M y P a a M MN y PPQ b Ly x O – + P M a y Nb a Q LM3 x –b a y N a – b2

1

2

2

2

2

=

2

2

2

4

=

y2 =

ρ =

Now,

2

2

2

3

2

2

2

2

4

2

3

2

o1 + t

3 y12 2

y2

+

F1 + b x I GH a x JK F– b I GH a y JK 4 2

=

ρ =

da y –

4

4

4

da y 4

Taking magnitude

ρ =

The tangent at (x0, y0) to the ellipse

x x0 y y0 + 2 = 1 a2 b

2

+ b4 x 2

2

a6 y3 2

3 2

4 2

=

2

da y −

OP Q

y2 =1 b2

i

3 2

i

3 2

+ b4 x 2

×

3

a2 y3 b4

a 4b 4

+ b4 x 2

a 4b 4

i

3 2

x2 y2 + = 1 is a 2 b2

...(1)

12

ENGINEERING MATHEMATICS—II

Length of perpendicular from (0, 0) upon this tangent

1

=

FG x IJ + FG y IJ H a K Hb K 2

0 2

2

0 2

a 2b2

=

a 4 y02 + b 4 x02 So, the length of perpendicular from the origin upon the tangent at (x, y) is a 2b 2

p =

a 4 y02 + b 4 x02

By replacing x0 by x and y0 by y

a 2b 2

p =

a4 y2 + b4 x2

Reciprocal and cube on both sides, we get, ⇒

1 = p3

=

da y 4

+ b4 x 2

2

i

3 2

i

3 2

a 6b 6

d

a 4 y 2 + b4 x2 4 4

a b

×

1 a b

2 2

By using eq. (1), we get ρ 1 2 2 3 = p a b

⇒

ρ =

a 2b 2 · p3

ax , 11. Show that, for the curve y = a+x Solution. Here,

y =

FG 2ρ IJ HaK

2 3

F xI F yI = G J + GH JK H yK x 2

ax a+x

Differentiating w.r.t. x

LM ba + xgb1g − x OP a MN ba + xg PQ = ba + xg 2

y1 = a Again Differentiating w.r.t. x

2 y2 = a

2

–2

=

– 2a 2

b a + x g ba + x g 3

3

2

2

·

13

DIFFERENTIAL CALCULUS—I

ρ =

Now,

d1 + y i

3 2 2 1

y2

Substituting y1 and y2, we get

R|1 + a U| S| ba + xg V| W T R| − 2a U| S| ba + xg V| W T {ba + xg + a } – 2a b a + x g FG x IJ + FG y IJ H yK H xK R| |S 2 {ba + xg + a } || 2a ba + xg T ba + x g + a a ba + x g ba + x g + a a ba + x g 3 2

4

4

=

2

3

3 4 2

4

ρ =

To show that

L.H.S.

FG 2ρIJ HaK

23

FG 2ρ IJ HaK

2

=

2 3

2

4

=

4

=

=

using (1)

2

2

FG x IJ + FG y IJ H yK H xK

R| x U| S| FG ax IJ V| TH a + xK W ba + x g + a2

∴ L.H.S. = R.H.S. using (2) and (3).

...(2)

2

2

=

2 3

2

2

2

R.H.S.

U| |V || W

4

2

=

3 2

3

3

4

=

...(1)

3

2

2

2

R| F ax I U| S GH a + x JK V| + | T x W a2

ba + x g

2

2

...(3)

14

ENGINEERING MATHEMATICS—II

12. Find ρ at any point on x = a (θ + sinθ) and y = a (1 – cosθ). Solution. Here x = a (θ + sinθ), y = a (1 – cosθ) Differentiating w.r.t. θ

dx dy = a (1 + cos θ), = a sin θ dθ dθ dy dy a sin θ = dθ = y1 = dx dx a 1 + cos θ dθ

b

g

θ θ cos 2 2 2 θ 2 cos 2

2 sin =

y 1 = tan Again differentiating w.r.t. θ y2 = = =

θ 2

FG IJ H K d F G tan θ2 IJK × ddxθ dθ H 1 F θI 1 sec G J × × H 2 K 2 a b1 + cos θg d θ tan dx 2

