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Engineering Mathematics – I Dr. V. Lokesha

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Engineering Mathematics – I

(10 MAT11)

LECTURE NOTES (FOR I SEMESTER B E OF VTU)

VTU-EDUSAT Programme-15

Dr. V. Lokesha Professor and Head

DEPARTMENT OF MATHEMATICS ACHARYA INSTITUTE OF TECNOLOGY

Soldevanahalli, Bangalore – 90


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ENGNEERING MATHEMATICS – I

Content

CHAPTER

UNIT I DIFFERENTIAL CALCULUS – I

UNIT II DIFFERENTIAL CALCULUS – II

UNIT III DIFFERENTIAL CALCULUS – III


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UNIT - I DIFFERENTIAL CALCULUS – I

Introduction: The mathematical study of change like motion, growth or decay is calculus. The Rate of change of given function is derivative or differential. The concept of derivative is essential in day to day life. Also applicable in Engineering, Science, Economics, Medicine etc.

Successive Differentiation: Let y = f (x) --(1) be a real valued function. The first order derivative of y denoted by or y’ or y1 or ∆1 The Second order derivative of y denoted by or y’’ or y2 or ∆2 Similarly differentiating the function (1) n-times, successively, the n th order derivative of y exists denoted by or yn or yn or ∆n The process of finding 2nd and higher order derivatives is known as Successive Differentiation.

nth derivative of some standard functions: 1. y = eax Sol : y1 = a eax y2 = a2 eax Differentiating Successively yn = an eax ie. Dn[eax] = an eax For, a =1 Dn[ex] = ex

dxdy

2

2

dxyd

n

n

dxyd


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Leibnitz’s Theorem : It provides a useful formula for computing the nth derivative of a product of two functions. Statement : If u and v are any two functions of x with un and vn as their nth derivative. Then the nth

derivative of uv is

(uv)n = u0vn + nC1 u1vn-1 + nC2u2vn-2 + …+nCn-1un-1v1+unv0 Note : We can interchange u & v (uv)n = (vu)n,

nC1 = n , nC2 = n(n-1) /2! , nC3= n(n-1)(n-2) /3! …

1. Find the nth derivations of eax cos(bx + c) Solution: y1 = eax – b sin (bx +c) + a eax cos (b x + c), by product rule. .i.e, y1 = eax ( ) ( )[ ]cbxsinbcbxcosa +−+

Let us put a = r cos θ , and b = r sin θ .

∴ 222 rba =+ and tan a/b=θ

.ie., 22 bar += and θ = tan-1 (b/a)

Now, [ ])cbxsin(sinr)cbxcos(cosrey ax1 +θ−+θ=

Ie., y1 = r eax cos ( )cbx ++θ

where we have used the formula cos A cos B – sin A sin B = cos (A + B) Differentiating again and simplifying as before, y2 = r2 eax cos ( )cbx2 ++θ .

Similarly y3 = r3 e ax cos ( )cbx3 ++θ .

………………………………………

Thus ( )cbxncosery axnn ++θ=

Where 22 bar += and θ = tan-1 (b/a).

Thus Dn [eax cos (b x + c)]

= ( )[ ][ ]cbxabneba axn +++ − /tancos)( 122


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2. Find the nth derivative of log 384 2 ++ xx

Solution : Let y = log 3x8x4 2 ++ = log (4x2 + 8x +3) ½

ie., 21y = log (4x2 + 8x +3) ∵ log xn = n log x

21y = log { (2x + 3) (2x+1)}, by factorization.

∵ 21y = {log (2x + 3) + log (2x + 1)}

Now ( ) ( )( )

( ) ( )( ) ⎭

⎬⎫

⎩⎨⎧

+−−

++

−−=

−−

n

n1n

n

n1n

n 1x22!1n1

3x22!1n1

21y

Ie., yn = 2n-1 (-1) n-1 (n-1) ! ( ) ( ) ⎭

⎬⎫

⎩⎨⎧

++

+ nn 1x21

3x21

3. Find the nth derivative of log 10 {(1-2x)3 (8x+1)5}

Solution : Let y = log 10 {1-2x)3 (8x+1)5} It is important to note that we have to convert the logarithm to the base e by the property:

10logxlogxlog

e

e10 =

Thus ( ) ( ){ }53e

e

1x8x21log10log

1y +−=

Ie., ( ) ( ){ }1x8log5x21log310log

1ye

++−=

( ) ( ) ( )( )

( ) ( )( ) ⎭

⎬⎫

⎩⎨⎧

+−−

+−

−−−=∴

−−

n

n1n

n

n1n

en 1x8

8!1n15x21

2!1n1.310log

1y

Ie., ( ) ( ) ( )( )

( )( ) ⎭

⎬⎫

⎩⎨⎧

++

−−−−

=−

n

n

n

n

e

n1n

n 1x845

x2113

10log2!1n1y


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4. Find the nth derivative of e2x cos2 x sin x

Solution : >> let y = e2x cos2 x sin x = e2x ⎥⎦⎤

⎢⎣⎡ +

2x2cos1 sin x

ie., 2

eyx2

= (sin x + sin x cos 2x)

= ( )[ ]⎭⎬⎫

⎩⎨⎧ −++ xsinx3sin

21xsin

2e x2

= ( )∵xsinx3sinxsin24

e x2

−+ sin (-x) = -sin x

4ey

x2

=∴ (sin x + sin 3x)

Now ( ) ( ){ }x3sineDxsineD41y x2nx2n

n +=

Thus ( ) ( )[ ] ( ) ( )[ ]{ }x323tannsine13x21tannsine541y 1x2n1x2n

n +++= −−

( ) ( )[ ] ( ) ( )[ ]{ }x323tannsin13x21tannsin54

ey 1n1nx2

n +++=∴ −−

5. Find the nth derivative of e2x cos 3x

Solution : Let y=e2x cos3 x = e 2x. 41 (3 cos x + cos 3x)

Ie., y = 41 (3 e2x cos x + e2x cos 3x)

41yn =∴ {3Dn (e2x cos x) + Dn (e2x cos 3x)}

( ) ( )[ ] ( ) ( )[ ]{ }xnexney xnxn

n 323tancos1321tancos5341 1212 +++= −−

Thus ( ) ( )[ ] ( ) ( )[ ]{ }x323tanncos13x21tanncos534

ey 1n1nx2

n +++= −−


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6. Find the nth derivative of ( )( )3x21x2x2

++

Solution : ( )( )3x21x2xy

2

++= is an improper fraction because; the degree of the

numerator being 2 is equal to the degree of the denominator. Hence we must divide and rewrite the fraction.

3x8x4

x4.41

3x8x4xy 2

2

2

2

++=

++= for convenience.

4x2 +8x +3

1

2

2

38384

4

−−++

xxx

x

∴ ⎥⎦⎤

⎢⎣⎡

++−−

+=3x8x4

3x8141y 2

Ie., ⎥⎦⎤

⎢⎣⎡

+++

−=3x8x4

3x841

41y 2

The algebraic fraction involved is a proper fraction.

Now ⋅⎥⎦⎤

⎢⎣⎡

+++

−=3x8x4

3x8D410y 2

nn

Let ( )( ) 3x2B

1x2A

3x21x23x8

++

+=

+++

Multiplying by (2x + 1) (2x + 3) we have, 8x + 3 = A (2x + 3) + B (2x + 1) ................(1)

By setting 2x + 1 = 0, 2x + 3 = 0 we get x = -1/2, x = -3/2. Put x = -1/2 in (1): -1 -1 + A (2) ⇒ A = -1/2

Put x = -3/2 in (1): -9 = B (-2) ⇒ B = 9/2

∴ ⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡

++⎥⎦

⎤⎢⎣⎡

+−−=

3x21D

29

1x21D

21

41y nn

n


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( ) ( )( )

( )( ) ⎭

⎬⎫

⎩⎨⎧

+−

⋅++

−⋅−−= ++ 1n

nn

1n

nn

3x22!n19

1x22!n11

81

ie., ( )( ) ⎭

⎬⎫

⎩⎨⎧

++

+−

= ++

+

1n1n

n1n

n 3x29

)1x2(1

82!n1y

7. Find the nth derivative of )2()1(

4

++ xxx

Solution : y = )2()1(

4

++ xxx is an improper fraction.

(deg of nr. = 4 > deg. of dr. = 2) On dividing x4 by x2 + 3 x + 2, We get

y = ( x2 – 3x + 7 ) + ⎥⎦⎤

⎢⎣⎡

++−−

231415

2 xxx

∴ yn = Dn (x2-3x+7)-Dn ⎥⎦⎤

⎢⎣⎡

++−

231415

2 xxx

But D = ( x2 – 3x + 7 ) = 2x – 3, D2 ( x2 – 3x + 7 ) = 2 D3( x2 – 3x + 7 ) = 0......... Dn ( x2 – 3x + 7 ) = 0 if n > 2

Hence yn = -Dn ⎥⎦

⎤⎢⎣

⎡++

+)2()1(

1415xx

x

Now, let Dn )2()1(23

14152 +

++

=++

+x

Bx

Axx

x

=> 15x+ 14 = A(x+2) + B(x+ 1 ) Put x = - 1 ; - 1 = A ( 1 ) or A = - 1 Put x = - 2 ; - 16 = B ( - 1 ) or B = 16

Yn = ⎭⎬⎫

⎩⎨⎧

⎥⎦⎤

⎢⎣⎡

++⎥⎦

⎤⎢⎣⎡

+−

2116

11

xD

xD nn


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1 1

1 1

( 1) ! 1 ( 1) ! 116( 1) ( 2)

1 16( 1) ! 2( 1) ( 2)

n n n n

n n

nn n n

n nx x

y n nx x

+ +

+ +

− −= −

+ +

⎧ ⎫= − − >⎨ ⎬

+ +⎩ ⎭ 8. Show that

⎭⎬⎫

⎩⎨⎧ −−−−

−=⎟

⎠⎞

⎜⎝⎛

+ nx

xn

xx

dxd

n

n

n

n 131

211log!)1(log

1

Solution : Let y = x

xx

x 1.loglog= and let u = log x, v =

x1

We have Leibnitz theorem,

(uv)n = uvn + vuvunvun nnCnC +++ −− ....2211 21 …… (1)

Now, u = log x ∴un = n

n

xn )!1()1( 1 −− −

xv 1

= ∴vn = 1

!)1(+

−n

n

xn

Using these in (1) by taking appropriate values for n we get,

Dn = n

n

n

n

xn

xn

xnx

xx )!1()1(.1!)1(..loglog 1

1

−−+

−=⎟

⎠⎞

⎜⎝⎛ −

+

( )

xxn

xn

xnn

n

n

n

n

1.!1)1(......

)!2()1(12.1

)1(

1

1

2

2

−−++

−−⎟⎠⎞

⎜⎝⎛−

−+

−

−

−

Ie.. = log x 1

1

1

!)1(!)1(+

−

+

−+

−n

n

n

n

xn

xn

1

1

1

2 )!1()1(....2

!)1(+

−

+

− −−++

−− n

n

n

n

xn

xn

⎥⎦

⎤⎢⎣

⎡ −−++

−−−

−−

−−−

+

−

1

121

1

2 )!1()1(....2)1()1(log!)1(

nnx

xn

n

n

Note : (-1)-1 = 1)1(

1)1(;11

12

2 =−

=−−=−

−


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Also nnn

nn

n 1)!1(

)!1(!

)!1(=

−−

=−

⎥⎦⎤

⎢⎣⎡ −−−−

−=⎥⎦

⎤⎢⎣⎡∴ + n

xx

nx

xdxd

n

n

n

n 1...31

211log!)1(log

1

9. If yn= Dn (xn logx) Prove that yn = n yn-1+(n-1)! and hence deduce that

yn = n ⎟⎠⎞

⎜⎝⎛ +++++

nx 1....

31

211log

Solution : yn = Dn(xn log x) = Dn-1 {D (xn log x}

= Dn-1 ⎭⎬⎫

⎩⎨⎧ + − xnx

xx nn log1. 1

= Dn-1(xn-1) + nDn-1 (xn-1 log x} ∴ yn = (n-1)! +nyn-1. This proves the first part.

Now Putting the values for n = 1, 2, 3...we get y1 = 0! + 1 y0 = 1 + log x = 1! (log x + 1 ) y2 = l! + 2y1 = l+2 (l + log x)

ie., y2 = 21og x + 3 = 2(log x + 3/2) = 2! ⎟⎠⎞

⎜⎝⎛ ++

211log x

y3 = 2! + 3y2 = 2 + 3(2 log x + 3)

ie., y3 = 61og x+ll = 6 (log x + ll/6) = 3! ⎟⎠⎞

⎜⎝⎛ +++

31

211log x

…………………………………………………………………………..

⎟⎠⎞

⎜⎝⎛ +++++=

nxnyn

1...31

211log!

10. If y = a cos (log x) + b sin ( log x), show that

x2y2 + xy1 + y = 0. Then apply Leibnitz theorem to differentiate this result n times. or If y = a cos (log x) + b sin (log x ), show that x2yn + 2 + (2n+l)xyn + l+(n2+1)yn = 0. [July-03]


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Solution : y = a cos (log x) + b sin (log x) Differentiate w.r.t x

∴ y1 = -a sin (log x) x1 + b cos (log x).

x1

(we avoid quotient rule to find y2) . => xy1 = - a sin (log x) + b cos (log x) Differentiating again w.r.t x we have,

xy2 + 1 y1 = - a cos (log x) + b sin ( log x) x1

or x2y2 + xy1 = - [ a cos (log x) + b sin (log x) ] = -y ∴ x2y2+xy1+y = 0 Now we have to differentiate this result n times. ie., Dn (x2y2) + Dn (xy1) + Dn (y) = 0 We have to employ Leibnitz theoreom for the first two terms. Hence we have,

⎭⎬⎫

⎩⎨⎧ −

++ −− )(.2.2.1

)1()(.2.)(. )222

12

2 yDnnyDxnyDx nnn

{ } 0)(.1.)(. 11

1 =++ −n

nn yyDnyDx

ie., {x2yn + 2 + 2n x yn + 1 + n (n – 1)yn} + {xyn+1+nyn}+yn = 0 ie., x2yn + 2 + 2n x yn + 1 + n2yn - nyn + xyn+1+nyn+yn = 0 ie., x2yn + 2 + (2n+l)xy n+l + (n2+l)yn = 0

11. If cos-1 (y/b ) = log (x/n)n, then show that

x2yn + 2 + (2n+l) xy n+l + 2n2yn = 0 Solution :By data, cos-1 (y/b) = n log (x/n) ∴log(am) = m log a

=> by = cos [n log (x/n )]

or y = b . cos [ n log (x/n)] Differentiating w.r.t x we get,


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y1 = -b sin [n log (x/n)] ( ) nnxn 1

/1

⋅⋅⋅

or xy1 = - n b sin [n log (x/n )] Differentiating w.r.t x again we get,

xy2 + 1. 1y = - n . b cos [ n log (x/n )] n nnx1.

)/(1

or x (xy2+y1) = n2b cos [n log (x/n) ] =-n2y, by using (1). or x2y2 +xyl + n2y = 0 Differentiating each term n times we have,

D(x2y2) + Dn(xy1) + n2Dn (y) = 0 Applying Leibnitz theorem to the product terms we have,

{ } 0.1.

.2.2.1

)1(.2.

21

122

=+++⎭⎬⎫

⎩⎨⎧ −

++

+

++

nnn

nnn

ynynxy

ynnyxnyx

ie x2yn+2 + 2 x yn+1 + n2yn + xy n+1+ nyn + n2yn=0 or x2 yn+2 + (2n + l) xyn+1 + 2n2yn = 0

12. If y = sin( log (x2 + 2 x + 1)),

or [Feb-03] If sin-1 y = 2 log (x + 1), show that (x+l)2yn + 2 + (2n+1)(x+1)yn+l + (n2 + 4)yn = 0

Solution : By data y = sin log (x2 + 2 x + 1 )

∴ y1 = cos log (x2 + 2 x + 1) 22)1(

12 +

+x

x

ie., y1 = cos log (x2 + 2 x + 1) 12

12 ++ xx

2 (x + 1)

ie., y1 = )1(

) 1 x 2 xlog(cos2 2

+++

x

or (x + 1) y1 = 2 cos log (x2 + 2 x + 1 ) Differentiating w.r.t x again we get


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(x+1)y2 + 1 y1 = -2 sin log (x2 + 2x + 1) )1(2.)1(

12 +

+x

x

or (x + 1)2y2 + (x+1) y1 = -4y or (x+l )2y2 + (x+l) y1 + 4y = 0 , Differentiating each term n times we have,

Dn [(x + 1)2y2] +Dn [(x+ 1)y1] + Dn [y] = 0 Applying Leibnitz theorem to the product terms we have,

⎭⎬⎫

⎩⎨⎧ −

++++ ++ nnn ynnyxnyx .2.2.1

)1().1(2.)1( 122

+ {(x+l) yn + 1+n. 1 .yn} + 4yn = 0 ie., (x+l)2yn + 2 + 2n (x+1)yn+1

+ n2yn-nyn + (x+l)yn+l + nyn + 4yn = 0 ie., (x+l)2yn + 2 + (2n + l) (x + l) y n+ 1 + (n2 + 4)yn = 0

13. If = log ( )21 xx ++ prove that

(1 + x2) yn+2 + (2n + 1) xyn+1 + n2yn = 0

>> By data, y = log ( )21 xx ++

∴y1 = ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

++

++x

xxx2.

1211

)1(1

22

Ie., 22

2

211

11

1)1(

1xx

xxxx

y+

=+

++

++

or 11 12 =+ yx

Differentiating w.r.t.x again we get

0.2.)12

11 1222 =

+++ yx

xyx

or (1+x2)y2 + xy1 = 0 Now Dn [(l+x2)y2] + Dn[xy1] = 0 Applying Leibnitz theorem to each term we get,


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⎭⎬⎫

⎩⎨⎧ −

+++ ++ nnn ynnyxnyx .2.2.1

)1(.2.)1( 122

+ [x . yn + 1+n .1 yn] = 0 Ie., (1 + x2) yn +2 + 2 n x yn + 1 + n2yn – nyn + xyn+l+ nyn = 0 or (l+x2)yn + 2 + (2n + l)xyn+1+n2yn = 0

14. If x = sin t and y = cos mt, prove that

(l-x2)yn + 2-(2n+1)xyn+l + (m2-n2)yn = 0. [Feb-04] Solution : By data x = sin t and y = cos mt

x = sin t => t = sin-1 x and y = cos mt becomes y = cos [ m sin-1x) Differentiating w.r.t.x we get

y1 = - sin (m sin-1x) 21 x

m−

or 121 yx+ = - m sin (m sin-1x)

Differentiating again w.r.f .x we get,

2

1122

2

1).sin(cos)2(

1211

xmxmmyx

xyx

−−=−

−+− −

or (1 -x2)y2-xyl = -m2y or (1 -x2)y2 –xy1 +m2y = 0

Thus (1-x2)yn+2-(2n+1)xyn+1+(m2-n2)yn=0 15. If x = tan ( log y), find the value of (l+x2)yn+1 + (2nx-l) yn+n(n-1)yn-1 [July-04] Solution : By data x = tan(log y) => tan-1 x = log y or y = etan-1 x Since the desired relation involves yn+1, yn and yn-1 we can find y1 and differentiate n times the result associated with y1 and y.