2

θ 2

sec 2 =

2a × 2 cos2 y2 =

ρ =

1 4a cos4

θ 2

o1 + y t

3 2 2 1

y2

RS1 + tan θ UV 2W T R| U| 1 S| θV 4a cos | 2W T 2

=

θ 2

4

3 2

15

DIFFERENTIAL CALCULUS—I

= =

RSsec FG θ IJ UV T H 2K W 2

3 2

1

× 4a cos θI F cos G J H 2K F θI 4 a cos G J · H 2K 3

ρ =

FG θ IJ H 2K FG θ IJ H 2K

× 4a cos4 4

13. Find the radius of curvature at the point ′θ′ on the curve x = a log sec θ, y = a (tan θ – θ). Solution x = a log sec θ, y = a (tan θ – θ) Differentiating w.r.t. θ dy dx 1 ⋅ sec θ ⋅ tan θ , = a = a (sec2 θ – 1) dθ dθ sec θ = a tan θ = a tan2 θ

∴

dy dy = dθ y1 = dx dx dθ a tan 2 θ = a tan θ y 1 = tan θ

b g

d2y d tanθ = dx 2 dx d dθ = tan θ ⋅ dθ dx

y2 =

b g

2 = sec θ ×

=

Now,

ρ =

sec 2 θ a tan θ

o1 + y t

3 2 2 1

y2

d1 + tan θi F sec θ I GH tan θ JK 2

=

1 a tan θ

2

3 2

16

ENGINEERING MATHEMATICS—II

sec 3 θ × a tan θ sec 2 θ ρ = a sec θ tan θ. =

14. For the curve x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), show that the radius of curvature at ′θ′ varies as θ. Solution x = a (cos θ + θ sin θ)

dx = a (– sin θ + θ cos θ + sin θ) = a θ cos θ dθ y = a (sin θ – θ cos θ)

⇒

dy = a (cos θ + θ sin θ – cos θ) = a θ sin θ dθ

⇒

y1 = y2 = =

dy a θ sin θ = = tan θ dx a θ cos θ

FG IJ b H K d dθ tan θg ⋅ b dθ dx

d dy d = tan θ dx dx dx

2 = sec θ ×

=

ρ =

Now,

=

g

1 a θ cos θ

1 a θ cos3 θ

d

1 + y12

i

3 2

y2

d i F 1 I GH a θ cos θ JK 1 + tan 2 θ

3 2

3

= sec3 θ × a θ cos3 θ = aθ ρ ∝ θ.

i.e.,

15. If ρ1 and ρ2 are the radii of curvatures at the extremities of a focal chord of the parabola

b g . Solution. If dat , 2at i and dat , 2at i are the extremities of a focal chord of the parabola

y2 = 4ax, then show that ρ 1 2 1

y2 = 4ax. Then

1

–2 3

+ ρ2

–2 3

= 2a

2 2

t1 · t2 = – 1

2

−2 3

17

DIFFERENTIAL CALCULUS—I

The parametric equations to x x′ x″

∴

the parabola are y = 2at = at2, = 2at y′= 2a = 2a y″ = 0

ρ =

ox ′

2

+ y′2

t

3 2

x′ y ″ – x″ y′

{b2at g + b2ag } d – 4a i

3 2 2

2

=

2

= –

d

8a 3 1 + t 2 4a

2

d

i

2 ρ = − 2a 1 + t

ρ

–2 3

–2 ρ3

= =

i

3 2

3 2

b2ag d1 + t i –2 3

1

2

−1

1

×

b2ag d1 + t i 2 3

2

Let t1 and t2 be extremities of a focal chord. Then t2 = – –2

–2

ρ1 3

Now,

3 = ρt = t1 =

–2

ρ 23

= ρ

= –2

Adding

–2

ρ1 3 + ρ23

=

=

i.e.,

–2 3 1

ρ

–2

+ ρ 23

=

–2 3 t=

–1 t1

1

1

×

1

b2ag d1 + t i 2 3

=

×

1

2 1

1

×

b2ag d1 + t i 2 3

2 1

t12

b2ag d1 + t i 1 R 1 ST1 + t + 1 +t t UVW b2 a g b2ag × d11 ++ tt i d i 2 3

2 1

2 1

2 3

–2 3

b2 a g

–2 3

2 1

2 1

2 1 2 1

· Hence proved.

1 · t1

18

ENGINEERING MATHEMATICS—II

EXERCISE 1.1 1. Find ρ at any point on y = log sin x. 2. Find ρ at x = 1 on y =

Ans. cosec x

log x . x

FG x IJ · H 2K

π 3. Find the radius of curvature at x = on y = log tan 4

x2 y2 + 4. Find the radius of curvature of (3, 4) on = 2. 9 16

5. Find the radius of curvature at (x1, y1) on b2x2 – a2y2 = a2b2.

LMAns. N

OP 2 2Q LM O 3I P F MNAns. 2 × GH 2 JK PQ LMAns. 125OP N 12 Q LM b x + a y OP MNAns. e a b j PQ 3 2

2 2 1

Ans. 4 2

d4 x

7. Show that ρ at any point on 2xy = a2 is

4

+ a4

i

3 2

8a 2 x 3

8. Show that ρ at (0, 0) on y2 = 12x is 6. 9. Find radius of curvature at x = 2 on y2 =

b

·

g·

LMAns. 1 OP N 3Q LMAns. a OP 2Q N LMAns. 5 5 a OP 6 PQ MN

x x−2 x −5

10. Find ρ at (a, a) on x3 + y3 = 2a3.

11. Find the radius of curvature at (a, 2a) on x2y = a(x2 + a2). 12. Find the radius of curvature at (–2a, 2a) on x2y = a (x2 + y2). 13. Show that ρ at (a

a

sin3θ)

on

3 4 2 2 1

4 4

6. Find ρ at (4, 2) on y2 = 4 (x – 3).

cos3θ,

3

2 x3

+

2 y3

=

2 a3

Ans. 2a

is 3a sin θ cos θ.

π 14. Find radius of curvature at θ = on x = a sin θ, y = b cos 2θ. 3

LM a MNAns. e

15. Find ρ for x = t – sin ht cos ht, y = 2cos ht.

Ans. 2 cos h 2 t sin ht

2

+ 12b 2 4ab

O j PP Q 3 2

19

DIFFERENTIAL CALCULUS—I

1.2.3 Radius of Curvature in Pedal Form Let polar form of the equation of a curve be r = f (θ) and P(r, θ) be a given point on it. Let the tangent to the curve at P subtend an angle ψ with the initial side. If the angle between the radius vector OP and the tangent at P is φ then we have ψ = θ + φ (see figure). Let p denote the length of the perpendicular from the pole O to the tangent at P. Then from the figure,

r = f (G)

OM p = OP r p = r sin φ

B

r

O

P (r, G)

O

G

X

sin φ = Hence, ∴

dψ dθ dφ dθ dφ dr 1 + = + ⋅ = = ds ds ds ds dr ds ρ

We know that tan φ = r ⋅

Hence,

sin φ = r ⋅

From (2),

dθ ds

dr ds

LM N

1 dφ sin φ + r cos φ ⋅ r dr

b

1 d ⋅ r sin φ r dr r sin φ = p

=

Since,

Fig. 1.2

dθ dr

sin φ dφ 1 + cos φ ⋅ = r dr ρ

=

g

OP Q

dr dp This is the Pedal form of the radius of curvature.

Therefore,

M

dθ ds dr ds

i.e.,

cos φ =

...(2)

p

r⋅

sin φ = cos φ

and

...(1)

ρ = r⋅

...(3)

1.2.4 Radius of Curvature in Polar Form Let r = f (θ) be the equation of a curve in the polar form and p(r, θ) be a point on it. Then we know that 1 1 1 Differentiating w.r.t. r, we2 get = 2 + 4 p r r

FG dr IJ H dθ K

2

...(4)

20

ENGINEERING MATHEMATICS—II

FG IJ d– 4r i + 1 ⋅ 2 ⋅ dr ⋅ d FG dr IJ H K dθ dr H dθ K r – 2 4 F dr I 2 dr d r dθ ⋅ ⋅ ⋅ − .G J + r r H dθ K r dθ dθ dr – 2 4 F dr I 2 d r + ⋅ − G J H K r r dθ r dθ L 1 2 FG dr IJ – 1 d r OP p M + MN r r H dθ K r dθ PQ 2