Consider y = ⋅− xe

1tan ∴ y. = ⋅− xe

1tan21

1x+

or (1 +x2)y1 = y Differentiating n times we have


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Dn[(l+x2)y1]=Dn[y] Anplying Leibnitz theorem onto L.H.S, we have, {(l+x2)Dn(y1) + n .2x .Dn-1 (y1)

nn yyDnn

=−

+ − )}(.2.2.1

)1(1

2

Ie., (1+x2)yn+1+2n x yn + n (n-1) yn-1-yn=0 Or (l+x2)yn + 1 + (2nx-l)yn + n(n-l)yn-1 = 0


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Continuity & Differentiability Some Fundamental Definitions A function f (x) is defined in the interval I, then it is said to be continuous at a point x = a if A function f (x) is said to be differentiable at x = a if Ex : Consider a function f (x) is defined in the interval [-1,1] by f (x) =

It is continuous at x = 0 But not differentiable at x = 0 Note : If a function f (x) is differentiable then it is continuous, but converse need not be true. Geometrically :

(1) If f (x) is Continuous at x =a means, f (x) has no breaks or jumps at the point x = a Ex : Is discontinuous at x=0 (2) If f (x) is differentiable at x = a means, the graph of f (x) has a unique tangent at the point or graph is smooth at x = a

1. Give the definitions of Continuity & Differentiability: Solution: A function f (x) is said to be continuous at x = a, if corresponding to an arbitrary positive number ε, however small, their exists another positive number δ such that. ⏐f (x) – f (a)⏐ < ε, where ⏐x - a⏐ < δ It is clear from the above definition that a function f (x) is continuous at a point ‘a’. If (i) it exists at x = a (ii) f (x) = f (a)

i.e, limiting value of the function at x = a is to the value of the function at x = a

axLt→

⎩⎨⎧

≤≤≤≤−−

=10 01

xxxx

x

0

( ) ( )lim '( ) Ih

f x h f a f a exists ah→

+ −= ∈

)()(lim afxfax

=→

1 1 0( )

0 1x

f xx x

− − ≤ ≤⎧= ⎨ < ≤⎩


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Differentiability: A function f (x) is said to be differentiable in the interval (a, b), if it is differentiable at every point in the interval. In Case [a,b] the function should posses derivatives at every point and at the end points a & b i.e., Rf1 (a) and Lf1 (a) exists.

2. State Rolle’s Theorem with Geometric Interpretation. Statement: Let f (x) be a function is defined on [a,b] & it satisfies the following Conditions.

(i) f (x) is continuous in [a,b]

(ii) f (x) is differentiable in (a,b) (iii) f (a) = f (b) Then there exists at least a point C∈ (a,b), Here a < b such that f1 ( c ) = 0

Proof: Geometrical Interpretation of Rolle’s Theorem: Y y = f (x) P R A B A B f(a) Q S →f(a) → f(b) O x = a x = c x = b x =a c1 c2 c3 c4 x = b Let us consider the graph of the function y = f (x) in xy – plane. A (a,.f(a)) and B (b, f( b ) ) be the two points in the curve f (x) and a, b are the corresponding end points of A & B respectively. Now, explained the conditions of Rolle’s theorem as follows.

(i) f (x) is continuous function in [a,b], Because from figure without breaks or jumps in between A & B on y = f (x).

(ii) f (x) is a differentiable in (a,b), that means let us joining the points A & B, we

get a line AB. ∴ Slope of the line AB = 0 then ∃ a point C at P and also the tangent at P (or Q or R or S) is Parallel to x –axis. ∴ Slope of the tangent at P (or Q or R or S) to be Zero even the curve y = f (x) decreases or increases, i.e., f (x) is Constant.


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f1 (x) = 0 ∴f1 (c) = 0 (iii) The Slope of the line AB is equal to Zero, i.e., the line AB is parallel to x – axis. ∴ f (a) = f (b)

3. Verify Rolle’s Theorem for the function f (x) = x2 – 4x + 8 in the internal [1,3] Solution: We know that every Poly nominal is continuous and differentiable for all points and hence f (x) is continuous and differentiable in the internal [1,3]. Also f (1) = 1 – 4 + 8 = 5, f (3) = 32 – 43 + 8 = 5 Hence f (1) = f (3) Thus f (x) satisfies all the conditions of the Rolle’s Theorem. Now f1 (x) = 2x – 4 and f1 (x) = 0 ⇒ 2x – 4 = 0 or x = 2. Clearly 1 < 2 < 3. Hence there exists 2t (1,3) such that f1 (2) = 0. This shows that Rolle’s Theorem holds good for the given function f (x) in the given interval.

4. Verify Rolle’s Theorem for the function f (x) = x (x + 3) in the interval [-3, 0] Solution: Differentiating the given function W.r.t ‘x’ we get

=

∴ f1(x) exists (i.e finite) for all x and hence continuous for all x. Also f (-3) = 0, f (0) = 0 so that f (-3) = f (0) so that f (-3) = f (0). Thus f (x) satisfies all the conditions of the Rolle’s Theorem. Now, f1 (x) = 0

⇒ = 0

Solving this equation we get x = 3 or x = -2 Clearly –3 < -2 < 0. Hence there exists –2∈ (-3,0) such that f1 (-2) = 0 This proves that Rolle’s Theorem is true for the given function.

2x

e−

2221 )32(21)3()(

xxexexxxf −− ++⎟

⎠⎞

⎜⎝⎛−+=

22 )6(21 x

exx −−−−

22 )6(21 x

exx −−−−


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5. Verify the Rolle’s Theorem for the function Sin x in [-π, π] Solution: Let f (x) = Sin x Clearly Sinx is continuous for all x. Also f1 (x) = Cos x exists for all x in (-π, π) and f (-π) = Sin (-π) = 0; f (π) = Sin (π) = 0 so that f (-π) = f (π) Thus f (x) satisfies all the conditions of the Rolle’s Theorem . Now f1 (x) = 0 ⇒ Cos x = 0 so that

X = ±

Both these values lie in (-π,π). These exists C = ±

Such that f1 ( c ) = 0 Hence Rolle’s theorem is vertified.

6. Discuss the applicability of Rolle’s Theorem for the function f (x) = ІxІ in [-1,1]. Solution: Now f (x) = ⏐x⏐= x for 0 ≤ x ≤ 1 -x for –1 ≤ x ≤ 0 f (x) being a linear function is continuous for all x in [-1, 1]. f(x) is differentiable for all x in (-1,1) except at x = 0. Therefore Rolle’s Theorem does not hold good for the function f (x) in [-1,1]. Graph of this function is shown in figure. From which we observe that we cannot draw a tangent to the curve at any point in (-1,1) parallel to the x – axis. Y y = ⏐x⏐ x -1 0 1

2π

2π


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Exercise: 7. Verify Rolle’s Theorem for the following functions in the given intervals.

a) x2 – 6x + 8 in [2,4]

b) (x – a)3 (x – b)3 in [a,b]

c) log in [a,b]

8. Find whether Rolle’s Theorem is applicable to the following functions. Justify your answer.

a) f (x) = ⏐x – 1 ⏐ in [0,2]

b) f (x) = tan x in [0, π] .

9. State & prove Lagrange’s (1st) Mean Value Theorem with Geometric meaning. Statement: Let f (x) be a function of x such that (i) If is continuous in [a,b] (ii) If is differentiable in (a,b) Then there exists atleast a point (or value) C∈ (a,b) such that.

i.e., f (b) = f (a) + (b – a) f1 (c) Proof:

⎭⎬⎫

⎩⎨⎧

++

xbaabx)(

2

abafbfcf

−−

=)()()(1

[b,f(b)]

[a,f(a)]

x

y

a c b


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Define a function g (x) so that g (x) = f (x) – Ax ---------- (1) Where A is a Constant to be determined. So that g (a) = g (b) Now, g (a) = f (a) – Aa G (b) = f (b) – Ab ∴ g (a) = g (b) ⇒ f (a) – Aa = f (b) – Ab.

i.e., ---------------- (2)

Now, g (x) is continuous in [a,b] as rhs of (1) is continuous in [a,b] G(x) is differentiable in (a,b) as r.h.s of (1) is differentiable in (a,b). Further g (a) = g (b), because of the choice oif A. Thus g (x) satisfies the conditions of the Rolle’s Theorem. ∴ These exists a value x = c sothat a < c < b at which g1 ( c ) = 0 ∴Differentiate (1) W.r.t ‘x’ we get g1 (x) = f 1 (x) – A ∴ g1 ( c ) = f1 ( c )- A (∵ x =c) ⇒ f 1 ( c ) - A = 0 (∵g1 ( c ) = 0) ∴f1 ( c ) = A -------------- (3) From (2) and (3) we get

(or) f (b) = f (a) + (b – a) f1 (c) For a < c < b

Corollary: Put b – a = h i.e., b = a + h and c = a + θ h Where 0 < θ < 1 Substituting in f (b) = f (a) + (b – a) f1 ( c ) ∴ f (a + h) = f (a) + h f (a + θ h), where 0 < θ < 1.

abafbfA

−−

=)()(

abafbfcf

−−

=)()()(1


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2011 Geometrical Interpretation:- Since y = f (x) is continuous in [a,b], it has a graph as shown in the figure below, At x = a, y = f (a) At x = b, y = f (b) Y B Y B P α A C A Q x X 0 a c b 0 Figure (ii) Figure (i) Slope of the line joining the points A (a,f(a)) and B ( (b,f (b))

Is (∴ Slope = m = tan θ)

= tan α Where α is the angle mode by the line AB with x – axis = Slope of the tangent at x = c = f1 ( c ), where a < c < b Geometrically, it means that there exists at least are value of x = c, where a < c < b at which the tangent will be parallel to the line joining the end points at x = a & x = b. Note: These can be more than are value at which the tangents are parallel to the line joining points A & B (from Fig (ii)).

abbfbf

−− )()(


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10. Verify Lagrange’s Mean value theorem for f(x) = (x – 1) (x – 2) (x – 3) in [0,4].

Solution: Clearly given function is continuous in [0,4] and differentiable in (0,4), because f (x) is in polynomial. f (x) = (x – 1) (x – 2) (x – 3) f (x) = x3 – 6x2 + 11x – 6 and f (0) = 03 – 6(0)2 + 11 (0) – 6 = -6 f(4) = 43 – 6 (4)2 + 11 (4) – 6 = 6 Differentiate f (x) W.r.t x, we get F1 (x) = 3x2 – 6x + 11 Let x = c, f1 ( c) = 3c2 – 6c + 11 By Lagrange’s Mean value theorem, we have

∴3c2 – 6c + 11 = 3 ⇒ 3c2 – 6c + 8 = 0 Solving this equation, we get

C = 2 ± ∈ (0,4)

Hence the function is verified.

11. Verify the Lagrange’s Mean value theorem for f (x) = logx in [1,e]. Solution: Now Logx is continuous for all x > 0 and hence [1,e].

Also which exists for all x in (1,e)

Hence f (x) is differentiable in (1,e) ∴by Lagrange’s Mean Value theorem, we get

)04()0()4()()()(1

−−

=−−

=ff

abafbfcf

34

)6(6=

−−=

32

xxf 1)(1 =


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2011

⇒ C = e – 1 ⇒ 1 < e - < 2 < e

Since e∈ (2,3) ∴ So that c = e – 1 lies between 1 & e Hence the Theorem.

12. Find θ for f (x) = Lx2 + mx + n by Lagrange’s Mean Value theorem. Solution: f (x) = Lx2 + mx + n ∴f1 (x) = 2 Lx + m We have f (a + h) = f (a) + hf1 (a + θ h) Or f (a + h) – f (a) = hf1 (a + θ h) i.e., { (a + h)2 + m (a + h) + n} – { a2 + ma + n} = h {2 (a + θ h) + m} Comparing the Co-efficient of h2, we get

1 = 2θ ∴θ = ∈ (0,1)

Exercise:

13. Verify the Lagrange’s Mean Value theorem for f (x) = Sin2x in

14. Prove that, < tan-1 b – tan-1 a < if 0 < a < b and reduce that

15. Show that in

16. Prove that Where a < b. Hence reduce

.

ceceLogLoge 1

111

11

=−

⇒=−−

21

⎥⎦⎤

⎢⎣⎡

2,0 π

21 bab

+−

21 aab

+−

61

434tan

253

41 +<<+ − ππ

12<<

xSinx

π⎟⎠⎞

⎜⎝⎛

2,0 π

2

11

2 11 babaSinbSin

aab

−

−<−<

−

− −−

151

641

321

61 −<<− − ππ Sin


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17. State & prove Cauchy’s Mean Value Theorem with Geometric meaning. Proof: The ratio of the increments of two functions called Cauchy’s Theorem. Statement: Let g (x) and f (x) be two functions of x such that, (i) Both f (x) and g (x) are continuous in [a,b] (ii) Both f (x) and g (x) are differentiable in (a,b) (iii) g1 (x) ≠ 0 for any x ε (a,b) These three exists at least are value x = c ε (a,b) at which

Proof: Define a function, φ (x) = f (x) – A. g (x) ------------------ (1) So that φ (a) = φ (b) and A is a Constant to be determined. Now, φ (a) = f (a) – A g (a) φ (b) = f (b) – A. g (b) ∴ f (a) – A g (a) = f (b) – A. g (b)

-------------------- (2)

Now, φ is continuous in [a,b] as r.h.s of (1) is continuous in [a,b] and φ (x) is differentiable in (a,b) as r.h.s of (1) is differentiable in [a,b]. Also φ (a) = φ (b) Hence all the conditions of Rolle’s Theorem are satisfied then there exists a value x = c ∈ (a,b) such that φ1 ( c ) = 0. Now, Differentiating (1) W.r.t x, we get φ1 (x) = f1 (x) – A.g1 (x) at x = c ∈ (a,b)

)()()()(

)()(

1

1

agbgafbf

cgcf

−−

=

)()()()(

agbgbfbfA

−−

=⇒


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2011

∴φ1 ( c ) = f1 ( c ) – A g1 (c) 0 = f1 ( c) – A g1 ( c ) (∵ g1 (x) ≠ 0)

--------------- (3)

Substituting (3) in (2), we get

, where a < c < b

Hence the proof. 18. Verify Cauchy’s Mean Value Theorem for the function f (x) = x2 + 3, g (x) = x3 + 1 in [1,3] Solution: Here f (x) = x2 + 3, g (x) = x3 + 1 Both f (x) and g (x) are Polynomial in x. Hence they are continuous and differentiable for all x and in particular in [1,3] Now, f1 (x) = 2x, g1 (x) = 3x2 Also g 1 (x) ≠ 0 for all x ∈ (1,3) Hence f (x) and g (x) satisfy all the conditions of the cauchy’s mean value theorem. Therefore

, for some c : 1 < c < 3

i.e.,

i.e.,

Clearly C = lies between 1 and 3.

Hence Cauchy’s theorem holds good for the given function.

)()(

1

1

cgcfA =⇒

)()()()(

)()(

1

1

agbgafbf

cgcf

−−

=

)()(

)1()3()1()3(

1

1

cgcf

ggff

=−−

2326

228412

c=

−−

612

613

31

132

==⇒= Cc

612


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19. Verify Cauchy’s Mean Value Theorem for the functions f (x) Sin x, g (x) = Cos x in

Solution: Here f (x) = Sin x, g (x) = Cos x so that f1 (x) = Cos x ,g1 (x) = - Sinx

Clearly both f (x) and g (x) are continuous in , and differentiable in

Also g1 (x) = -Sin x ≠ 0 for all x ∈ ∴ From cauchy’s mean value theorem we obtain

for some C : 0 < C <

i.e., i.e., -1 = - Cot c (or) Cot c = 1

∴ C = , clearly C = lies between 0 and

Thus Cauchy’s Theorem is verified. Exercises: 20. Find C by Cauchy’s Mean Value Theorem for

a) f (x) = ex, g (x) = e-x in [0,1]

b) f (x) = x2, g (x) = x in [2,3] 21. Verify Cauchy’s Mean Value theorem for

a) f (x) = tan-1 x, g (x) = x in

b) f(x) = log x, g (x) = in [1,e]

⎥⎦⎤

⎢⎣⎡

2,0 π

⎥⎦⎤

⎢⎣⎡

2,0 π ( )2,0 π

( )2,0 π

)()(

)0(2

)0(2

1

1

cgcf

gg

ff=

−⎟⎠⎞

⎜⎝⎛

−⎟⎠⎞

⎜⎝⎛

π

π

2π

SincCosc−

=−−

1001

4π

4π

2π

⎥⎦

⎤⎢⎣

⎡ 1,3

1

x1


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Generalized Mean Value Theorem: 22. State Taylor’s Theorem and hence obtain Maclaurin’s expansion (series) Statement: If f (x) and its first (n – 1) derivatives are continuous in [a,b] and its nth derivative exists in (a,b) then

f(b) = f(a) + (b – a) f1 (a) + f11 (a) + ---------+ fn-1(a) +

Where a < c < b Remainder in Taylor’s Theorem: We have

f (x) = f (a) + (x – a) f 1 (a) + f 11 (a) + --------- + f n-1 (a) +

f n [a + (x – a) θ}

f (x) = S n (x) + R n (x)

Where R n (x) = f n [a + (x – a) θ] is called the Largrange’s form of the Remainder.

Where a = 0, R n (x) = f n (θx), 0 < θ < 1

Taylor’s and Maclaurin’s Series: We have f (x) = Sn (x) + Rn (x) ∴

If

Thus converges and its sum is f (x).

This implies that f (x) can be expressed as an infinite series.

i.e., f (x) = f (a) + (x – a) f 1 (a) + f 11 (a) + ---------- to ∞

This is called Taylor’s Series.

!2)( 2ab −

)!1()( 1

−− −

nab n

)(!

)( cfn

ab nn−

!2)( 2ax −

)!1()( 1

−− −

nax n

nax n

∠− )(

!)(

nax n−

!nxn

)()]()([ xRLimxSxfLim nnnn ∞→∞→=−

)()(0)( xSLimxthenfxRLim nnnn ∞→∞→=

)(xSLim nn ∞→

!2)( 2ax −


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Putting a = 0, in the above series, we get

F (x) = f (0) + x f 1 (0) + f 11 (0) + --------- to ∞

This is called Maclaurin’s Series. This can also denoted as

Y = y (0) + x y 1 (0) + y 2 (0) + -------- + y n (0) ---------- to ∞

Where y = f (x), y1 = f 1 (x), -------------- y n = f n (x) 23. By using Taylor’s Theorem expand the function e x in ascending powers of (x – 1) Solution: The Taylor’s Theorem for the function f (x) is ascending powers of (x – a) is

f (x) = f (a) + (x – a) f 1 (a) + f 11 (a) + ------------ (1)

Here f (x) = e x and a = 1 f 1 (x) = e x ⇒ f 1 (a) = e f 11 (x) = e x ⇒ f 11 (a) = e ∴ (1) becomes

e x = e + (x – 1) e + e + -------------

= e { 1 + (x – 1) + + --------}

24. By using Taylor’s Theorem expand log sinx in ascending powers of (x – 3) Solution: f (x) = Log Sin x, a = 3 and f (3) = log sin3

Now f 1 (x) = = Cotx, f 1 (3) = Cot3

f 11 (x) = - Cosec 2x, f 11 (3) = - Cosec 23

f 111 (x) = - 2Cosecx (-Cosecx Cotx) = 2Cosec 3x Cotx ∴ f 111 (3) = 2Cosec 33 Cot3

∴ f (x) = f (a) + (x – a) f 1 (a) + f 11 (a) + f 111 (a) + ------------

∴Log Sinx = f (3) + (x – 3) f 1(3) + f 11 (3) + f 111 (3) + -----------

= logsin3 + (x – 3) Cot3 + (-Cosec23) + 2 Cosec 33Cot3 + ---

!2)( 2x

!2)( 2x

!)(

nx n

2)( 2

∠− ax

2)1( 2−x

2)1( 2−x

SinxCosx

!2)( 2ax −

!3)( 3ax −

!2)3( 2−x

!3)3( 3−x

!2)3( 2−x

!3)3( 3−x


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Exercise:

25. Expand Sinx is ascending powers of

26. Express tan –1 x in powers of (x – 1) up to the term containing (x – 1) 3 27. Apply Taylor’s Theorem to prove

e x + h = e x

Problems on Maclaurin’s Expansion:

28. Expand the log (1 + x) as a power series by using Maclaurin’s theorem. Solution: Here f (x) = log (1 + x), Hence f (0) = log 1 = 0 We know that

= , n = 1,2,---------

Hence f n (0) = (-1) n-1 (n – 1) ! f 1 (0) = 1, f 11 (0) = -1, f 111 (0) = 3!, f 1v (0) = - 3! Substituting these values in

f (x) = f (0) + x f 1 (0) + f 11 (0) + -------- + f n (0) + ------------

∴ log (1+x) = 0 + x . 1 + (-1) + 2! + - 3! + ---------------

= x - + - + ---------

This series is called Logarithmic Series.

⎟⎠⎞

⎜⎝⎛ −

2πx

⎥⎦

⎤⎢⎣

⎡−−−−−−−−−++++

!3!21

32 hhh

{ }⎭⎬⎫

⎩⎨⎧

+=+= −

−

xdxdx

dxdxf n

n

n

nn

11)1log()( 1

1

n

n

nn)1(

)!1()1( 1

+−− −

!2

2x!n

xn

!2

2x!3

3x!4

4x

2

2x3

3x4

4x


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29. Expand tan –1 x by using Macluarin’s Theorem up to the term containing x 5 Solution: let y = tan –1 x, Hence y (0) = 0

We find that y 1 = which gives y 1 (0) = 1

Further y 1 (1 + x 2) = 1, Differentiating we get Y 1 . 2x + (1 + x 2) y 2 = 0 (or) (1 + x2) y 2 + 2xy 1 = 0 Hence y 2 (0) = 0 Taking n th derivative an both sides by using Leibniz’s Theorem, we get

(1 + x 2) y n + 2 + n . 2xy n +1 + . 2. y n + 2xy n – 1 + n.2.y n = 0

i.e., (1 + x 2) y n + 2 + 2 (n +1) x y n + 1 + n (n + 1) y n = 0 Substituting x = 0, we get, y n + 2 (0) = -n (n + 1) y n (0) For n = 1, we get y 3 (0) = - 2y 1 (0) = - 2 For n = 2, we get y 4 (0) = - 2 .3.y 2 (0) = 0 For n = 3, we get y 5 (0) = - 3.4.y 3 (0) = 24 Using the formula

Y = y (0) + x y 1 (0) + y 2 (0) + y 3 (0) + ---------

We get tan –1 x = x - + - -------------

Exercise: 30. Using Maclaurin’s Theorem prove the following:

a) Secx = 1 + + + --------

b) Sin –1 x = x + + + --------------

c) e x Cos x = 1 + x - + ---------------

d) Expand e ax Cos bx by Maclaurin’s Theorem as far as the term containing x 3

211x+

2.1)1( −nn

!2

2x!3

3x

3

3x5

5x

!2

2x!4

5 4x

6

3x40

3 5x

3

3x


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Exercise : Verify Rolle’s Theorem for

(i) in ,

(ii) in

(iii) in .

Exercise : Verify the Lagrange’s Mean Value Theorem for

(i) in

(ii) in

Exercise : Verify the Cauchy’s Mean Value Theorem for

(i) and in

(ii) and in

(iii) and in

)cos(sin)( xxexf x −= ⎥⎦⎤

⎢⎣⎡

45,

4ππ

2/)2()( xexxxf −= [ ]2,0

2sin 2( ) x

xf xe

= ⎥⎦⎤

⎢⎣⎡

45,

4ππ

( ) ( 1)( 2)f x x x x= − −10,2

⎡ ⎤⎢ ⎥⎣ ⎦

1( )f x Tan x−= [ ]0,1

( )f x x=1( )g xx

= 1 ,14

⎡ ⎤⎢ ⎥⎣ ⎦

21( )f xx

=1( )g xx

= [ ],a b

( )f x Sin x= ( )g x Cos x= [ ],a b


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UNIT – II

DIFFERENTIAL CALCULUS-II

Give different types of Indeterminate Forms. If f (x) and g (x) be two functions such that and both exists, then

If = 0 and = 0 then

Which do not have any definite value, such an expression is called

indeterminate form. The other indeterminate forms are , 00, ∞0 and 1 ∞

1. State & prove L’ Hospital’s Theorem (rule) for Indeterminate Forms.

L’Hospital rule is applicable when the given expression is of the form or

Statement: Let f (x) and g (x) be two functions such that (1) = 0 and = 0

(2) f1 (a) and g1 (a) exist and g1 (a) ≠ 0

Then

Proof: Now , which takes the indeterminate form . Hence applying the

L’ Hospitals theorem, we get

)(xfLimax→

)(xgLimax→

)(

)(

)()(

xgLim

xfLim

xgxfLim

ax

ax

ax→

→

→=

)(xfLimax→

)(xgLimax→

00

)()(

=→ xg

xfLimax

∞−∞∞×∞∞ ,0,

00

∞∞

)(xfLimax→

)(xgLimax→

)(

)(

)()(

1

1

xgLim

xfLim

xgxfLim

ax

ax

ax→

→

→=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

→→)(

1

1

)()(

xf

xgLimxgxfLim

axax 00


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2011

If then

1

i.e

If = 0 or ∞ the above theorem still holds good.

2. Evaluate form

Solution: Apply L’Hospital rule, we get

= 1

3. Evaluate

Solution: = form

Apply L’ Hospital rule

=

2

1

1

2

1

2

1

)()(

)()(

)]([)(

)]([)(

)()(

⎥⎦

⎤⎢⎣

⎡=

−

−

=→→→ xg

xfxfxgLim

xfxf

xgxg

LimxgxfLim

axaxax

2

1

1

)()(

)()(

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=

→→ xgxfLim

xfxgLim

axax

∞≠≠→

andxgxfLim

ax0

)()(

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=

→→ )()(

)()(

1

1

xgxfLim

xfxgLim

axax

)()(

)()(

1

1

xgxfLim

xgxfLim

axax →→=

)()(

xgxfLim

ax→

00

=→ x

SinxLimax

11

11==

→

θCosCosxLimax

1=∴→ x

SinxLimax

CotxSinxLim

ax

log→

CotxSinxLim

ax

log→ ∞

∞−=

∞=

0log0

0logCot

Sin

xCotxxCoCoxCosceLim

ax secsec2

2

−−

→


10 MAT11

39

2011

Exercise: 1

Evaluate

a)

b)

c)

d )

4. Explain ∞ - ∞ and 0 × ∞ Forms:

Solution: Suppose = 0 and in this case

- g(x) = 0 × ∞, reduce this to or form

Let form

Or form

L’ Hospitals rule can be applied in either case to get the limit. Suppose = ∞ and in this case form, reduce

this or form and then apply L’Hospitals rule to get the limit

02

1==

→ CotxLim

ax

0log=∴

→ CotxSinxLim

ax

xxLim

x

tan0→

xx

xLim1

0)1( +

→

xaLim

x

x

1−∞→

axaxLim

nn

x −−

→0

)(xfLimax→

∞=→

)(xgLimax

)(xfLimax→ 0

0∞∞

[ ]00

)(1

)()().( =⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=→→

xg

xfLimxgxfLimaxax

[ ]∞∞

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=→→

)(1

)()().(xf

xgLimxgxfLimaxax

)(xfLimax→

∞=→

)(xgLimax

[ ] ∞−∞=→

)().( xgxfLimax

00

∞∞


10 MAT11

40

2011

5. Evaluate

Solution: Given = ∞ - ∞ form

∴ Required limit = form

Apply L’Hospital rule.

=

6. Evaluate

Solution: Given limit is ∞ - ∞ form at x = 0. Hence we have

Required limit =

= form

Apply L’ Hospital’s rule

=

= form

Apply L’ Hospitals rule

⎭⎬⎫

⎩⎨⎧ +

−→ 20

)1log(1x

xx

Limx

⎭⎬⎫

⎩⎨⎧ +

−→ 20

)1log(1x

xx

Limx

001log(

20=

⎭⎬⎫

⎩⎨⎧ +−

→ xxxLim

x

xxLim

x 21

11

0

+−

→

21

)1(21

21

00=

+=+=

→→ xLim

xx

x

Limxx

⎭⎬⎫

⎩⎨⎧ −

→Cotx

xLimx

10

⎭⎬⎫

⎩⎨⎧ −

→ SinxCosx

xLimx

10

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛ −

→ 00

0 xSinxxCosxSinxLim

x

SinxxCosxxSinxCosxCosxLim

x ++−

→0

⎟⎠⎞

⎜⎝⎛=

+→ 00

0 SinxxCosxxSinxLim

x


10 MAT11

41

2011

=

=

7. Evaluate tan x log x

Solution: Given limit is (0 × - ∞) form at x = 0



=

= form

Apply L Hospitals rule

=

8. Evaluate

Solution: Given limit is (∞ × 0) form at x = 1



CosxxSinxCosxSinxxCosxLim

x +−+

→0

020

020

==−

0→xLim

⎥⎦⎤

⎢⎣⎡

∞∞−

=→ Cotx

xLimx

log0

xCoxLim

x 20 sec

1

−→

⎟⎠⎞

⎜⎝⎛−

→ 002

0 xxSinLim

x

01

20

=−

→

SinxCosxLimx

xxSecLimx

log.21

⎟⎠⎞

⎜⎝⎛

→

π

00

2

log1 xCos

xLimx π→

πππ2

2.

2

1

1−=

−=

→ xSin

xLimx


10 MAT11

42

2011

Exercise: 2

Evaluate

a)

b)

c)

d)

9. Explain Indeterminate Forms , , , Solution: At x = a, takes the indeterminate form (i) if f(x) = 0 and g (x) = 0 (ii) if f(x) = 1 and g (x) = ∞

(iii) if f(x) = ∞ and g(x) = 0 and

(iv) if f(x) = 0 and f (x) = ∞

In all these cases the following method is adopted to evaluate

Let L = so that

Log L = [g(x) log f (x)] = 0 × ∞

Reducing this to and applying L’ Hospitals rule, we get Log L = a Or

L = ea

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−→ xxxLim

x log1

11

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−

→ axCot

xaLim

x 0

⎟⎠⎞

⎜⎝⎛

−−

→ SinxSecxLim

x 11

2π

xaLim xx

⎟⎠⎞⎜

⎝⎛ −

∞→1

1

00 ∞1 0∞ ∞0

[ ] )()( xgxf

00ax

Lim→ ax

Lim→

∞1ax

Lim→ ax

Lim→

0∞ax

Lim→ ax

Lim→

∞0ax

Lim→ ax

Lim→

axLim

→[ ] )()( xgxf

axLim

→[ ] )()( xgxf

axLim

→

∞∞or

00


10 MAT11

43

2011

10. Evaluate x Sinx

Solution: let L = x Sinx ⇒ 00 form at x = 0

Hence Log L = Sinx log x ⇒ 0 × ∞ form

∴ form

Apply L’ Hospital rule,

= form

Apply L’ Hospitals rule we get

=

∴ L = 0

11. Evaluate

Solution: let L = is 1∞ form

form


=

0→xLim

0→xLim

0→xLim

⎟⎠⎞

⎜⎝⎛

∞∞

==→→ Coscex

xLimSinx

xLimLogLxx

log1log

00

( )00tan.1

00 xxSinxLim

CoscexCotxxLim

xx

−=

− →→

−∞=−

→ 1tansin 2

0

xCosxxxSecLimx

011=

∞===⇒−∞=∴ ∞

∞−

eeLLogL

xx

xLim −

→

11

1)(

x

xxLim −

→

11

1)(

⎟⎠⎞

⎜⎝⎛≡⎟

⎠⎞

⎜⎝⎛

−=∴

→ 00log

11

1x

xLimLogLx

111

1

11−=

−=

− →→ xLimxLimxx


10 MAT11

44

2011

∴ Log L = -1

⇒ L =

12. Evaluate

Solution: let L = form

form

form


=

form

Apply L’ Hospital rule, we get

Log L =

ee 11 =−

21

0

tan x

x xxLim ⎟

⎠⎞

⎜⎝⎛

→

∞

→≡⎟

⎠⎞

⎜⎝⎛ 1tan 2

1

0

x

x xxLim

)0(tanlog120

×∞≡⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛=∴

→ xx

xLimLogLx

( )⎟⎠⎞

⎜⎝⎛≡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=∴→ 0

0tanlog20 x

xx

LimLogLx

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧ −

→ xx

xxxSec

xx

Limx 2

tan

tan1 2

2

0

⎟⎠⎞

⎜⎝⎛=⎥

⎦

⎤⎢⎣

⎡ −=

→ 00tan

21

3

2

0 xxxxSecLimLogL

x

2

222

0 3tan2

21

xxSecxxxSecxSecLim

x

−+=

→

⎟⎠⎞

⎜⎝⎛=

→ xxxSecLim

x

tan)(31 2

0

31

31

eL =∴


10 MAT11

45

2011

Exercise: 3 Evaluate the following limits.

a) b)

c) d)

Evaluate the following limits.

Cotx

xSecxLim )(

0→

⎟⎠⎞

⎜⎝⎛

→⎟⎠⎞

⎜⎝⎛ −

ax

ax axLim

2tan

2π

xexLim

x

x

−+→

1

0

)1( 2)(0

xx

CosaxLim∞

→

3

30 0

2 log(1 ) log(1 )( ) lim ( ) limsin sin

x x

x x

e e x xi iix x x

−

→ →

− − + +

20 22

1 sin cos log(1 ) log sin( ) lim ( ) limtan ( )

2x x

x x x xiii ivx x x

π π→ →

+ − + −

−

1

2 202

cosh log(1 ) 1 sin sin( ) lim ( ) limx x

x x x x xv vix xπ

−

→ →

+ − − +

2 2

0

(1 )( ) limlog(1 )

x

x

e xviix x→

− ++

( )0

( ) lim 2 cot( ) ( ) lim cos cotx a x

xi x a ii ecx xa→ →

⎛ ⎞− − −⎜ ⎟⎝ ⎠

220

2

1( ) lim tan sec ( ) lim cot2 xx

iii x x x iv xxπ

π→→

⎡ ⎤ ⎛ ⎞− −⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

[ ]202

1 1( ) lim ( ) lim 2 tan sectanx x

v vi x x xx x x π

π→ →

⎡ ⎤− −⎢ ⎥⎣ ⎦

22

1

0 0

1 cos( ) lim(cos ) ( ) lim2

b xx

x x

xi ax ii→ →

⎛ ⎞+⎜ ⎟⎝ ⎠


10 MAT11

46

2011

Evaluate the following limits.