– 2 dp –2 dr ⋅ + = 3 3 p dr dθ r

–5

4

2

=

3

2

5

4

2

2

=

Hence, Now,

dp = dr

3

2

5

4

2

2

2

3

3

ρ = r⋅

5

dr = dp

=

r2 + 2

4

2

r

p3

R| 1 + 2 FG dr IJ S| r r H dθ K T

r6 ⋅

1 p3

3

FG dr IJ H dθ K

5

2

−r⋅

2

−

1 d 2r r 4 dθ 2

U| V| W

d 2r dθ 2

By using equation (4),

R| 1 1 FG dr IJ U| r ⋅S + V T| r r H dθ K W| F dr I d r r +2G J –r⋅ H dθ K dθ

3 2

2

6

2

=

4

2

2

2

2

R|r + FG dr IJ U| S| H dθ K V| T W F dr I d r +2G J −r⋅ H dθ K dθ 2

2

ρ=

2

r

where

∴

r1 =

ρ=

2

3 2

2

2

dr d 2r , r = · dθ 2 dθ 2

o

r 2 + r12

r + 2

2r12

t

3 2

− r r2

This is the formula for the radius of curvature in the polar form.

...(5)

21

DIFFERENTIAL CALCULUS—I

WORKED OUT EXAMPLES 1. Find the radius of the curvature of each of the following curves: (i) r3 = 2ap2 (Cardiod) (ii) p2 = ar

1 1 1 r2 = + − (Ellipse). p 2 a 2 b 2 a 2b 2

(iii)

Solution. (i) Here r3 = 2ap2 Differentiating w.r.t. p, we get

3r 2 ⋅

dr = 4ap dp 4ap dr = dp 3r 2

⇒

ρ = r⋅

Hence,

Fr I where p = G J H 2a K 3

dr 4ap 4ap =r⋅ 2 = dp 3r 3r

1 2

Fr I 4a ⋅ G J H 2a K 3

ρ =

1 2

3r 3

4a r 2 2 2ar = = 3 3r 2a (ii) Here

p2 = ar

Differentiating w.r.t. p, we get

dr dp

Then

2p = a ⋅

⇒

2p dr = a dp

where p =

ar · 3

dr 2 ⋅ ar 2r 2 =r⋅ = ρ = r dp a a (iii) Given

1 p2

=

1 1 r2 + – a2 b 2 a 2b 2

Differentiating w.r.t. p, we get –2 –1 dr = 2 2 2r ⋅ p3 dp a b

22

ENGINEERING MATHEMATICS—II

Hence Therefore,

dr a 2b 2 = dp p 3r ρ = r.

dr a 2b 2 a 2b 2 =r. 3 = 3 dp p r p

2. Find the radius of curvature of the cardiod r = a (1 + cos θ) at any point (r, θ) on it. Also

ρ2 is a constant. r Solution. Given r = a (1 + cos θ) Differentiating w.r.t. θ dr = – a sin θ r1 = dθ

prove that

d 2r = – a cos θ and r2 = dθ 2 ∴ The radius of curvature in the polar form ρ =

or

2

r + 2

+

t

3 2 2 r1

2r12

− r r2

g + a sin θ} { b a b1 + cos θg + 2a sin θ − ab1 + cos θgb – a cos θg a o1 + 2 cos θ + cos θ + sin θt a o1 + 2 cos θ + cos θ + 2 sin θ + cos θ + cos θt a m2 b1 + cos θgr 3 b1 + cos θg 2 2 a b1 + cos θg a 2 1 + cos θ

=

2

2

2

2

2

2

3 2

2

2

3

=

2

2

2

3 2

2

2

3 2

=

1 2

=

3 2

=

F θI 2 a G 2 cos J H 2K 2

3 4 θ a cos ρ = 3 2 Squaring on both sides, we get ρ2 =

16 2 θ a cos2 9 2

b

8 a2 1 + cos θ = 9

1 2

LM where cos θ = 1 b1 + cos θgOP 2 2 Q N LM where 1 + cos θ = r OP aQ N 2

g

23

DIFFERENTIAL CALCULUS—I

ρ2

8a 2 r 8ar ⋅ = = 9 a 9

8a ρ2 = which is constant. 9 r

Hence,

3. Show that for the curve rn = an cos nθ the radius of curvature is rn

an

Solution. Here = cos nθ Taking logarithms on both sides, we get n log r = n log a + log cos nθ Differentiating w.r.t. θ, we have