21

12 log(1 )

1 0

sin( ) lim(1 ) ( ) limx

x

x x

xiii x ivx

−

→ →

⎛ ⎞− ⎜ ⎟

⎝ ⎠

( )cottan

0 0( ) lim(sin ) ( ) lim 1 sin xx

x xv x iv x

→ →+

( )2 tan 2cos

04

( ) lim(cos ) ( ) lim tan xec x

x xvii x viii x

π→ →

1

0

1( ) lim( ) ( ) lim1 3

x x x xx

x x

ax a b cix xax→∞ →

⎛ ⎞+ + +⎜ ⎟− ⎝ ⎠

22

1

0 0

1 cos( ) lim(cos ) ( ) lim2

b xx

x x

xi ax ii→ →

⎛ ⎞+⎜ ⎟⎝ ⎠

21

12 log(1 )

1 0

sin( ) lim(1 ) ( ) limx

x

x x

xiii x ivx

−

→ →

⎛ ⎞− ⎜ ⎟

⎝ ⎠

( )cottan

0 0( ) lim(sin ) ( ) lim 1 sin xx

x xv x iv x

→ →+

( )2 tan 2cos

04

( ) lim(cos ) ( ) lim tan xec x

x xvii x viii x

π→ →

1

0

1( ) lim( ) ( ) lim1 3

x x x xx

x x

ax a b cix xax→∞ →

⎛ ⎞+ + +⎜ ⎟− ⎝ ⎠


10 MAT11

47

2011

Polar Curves If we traverse in a hill section where the road is not straight, we often see caution boards hairpin bend ahead, sharp bend ahead etc. This gives an indication of the difference in the amount of bending of a road at various points which is the curvature at various points. In this chapter we discuss about the curvature, radius of curvature etc. Consider a point P in the xy-Plane.

r = length of OP= radial distance θ = Polar angle ( r, θ)→ Polar co-ordinates

Let r = f (θ) be the polar curve x = r Cos θ y = r Sin θ

Relation (1) enables us to find the polar co-ordinates ( r, θ) when the Cartesian co-ordinates ( x, y) are known. Expression for arc length in Cartesian form. Proof: Let P (x,y) and Q (x + δx, Y + δy) be two neighboring points on the graph of the function y = f (x). So that they are at length S and S + δs measured from a fixed point A on the curve. Y = f (x) Q T (Tangent) δs δy A δx N O δx X

From figure, ,

∠TPR = ψ and PR =δx, RQ = δy

SPQ δ=∩

SAP =∩

( )2 2 1, tan (1)yr x y xθ − −−−−−−= + =


10 MAT11

48

2011

∴ Arc PQ = δS From ∆le PQR, we have [Chord PQ]2 = PR2 + QR2 [Chord PQ]2 = (δx)2 + (δy)2 (∵ from figure) When Q is very close to point P, the length of arc PQ is equal to the length of Chord PQ. i.e arc PQ = Chord PQ = δs ∴ (δs)2 = (δx)2 + (δy)2 -------- (1) ÷ (δx)2, we get

When Q → P along the curve, δx → 0, δs → 0

i.e.,

∴ --------------(2)

Similarly, dividing (1) by δy and taking the limit as δy → 0, we get

------------(3)

22

1 ⎟⎠⎞

⎜⎝⎛+=⎟

⎠⎞

⎜⎝⎛

xy

xs

δδ

δδ

2

0

2

01 ⎟

⎠⎞

⎜⎝⎛+=⎟

⎠⎞

⎜⎝⎛∴

→→ xyLim

xsLim

xx δδ

δδ

δδ

22

1 ⎟⎠⎞

⎜⎝⎛+=⎟

⎠⎞

⎜⎝⎛

dxdy

dxds

2

1 ⎟⎠⎞

⎜⎝⎛+=

dxdy

dxds

2

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

dydx

dyds


10 MAT11

49

2011

Expressions for &

Trace a tangent to the curve at the point P, it makes an angle ψ with the x – axis. From ∆le PRT, we have

tan ψ =

Q Equation (2) becomes, Y Tangent T T A P ψ R ψ P R ψ O M N X

=

∴ = Sec ψ ( Or) = Cos ψ

and equation (3) becomes

= = Cosec ψ

∴ = Cosecψ or = Sin ψ

Derive an expression for arc length in parametric form. Solution: Let the equation of the curve in Parametric from be x = f (t) and y = g (t). We have,

dxds

dyds

dxdy

dxds ψ2tan1+

ψψ sec2 == Sec

dxds

dsdx

dyds ψ21 Cot+ ψ2secCo=

dyds

dsdy


10 MAT11

50

2011

(δs)2 = (δx)2 + (δy)2 ÷ by (δt)2, we get

Taking the limit as δt → 0 on both sides, we get

(Or) ------------ (4)

Derive an expression for arc length in Cartesian form. Solution: Let P (r,θ) and Q (r + δr, θ + δθ) be two neighboring points on the graph of the function r = f (θ). So that they are at lengths S and s + δs from a fixed point A on the curve. ∴ PQ = (S + δs) – s = δs Draw PN ⊥ OQ From ∆le OPN, Q Tangent N r + δr P φ r δθ θ X O

222

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

ty

tx

ts

δδ

δδ

δδ

2

0

2

0

2

0⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

→→→ tyLim

txLim

tsLim

ttt δδ

δδ

δδ

δδδ

222

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛∴

dtdy

dtdx

dtds

22

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=

dtdy

dtdx

dtds


10 MAT11

51

2011

i.e., = Sin δθ (or) PN = r sinδθ

and = Cos δθ

i.e = Cos δθ (or) ON = r Cos δθ

When Q is very close to P, the length of arc PQ as equal to δs, where δs as the length of chord PQ. In ∆le PQN, (PQ)2 = (PN)2 + (QN)2 but PN = r sin δθ (∵Sin δθ ≈ δθ) And QN = OQ – ON = (r + δr) – r Cos δθ = r + δr – r ( ∵Cos δθ ≈ 1) ∴ QN = δr And (PQ)2 = (PN)2 + (QN)201 (δS)2 + (rδθ)2 + (δr)2 ---------------- (5) ÷ by (δθ)2 we get

When Q → P along the curve δθ → 0 as δS → 0 and δr → 0

δθSinOPPN

=

rPN

OPON

rON

22

2

⎟⎠⎞

⎜⎝⎛+=⎟

⎠⎞

⎜⎝⎛

δθδ

δθδ rrS


10 MAT11

52

2011

i.e

i.e., --------------- (6)

Similarly equation (5) ÷ by (δr)2 and taking the limits as δr → 0 we get

+ 1

i.e + 1

∴ --------------- (7)

Note: Angle between Tangent and Radius Vector:- We have,

Tan φ = r

i.e., r

= r .

∴Sinφ = r and Cos φ =

2

0

22

0⎟⎠⎞

⎜⎝⎛+=⎟

⎠⎞

⎜⎝⎛

→→ δθδ

δθδ

δθδθ

rLimrSLim

22

2

⎟⎠⎞

⎜⎝⎛+=⎟

⎠⎞

⎜⎝⎛

θθ ddrr

dds

22 ⎟

⎠⎞

⎜⎝⎛+=

θθ ddrr

dds

2

0

22

0⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

→→ rrLim

rSLim

rr δδθ

δδ

δδ

22

2

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

drdr

drds θ

12

2 +⎟⎠⎞

⎜⎝⎛=

drdr

drds θ

drdθ

=φφ

CosSin

drdθ

dsdθ

drds

=φφ

CosSin

dsdr

dsdr θ

dsdθ

dsdr


10 MAT11

53

2011

Find and for the following curves:-

1) y = C Cos h

Solution: y = C Cos h

Differentiating y w.r.t x. we get

= Sin h

= Cosh

Again Differentiating y w.r.t y we get

1 = C Sin h

i.e = Cosech

2) x3 = ay2 Solution : x3 = ay2 Differentiating w.r.t y and x separately we get

dxds

dyds

⎟⎠⎞

⎜⎝⎛

cx

⎟⎠⎞

⎜⎝⎛

cx

dxdy

⎟⎠⎞

⎜⎝⎛

cx

2

1 ⎟⎠⎞

⎜⎝⎛+=∴

dxdy

dxds

( ) ( )cxCoshc

x 22sinh1 =+≡

dxds

⎟⎠⎞

⎜⎝⎛

cx

⎟⎠⎞

⎜⎝⎛

cx

C1

dydx

dydx

⎟⎠⎞

⎜⎝⎛

cx

( )cxhCo

dyds 2sec1 +=∴


10 MAT11

54

2011

3x2 = 2 ay and 3 x2 = 2 ay

i.e., = and =

We know that

=

=

=

= and

=

(∴x3 = ay2)

3. y = log cos x Solution . y= log cos x Differentiating w.r.t x and y separately, we get

= ( – Sin x) = - tan x

i.e., = - tan x

and 1 = (- Sin x)

dydx

dxdy

dydx

232

xay

dxdy

ayx

23 2

dxds 2

1 ⎟⎠⎞

⎜⎝⎛+

dxdy

22

231 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+

ayx

22

2

22

3

491

491

yaxay

yaxx

+=+

dxds

ax

491+

2

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

dydx

dyds 2

2321 ⎟

⎠⎞

⎜⎝⎛+

xay

21

21

4

22

941

941 ⎟

⎠⎞

⎜⎝⎛ +=⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

xa

xya

dyds

dxdy

Cosx1

dxdy

Cosx1

dydx


10 MAT11

55

2011

i.e., 1 = - tan x (Or) = - Cot x

We have

= and

& =

=

∴ = Sec x and

Find for the following Curves:-

1. x = a (Cos t + t Sin t), y = a (Sin t = t Cos t) 2. x = a Sec t , y = b tan t 3. x = a , y = a Sin t Solution of 1 Given x = a (Cos t + t Sin t), y = a (Sin t – t Cos t ) Differentiating x & y W.r.t ‘t’, we get

= a (-Sin t + Sin t + t Cos t)

= a t cos t

and = a (Cos t – Cos t + t Sin t)

= at Sin t

dydx

dydx

dxds 2

1 ⎟⎠⎞

⎜⎝⎛+

dxdy

2

1 ⎟⎟⎠

⎞⎜⎜⎝

⎛+=

dydx

dyds

xdxds 2tan1+=∴

dyds xCot 21+

xSec2

dxds 2

12 )1( xCot

dyds

+=

dtds

( )2tanlog tCost +

dtdx

dtdx

dtdy

dtdy

22

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=∴

dtdy

dtdx

dtds


10 MAT11

56

2011

=

= at

Solution of 2 x = a Sec t, x = b tan t

∴ = a Sec t tan t, = b Sec2t

We have

= (a2Sec2 t tan2t + b2 Sec4t)

= [a2 Sec2 t (Sec2t – 1) + b2 Sec4 t]

= [a2 Sec4 t – a2 Sec2 t + b2 Sec4 t]

= [(a2 + b2) Sec4 t – a2 Sec2t]

Solution of 3

x = a (Cos t + log tan ), y = a Sin t

Differentiating x and y w.r.t ‘t’ we get

= a , = a Cos t

= , =

= a Cot t

tSintatCosta 222222 +

dtds

dtdx

dtdy

22

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=∴

dtdy

dtdx

dtds

21

21

21

dtds 2

1

2t

dtdx

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+−

21.2

2tan1int 2 tSec

tS

dtdy

22

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛=∴

dtdy

dtdx

dtds

21

22

22

2

2tan22int

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+− tCosa

t

tSecSa

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+−

2tan42

2tan2int

2

42222

t

tSecat

tSecSaa


10 MAT11

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2011

Find and for the following curves

1. r = a (1 – Cos θ)

2 .r2 = a2 Cos 2θ

3. r = a eθCot α Solution of 1

r = a (1 – Cos θ)

Differentiating r w.r.t θ we get

= a Sin θ

Hence

=

= {a2 (1 – Cos θ)2 + a2 Sin2 θ}

= {a2 (1 – 2 Cos θ + Cos2θ) + a2 Sin2 θ}

= {a2 – 2a2 Cos θ + a2}

= {2a2 – 2a2 Cos θ}

= a{ 2(1 – Cos θ)}

= a {2 (2 Sin2 )}

= 2 a Sin

And =

=

θdds

θddr

θddr

θdds 2

12

2

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

θddrr

21

21

21

21

21

2θ 2

1

θdds

2θ

drds 2

12

21⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

drdr θ

21

22

22 )1(1⎭⎬⎫

⎩⎨⎧ −

+θθ

SinaCosa


10 MAT11

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2011

=

=

=

Solution of 2

r2 = a2 Cos 2θ

Differentiating W.r.t ‘θ’ we get

2r = -a2 Sin 2θ . 2

r = -a2 Sin 2θ

=

Hence

=

=

=

=

=

21

2

22 )1(

⎭⎬⎫

⎩⎨⎧ −+

θθθ

SinCosSin

{ }

222222

2)1(2 2

1

θθ

θ

θ

θ

θθ

CosSin

Sin

Sin

Sin

SinCos

=⎟⎠⎞

⎜⎝⎛

=−

drds

2

1θCos

θddr

θddr

θddr

rSina βθ22

θdds 2

12

2

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

θddrr

21

222 2

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+

rSinar θ

21

2

242 2

⎭⎬⎫

⎩⎨⎧

+r

Sinar θ

21

2

242

22

⎭⎬⎫

⎩⎨⎧

+θθθ

CosaSinaCosa

21

2

2424

222

⎭⎬⎫

⎩⎨⎧ +

θθθ

CosaSinaCosa


10 MAT11

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2011

=

Or = (∵r2 = a2 Cos 2θ)

And =

=

=

=

= = = Cosec 2θ

Solution 3

r = aeθCot α , here α is constant

Differentiating w.r.t ‘θ’ we get

= a eθCot α. Cot α

Hence

=

= {a2 e2θCotα + a2 e2θCot α cot2 α}

= a eθCot α {1 + Cot2α}

= a eθCot α {Cosce2α}

θθ 22

21

2

4

Cosa

Cosaa

=⎭⎬⎫

⎩⎨⎧

θdds

ra

ra 22

1

2

4

=⎭⎬⎫

⎩⎨⎧

drds 2

12

21⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

drdr θ

21

2

22

21

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛

−+

θSinarr

21

24

4

21

⎭⎬⎫

⎩⎨⎧

+θSina

r

21

24

2424

2sin22

⎭⎬⎫

⎩⎨⎧ +

θθθ

aCosaSina

21

24

4

2sin ⎭⎬⎫

⎩⎨⎧

θaa

θ21

Sin

θddr

θdds 2

12

2

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

θddrr

21

21

21


10 MAT11

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2011

= a eθCot α Cosce α

and =

=

= {1 + tan2 α}

= {Sec2 α} = Sec α

Exercises:

Find and to the following curves.

1. rn = an Cos nθ

2. r (1 + Cos θ) = a

3. rθ = a

Note:

We have Sin φ = r and

Cos φ =

∴ = Cosφ = (1 – Sin2φ)

= Since P = r Sin φ.

=

θdds

drds 2

12

21⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

drdr θ

21

2222

2

11⎭⎬⎫

⎩⎨⎧ +

ααθαθ

CotCotCot

eaea

21

21

drds

θdds

dsdθ

dsdr

dsdr 2

1

21

2

2

1 ⎥⎦

⎤⎢⎣

⎡−

rp

dsdr

rpr 22 −


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2011

=

Q

r+δr P(x,y) φ OR = P r

O X P

R Sin φ =

∴ P = r Sin φ

Prove that with usual notations ⎟⎠⎞

⎜⎝⎛=

drdr θφtan

Let P (r, θ) be any point on the curve r = f (θ)

r OP and POX =θ=∴∧

Let PL be the tangent to the curve at P subtending an angle ψ with the positive direction of

the initial line (x – axis) and φ be the angle between the radius vector OP and the tangent PL.

That is φ=∧

L P O

From the figure we have

drds

∴22 pr

r−

rp

OPOR

=


10 MAT11

62

2011 θ+φ=ψ

(Recall from geometry that an exterior angle is equal to the sum of the interior opposite angles) ( )θ+φ=ψ⇒ tan tan

θφφ+φ

ψ tan tan - 1 tan tan tan or …(1)

Let (x, y) be the Cartesian coordinates of P so that we have, X = r cos θ , y = r sin θ

Since r is a function of θ , we can as well regard these as parametric equations in terms of θ .

We also know from the geometrical meaning of the derivative that

PL tangent theof slope dxdy tan ==ψ

ie., ddx

ddy tan

θθ=ψ since x and y are function of θ

ie., ( )

( ) θ=′

θ′+θθ′+θ

=θ

θ

θθ=ψ

ddr r where

cos r sin r -sin r cosr

cosr dd

sin r dd

tan

Dividing both the numerator and denominator by θ′ cos r we have,

θ′θ′

+θ′θ−

θ′θ′

+θ′θ

=ψ

cos r cos r

cos rsin r

cos rsin r

cos r cosr

tan

Or θ

′

θ+′=ψ

tan . rr - 1

an t rr

tan …(2)

Comparing equations (1) and (2) we get

⎟⎠⎞

⎜⎝⎛ θ

=φ⎟⎠⎞

⎜⎝⎛

θ

=′

=φdrdr or tan

ddrr

rr tan


10 MAT11

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2011

Prove with usual notations 2

422 ddr

r1

r1

p1

⎟⎠⎞

⎜⎝⎛

θ+= or

r1 u where

ddu u 1 2

22 =⎟

⎠⎞

⎜⎝⎛+=

θp

Proof : Let O be the pole and OL be the initial line. Let P (r, θ ) be any point on the curve and hence

we have OP = r and θ=∧

POL

Draw ON = p (say) perpendicular from the pole on the tangent at P and let φ be the angle

made by the radius vector with the tangent.

From the figure θ==∧∧

POL 90 PNO

Now from the right angled triangle ONP

OPON sin =φ

ie., φ==φ sin r por rP sin

we have p = r sin φ …(1)

and θ

=φddr

r1 cot …(2)

Squaring equation (1) and taking the reciprocal we get,


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2011

φ=φ

= cosec r1

p1 ie.,

sin1 .

r1

p 1 2

22222

Or ( )φ+= cot 1 r1

p 1 2

22

Now using (2) we get,

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

θ+=

2

222 ddr

r11

r1

p 1

Or 2

422 ddr

r1

r1

p 1

⎟⎠⎞

⎜⎝⎛

θ+= …(3)

Further, let u r1

=

Differentiating w.r.t. θ we get, 22

42 ddu

ddr

r1

ddu

ddr

r 1

⎟⎠⎞

⎜⎝⎛

θ=⎟

⎠⎞

⎜⎝⎛

θ⇒

θ=⎟

⎠⎞

⎜⎝⎛

θ− , by squaring

Thus (3) now becomes 2

22 d

du u p 1

⎟⎠⎞

⎜⎝⎛

θ+= …(4)

1. Find the angle of intersection of the curves:

( ) ( )θ=θ+= cos-1b r & cos 1 a r

Solution : r = a (1 + cos θ) : r = b(1 – cos θ)

) cos (1 log a log r log θ++=⇒ : log r = log b + log (1 – cos θ)

Differentiating these w.r.t. θ we get

cos1

sin - 0 ddr

r1

θ+θ

+=θ

: cos1

sin 0 ddr

r1

θ−θ

+=θ

( ) ( )( )/2 cos 2

/2 cos /2sin 2 cot 21 θθθ−

=φ : ( ) ( )( )/2 sin 2

/2 cos /2sin 2 cot 22 θθθ

=φ

ie., ( ) ( )/2 /2cot /2 tan - cot 1 θ+π=θ=φ : ( )/2cot cot 2 θ=φ

/2 /2 1 θ+π=φ⇒ : /2 2 θ=φ


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2011

∴ angle of intersection = /2 /2 - /2 /2 - 21 πθθπφφ =+=

Hence the curves intersect orthogonally. 2. S.T. the curves ( ) ( )θ=θ+= sin-1a r & sin 1 a r cut each other orthogonally

Solution : log r = log a + log (1 + sin θ) : log r = log a + log (1 – sin θ)

Differentiating these w.r.t θ we get

θ+

θ=

θ sin 1 cos

ddr

r1 :

θ−θ

=θ sin 1

cos- ddr

r1

ie., sin 1

cos cot 1 θ+θ

=φ : sin 1

cos- cot 2 θ−θ

=φ

We have θθ

=φθ

θ+=φ

cos -sin - 1 tan and

cossin 1 tan 21

1- cos-

cos cossin-1 tan . tan 2

2

2

2

21 =θθ

=θ−θ

=φφ∴

Hence the curves intersect orthogonally.


r = sin θ + cos θ, r = 2 sin θ

Solution : log r = log (sin θ + cos θ) : r = 2 sin θ

( )θ+θ=⇒ cos sin log r log : log r = log 2 + log (sin θ)

Differentiating these w.r.t θ we get

θ+θθθ

=θ cos sin

sin - cos ddr

r1 :

sin cos

ddr

r1

θθ

=θ

ie., ( )( )θ+θ

θθ=φ

tan 1 cos tan -1 cos cot 1 : θ=φ⇒θ=φ cot cot 22

ie., ( ) θ+π=φ⇒θ+π=φ /4 /4cot cot 11

/4 - /4 - 21 π=θθ+π=φφ∴

The angle of intersection is 4/π


10 MAT11

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2011

4. Find the angle of the curves: r = a log θ and r = a/ log θ

Solution : r = a log θ : r = a/ log θ

( )θ+=⇒ log log a log r log : ( )θ= log log - a log r log

Differentiating these w.r.t q, we get,

θθ=

θ . log1

ddr

r1 :

θθ−=

θ . log1

ddr

r1

ie., log

1 cot 1 θθ=φ :

log 1 - cot 2 θθ

=φ

Note : We can not find 21 and φφ explicitly.

θθ=φ∴ log tan 1 : θθ=φ log tan 2

Now consider, θθ=φ log tan 1 : θθ=φ log - tan 2

Now consider, ( )21

21 tan tan 1 log 2 - tan

φφθθ

φφ+

=( )2log1

log2θθ−

θθ= ......…(1)

We have to find θ by solving the given pair of equations :

θ=θ= a/log r and log a r

Equating the R.H.S we have a θ

=θ log

a log

ie., ( ) e logor 1 1 log 2 =θ⇒θ==θ

Substituting ( ) get we1in e =θ

( ) 221 e - 12e tan =φ−φ ( )1 e log =∵

∴ angle of intersection e tan2 e - 1

2e tan - 1-2

1-21 =⎟

⎠⎞

⎜⎝⎛=φφ


r = a (1 – cos θ ) and r = 2a cos θ

Solution : r = a (1 – cos θ) : r = 2a cos θ

Taking logarithms we have,

Log r = log a + log (1 – cos θ) : log r = log 2a + log (cos θ)

Differentiating these w.r.t θ, we get,


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2011

cos - 1

sin ddr

r1

θθ

=θ

: cos

sin - ddr

r1

θθ

=θ

ie., ( ) ( )( ) θ=φθ

θθ=φ tan - cot :

/2 sin 2/2 cos /2sin 2 cot 221

ie., ( )/2cot cot 1 θ=φ : ( )θ+π=φ /2cot cot 2

/2 1 θ=φ⇒ : θ+π=φ /2 2

2 2 / -/2 - /2 21 θπθπθφφ +==−∴ …(1)

Now consider ( ) θ=θ= cos 2a cos - 1 a r

Or ( )1/3 cos or 1 cos 3 -1=θ=θ

Substituting this value in (1) we get,

The angle of intersection ( )1/3 cos . 1/2 2/ -1+π

6. Find the angle of intersection of the curves : θ= ar and θ= / ar

Solution : θ= ar : θ= / ar

θ+=⇒ log a log r log : θ= log - a log r log

Differentiating these w.r.t θ, we get,

θ

=θ

1 ddr

r1 :

θ=

θ1 -

ddr

r1

ie., θ

=φ1 cot 1 :

θ=φ

1 - cot 2

or θ=φ tan 1 : θ=φ - tan 2

Also by equating the R.H.S of the given equations we have

1 1 or a/ a 2 ±=θ⇒=θθ=θ

When and 1- tan 1, tan 1, 21 =φ=φ=θ

When 1 tan 1,- tan 1,- 21 =φ=φ=θ .

/2 - 1 - tan . tan 2121 π=φφ⇒=φφ∴

The curves intersect at right angles.


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2011

7. Find the pedal equation of the curve: ( ) 2a cos - 1 r =θ

Solution : ( ) 2a cos - 1 r =θ

( ) 2a log cos - 1 log r log =θ+⇒

Differentiating w.r.t θ, we get

θθ

=θ

=θ

θ+

θ cos - 1sin -

ddr

r1or 0

cos - 1sin

ddr

r1

( ) ( )( )

( )/2cot - /2 sin 2

/2 cos /2sin 2- cot 2 θ=θ

θθ=φ∴

ie., ( ) ( )/2 - /2-cot cot θ=φ⇒θ=φ∴

Consider φ= sin r p

( ) ( )/2sin r - por /2-sin r p θ=θ=∴

Now we have, ( ) 2a cos - 1 r =θ …(1)

( )/2sin r - p θ= …(2)

We have to eliminate θ from (1) and (2)

(1) can be put in form ( ) 2a 2/sin 2 . 2 =θr

ie., ( ) a /2 sin r 2 =θ

But p/-r = sin (θ/2), from (2)

ar por a rpr 2

2

2

==⎟⎟⎠

⎞⎜⎜⎝

⎛∴

Thus p2 = ar is the required pedal equation.

8. Find the pedal equation of the curve: r2 = a2 sec 2θ

Solution : r2 = a2 sec 2θ

)2 (sec log a log 2 r log 2 θ+=⇒

Differentiating w.r.t θ , we get,

θ=θθ

θθ=

θ2tan

ddr

r1 ie.,

2 sec2 tan 2 sec 2

ddr

r2

ie., ( ) θπ=φ⇒θπ=φ 2 - /2 2-/2cot cot


10 MAT11

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2011

Consider ( ) θ=θπ=∴φ= 2 cosr p ie., 2-/2sin r p sin r p

Now we have, θ= 2 sec a r 22 …(1)

θ= 2 cosr p …(2)

From (2) p/r = cos 2θ or r/p = sec 2 θ

Substituting in (1) we get, ( )r/p a r 22 = or pr = a2

Thus pr = a2 is the required pedal equation.

9. Find the pedal equation of the curve: rn = an cos nθ

Solution : rn = an cos nθ

( )θ+=⇒ n cos log a logn r logn

Differentiating w.r.t q we get

θ=θθ

θ=

θn tan -

ddr

r1 ie.,

n cosnsin n -

ddr

rn

( ) θ+π=φ⇒θ+π=φ∴ n /2 n /2cot cot


( ) θ=θ=π=∴ n cosr p ie., n /2sin r p

Now we have, θ= n cos a r nn …(1)

θ= n cosr p …(2)

( )p/r a r is (2) of econsequenc a as (1) nn =∴

Thus rn + 1 = pan is the required pedal equation.

10. Find the pedal equation of the curve: rm = am (cos mθ + sin mθ)

Solution : rm = am (cos mθ + sin mθ)

Differentiating w.r.t θ, we get,

θ+θθ+θ

=θ msin m cos

m cos m msin m- ddr

rm

ie., ( )( )θ+θ

θθ=

θ+θθθ

=θ mtan 1 m cos

m tan -1 m cos msin m cosmsin -m cos

ddr

r1

( ) θ+π=φ⇒θ+π=φ∴ m /4 m /4cot cot


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2011


( )θ+π=∴ m /4sin r p

ie., ( ) ( )[ ]θπ+θπ= msin /4 cos m cos /4sin r p

ie., ( )θ+θ= msin m cos 2r p

(we have used the formula of sin (A + B) and also the values ( ) ( ) )21/ /4 cos /4 sin =π=π

Now we have, ( )θ+θ= msin m cos a r mm …(1)

( )θ+θ= msin m cos 2r p …(2)

Using (2) in (1) we get,

pa 2 ror r

2p . a r m1mmm == +

Thus pa 2 r m1 m =+ is the required pedal equation.

11. Establish the pedal equation of the curve:

θ+θ= n cos bnsin a r nnn in the form ( ) 2 n 2n22n 2 r ba p +=+

Solution : We have θ+θ= n cos bnsin a r nnn

( )θ+θ=⇒ n cos b nsin a log r logn nn

Differentiating w.r.t θ we get

θ+θθθ

=θ ncosb nsin a

nsin nb - n cos na ddr

rn

nn

nn

Dividing by n, θ+θθθ

=φncosbnsin ansin b - n cosa cot nn

nn


Since φ cannot be found, squaring and taking the reciprocal we get,

( )φ+=φ= cot 1 r1

p1or cosec

r1

p1 2

222

22

( )( ) ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

θ+θ

θθ+=∴ 2nn

2nn

22n cos bnsin a

nsin b -n cos a1 r1

p1


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2011

( ) ( )( ) ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

θ+θ

θθ+θ+θ= 2nn

2nn2nn

22n cos bnsin a

nsin b - n cos a n cos b nsin a r1

p1 e.,i

( ) ( )( ) ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

θ+θ

θ+θ+θ+θ= 2nn

22n222n2

22n cos bnsin a

n sin n cos b n cos n sina r1

p1 e.,i

(product terms cancels out in the numerator)

( )2nn

n 2n 2

22n cos b nsin a

b a . r1

p1 .,ie

θ+θ

+=

( ) ,r

ba . r1

p1 or 2n

n2n 2

22+

= by using the given equation.

( ) 2n 2n 2n 22 r b ap +=+∴ is the required pedal equation.

12. Define Curvature and Radius of curvature Solution: A Curve Cuts at every point on it. Which is determined by the tangent drawn. Y y = f (x) Tangent s P (x,y) A O ψ X Let P be a point on the curve y= f (x) at the length ‘s’ from a fixed point A on it. Let the tangent at ‘P’ makes are angle ψ with positive direction of x – axis. As the point ‘P’ moves along curve, both s and ψ vary.

The rate of change ψ w.r.t s, i.e., as called the Curvature of the curve at ‘P’.

The reciprocal of the Curvature at P is called the radius of curvature at P and is denoted by ρ.

dsdψ

ψψρdds

dsd ==∴1


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2011

(or)

Also denoted ∴K =

K is read it as Kappa.

13. Derive an expression for radius of curvature in Cartesian form. Solution :(1) Cartesian Form: Y = f (x) P T A S ψ dy ψ ψ P dx R X O Let y = f (x) be the curve in Cartesian form.

We know that, tan ψ = (From Figure) ----------- (1)

Where ψ is the angle made by the tangent at P with x – axis. Differentiating (1) W.r.t x, we get

Sec2ψ . =

i.e., = Sec2 ψ

= (1 + tan2 ψ)

= (1 + tan2 ψ) .

= (1 + tan2 ψ)

= . . from eqn (1)

ψρ

dds

=∴dsdψ

ρ=

1

K1

=ρdsdψ

dxdy

dxdψ

2

2

dxyd

2

2

dxyd

dxdψ

dxdψ

dxdψ

dxds

ρ1 2

12

1⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

dxdy

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

2

1dxdy

ρ1 2

12

1⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

dxdy


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2011

=

∴ ρ =

∴ ρ = ( 1+ y1 2 )3/2 ……………(1) y2

Where y1 = , y2 =

This is the formula for Radius of Curvature in Cartesian Form. 14. Show that the Curvature of a Circle at any point on it, is a Constant

Tangent

PP P

A ψ X T Solution: Consider a Circle of radius r. Let A be a fixed point and ‘P’ be a given point on the circle such that arc AP = S. Let the angle between the tangent to the Circle at A and P be ψ. Then clearly AOP = ψ. ∴ AP = rψ i.e., S = rψ This is the intrinsic equation of the circle. Differentiating w.r.t S we get

1 = r Or K = =

∴ K = which is Constant

Hence the Curvature of the Circle at any point on it is constant.

ρ1 2

32

1⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

dxdy

23

2

2

2

1

dxyd

dxdy

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛+

dxdy

2

2

dxyd

dsdψ

dsdψ

r1

r1

O r


10 MAT11

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2011

15. Derive an expression for radius of curvature in parametric form.

Solution: We have ρ = (1 + y 12 )3/2

y2 Let x =f (t), y = g (t) be the curve in Parametric Form.

Then y1 =

Y2 = = =

= . .

=

∴ Y2 =

Substituting Y1 and Y2 in equation 1.

ρ = , we get

ρ =

-------------------- (2)

Equation (2) is called the Radius of Curvature in Parametric Form.

•

•

==X

Y

dtdx

dtdy

dxdy

2

2

dxyd

dxd

⎟⎠⎞

⎜⎝⎛

dxdy

dxd

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛•

•

X

Ydtd

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛•

•

X

Ydxdt

dtd

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛•

•

X

Y

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

dtdx1

dtd

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛•

•

X

Y•

X

1

( )2xxyyx −

•

)(

1

X

3

⎟⎠⎞

⎜⎝⎛

−•

•

x

xyyX

( )2

232

11yY+

( )( )

( )3

23

21

xxyyx

xY

−⎥⎦⎤

⎢⎣⎡ +

( ) ( )[ ]xyyx

yx−

+=

2322

ρ


10 MAT11

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2011

16. Derive an expression for radius of curvature in polar form.

Solution: Let r = f (θ) be the curve in the Polar Form. We know that, Angle between the tangent and radius vector,

Tan φ = r

= r.

i.e., tan φ =

Differentiate w.r.t ‘θ’ we get

Sec2 φ . =

=

=

=

From figure ψ = θ + φ Differentiating w.r.t θ, we get

drdθ

θddr

1

( )θddr

r

θφ

dd

2

2

2

.

⎟⎠⎞

⎜⎝⎛

−

θ

θθθ

ddr

drdr

ddr

ddr

( )( ) ⎪

⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=∴ 2

222

2

1

θ

θθφθ

φ

ddr

drdrd

dr

Secdd

( )( ) ⎪

⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧ −

+ 2

222

2tan11

θ

θθφ

ddr

drdrd

dr

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎠⎞

⎜⎝⎛

−⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

+2

2

22

2

2

1

1

θ

θθ

θddr

drdr

ddr

ddrr

θφ

dd

22

2

22

rddr

drdr

ddr

+⎟⎠⎞

⎜⎝⎛

−⎟⎠⎞

⎜⎝⎛

θ

θθ


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2011

= 1 +

∴ = 1 +

=

Also we know that =

Now, ρ= = .

= .

ρ ----------- (3)

Where r1 = , r2 =

Equation (3) as called the radius of curvature in Polar form.

17. Derive an expression for radius of curvature in pedal form.

Solution: Let p = r Sin φ be the curve in Polar Form. We have p = r Sin φ Differentiating p W.r.t r, we get

= Sin φ + r Cos φ

θψ

dd

θφ

dd

θψ

dd

22

2

22

rddr

drdr

ddr

+⎟⎠⎞

⎜⎝⎛

−⎟⎠⎞

⎜⎝⎛

θ

θθ

θψ

dd

22

2

222 2

rddr

drdr

ddrr

+⎟⎠⎞

⎜⎝⎛

−⎟⎠⎞

⎜⎝⎛+

θ

θθ

θdds 2

12

2

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

θddrr

ψdds

θdds

ψθ

dd

21

22

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛+

θddrr

2

222

22

2θθ

θ

drdr

ddrr

rddr

−⎟⎠⎞

⎜⎝⎛+

+⎟⎠⎞

⎜⎝⎛

( )2

21

2

232

12

2 rrrrrr

−++

=

θddr

2

2

θdrd

drdp

drdφ


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2011

But Sin φ = r , Cos φ=

∴ = r + r

= r + .

= r + r

= r (θ + φ)

= r

∴ = r

= r. ∵ =

∴ ρ= r --------------- (4)

Equation (4) is called Radius of Curvature of the Curve in Polar Form.

18. Find the radius of curvature at (x, y) for the curve ay2 = x3.

Solution: Given ay2 = x3 -------- (1) is in Cartesian form. We have, Radius of curvature in Cartesian form.

------------ (2)

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

dsdθ

dsdr

drdp

dsdθ

dsdr

drdφ

dsdθ

drdφ

dsdr

dsdθ

dsdφ

dsd

dsdψ

drdp

dsdψ

drdp

ρ1

dsdψ

ρ1

dpdr

( )2

232

11yy+

=ρ


10 MAT11

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2011

a 2y

Differentiating y1 w.r.t ‘x’ we get

Substitute Y1 and Y2 in (2), we get.

19. Find the radius of curvature at (x,y) for the curve y = c log Sec

Solution: Given y = c log Sec -------- (1)

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

= c tan . = tan

Differentiating y1 W.r.t x we get

=

Substitute y1 and y2 in

∴

ρ =

ax

axa

xayxy

dxdyx

dxdy

23

2

3233

21

3

22

12 =

⎟⎟⎠

⎞⎜⎜⎝

⎛===⇒=

axy

23

1 =∴

xaxadxydy

43

21

23

2

2

2 ===

( )a

xax6

94 23

+=ρ

⎟⎠⎞

⎜⎝⎛

cx

⎟⎠⎞

⎜⎝⎛

cx

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

=cx

cxSec

cxSec

cy tan.1.1 ⎟⎠⎞

⎜⎝⎛

cx

c1

⎟⎠⎞

⎜⎝⎛

cx

⎟⎠⎞

⎜⎝⎛=

cxSecy 2

2 c1

c1

⎟⎠⎞

⎜⎝⎛

cxSec 2

( )2

232

11yy+

=δ

( )[ ]( )

( )[ ]( )cxSec

c

cxSec

cxSec

c

cx

2

23

2

2

23

2

11tan1

=+

=δ

( )cxcSec


10 MAT11

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2011

20. Find the radius of curvature at the point ‘t’ on the curve x = a (t + Sint), y = a (1 – Cost).

Solution: Given Curves are in Parametric Form

∴ Radius of Curvature, ------ (1)

Differentiating the given Curves W.r.t t, we get

= a (1 + Cost) = = a Sint

Differentiating W.r.t ‘t’ we get

= - a Sint, = a Cost Substitute , , and in (1), we get

=

=

21. Find the Radius of Curvature to x = , y = a Sint at t. Solution: Here x = , y = a Sin t

=a { -sint + }

=a { -sint + } = a { (1-sin 2t) / sint } = a Cos2t / sint

( ) ( )( )xyyx

yx−

+=

2322

ρ

dtdxx = y

dtdy

x y

x y y x

( )int)int()1(

int)1( 23222

aSaSaCostCostaSaCosta

−−+++

=ρ

{ }{ }tSintCosCosta

tSintCosCosta222

23223 21

+++++

{ } { }222.8

2222.2

)1()1(2

2

3

2

23

22

3

tCos

tCosatCos

tCosa

CostCosta

==++

24 taCos=ρ

( ){ }2tanlog tCosta +

( ){ }2tanlog tCosta +

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+−=21.2.

2tan1int 2 tSec

tSa

dtdx

2/cos2/sin21

tt

2/sin21

t

dtdx


10 MAT11

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2011

= = tant

Differentiating W.r.t ‘x’ we get

Substitute & in , we get

i.e.,

∴ρ = a Cot t.

22. Find the Radius of Curvature to at

Solution: Given -------------- (1) Differentiating (1) w.r.t to ‘x’ we get

i.e., (From (1))

----------- (2)

aCostdxdyand =

dtdxdtdy

dxdy

//

=∴ttta

aCostsin/coscos

tdxdy tan=∴

int

1.12

2222

2

StaCos

tSec

dtdx

tSecdxdttSec

dxyd

===

int1 42

2

tSSecadx

yd==

dxdy

2

2

dxyd ( )

2

232

11yy+

=ρ

( ) aCottSCosta

tSSectaSec

tSSeca

t===

+=

intintint1tan1

4

3

4

232

ρ

ayx =+ ⎟⎠⎞

⎜⎝⎛

4,

4aa

ayx =+

02

12

1=+

dxdy

yx

( )x

xaxy

dxdyy −

−=−==1

xay −= 11


10 MAT11

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2011

Also -------------- (3)

At the given point

Then = - 1 &

∴ Substitute y1 and y2 in

ρ =

ρ

23. Show that for the Cardioids r = a (1 + Cosθ), ρ2 / r 2 is a constant Solution: r = a (1 + Cosθ)

= - a Sinθ

We have,

, is Pedal Equation.