an · n + 1 rn –1

b g

n dr n sin nθ = 0– r dθ cos nθ

dr = – r tan nθ dθ Differentiating w.r.t. θ again, we obtain r1 =

r2 =

RS T – onr sec

dr d 2r − rn sec 2 nθ + tan nθ ⋅ 2 = dθ dθ

=

t

nθ − r tan 2 nθ

2

= r tan2 nθ – nr sec2 nθ Using the polar form of ρ, we get ρ =

=

=

= = =

o

r 2 + r12

r + 2

2r12

t

3 2

− rr2

or

b

2

r 2 + 2 – r tan nθ

t

+ r 2 tan 2 nθ

g

2

d

3 2

− r r tan 2 nθ − nr sec 2 nθ

i

r 3 sec 3 nθ r 2 1 + 2 tan 2 nθ − tan 2 nθ + n sec 2 nθ

r sec 3 nθ n + 1 sec 2 nθ

b g

r n + 1 cos nθ

b g

r

bn + 1g FGH ar IJK n

n

=

UV W

an · n + 1 rn–1

b g

LMwhere cos nθ = r OP a Q N n

n

24

ENGINEERING MATHEMATICS—II

4. Find the radii of curvature of the following curves: (ii) r (1 + cos θ) = a (i) r = aeθ cot α (iii) θ =

r2 − a2 a

– cos –1

FG a IJ · H rK

Solution. (i) Here r = aeθ Differentiating w.r.t θ

cot α

dr = aeθ cot α · cot α dθ = r · cot α

So that,

tan φ =

=

r dr dθ r = tan α r cot α

Hence, φ = α, since p = r sin φ We get, p = r sin α. This is the Pedal equation of the given curve. From which, we get

dr 1 = dp sin α Hence,

p = r⋅

dr = r cosec α. dp

(ii) Given equation of the curve is r (1 + cos θ) = a Differentiating w.r.t. θ, we get r (– sin θ) + (1 + cos θ) ·

dr =0 dθ

dr r sin θ = dθ 1 + cos θ

or

We have,

1 p2

1 1 = 2 + 4 r r

=

FG dr IJ H dθ K

2

r 2 sin 2 θ 1 1 + ⋅ r 2 r 4 1 + cos θ 2

b

g 1 L sin θ O M P 1+ r MN b1 + cos θg P Q 1 L b1 + cos θg + sin M r MN b1 + cos θg 2

=

2

2

2

=

2

2

2

θ

OP PQ

25

DIFFERENTIAL CALCULUS—I

= = where 1 + cos θ =

LM 2 b1 + cos θg OP MN b1 + cos θg PQ

1 r2

r

2

2 1 + cos θ

b

2

g

a r 1 p2

=

2 a r2 ⋅ r

=

2 ar

ar which is the pedal equation of the curve. 2 Differentiating w.r.t. p, we get Hence,

p2 =

2p =

a dr ⋅ 2 dp

⇒

4p dr = a dp

∴

ρ = r⋅

dr dp

= r⋅

4p where p = a

= r.

4 a

ar 2

ar 2 3

a r2 .

= 2 2 (iii) Here,

Then,

θ =

r2 − a2

− cos–1

a

dθ 2r + = dr a ⋅ 2 ⋅ r2 − a2

=

=

r a r −a 2

2

–

r2 − a2 ar r 2 − a 2

FG a IJ H rK 1

F1 − a GH r a

r r − a2 2

2 2

FG – a IJ I HrK JK 1 2

26

ENGINEERING MATHEMATICS—II

so that

dθ = dr

r2 − a2 ar

dr = dθ

ar

r2 − a2 We have the Pedal equation, we get 1 p2

1 p2

FG dr IJ H dθ K

2

=

1 1 + 4 2 r r

=

a 2r 2 1 1 + ⋅ r2 r4 r2 − a2

=

a2 1 1 + r2 r2 − a2

=

1 r − a2

RS T

d

UV W

i

2

p2 = r2 – a2

Hence From this we get

p dr = r dp ρ = r⋅

∴

p = p = r2 − a2 ⋅ r

EXERCISE 1.2 1. Find the radius curvature at the point ( p, r) on each of the following curves:

LMAns. r OP N aQ LMAns. a OP N 3r Q LM Ans. a OP MN bn + 1g r PQ LM a + r OP MMAns. d r + 2ai PP MN PQ 3

(i) pr =

a2

(ii)

r3

a2p

(iii)

pan

(Hyperbola)

2

2

=

(Lemniscate)

n

(iv) p =

=

r n+1

(Sine spiral)

r4 (Archimedian spiral) r2 + a2

n −1

2

3 2 2

2

2

27

DIFFERENTIAL CALCULUS—I

2. Find the radius of curvature at (r, θ) on each of the following curves:

LM r a + r OP MM Ans. d a i PP MN PQ LMAns. a OP N 3rQ LMAns. r OP N aQ LMAns. r OP N 3p Q 3 2 2

2

(i) r =

a θ

2

(iii)

r2

=

a2

cos 2θ

3

(v) r2 cos 2θ = a2

2

4

(vii) r = a sec 2θ

2

LMAns. a OP N 2Q

(ii) r = a cos θ

3

LM Ans. a OP (iv) = sin nθ MN bn + 1g r PQ LMAns. 2 ar OP a (vi) r = b1 − cos θg 3 PQ 2 MN LM Ans. na OP (viii) r = a sin nθ N 2Q n

rn

an

n −1

3. If ρ1 and ρ2 are the radii of curvature at the extremities of any chord of the cardiode 2 2 r = a (1 + cos θ) which passes through the pole. Prove that ρ1 + ρ2

=

16a 2 · 9

1.3 SOME FUNDAMENTAL THEOREM 1.3.1 Rolle’s Theorem If a function f (x) is 1. continuous in a closed interval [a, b], 2. differentiable in the open interval (a, b) and 3. f (a) = f (b). Then there exists at least one value c of x in (a, b) such that f ′ (c) = 0 (No proof).

1.3.2 Lagrange’s Mean Value Theorem Suppose a function f (x) satisfies the following two conditions. 1. f (x) is continuous in the closed interval [a, b]. 2. f (x) is differentiable in the open interval (a, b). Then there exists at least one value c of x in the open interval (a, b), such that

bg

f (b) – f a b–a

= f ′ (c)

28

ENGINEERING MATHEMATICS—II

Proof. Let us define a new function φ(x) = f (x) – k·x

...(1)

where k is a constant. Since f (x), kx and φ (x) is continuous in [a, b], differentiable in (a, b). From (1) we have,

φ (a) = f (a) – k·a φ (b) = f (b) – k·b

∴

φ (a) = φ (b) holds good if f (a) – k·a = f (b) = k·b

i.e.,

k (b – a) = f (b) – f (a)

or

k =

bg bg

f b − f a b−a

...(2)

Hence, if k is chosen as given by (2), then φ (x) satisfy all the conditions of Rolle’s theorem. Therefore, by Rolle’s theorem there exists at least one point c in (a, b) such that φ′(c) = 0. Differentiating (1) w.r.t. x we have, φ′(x) = f ′(x) – k φ′(c) = 0 gives f ′(c) – k = 0

and

k = f ′(c)

i.e.,

...(3)

Equating the R.H.S. of (2) and (3) we have

bg bg

f b − f a b−a

= f ′(c)

...(4)

This proves Lagrange’s mean value theorem.

1.3.3 Cauchy’s Mean Value Theorem If two functions f (x) and g (x) are such that 1. f (x) and g (x) are continuous in the closed interval [a, b]. 2. f (x) and g (x) are differentiable in the open interval (a, b). 3. g′ (x) ≠ 0 for all x ∈ (a, b). Then there exists at least one value c ∈ (a, b) such that

bg g bbg – g b a g

f (b) – f a

=

b g· g ′ b cg f′ c

Proof: Let us define a new function φ (x) = f (x) – kg (x)

...(1)

where k is a constant. From the given conditions it is evident that φ (x) is also continuous in [a, b], differentiable in (a, b). Further (1), we have φ (a) = f (a) – k g (a); φ (b) = f (b) – k g (b)