= a2Sin2θ

=

=

23

23

2

2

222

1

x

axadx

ydy =⎟⎠⎞

⎜⎝⎛−−== −

⎟⎠⎞

⎜⎝⎛

4,

4aa

2

11 aay −=

( ) aa

ay 4

421

232 =−=

( )2

232

11yy+

=ρ

( )24

224

24

)1(1 23

232 aaa

a===

−+

2a

=

θddr

2

422

111⎟⎠⎞

⎜⎝⎛+=

θddr

rrP

42

11rr

+

4

2222

4

222 )1(r

SinaCosar

Sinar θθθ ++=

+

4

2 )1(2r

Cosa θ+


10 MAT11

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2011

(∵ r = a (1 + Cosθ)

Differentiating w.r.t ‘P’ we get

2P =

Now,

And

24. Find the Radius of Curvature of the Curve

Solution: Given

Differentiating w.r.t to P, we get

⎟⎠⎞

⎜⎝⎛=∴

ar

ra

P 4

2

2

21

θCosar

+=∴ 1

32

21ra

P=

arP2

32 =∴

dpdrr

a23

21

234

rap

dpdr

=

rap

dpdrr

34

==ρ

98

2.

916

9161 3

3

2

2

22

2

2 aa

rra

rpa

rr==⎥

⎦

⎤⎢⎣

⎡=

ρ

22

2

222

111ba

rbaP

−+=

22

2

222

111ba

rbaP

−+=

dpdrr

baP212

223

−=

−

rpba

dpdr

3

22

=∴

3

22

3

22

..pba

rpbar

dpdrr ===∴ ρ


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2011

25. Find the Radius of Curvature at (r,θ) on r =

Solution: Given r =

Differentiating w.r.t to ‘θ’ we get

We have

Differentiating above result w.r.t to‘P’ we get

θa

θa

θθθθθraa

ddr −

=−

=−

=1.2

θθr

ddr −

=

2

422

111⎟⎠⎞

⎜⎝⎛+=

θddr

rrP

2222

2

42

1111θθ rr

rrr

+=+=

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=⎟

⎠⎞

⎜⎝⎛ += 2

2

2222

111111θ

θθ rrP

12 +=∴

θ

θrP

22

2

22

.

1

.

1 rara

ra

rarrP

+=

+

=+

=θ

θ

( )22

22

22 22

1..1

radpdrr

raar

dpdrara

+

/+/

−+=

dpdr

raararara ⎟⎟

⎠

⎞⎜⎜⎝

⎛

+−+=+

22

22222 .

dpdr

raararara

22

22222 ).(

+

−+=+


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2011

Thus,

Exercises:

(1) Find the Radius of the Curvature at the point (s, ψ) on S = a log tan

(2) Find the Radius of the Curvature of xy = C2 at (x,y) (3) Find the Radius of the Curvature of xy3 = a4 at (a,a) (4) Find the Radius of Curvature at the point θ on x = C Sin 2θ (1 + Cos 2θ), y = C Cos 2θ (1 – Cos 2θ) (5) If ρ1 and ρ2 are the radii of curvature at the extremities of any chord of the cardiode

r = a (1 + Cosθ) which posses through the Pole prove that

( )dpdrarara 32222 =++

( )dpdr

ara

=+

3

2322

( )3

2322..

arar

dpdrr +

==ρ

( ) 2322

3 raar

+=∴ ρ

⎟⎠⎞

⎜⎝⎛ +

24ψπ

916 2

22

21

a=+ ρρ


10 MAT11

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2011

DIFFERENTIAL CALCULUS – III PARTIAL DIFFERENTIATION Introduction :- Partial differential equations abound in all branches of science and engineering and many areas of business. The number of applications is endless. Partial derivatives have many important uses in math and science. We shall see that a partial derivative is not much more or less than a particular sort of directional derivative. The only trick is to have a reliable way of specifying directions ... so most of this note is concerned with formalizing the idea of direction So far, we had been dealing with functions of a single independent variable. We will now consider functions which depend on more than one independent variable; Such functions are called functions of several variables.

Geometrical Meaning Suppose the graph of z = f (x,y) is the surface shown. Consider the partial derivative of f with respect to x at a point (x0, y0). Holding y constant and varying x, we trace out a curve that is the intersection of the surface with the vertical plane y = y0. The partial derivative fx(x0,y0). measures the change in z per unit increase in x along this curve. That is, fx(x0, y0) is just the slope of the curve at (x0, y0). The geometrical interpretation of fy(x0, y0). is analogous.


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2011

Real-World Applications: Rates of Change: In the Java applet we saw how the concept of partial derivative could be applied geometrically to find the slope of the surface in the x and y directions. In the following two examples we present partial derivatives as rates of change. Specifically we explore an application to a temperature function ( this example does have a geometric aspect in terms of the physical model itself) and a second application to electrical circuits, where no geometry is involved. I. Temperature on a Metal Plate The screen capture below shows a current website illustrating thermal flow for chemical engineering. Our first application will deal with a similar flat plate where temperature varies with position. * The example following the picture below is taken from the current text in SM221,223: Multivariable Calculus by James Stewart.


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2011

Suppose we have a flat metal plate where the temperature at a point (x,y) varies according to position. In particular, let the temperature at a point (x,y) be given by, where T is measured in oC and x and y in meters. Question: what is the rate of change of temperature with respect to distance at the point (2,1) in (a) the x-direction? and (b) in the y-direction ? Let's take (a) first. What is the rate of change of temperature with respect to distance at the point (2,1) in (a) the x-direction? What observations and translations can we make here? Rate of change of temperature indicates that we will be computing a type of derivative. Since the temperature function is defined on two variables we will be computing a partial derivative. Since the question asks for the rate of change in the x-direction, we will be holding y constant. Thus, our question now becomes: What is at the point (2,1)? Conclusion : The rate of change of temperature in the x-direction at (2,1) is degrees per meter; note this means that the temperature is decreasing !

2 2( , ) 60 /1T x y x y= + +

dTdx

2 2 2 2 1

2 2 2

2

( , ) 60 /1 60(1 )

60(2 )(1 )

20(2,1) 60(4)(1 4 1) 3

T x y x y x yT x x yxT

x

−

−

−

= + + = + +∂ = − + +∂∂ −= − + + =∂

203

−


10 MAT11

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2011

Part (b): The rate of change of temperature in the y-direction at (2,1) is computed in a similar manner. Conclusion : The rate of change of temperature in the y-direction at (2,1) is degrees per meter; note this means that the temperature is decreasing !

II. Electrical Circuits: Changes in Current The following is adapted from an example in a former text for SM221,223 Multivariable

Calculus by Bradley and Smith. * In an electrical circuit with electromotive force (EMF) of E volts and resistance R ohms, the current, I, is I=E/R amperes. Question: (a) At the instant when E=120 and R=15 , what is the rate of change of current with respect to voltage. (b) What is the rate of change of current with respect to resistance? (a) Even though no geometry is involved in this example, the rate of change questions can be answered with partial derivatives. we first note that I is a function of E and R, namely, I(E,R) = ER-1

2 2 2 2 1

2 2 2

2

( , ) 60 /1 60(1 )

60(2 )(1 )

10(2,1) 60(2)(1 4 1) 3

T x y x y x yT y x yxT

x

−

−

−

= + + = + +∂ = − + +∂∂ −= − + + =∂

103

−


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2011

The rate of change of current with respect to voltage = the partial derivative of I with respect to voltage, holding resistance constant is when E=120 and R=15 , we have verbal conclusion : If the resistance is fixed at 15 ohms, the current is increasing with respect to voltage at the rate of 0.0667 amperes per volt when the EMF is 120 volts. Part (b): What is the rate of change of current with respect to resistance? Using similar observations to part (a) we conclude: The partial derivative of I with respect to resistance, holding voltage constant = when E=120 and R=15 , we have Conclusion : If the EMF is fixed at 120 volts, the current is decreasing with respect to resistance at the rate of 0.5333 amperes per ohm when the resistance is 15 ohms.

Key Words :- Then the partial derivative of z w.r.t x is given by The partial derivative of z w.r.t y is given by

1I RE−∂ =∂

115 0.0667IE

−∂ = ≈∂

1I ERE−∂ =∂

1(120,15) 120(15) 0.5333IE

−∂ = − ≈ −∂

0

( , ) ( , )limxx

z f x x y f x yzx xδ

δδ→

∂ + −= =

∂

0

( , ) ( , )limyy

z f x y y f x yzy yδ

δδ→

∂ + −= =

∂


10 MAT11

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2011

1. If ( )byaxsin e u by-ax += show that u ab 2a - =∂∂

∂∂

yu

xub

Solution : ( )by ax sin e u by-ax +=

( ) ( )by ax sin a.e a . by ax cos e xu by-axby -ax +++=

∂∂

∴

( ) ua by ax cos e a xu e.,i by -ax ++=

∂∂ …(1)

Also ( ) ( ) ( )by ax sin e b- b . by ax cos e by-axby -ax +++=∂∂

xu

( ) bu by ax cos e b yu .,ie by -ax −+=

∂∂ …(2)

Now yu a

xu b

∂∂

=∂∂ by using (1) and (2) becomes

( ) ( ) abu byax cos abe -abu by ax cos abe by-axbyax ++++= −

= 2 abu

Thus abu 2 yu a -

xu b =

∂∂

∂∂

2. If ( ) that prove ,by-ax f e u byax +=

abu2yu a

xu b =

∂∂

=∂∂

Solution : ( )by -ax f eu byax+= , by data

( ) ( )by -ax fae a by-ax f . e xu byaxby ax ++ +′=

∂∂

Or ( ) u a by-ax f . e a by ax +′=∂∂ +

xu …(1)

Next, ( ) ( ) ( )by -ax f e b b- . by-ax f e yu by ax by ax ++ +′=

∂∂

Or ( ) ba by -ax f e b yu byax +′−=

∂∂ + …(2)

Now consider L.H.S yua

xu b

∂∂

+∂∂

=


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2011

( ){ } ( ){ }bu by -ax f be- a au by -ax f ae b byaxbyax +′++′= ++

( ) ( ) abu by -ax fe ab-abu by -ax f e ab by ax byax +′+′= ++

= 2abu = R.H.S

Thus u b a 2 yua

xu b =

∂∂

+∂∂

3. If 222 z y x log u ++= , show that ( ) 1 zu

yu

xu z y x 2

2

2

2

2

2222 =⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

++

Solution : By data ( )222222 zyx log 21 zy x log u ++=++=

The given u is a symmetric function of x, y, z, (It is enough if we compute only one of the required partial derivative)

222222 y 2 . 1 .

21

zxxx

zyxxu

++=

++=

∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛

++∂∂

=⎟⎠⎞

⎜⎝⎛

∂∂

∂∂

=∂∂

222

2

zy u u

xx

xxxx2

ie., ( )

( ) ( )2222

222

2222

222

zy

zy zy

2 . - 1 zy

++

−+=

++

++

x

x

x

xxx

( )2222

222

2

2

zy

zy u ++

−+=

∂∂

∴x

xx

…(1)

Similarly ( )2222

222

2

2

zy

xz u ++

−+=

∂∂

∴x

yy

…(2)

( )2222

222

2

2

zy

yx u ++

−+=

∂∂

x

zz

…(3)

Adding (1), (2) and (3) we get,

( ) 2222222

222

2

2

2

2

2

2

zy1

zy

zy uu u ++

=++

++=

∂∂

+∂∂

+∂∂

xx

xzyx

Thus ( ) 1 zu

yu u zy 2

2

2

2

2

2222 =⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

+∂∂

++x

x


10 MAT11

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4. If u = log (tan x + tan y + tan z), show that, sin 2x ux + sin 2y uy + sin 2z uz = 2

Solution : u = log (tan x + tan y + tan z) is a symmetric function.

z tan y tan tan

sec u2

++=

xx

x

( )z tan y tan tan .sec cos sin 2 u 2 sin

2

++=

xxxxx x

Or z tan y tan tan

tan 2 u 2 sin++

=x

xx x …(1)

Similarly z tan y tan tan

tan 2 u 2 sin++

=x

yy y …(2)

z tan y tan tan

tan 2 u 2 sin++

=x

zz y …(3)


( )( ) 2

z tan y tan x tanz tan y tan tan x 2 u 2zsin u2y sin u 2 sin zy =

++++

=++xx

Thus 2 u 2zsin u2y sin u2x sin zyx =++

5. If u = log (x3 + y3 + z3 – 3xyz) then prove that zyx

3 zu

yu

xu

++=

∂∂

+∂∂

+∂∂ and hence show

that ( )2

2

zy9- u ++

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

xzyx

Solution : u = log (x3 + y3 + z3 – 3xyz) is a symmetric function

yz3 - zy3yz - 3 u333

2

xxx

x ++=

∂∂ …(1)

yz3 - zy3zx - 3 u333

2

xxy

y ++=

∂∂ …(2)

yz3 - zy3xy - 3 u

333

2

xxz

z ++=

∂∂ …(3)



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( )( )yz3zy

zyzyzy 3 zu u u

333

222

xxxxx

yx −++−−−++

=∂∂

+∂∂

+∂∂

Recalling a standard elementary result,

( ) ( )cabcabcba cba abc3cba 222333 −−−++++=−++

We have,

( )( ) ( )xxxx

xxxx zyzyzy zy

zyzyzy 3 zu

yu u

222

222

−−−++++−−−++

=∂∂

+∂∂

+∂∂

Thus zy

3 zu

yu u

++=

∂∂

+∂∂

+∂∂

xx

Further u z

y

2

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂x

u z

y

z

y

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=xx

zu

yu u

z

y ⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=xx

z y

3 z

y

⎟⎟⎠

⎞⎜⎜⎝

⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

=xx

, by using the earlier result.

⎟⎟⎠

⎞⎜⎜⎝

⎛++∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛++∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛++∂

∂=

z y 3

z

z y 3

y

zyx3

x

xx

( ) ( ) ( ) ( )2222 zy

9 zy

3 zy

3 zy

3 ++

−=

++−

+++

−+

++−

=xxxx

Thus ( )2

2

zy9- u

z

y

x ++=⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+∂∂

x


10 MAT11

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2011

6. If ( ) θ=θ== sin r y , cosr and rf u x ,

prove that ( ) ( )r r1 r 2

2

2

2

ffyu

xu ′+′′=

∂∂

+∂∂

Solution :Observing the required partial derivative we conclude that u must be a function of x, y. But

u = f( r) by data and hence we need to have r as a function of x, y. Since x = r cos θ, y = r

sin θ we have x2 + y2 = r2.

( ) 22 yx r wherer f u have we +==∴

( ) ( ) ( ) and .

rr - r

rr u 2

222

32

2

xfxfx

′′+

′=

∂∂

( ) ( ) ( ) .

rr - r

rr u 2

222

32

2yfyf

y′′

+′

=∂∂

Adding these results we get,

( ) ( ){ } ( ) ( )222

22232

2

2

2y

rrf y - 2

rrf u u

+′′

++′

=∂∂

+∂∂ xxx

yx

( ) ( ) ( ) ( )r r r1 r .

rr r .

rr 2

22

3 ffff ′′+′=′′

+′

=

Thus ( ) ( )r r1 r 2

2

2

2

ffyu

xu ′+′′=

∂∂

+∂∂

7. Prove that nuyuy

xux =

∂∂

+∂∂

Proof : Since u = f (x, y) is a homogeneous function of degree n we have by the definition,

( )y/x g x u n= …(1)

Let us differentiate this w.r.t x and also w.r.t.y

( ) ( )xxx

xx y/ g n y- . y/ g . xu 1 -n

2n +⎟

⎠⎞

⎜⎝⎛′=

∂∂

∴

ie., ( ) ( )xxxxx

y / g n y/ gy u 1 -n 2 -n +′=∂∂ …(2)

Also ( ) ⎟⎠⎞

⎜⎝⎛′=

∂∂

xxx

y1 . y/ g . u n


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ie., ( ) y/ g . u 1 -n xxy

′=∂∂ …(3)

Now consider yx

x∂∂

+∂∂ uy u as a consequence of (2) and (3)

( ) ( )[ ] ( )[ ]xxxxxxx y / g y y/ g n y/ gy - 1 -n 1 -n 2 -n ′++′=

( ) ( ) ( )xxxxxx y/ gy y/ g n y/ gy 1 -n n1n ′++′−= −

( ) y/ g x. n n x=

= n u, by using (1) Thus we have proved Euler’s theorem

un yuu ;u n yuy u y =+=

∂∂

+∂∂

xxx

x

8. Prove that ( )u 1 -n n

y 2 22

=∂∂

∂+

∂∂

yxux

xux 2

2

Proof : Since u = f (x, y) is a homogeneous function of degree n, we have Euler’s theorem

un

yuy u =

∂∂

+∂∂

xx …(1)

Differentiating (1) partially w.r.t. x and also w.r.t y we get,

xxxxx

∂∂

=∂∂

∂+⎟⎟

⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂ un

y uy u 1. u

2

2

2 …(2)

Also, yun

yu . 1

yuy

yu 2

22

∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

+∂∂

∂x

x …(3)

We shall now multiply (2) by x and (3) by y.

and u n yuy u x u

22

xx

xx

xxx 2

2

∂∂

=∂∂

∂+

∂∂

+∂∂

yuny

yuy

yu y

yuy 2

22

2

∂∂

=∂∂

+∂∂

+∂∂

∂x

x

Adding these using the fact that x∂∂

∂ yu2

= x∂∂

∂ yu2

we get,


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⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛

∂∂

+∂∂

∂+

∂∂

yuy u n

yuy u

yuy

yuy 2 u

2

22

2

2

22

xx

xx

xx

xx

ie., ( ) ( )1 usingby ,un n u n yuy

yuy 2 u

2

22

2

2

22 =+

∂∂

+∂∂

∂+

∂∂

x x

xx

or ( ) ( ) u 1 -n n nu -nun yuy

yuy 2 u

2

22

2

2

22 =+

∂∂

+∂∂

∂+

∂∂

x x

xx

Thus ( ) u 1 -n n yuy

yuy 2 u

2

22

2

2

22 +

∂∂

+∂∂

∂+

∂∂

x x

xx

ie., ( ) u 1 -n n uy uy 2 u yy2

y2 =++ xxx xx

9. If y

z z

y z y

+

++

++

=xx

xu show that 0 zu z

yuy

xu x =

∂∂

+∂∂

+∂∂

Solution : (Observe that the degree is 0 in every term)

y z

zy

z y u

++

++

+=

xxx

We shall divide both numerator and denominator of every term by x.

( ){ }xxxxx

xxx

z/ ,y/ g /y1

z 1/z

/yz/ y/

1 u =+

++

++

=

⇒ u is homogeneous of degree 0. ∴ n = 0

We have Euler’s theorem, un zu z

yuy u =

∂∂

+∂∂

+∂∂

xx

Putting n = 0 we get, 0 zu z

yuy u =

∂∂

+∂∂

+∂∂

xx

10. If 3 y that show y log 44

=∂∂

+∂∂

⎟⎟⎠

⎞⎜⎜⎝

⎛++

=yu

xux

yxxu

Solution : we cannot put the given u in the form xn g (y/x)

( )( )

( )( )⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

++

=+

+=

++

=∴x

xxxxxx

xx

/y1y/1

y/1 /y1

yy e

43

44444u

ie., eu = x3 g (y/x) ⇒ eu is homogeneous of degree 3 ∴ n = 3

Now applying Euler’s theorem for the homogeneous function eu


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We have ( ) ( ) uuu

en yey e =

∂∂

+∂

∂x

x

ie., uuu 3e yu ey u e =

∂∂

+∂∂

xx

Dividing by eu we get 3 yuy u =

∂∂

+∂∂

xx

11. If ⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=yxyxu

331- tan show that

(i) x ux + y uy = sin 2 u (ii) x2uxx + 2 x y uxy + y2uyy = sin 4 u – sin 2 u

Solution : (i) databy y - y tan u

331-

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=

xx

( )( )

( )( )⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−+

=+

=+

=⇒x/y1

y/1 y/ - 1

/y1 y - y u tan

32

33333 xxxxxx

xx

ie., tan u = x2/g (y/x) ⇒ tan u is homogeneous of degree 2.

Applying Euler’s theorem for the function tan u we have, ( ) ( )

2 n ;u tan .n y

utany utan ==∂

∂+

∂∂

xx

ie., u tan 2 yuu secy uu sec 22 =

∂∂

+∂∂

xx

or u 2sin u sin u cos 2 u cosusin u cos 2

u secu tan 2

yuy u 2

2 ====∂∂

+∂∂

xx

2usin yu u y =+∴ xx

(ii) We have xux + y uy = sin 2 u …(1) Differentiating (1) w.r.t x and also w.r.t y partially we get

xxxxxx u .2u cos 2 yuu . 1 u y =++ …(2)

And x uyx + yuyy + 1 . uy = 2 cos 2u . uy …(3) Multiplying (2) by x and (3) by y we get,

xxxxx xxxx u u. 2 cos 2 uy u u y2 =++


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yyyyyx yyx u u. 2 cos 2 uy u uy 2 =++

Adding these by using the fact that uyx = uxy, we get

( ) ( )yyyy2

y2 uy u2u cos 2 yu y uy uy 2 u +=++++ xxxxx xxxx

By using (1) we have,

2usin -u 2sin 2u cos 2 uy uy 2 u yy2

y2 =++ xxx xx

(since sin 2θ = 2 cos θ sin θ, first term in the R.H.S becomes sin 4u)

Thus x2uxx + 2 x y uxy + y2 uyy = sin 4 u – sin 2u

12. If ⎟⎟⎠

⎞⎜⎜⎝

⎛=

xz ,

zy ,

yx f u Prove that 0

zu z

yuy

xu x =

∂∂

+∂∂

+∂∂

>> here we need to convert the given function u into a composite function.

Let ( )x

x z r ,zy q ,

y p wherer q, p, f u ====

ie., ( ) ( ){ } z y, x,u z y, x, r q, p, u →⇒→→

xxxx ∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

=∂∂

∴r

ru q

qu p

pu u

ie., ⎟⎠⎞

⎜⎝⎛

∂∂

+∂∂

+∂∂

=∂∂

2z- .

ru 0 .

qu

y1 .

pu u

xx

ru z -

pu

y u

∂∂

∂∂

=∂∂

∴x

xx

x …(1)

Similarly by symmetry we can write,

pu x -

qu

z u y

∂∂

∂∂

=∂∂

yy

y …(2)

qu y -

ru

x u z

∂∂

∂∂

=∂∂

zz

z …(3)

Adding (1), (2) and (3) we get 0zu z

yuy u x =

∂∂

+∂∂

+∂∂

x


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2011

13. If u = f(x – y, y – z, z – x) show that 0 =∂∂

+∂∂

+∂∂

zu

yu

xu

>> Let u = f(p, q, r) where p = x – y, q = y – z, r = z – x

xxxx ∂∂

∂∂

+∂∂

∂∂

+∂∂

∂∂

=∂∂

∴r

ru q

qu p

pu u

ie., ( )1- ru 0.

qu 1 .

pu u

∂∂

+∂∂

+∂∂

=∂∂

x

ru -

pu u

∂∂

∂∂

=∂∂

∴x

…(1)

Similarly we have by symmetry

pu -

qu u

∂∂

∂∂

=∂∂

y …(2)

qu -

ru u

∂∂

∂∂

=∂∂

z …(3)

Adding (1), (2) and (3) we get, 0 zu

yu u

=∂∂

+∂∂

+∂∂

x

14. If z = f(x, y) where x = r cos θ and y = r sin θ

Show that 2

2

222 z r1

rz

yz z

⎟⎠⎞

⎜⎝⎛

θ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

x

Solution : ( ) ( ){ } ( )θ→⇒θ→→ r, z r, y ,z x

θ∂∂

∂∂

+θ∂

∂∂∂

∂∂

∂∂

+∂∂

∂∂

=∂∂

∴y

yz z ;

ry

yz

r z

rz x

xx

x

ie., θ∂∂

+θ∂∂

=∂∂ sin

yz cos

xz

rz …(1)

and ( ) ( ) ⎥⎦

⎤⎢⎣

⎡θ

∂∂

+θ∂∂

=θ∂∂

+θ∂∂

=θ∂

∂ cos yz sin z-r cosr

yz sin r - z z

xx

or θ∂∂

+θ∂∂−

=θ∂

∂= cos

yz in s z z

r1

x

squaring and adding (1), (2) and collecting suitable terms have,


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[ ]θ+θ⎟⎠⎞

⎜⎝⎛

∂∂

=⎟⎠⎞

⎜⎝⎛

θ∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂ 22

22

2

2

sin cos z z r1

rz

x

[ ] θθ∂∂

∂∂

θθ∂∂

∂∂

+θ+θ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+ cos sin yz z 2 - sin cos

yz z 2 cos sin

yz 22

2

xx

L.H.S R.H.S ie., yz z z

r1

rz

222

2

2

=⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎟⎠⎞

⎜⎝⎛

∂∂

=⎟⎠⎞

⎜⎝⎛

θ∂∂

⎟⎠⎞

⎜⎝⎛

∂∂

∴x

15. If ( ) vv eexfz −=+== − -uu e y ,e x wherey ,

Prove that vz

uz

yz

xz

∂∂

−∂∂

=∂∂

−∂∂ y x

Solution : ( ) ( ){ } ( ) vu, z vu, y x, z →⇒→→

vy

yz

v z

vz ;

uy

yz

u z

uz

∂∂

∂∂

+∂∂

∂∂

=∂∂

∂∂

∂∂

+∂∂

∂∂

=∂∂

∴x

xx

x

( )u-u e- e . .,yz

xz

uzie

∂∂

+∂∂

=∂∂ …(1)

( ) ( )v-v- e- yz e- z

vz

∂∂

+∂∂

=∂∂

x …(2)

Consider R.H.S = v∂

∂−

∂∂ z uz and (1) – (2) yields

( ) ( ) y . yz - . z ee

yz - e e z vu-v-u

∂∂

∂∂

=−∂∂

+∂∂ x

xx

Thus yzy z

yz -

uz

∂∂

−∂∂

=∂∂

∂∂

xx ie., R.H.S = L.H.S

16. Find ( )( )z,y,x

w,v,u∂∂ where u = x2 + y2 + z2, v = xy+yz+zx, w=x+y+z

Solution : The definition of J = ( )( )z,y,x

w,v,u∂∂

u u ux y zv v vx y zw w wx y z

∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂

=∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂

But u = x2 + y2 + z2, v = xy+yz+zx, w=x+y+z


10 MAT11

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Substituting for the partial derivatives we get

J = 111

xyzxzyz2y2x2

+++

Expanding by the first row, J = 2x {(x + z) – (y + x)} -2y {(y + z) – (y + x)} +2z {(y + z) – (x + z)} = 2x (z-y) – 2y(z-x) + 2z(y-x) = 2(xz – xy – yz + xy + yz – xz) = 0 Thus J = 0

17. If u = xyz , v =

yzx , w =

zxy , show that ( )

( )z,y,xw,v,u

∂∂ = 4

Solution : by data u = xyz , v =

yzx , w =

zxy

( )( )z,y,x

w,v,u∂∂ =

zw

yw

xw

zv

yv

xv

zu

yu

xu

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

=

2

2

2

zxy

zx

zy

yx

yzx

yz

xy

xz

xyz

−

−

−

= ⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛ −

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−yx

zx

zxy

yzx

xyz

222

- ⎭⎬⎫

⎩⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⋅+

⎭⎬⎫

⎩⎨⎧

⋅−⎟⎟⎠

⎞⎜⎜⎝

⎛ −22 yzx

zy

zx

yz

xy

yx

zy

yzx

yz

xz

= ⎭⎬⎫

⎩⎨⎧

++⎭⎬⎫

⎩⎨⎧ −

−−

⎭⎬⎫

⎩⎨⎧

−−

yx

yx

xy

zx

zx

xz

yzx

yzx

xyz 22

2

= 0+1+1+1+1=4

Thus ( )( )z,y,x

w,v,u∂∂ = 4


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2011

18. If u + v = ex cos y and u – v = ex sin y find the jacobian of the functions u and v w.r.t x and y.

Solution : we have to find ( )( )y,x

v,u∂∂ =

yv

xv

yu

xu

∂∂

∂∂

∂∂

∂∂

Using the given data we have to solve for u and v in terms of x and y. By data u + v = ex cos y ……(1) u – v = ex sin y ……(2) (1) + (2) gives : 2 u = ex (cos y + sin y) (2) – (2) gives : 2 v = ex (cos y – sin y)

Ie., u = 2

e x

(cos y + sin y) ; v = 2ex

(cos y – sin y)

∴xu

∂∂ =

2e x

(cos y + sin y), xv

∂∂ =

2e x

(- sin y - cos y)

Now ( )( )yx

vu,,

∂∂ =

)ycosy(sin2e)ysiny(cos

2e

)ycosysin(2

e)ysiny(cos2

e

xx

xx

+−

−

+−+

= 2

e x

. 2

e x

{ - ( cos y + sin y)2 – (cos y – sin y)2}

= 4

2xe− {1+sin 2y) + (1 – sin 2y)} = 2

2xe−

Thus ( )( )y,x

v,u∂∂ =

2

2xe−


10 MAT11

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19. (a) If x = r cos θ , y = r sin θ find the value of ( )( )y,x

,r∂

θ∂

(b) Further verify that ( )( )θ∂

∂,r

y,x . ( )( )y,x

,r∂

θ∂ = 1

(a) Solution : We shall first express r, θ in terms of x and y.

We have x = r cos θ , y = r sin θ by data.

∴x2 + y2 = r2 and xy = tan θ or θ = tan-1 (y/x)

Consider r2 = x2 + y2 Differentiating partially w.r.t x and also w.r.t y we get,

2r xr

∂∂ = 2x and 2r

yr

∂∂ = 2y

∴ xr

∂∂ =

rx and

yr

∂∂ =

ry

Also consider θ = ( )xy /tan 1−

∴ x∂θ∂ = 2)x/y(1

1+

. 2xy− and

y∂θ∂ = 2)x/y(1

1+

. x1

i.e., x∂θ∂ = 22 yx

y+

− and y∂θ∂ = 22 yx

x+

Now 2222

),(),(

yxx

yxy

ry

rx

dydx

dyr

rr

yxr

++−

=∂∂

∂∂∂

=∂∂

θθθ

ryxryx

yxry

yxrxei 1

)()(

)()(.,. 22

22

22

2

22

2

=++

=+

++

=

ryxr 1

),(),(

=∂∂

∴θ


10 MAT11

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2011

Solution (b) : Consider θθ sin,cos ryrx ==

rrrr

dy

ry

xrx

ryx

=+=−

=∂

∂∂

∂∂

∂∂

=∂∂ )sin(cos

cossinsincos

),(),( 22 θθ

θθθθ

θ

θθ

rr

yx=

∂∂

∴),(),(

θ

From (1) and (2) : 11

),(),(

),(),(

=⋅=∂

∂⋅

∂

∂

rr

yxr

xyx θθ

20. If ( ) uvyvux =−= ,1 then show that 1/ =JJ

Solution : uvyvux =−= );1(

uvyu

vxv

uyv

ux

=∂∂

−=∂∂

=∂∂

−=∂∂ ,),1(

uvuv

vy

uy

vx

ux

vuyxJ

−−=

∂∂

∂∂

∂∂

∂∂

=∂∂

=)1(

),(),(

uJuuvuv =∴=+−= )1(

Next we shall express u and v in terms of x and y. By data x = u - uv and y = uv

Hence .uyx =+ Also yx

yuyv

+==

Now we have, ,1,1; =∂∂

=∂∂

∴+

=+=yu

xu

yxyvyxu


10 MAT11

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2011

2)(2)(10)(

yxx

yxyyx

xv

+=

+⋅−⋅+

=∂∂

22

/

)()(

11

),(),(

yxx

yxy

yv

xv

yu

xu

yxvuJ

++−=

∂∂

∂∂

∂∂

∂∂

=∂∂

=∴

uyxyxyx

yxy

yxx 1

)(1

)()()( 222 =+

=++

=+

++

=

Thus u

J 1/ = Hence u

uJJ 1/ ⋅=⋅ Thus 1/ =JJ

21. State Taylor’s Theorem for Functions of Two Variables. Statement: Considering f (x + h, y + k) as a function of a single variable x, we have by Taylor’s Theorem

f (x + h, y + k) = f (x, y + k) + h + ---------(1)

Now expanding f (x, y + k) as function of y only,

f (x, y + k) = f (x, y) + k + --------

∴ (i) takes the form f (x + h, y + k) = f (x,y) + k +-------------------------------- +

h

+

Hence f (x + h, y + k) = f (x , y) +h

In symbol we write it as

2

22 ),(!2

),(x

kyxfhx

kyxf∂

+∂+

∂+∂

2

22 ),(!2

),(y

yxfky

yxf∂

∂+

∂∂

2

22 ),(!2

),(y

yxfky

yxf∂

∂+

∂∂

⎭⎬⎫

⎩⎨⎧

−−−−−+∂

∂+

∂∂

+∂∂

2

22 ),(!2

),(),(y

yxfky

yxfkyxfx

⎭⎬⎫

⎩⎨⎧

−−−−−−−−+∂

∂+

∂∂

yyxkyxf

xh ),(),(

!2 2

22

)1(2!2

12

22

2

2

22 −−+⎟⎟

⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

∂+

∂∂

+∂∂

+∂∂

yfk

yxfhk

xfh

yfk

xf


10 MAT11

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2011

F (x + h, y + k) = f (x,y) +

Taking x = a and y = b, (1) becomes

f (a + h, b + k) = f (a,b) + [h f x (a,b) + kf y (a,b)] + [h 2 f xx (a,b)+ 2hkf xy (a,b)

+ k 2 f yy (a,b)] + --------- Putting a + h = x and b + k = y so that h = x – a, k = y – b, we get F (x,y) = f (a,b) + [(x – a) f x (a,b) + (y – b) f y (a,b)]

= [(x – a)2 f xx (a,b) + 2 (x – a) (y – b) f xy (a,b) + (y – b)2 f yy (a,b)] +------ (2)

This is Taylor’s expansion of f (x,y) in powers of (x – a) and (y – b). It is used to expand f (x,y) in the neighborhood of (a,b) corollary, putting a = 0, b = 0 in (2), we get

f (x,y) = f (0,0) + [x f x (0,0) + y f y (0,0)] + [ x 2 f xx (0,0) + 2xy f xy (0,0)

+ y 2 f yy (0,0) ] + --------- (3) This is Maclaurin’s Expansion of f (x,y) 22. Expand e x log (1 + y) in powers of x and y up to terms of third degree. Solution: Here f (x,y) = e x log (1 + y) ∴ f (0,0) = 0 f x (x,y) = e x log (1 + y) ∴ f x (0,0) = 0

f y (x,y) = e x ∴ f y (0,0) = 1

f xx (x,y) = e x log (1 + y) ∴ f xx (0,0) = 0

f xy (x,y) = e x ∴ f xy (0,0) = 1

−−−−+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂

+⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+∂∂ f

yk

xhf

yk

xh

2

!21

!21

!21

!21

y+11

y+11


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2011

f yy (x,y) = - e x (1 + y) -2 ∴ f yy (0,0) = -1 f xxx (x,y) = e x log (1 + y) ∴ f xxx (0,0) = 0

f xxy (x,y) = e x ∴ f xxy (0,0) = 1

f xyy (x,y) = -e x (1 + y) 2 ∴ f xyy (0,0) = -1 f yyy (x,y) = 2e x (1 + y)-3 ∴ f yyy (0,0) = 2 Now, Maclaurin’s expansion of f (x,y) gives

f (x,y) = f (0,0) + x (f x (0,0) + y fy (0,0) + {x2 fxx (0,0) + 2xy fxy (0,0) +

y2 fyy (0,0)} + {x3 fxxx (0,0) + 3x2 y fxxy (0,0) + 3xy2 fxyy (0,0) + y3 fyyy (0,0)} + -----------

∴ex log (1 + y) = 0 + x.0 + y (1) + {x2.0 + 2xy (1) + y2 (-1)}

+ {x3.0 + 3x2y (1) + 3xy2 (-1) + y3(2)}+---------

= y + xy - y2 + (x2y = xy2) + y3 + ----------

23. Expand f (x,y) = ex Cosy by Taylor’s Theorem about the point up to the Second

degree terms.

Solution: f (x,y) = ex Cosy and a = 1, b = ∴ f = =

fx (x,y) = ex Cos y ∴ f =

fy (x,y) = -ex Sin y ∴ fy = -

fxx(x,y) = ex Cos y ∴ fxx =

fxy (x,y) = -ex Sin y ∴ fxy = -

fyy (x,y) = - ex Cos y ∴ fyy = -

Hence by Taylor’s Theorem, we obtain

y+11

!21

!21

!21

!21

21

21

21

⎟⎠⎞

⎜⎝⎛

4,1 π

4π

⎟⎠⎞

⎜⎝⎛

4,1 π

2e

⎟⎠⎞

⎜⎝⎛

4,1 π

2e

⎟⎠⎞

⎜⎝⎛

4,1 π

2e

⎟⎠⎞

⎜⎝⎛

4,1 π

2e

⎟⎠⎞

⎜⎝⎛

4,1 π

2e

⎟⎠⎞

⎜⎝⎛

4,1 π

2e


10 MAT11

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2011

f (x,y) = f + +

+ -------------

i.e., ex Cosy = + +

------------

ex Cosy = -------}

Exercise: 1) Expand exy up to Second degree terms by using Maclaurin’s theorem 2) Expand Log (1 – x – y ) up to Third degree terms by using Maclaurin’s theorem 3) Expand x2y about the point (1,-2) by Taylor’s expansion 4) Obtain the Taylor’s expansion of ex Siny about the point up to Second degree terms 5) Expand esinx up to the term containing x4

⎟⎠⎞

⎜⎝⎛

4,1 π

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −+− fyyfx x 4

)1( π

!21

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−+− yyx fyxfx

22

4)1(2)1( π

2e

⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛ −+−

242)1( eyex π

!21

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ −⎟⎠⎞

⎜⎝⎛ −+⎟

⎠

⎞⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛ −−+−

2424)1(2

2)1(

22 eyeyxex ππ

2e

+⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−⎟

⎠⎞

⎜⎝⎛ −−−−+⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −−−+

22

44)1(2)1(

!21

4)1(1 πππ yyxxyx

( )2,0 π


10 MAT11

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Maxima and Minima:- In mathematics, the maximum and minimum (plural: maxima and minima) of a function, known collectively as extrema (singular: extremum), are the largest and smallest value that the function takes at a point within a given neighborhood. A function f (x, y) is said to have a Maximum value at (a,b) if their exists a neighborhood point of (a,b) (say (a+h, b+k)) such that f (a, b) > f (a+h, b+k). Similarly, Minimum value at (a,b) if there exists a neighborhood point of (a,b) (say (a+h, b+k)) such that f (a, b) < f (a+h, b+k). A Minimum point on the graph (in red) 2 2 3( , ) (1 )f x y x y x= + −

A Maximum point on the graph is at the top (in red)

Engineeri

A sad Sadd

Nec

• • • • •

ing Mathem

ddle point on

le point betw

cessary an

If fx = Functi Functio Functio If AC

further ana

matics – I

n the graph o

ween two hills

nd Suffic

=0 and fy =ion will be on will be mon will be n

C-B2 = 0 walysis w

I

f z=x2−y2 (in

s.

cient Con

=0 (Necesminimummaximum neither ma

we cannot m

where A =

red)

ndition:-

ssary Condm if AC-B2

if AC-B2

axima nor mmake any

, Bxxf= =

dition) 2 > 0 and A > 0 and Aminima if conclusion

, Cxyf= =

Dr. V. Lo

10 MA

A > 0 A < 0

AC-B2 <n without a

yyf=

kesha

AT11

110

0 any

0

2011


10 MAT11

111

2011 Working Procedure:-

• First we find Stationary points by considering fx =0 and fy =0 .

• Function will be minimum if AC-B2 > 0 and A > 0 at that stationary point

• Function will be maximum if AC-B2 > 0 and A < 0 at that stationary point

• Function will be neither maximum nor minimum if AC-B2 < 0 at that stationary point and it is called as SADDLE POINT. 25. Explain Maxima & Minima for Functions of Two Variables& hence obtain the Necessary

Conditions for Maxima, Minima. Solution: Let Z = f (x,y) be a given function of two independent variables x & y. The above equation represents a surface in 3D. Z P(a,b) f (a,b) Z = f (x,y) A = f (a,b) Y a O N b M(a,b) X A given points (a,b) on the surface has Co-ordinates [a,b, f (a,b)]


10 MAT11

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Definition: The function Z = f (x,y) is said to be a maximum at the point (a,b) if f (x,y) < f (a,b) in the neighborhood of the point (a,b) Cap Z Cup O Y (a,b) X Definition: The function Z = f (x,y) is said to posses a minimum at the point (a,b) if f (x,y) > f (a,b) in the neighborhood of the point (a,b) Necessary Condition for Maxima, Minima: If Z = f (x,y) has a max or min at (a,b) then f x (a,b) = 0, fy (a,b) = 0 Sufficient Conditions for Maxima, Minima: Put R = fxx (a,b), S = fxy (a,b), T = fyy (a,b) (1) Suppose S2 – RT > 0 There is no maxima or minima at (a,b) (2) Suppose S2 – RT < 0 Thus there is maxima or minima at (a,b) according as R < 0 Or R > 0 (3) Suppose S2 – RT = 0, Then there is a saddle point at (a,b)


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26. Find the maxima and minima of the functions f (x,y) = x3 + y3 – 3axy, a > 0 is constant. Solution: Given f (x,y) = x3 + y3 – 3axy fx = 3x2 – 3ay, fy = 3y2 – 3ax fxx = 6x fyy = 6y. Put fx = 0, fy = 0 and solve i.e., 3x2 – 3ay = 0 & 3y2 – 3ax = 0 i.e., x2 = ay & y2 = ax

⇒ y = ∴ ⇒ = ax (∵x2 = ay)

∴ x4 = a3x i.e., x (x3 – a3) = 0 ∴ x = 0, x = a ⇒ y = 0, y = ± a ∴ The critical Or stationary points are (0,0), (a,a) and (a,-a) (1) At (0,0) R = fxx (0,0) = 0 S = fxy (0,0) = -3a T = fyy (0,0) = 0 ∴ S2 – RT = 9a2 – 0 = 9a2 > 0 ∴ There is neither a maximum or a minimum at (0,0)

ax 2 22

⎟⎟⎠

⎞⎜⎜⎝

⎛ax

axax

=∴ 2

4


10 MAT11

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2011

27. Examine the following functions for extreme values f = x4 + y4 – 2x2 + 4xy – 2y2 Solution: fx = 4x3 – 4x + 4y fy = 4y3 – 4x – 4y fxy = 4, fxx = 12 x2 – 4, fyy = 12y2 – 4 Put fx = 0, fy = 0 and solve i.e., 4x3 – 4x + 4y = 0 → (1) 4y3 + 4x – 4y = 0 → (2) Adding (1) & (2), we get 4 (x3 + y3) = 0 i.e., x3 + y3 = 0 i.e., y = - x Substitute y = -x in (1), we get 4x3 – 4x – 4x = 0 i.e., 4x3 – 8x = 0 i.e., x3 – 2x = 0 ⇒ x (x2 – 2) = 0 i.e., x = 0 & x2 – 2 =0 i.e., x = 0 & x = ± x = , - ∴x = 0, ,- and corresponding values of y are y = 0, - , ∴ The critical points are (0,0), , - , - , (1) at (0,0) R = fxx (0,0) = - 4

2

2 2

2 2 2 2

2 2 2 2


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S = fxy (0,0) = 4 T = fyy (0,0) = -4 ∴ S2 – RT = 16 – (-4) (-4) = 16 – 16 = 0 i.e., S2 – RT = 0, These is a saddle point at (0,0) (2) at , - R = fxx , - =24– 4 = 20 S = fxy ,- = 4 T = fyy , - = 20 ∴ S2 – RT = 16 – (20) (20) = 16 – 400 = -384 < 0 Thus these is neither maximum nor minimum according to R < 0 or R > 0 at , - Hence R = 20 > 0 ∴ There is a minimum at , - = 4 + 4 – 4 – 8 – 4 = - 8 (3) at - , R = fxx = 20 > 0 S = fxy = 4 T = fyy = 20 ∴ S2 – RT = 16 – 400 = -384 < 0

2 2

2 2

2 2

2 2

2 2

2 2

( ) ( ) ( ) ( ) ( )2244

min 222242222 −−−+−−+=∴ f

2 2

( )2,2−

( )2,2−

( )2,2−


10 MAT11

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Since R > 0, ∴ There is minima at ∴ fmin = - 8 at ∴ Extreme Value = - 8 at & Exercise: 1) Find the extreme values of f = x3 y2 (1 – x – y) 2) Determine the maxima or minima of the function Sin x + Sin y + Sin (x + y) 3) Examine the function f(x,y) = 1+ sin( x2 + y2) for extremum. 28. If PV2 = K and if the relative errors in P is 0.05 and in V is 0.025 show that the error in K

is 10%.

Solution : KPV =2 by data. Also 025.0and05.0 ==VV

PP δδ

KVP loglog2log =+⇒

)(log)(log2)(log KVP δδδ =+⇒

10100)1.0(100

1.0)025.0(205.0.,.

1121.,.

=×=×∴

==+

=⋅+

KK

KKor

KKei

KK

VV

Pp

ei

δ

δδ

δδδ

Thus the error in K is 10%.

( )2,2−

( )2,2−

( )2,2− ( )2,2−


10 MAT11

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29. The time T of a complete oscillation of a simple pendulum is given by the formula

⋅= glT /2π

(i) If g is a constant find the error in the calculated value of T due to an error of 3% in the value of l.

(ii) Find the maximum error in T due to possible errors upto 1% in l and 3% in g.

Solution :

( ) 2 / , Constant, 100 3

1log log 2 (log log )2

1(log ) (log 2 ) (log )2

1. ., 0 021 1100 100 (3) 1.52 2

li T l g gl

T l g

T g

T li eT lT lor

T l

δπ

π

δ δ π δ

δ δ

δ δ

= = × =

⇒ = + −

⇒ = +

= + −

⎛ ⎞× = × = =⎜ ⎟⎝ ⎠

∴ the error in T = 1.5%. (ii) If g is not a constant we have,

⎟⎟⎠

⎞⎜⎜⎝

⎛×−⎟

⎠⎞

⎜⎝⎛ ×=× 100

21100

21100

gg

ll

TT δδδ

The error in T will be maximum if the error in l is positive and the error in g is negative (or vice-versa) as the difference in errors converts in to a sum.

2)3(21)1(

21100max =−−+=⎟

⎠⎞

⎜⎝⎛ ×∴

TTδ

∴ the maximum error in T is 2%.


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30. The current measured by a tangent galvanometer is given by the relation

c = k tan θ where θ is the angle of deflection. Show that the relative error in c due to a

given error in θ is minimum when θ = .450

Solution : Consider c = k tan θ . K is taken as a constant.

)(tanlogloglog θ+=⇒ kc

)(tanlog)(log)(log θδδδ +=⇒ kc

δθθθδ

tansec01.,.

2

+=cc

ei

θθδθδθδθ

θθθδ

cossincos1

sincos.,. 2 =⋅=

cor

ccei

δθθ

δ2sin

2.,. =ccei

The relative error in c being cc /δ minimum when the denominator of the R.H.S. is maximum

and the maximum value of a sine function is1.

⋅==⇒=∴ 00 4590212sin θθθ or

Thus the relative error in c is minimum when 045=θ

31. 2

21 mvTIf = is the kinetic energy, find approximately the change in T as m changes

from 49 to 49.5 and v changes from 1600 to 1590. 6 Marks

Solution : We have by data 2

21 mvT = and

5.0m5.49mm,49m =δ∴=δ+=

10v1590vv,1600v −=δ∴=δ+=


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We have to find .Tδ (logarithm is not required)

)(21 2mvT δδ =∴

{ }2.)2(21 vmvvm δδ +

{ } 000,44,1)1600()5.0()10()1600()2()49(21,. 2 =+−=ei

Thus the change in 000,44,1−== TT δ

32. The pressure p and the volume v of a gas are concentrated by the relation

.tan4.1 tconspv = Find the percentage increase in pressure corresponding to a

diminution of ½% in volume.

Solution : 1.4 Constant ( ),pv c say= = by data.

cvp loglog4.1log =+⇒

)(log)(log4.1)(log cvp δδδ =+⇒

,21100But;04.1,. −=×=⎟

⎠⎞

⎜⎝⎛+

vv

vv

ppei δδδ by data.

7.0100or01004.1100 +=×=⎟⎠⎞

⎜⎝⎛ ×+×∴

pp

vv

pp δδδ .

Thus the percentage increase in pressure = 0.7 33. Find the percentage error in the area of an ellipse when an error of +1% is made in

measuring the major and minor axis.

Solution : For the ellipse 1// 2222 =+ byax the area (A) is given by π ab where 2a and 2b are

the lengths of the major and minor axis. Let 2a = x and 2b = y.

By data .1100,1100 =×=×yy

xx δδ


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xyyxabA422πππ =⋅⋅==

yxA loglog)4/(loglog ++=∴ π

)(log)(log)4/(log)(log yxA δδπδδ ++=⇒

1001001000.,. ×+×=×++=yy

xx

AAor

yy

xx

AAei δδδδδδ

211100 =+=×∴AAδ

Thus error in the area = 2% 27. If the sides and angles of a triangle ABC vary in such way that the circum radius

remains constant, prove that

0coscoscos

=++c

cB

bA

a δδδ

Solution : If the triangle ABC is inscribed in a circle of radius r and if a,b,c respectively

denotes the sides opposite to the angles A,B,C we have the sine rule (formula) given by

rC

cB

bA

a 2sinsinsin

===

or CrcBrbAra sin2,sin2,sin2 ===

)(sin2),(sin2),(sin2 CrcBrbAra δδδδδδ ===⇒

CCrcBBrbAAraei δδδδδδ cos2,cos2,cos2.,. ===

or CrC

cBrB

bArA

a δδδδδδ 2cos

,2cos

,2cos

===

Adding all these results we get,

)(2)(2coscoscos

CBArCBArC

cB

bA

a++=++=++ δδδδδδδ

But A+B+C = 180 = π radians = constant.

δδ =++⇒ )( CBA (constant) = 0

Thus 0coscoscos

=++C

cB

bA

a δδδ


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Multiple choice Questions:

DIFFERENTIAL CALCULUS

1) The radius of curvature at any point of catenary S=C tan Φ is a) c sec2 Φ b) c cos2 Φ c) c tan-2 Φ d)none 2) Stationary points of f(x,y) = sin x+ sin y+ sin ( x + y ) is a) (п/3, п/3) b) (п/6, п/6) c) (п/4, п/4) d) none 3) If the curvature is zero, that point is known as-------------------- a) Point of inflection b) Stationary point c) (a) or (b) d) none 4) The radius of curvature of the curve y = 4 sin x- sin 2x at x=п/2 is a) 5√5/4 b) -5√5/4 c)5√5/2 d) none 5) The function x2+2xy+ 2y2+2x+2y has a minimum value at

a) (-3/2,1/2) b)(3/2, 1/2 ) c) (3/2,-1/2) d)none 6) The stationary point of f(x,y,z)=x2+y2+z2 where x,y,z are connected by x+y+z=a is a) (a,a) b) (-a,-a) c) (2a, 0) d) none 7) The radius of curvature of the curve √x+√y =1 at the point (1/4,1/4) is ------------ a) ρ =1/√2 b) ρ=√2 c) ρ= -1/√2 d) none ) 8) The expression for the derivative of arc length in Cartesian form is given by a) ds/d�= b) ds/dx=

c) ds/dr= d) ds/dt=

9)The formulae for radius of curvature in Cartesian form is

a) ρ = (x1) 2 + (y1) 2)3/2 /x1y11+x11y1

b) ρ =

c) ρ = [(x1) 2 + (y1) 2)3/2 ]

)dr/d (r 22 θ+ 2(dy/dx)1+2d?/dr) (1+ 2(dy/dx)1+

( ) 2/3

2

211

yy+


10 MAT11

122

2011 d) ρ = (1+(y1) 2 ) 3/2 10) The function for which Rolle’s theorem is true is: a) f(x)=log x in the interval [1/2,2] b) f(x)=|x+1| in the interval [-2,2] c) f(x)=| x |in the interval [-1,1] d) Non of the above 11) The value of ‘c’in Rolle’s theorem ,where and f(x)= cosx is equal to: a b) c) d)0 12).The expansion of tanx in powers of x by Maclaurin’s theorem is valid in the interval : a) b) c) d) 13) The value of ‘c’in Lagrange,s mean value theorem ,where [1,2] and f(x)= x(x-1) is : a)5/4 b)3/2 c)7/4 d) 11/6 14) The value of ‘c’in Rolle’s theorem, where [0, ] and f(x)= sinx is equal to: a) b) c) d) Non of these 15) The maximum value of logx/x is: a) 1 b)e c)2/e d) 1/e

16) The maximum value of (l/x ) x is equal to: a)e b)1 c) e 1/e d) (1/e)e

17) The difference bwtween the maximum and minimum values of the function a sinx+bcosx is: a) b) c) d) 18) Which one of the following statement is correct for the function f(x)=x3

a) f(x) has a maximum value at x=0 b) f(x) has a manimum value at x=0 c) f(x) has a neither a maximum nor a manimum value at x=0 d) f(x) has no point of inflexion 19) which one of the following is not an indeterminate form a) b) c) d)

20) In Lagrange,s mean value theorem , f1(c)= a) f(a) – f(b)/ b -a b) f(b) – f(a) / a -b c) f(b) – f(a) / b -a d) none

2/2/ ππ <<− c4/π 3/π π

( )∞∞− , ( )2/3,2/3 ππ− ( )ππ ,− ( )2/,2/ ππ−

π

6/π 3/π 2/π

222 ba + )(2 22 ba + 22 ba + 22 ba +−

∞+∞ ∞−∞ ∞∞ / ∞X0


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2011

21) In Cauchy’s mean value theorem , f1(c) / g1(c) = a) f(a) – f(b)/ g(b) – g(a) b) f(b) – f(a) / g(b) – g(a) .c) f(b) +f(a) / g(b) +g(a) d) none

22) is equal to :

a) 0 b) ∞ c) log (a/b) d) log (a-b) 23) The value of is equal to:

a) -1 b) ∞ c) 1 d) 0 24) The value of (1 – cos x / 3 x2 )is equal to:

a) 3 b) 1/3 c) 1/6 d) 1/9

25) ( ex + e-x – 20) / x2 is equal to : a) 1 b) -1 c) 1/2 d) -1/2 26) The value of ( 1 + x )1/x is : a) 1 b) -1 c) 1/e d) e

27) The value of =….

a) 1 b) 2 c) 3 d)0 28) The formulae for radius of curvature in polar form is

a) ρ = rdr/dp b) ρ = [ (r 2 + (r1) 2)3/2] / [r2+2 ( r1) 2- r r2]

c) ρ =(r 2 + (r1) 2)3/2 d) None

29) The value of 1 1/x =……

a) 1 b) ∞ c) -∞ d) 0

0lim

→x x

x xb -a

0lim

→x 1log

−xx

0lim

→x

0lim

→x

0lim

→x

xx

x

sinlim0→

0lim

→x


10 MAT11

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2011

KEY ANSWERS:

1-a 2-a 3-a 4-a 5-a 6-a 7-a 8-b 9-b 10-a

11-d 12-d 13-b 14-c 15-d 16-c 17-a 18-c 19-a 20-c

21-b 22-c 23-c 24-c 25-a 26-c 27-a 28-b 29-b --

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