1. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
1
2011
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

2. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
2
2011
ENGNEERING MATHEMATICS – I
Content
CHAPTER
UNIT I DIFFERENTIAL CALCULUS – I
UNIT II DIFFERENTIAL CALCULUS – II
UNIT III DIFFERENTIAL CALCULUS – III

3. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
3
2011
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
dx
dy
2
2
dx
y
d
n
n
dx
y
d

4. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
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10 MAT11
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6. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
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7. Engineering Mathematics – I Dr. V. Lokesha
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2011

8. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
8
2011
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 + n
C1 u1vn-1 + n
C2u2vn-2 + …+n
Cn-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 ( ) ( )
[ ]
c
bx
sin
b
c
bx
cos
a +
−
+
Let us put a = r cos θ , and b = r sin θ .
∴ 2
2
2
r
b
a =
+ and tan a
/
b
=
θ
.ie., 2
2
b
a
r +
= and θ = tan-1 (b/a)
Now, [ ]
)
c
bx
sin(
sin
r
)
c
bx
cos(
cos
r
e
y ax
1 +
θ
−
+
θ
=
Ie., y1 = r eax cos ( )
c
bx +
+
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 ( )
c
bx
2 +
+
θ .
Similarly y3 = r3 e ax cos ( )
c
bx
3 +
+
θ .
………………………………………
Thus ( )
c
bx
n
cos
e
r
y ax
n
n +
+
θ
=
Where 2
2
b
a
r +
= and θ = tan-1 (b/a).
Thus Dn [eax cos (b x + c)]
= ( )
[ ]
[ ]
c
bx
a
b
n
e
b
a ax
n
+
+
+ −
/
tan
cos
)
( 1
2
2

9. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
9
2011
2. Find the nth derivative of log 3
8
4 2
+
+ x
x
Solution : Let y = log 3
x
8
x
4 2
+
+ = log (4x2 + 8x +3) ½
ie.,
2
1
y = log (4x2 + 8x +3) ∵ log xn = n log x
2
1
y = log { (2x + 3) (2x+1)}, by factorization.
∵
2
1
y = {log (2x + 3) + log (2x + 1)}
Now
( ) ( )
( )
( ) ( )
( ) ⎭
⎬
⎫
⎩
⎨
⎧
+
−
−
+
+
−
−
=
−
−
n
n
1
n
n
n
1
n
n
1
x
2
2
!
1
n
1
3
x
2
2
!
1
n
1
2
1
y
Ie., yn = 2n-1 (-1) n-1 (n-1) !
( ) ( ) ⎭
⎬
⎫
⎩
⎨
⎧
+
+
+
n
n
1
x
2
1
3
x
2
1
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:
10
log
x
log
x
log
e
e
10 =
Thus ( ) ( )
{ }
5
3
e
e
1
x
8
x
2
1
log
10
log
1
y +
−
=
Ie., ( ) ( )
{ }
1
x
8
log
5
x
2
1
log
3
10
log
1
y
e
+
+
−
=
( ) ( )( )
( )
( ) ( )
( ) ⎭
⎬
⎫
⎩
⎨
⎧
+
−
−
+
−
−
−
−
=
∴
−
−
n
n
1
n
n
n
1
n
e
n
1
x
8
8
!
1
n
1
5
x
2
1
2
!
1
n
1
.
3
10
log
1
y
Ie.,
( ) ( ) ( )
( )
( )
( ) ⎭
⎬
⎫
⎩
⎨
⎧
+
+
−
−
−
−
=
−
n
n
n
n
e
n
1
n
n
1
x
8
4
5
x
2
1
1
3
10
log
2
!
1
n
1
y

10. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
10
2011
4. Find the nth derivative of e2x cos2 x sin x
Solution : >> let y = e2x cos2 x sin x = e2x
⎥
⎦
⎤
⎢
⎣
⎡ +
2
x
2
cos
1
sin x
ie.,
2
e
y
x
2
= (sin x + sin x cos 2x)
= ( )
[ ]
⎭
⎬
⎫
⎩
⎨
⎧
−
+
+ x
sin
x
3
sin
2
1
x
sin
2
e x
2
= ( )∵
x
sin
x
3
sin
x
sin
2
4
e x
2
−
+ sin (-x) = -sin x
4
e
y
x
2
=
∴ (sin x + sin 3x)
Now ( ) ( )
{ }
x
3
sin
e
D
x
sin
e
D
4
1
y x
2
n
x
2
n
n +
=
Thus ( ) ( )
[ ] ( ) ( )
[ ]
{ }
x
3
2
3
tan
n
sin
e
13
x
2
1
tan
n
sin
e
5
4
1
y 1
x
2
n
1
x
2
n
n +
+
+
= −
−
( ) ( )
[ ] ( ) ( )
[ ]
{ }
x
3
2
3
tan
n
sin
13
x
2
1
tan
n
sin
5
4
e
y 1
n
1
n
x
2
n +
+
+
=
∴ −
−
5. Find the nth derivative of e2x cos 3x
Solution : Let y=e2x cos3 x = e 2x.
4
1
(3 cos x + cos 3x)
Ie., y =
4
1
(3 e2x cos x + e2x cos 3x)
4
1
yn =
∴ {3Dn (e2x cos x) + Dn (e2x cos 3x)}
( ) ( )
[ ] ( ) ( )
[ ]
{ }
x
n
e
x
n
e
y x
n
x
n
n 3
2
3
tan
cos
13
2
1
tan
cos
5
3
4
1 1
2
1
2
+
+
+
= −
−
Thus ( ) ( )
[ ] ( ) ( )
[ ]
{ }
x
3
2
3
tan
n
cos
13
x
2
1
tan
n
cos
5
3
4
e
y 1
n
1
n
x
2
n +
+
+
= −
−

11. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
11
2011
6. Find the nth derivative of
( )( )
3
x
2
1
x
2
x2
+
+
Solution :
( )( )
3
x
2
1
x
2
x
y
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.
3
x
8
x
4
x
4
.
4
1
3
x
8
x
4
x
y 2
2
2
2
+
+
=
+
+
= for convenience.
4x2 +8x +3
1
2
2
3
8
3
8
4
4
−
−
+
+
x
x
x
x
∴ ⎥
⎦
⎤
⎢
⎣
⎡
+
+
−
−
+
=
3
x
8
x
4
3
x
8
1
4
1
y 2
Ie., ⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
−
=
3
x
8
x
4
3
x
8
4
1
4
1
y 2
The algebraic fraction involved is a proper fraction.
Now ⋅
⎥
⎦
⎤
⎢
⎣
⎡
+
+
+
−
=
3
x
8
x
4
3
x
8
D
4
1
0
y 2
n
n
Let
( )( ) 3
x
2
B
1
x
2
A
3
x
2
1
x
2
3
x
8
+
+
+
=
+
+
+
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
∴
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
+
+
⎥
⎦
⎤
⎢
⎣
⎡
+
−
−
=
3
x
2
1
D
2
9
1
x
2
1
D
2
1
4
1
y n
n
n

12. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
12
2011
( ) ( )
( )
( )
( ) ⎭
⎬
⎫
⎩
⎨
⎧
+
−
⋅
+
+
−
⋅
−
−
= +
+ 1
n
n
n
1
n
n
n
3
x
2
2
!
n
1
9
1
x
2
2
!
n
1
1
8
1
ie.,
( )
( ) ⎭
⎬
⎫
⎩
⎨
⎧
+
+
+
−
= +
+
+
1
n
1
n
n
1
n
n
3
x
2
9
)
1
x
2
(
1
8
2
!
n
1
y
7. Find the nth derivative of
)
2
(
)
1
(
4
+
+ x
x
x
Solution : y =
)
2
(
)
1
(
4
+
+ x
x
x
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 ) + ⎥
⎦
⎤
⎢
⎣
⎡
+
+
−
−
2
3
14
15
2
x
x
x
∴ yn = Dn (x2-3x+7)-Dn
⎥
⎦
⎤
⎢
⎣
⎡
+
+
−
2
3
14
15
2
x
x
x
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
(
14
15
x
x
x
Now, let Dn
)
2
(
)
1
(
2
3
14
15
2
+
+
+
=
+
+
+
x
B
x
A
x
x
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 =
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
+
+
⎥
⎦
⎤
⎢
⎣
⎡
+
−
2
1
16
1
1
x
D
x
D n
n

13. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
13
2011
1 1
1 1
( 1) ! 1 ( 1) ! 1
16
( 1) ( 2)
1 16
( 1) ! 2
( 1) ( 2)
n n n n
n n
n
n n n
n n
x x
y n n
x x
+ +
+ +
− −
= −
+ +
⎧ ⎫
= − − >
⎨ ⎬
+ +
⎩ ⎭
8. Show that
⎭
⎬
⎫
⎩
⎨
⎧
−
−
−
−
−
=
⎟
⎠
⎞
⎜
⎝
⎛
+
n
x
x
n
x
x
dx
d
n
n
n
n
1
3
1
2
1
1
log
!
)
1
(
log
1
Solution : Let y =
x
x
x
x 1
.
log
log
= and let u = log x, v =
x
1
We have Leibnitz theorem,
(uv)n = uvn + v
u
v
u
n
v
u
n n
n
C
n
C +
+
+ −
− ....
2
2
1
1 2
1
…… (1)
Now, u = log x ∴un = n
n
x
n )!
1
(
)
1
( 1
−
− −
x
v
1
= ∴vn = 1
!
)
1
(
+
−
n
n
x
n
Using these in (1) by taking appropriate values for n we get,
Dn = n
n
n
n
x
n
x
n
x
n
x
x
x )!
1
(
)
1
(
.
1
!
)
1
(
..
log
log 1
1
−
−
+
−
=
⎟
⎠
⎞
⎜
⎝
⎛
−
+
( )
x
x
n
x
n
x
n
n
n
n
n
n
1
.
!
1
)
1
(
......
)!
2
(
)
1
(
1
2
.
1
)
1
(
1
1
2
2
−
−
+
+
−
−
⎟
⎠
⎞
⎜
⎝
⎛
−
−
+
−
−
−
Ie.. = log x 1
1
1
!
)
1
(
!
)
1
(
+
−
+
−
+
−
n
n
n
n
x
n
x
n
1
1
1
2
)!
1
(
)
1
(
....
2
!
)
1
(
+
−
+
−
−
−
+
+
−
− n
n
n
n
x
n
x
n
⎥
⎦
⎤
⎢
⎣
⎡ −
−
+
+
−
−
−
−
−
−
−
−
+
−
1
1
2
1
1
2
)!
1
(
)
1
(
....
2
)
1
(
)
1
(
log
!
)
1
(
n
n
x
x
n
n
n
Note : (-1)-1 = 1
)
1
(
1
)
1
(
;
1
1
1
2
2
=
−
=
−
−
=
−
−

14. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
14
2011
Also
n
n
n
n
n
n 1
)!
1
(
)!
1
(
!
)!
1
(
=
−
−
=
−
⎥
⎦
⎤
⎢
⎣
⎡
−
−
−
−
−
=
⎥
⎦
⎤
⎢
⎣
⎡
∴ +
n
x
x
n
x
x
dx
d
n
n
n
n
1
...
3
1
2
1
1
log
!
)
1
(
log
1
9. If yn= Dn (xn logx)
Prove that yn = n yn-1+(n-1)! and hence deduce that
yn = n ⎟
⎠
⎞
⎜
⎝
⎛
+
+
+
+
+
n
x
1
....
3
1
2
1
1
log
Solution : yn = Dn(xn log x) = Dn-1 {D (xn log x}
= Dn-1
⎭
⎬
⎫
⎩
⎨
⎧
+ −
x
nx
x
x n
n
log
1
. 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! ⎟
⎠
⎞
⎜
⎝
⎛
+
+
2
1
1
log x
y3 = 2! + 3y2 = 2 + 3(2 log x + 3)
ie., y3 = 61og x+ll = 6 (log x + ll/6) = 3! ⎟
⎠
⎞
⎜
⎝
⎛
+
+
+
3
1
2
1
1
log x
…………………………………………………………………………..
⎟
⎠
⎞
⎜
⎝
⎛
+
+
+
+
+
=
n
x
n
yn
1
...
3
1
2
1
1
log
!
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]

15. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
15
2011
Solution : y = a cos (log x) + b sin (log x)
Differentiate w.r.t x
∴ y1 = -a sin (log x)
x
1
+ b cos (log x).
x
1
(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)
x
1
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
.
)
(
. )
2
2
2
1
2
2
y
D
n
n
y
D
x
n
y
D
x n
n
n
{ } 0
)
(
.
1
.
)
(
. 1
1
1 =
+
+ −
n
n
n
y
y
D
n
y
D
x
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) xyn+l + 2n2yn = 0
Solution :By data, cos-1 (y/b) = n log (x/n) ∴log(am) = m log a
=>
b
y
= cos [n log (x/n )]
or y = b . cos [ n log (x/n)]
Differentiating w.r.t x we get,

16. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
16
2011
y1 = -b sin [n log (x/n)]
( ) n
n
x
n
1
/
1
⋅
⋅
⋅
or xy1 = - n b sin [n log (x/n )]
Differentiating w.r.t x again we get,
xy2 + 1. 1
y = - n . b cos [ n log (x/n )] n
n
n
x
1
.
)
/
(
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
.
2
1
1
2
2
=
+
+
+
⎭
⎬
⎫
⎩
⎨
⎧ −
+
+
+
+
+
n
n
n
n
n
n
y
n
y
n
xy
y
n
n
y
x
n
y
x
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) 2
2
)
1
(
1
2
+
+
x
x
ie., y1 = cos log (x2 + 2 x + 1)
1
2
1
2
+
+ x
x
2 (x + 1)
ie., y1 =
)
1
(
)
1
x
2
x
log(
cos
2 2
+
+
+
x
or (x + 1) y1 = 2 cos log (x2 + 2 x + 1 )
Differentiating w.r.t x again we get

17. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
17
2011
(x+1)y2 + 1 y1 = -2 sin log (x2 + 2x + 1) )
1
(
2
.
)
1
(
1
2
+
+
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,
⎭
⎬
⎫
⎩
⎨
⎧ −
+
+
+
+ +
+ n
n
n y
n
n
y
x
n
y
x .
2
.
2
.
1
)
1
(
).
1
(
2
.
)
1
( 1
2
2
+ {(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 ( )
2
1 x
x +
+ prove that
(1 + x2) yn+2 + (2n + 1) xyn+1 + n2yn = 0
>> By data, y = log ( )
2
1 x
x +
+
∴y1 =
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
+
+
+
+
x
x
x
x
2
.
1
2
1
1
)
1
(
1
2
2
Ie.,
2
2
2
2
1
1
1
1
1
)
1
(
1
x
x
x
x
x
x
y
+
=
+
+
+
+
+
or 1
1 1
2
=
+ y
x
Differentiating w.r.t.x again we get
0
.
2
.
)
1
2
1
1 1
2
2
2
=
+
+
+ y
x
x
y
x
or (1+x2)y2 + xy1 = 0
Now Dn [(l+x2)y2] + Dn[xy1] = 0
Applying Leibnitz theorem to each term we get,

18. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
18
2011
⎭
⎬
⎫
⎩
⎨
⎧ −
+
+
+ +
+ n
n
n y
n
n
y
x
n
y
x .
2
.
2
.
1
)
1
(
.
2
.
)
1
( 1
2
2
+ [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)
2
1 x
m
−
or 1
2
1 y
x
+ = - m sin (m sin-1x)
Differentiating again w.r.f .x we get,
2
1
1
2
2
2
1
).
sin
(
cos
)
2
(
1
2
1
1
x
m
x
m
m
y
x
x
y
x
−
−
=
−
−
+
− −
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 = ⋅
−
x
e
1
tan
∴ y. = ⋅
−
x
e
1
tan
2
1
1
x
+
or (1 +x2)y1 = y
Differentiating n times we have

19. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
19
2011
Dn[(l+x2)y1]=Dn[y]
Anplying Leibnitz theorem onto L.H.S, we have,
{(l+x2)Dn(y1) + n .2x .Dn-1 (y1)
n
n
y
y
D
n
n
=
−
+ −
)}
(
.
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

20. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
20
2011
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
a
x
Lt
→
⎩
⎨
⎧
≤
≤
≤
≤
−
−
=
1
0
0
1
x
x
x
x
x
0
( ) ( )
lim '( ) I
h
f x h f a
f a exists a
h
→
+ −
= ∈
)
(
)
(
lim a
f
x
f
a
x
=
→
1 1 0
( )
0 1
x
f x
x x
− − ≤ ≤
⎧
= ⎨
< ≤
⎩

21. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
21
2011
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.

22. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
22
2011
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.
2
x
e
−
2
2
2
1
)
3
2
(
2
1
)
3
(
)
(
x
x
e
x
e
x
x
x
f
−
−
+
+
⎟
⎠
⎞
⎜
⎝
⎛
−
+
=
2
2
)
6
(
2
1 x
e
x
x
−
−
−
−
2
2
)
6
(
2
1 x
e
x
x
−
−
−
−

23. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
23
2011
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
2
π
2
π

24. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
24
2011
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:
⎭
⎬
⎫
⎩
⎨
⎧
+
+
x
b
a
ab
x
)
(
2
a
b
a
f
b
f
c
f
−
−
=
)
(
)
(
)
(
1
[b,f(b)]
[a,f(a)]
x
y
a c b

25. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
25
2011
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.
a
b
a
f
b
f
A
−
−
=
)
(
)
(
a
b
a
f
b
f
c
f
−
−
=
)
(
)
(
)
(
1

26. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
26
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)).
a
b
b
f
b
f
−
− )
(
)
(

27. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
27
2011
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
)
0
4
(
)
0
(
)
4
(
)
(
)
(
)
(
1
−
−
=
−
−
=
f
f
a
b
a
f
b
f
c
f
3
4
)
6
(
6
=
−
−
=
3
2
x
x
f
1
)
(
1
=

28. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
28
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
.
c
e
c
e
Log
Loge 1
1
1
1
1
1
=
−
⇒
=
−
−
2
1
⎥
⎦
⎤
⎢
⎣
⎡
2
,
0
π
2
1 b
a
b
+
−
2
1 a
a
b
+
−
6
1
4
3
4
tan
25
3
4
1
+
<
<
+ − π
π
1
2
<
<
x
Sinx
π
⎟
⎠
⎞
⎜
⎝
⎛
2
,
0
π
2
1
1
2
1
1 b
a
b
a
Sin
b
Sin
a
a
b
−
−
<
−
<
−
− −
−
15
1
6
4
1
3
2
1
6
1
−
<
<
− − π
π
Sin

29. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
29
2011
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
a
g
b
g
a
f
b
f
c
g
c
f
−
−
=
)
(
)
(
)
(
)
(
a
g
b
g
b
f
b
f
A
−
−
=
⇒

30. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
30
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
c
g
c
f
A =
⇒
)
(
)
(
)
(
)
(
)
(
)
(
1
1
a
g
b
g
a
f
b
f
c
g
c
f
−
−
=
)
(
)
(
)
1
(
)
3
(
)
1
(
)
3
(
1
1
c
g
c
f
g
g
f
f
=
−
−
2
3
26
2
28
4
12
c
=
−
−
6
1
2
6
13
3
1
13
2
=
=
⇒
= C
c
6
1
2

31. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
31
2011
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
c
g
c
f
g
g
f
f
=
−
⎟
⎠
⎞
⎜
⎝
⎛
−
⎟
⎠
⎞
⎜
⎝
⎛
π
π
2
π
Sinc
Cosc
−
=
−
−
1
0
0
1
4
π
4
π
2
π
⎥
⎦
⎤
⎢
⎣
⎡
1
,
3
1
x
1

32. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
32
2011
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
)
( 2
a
b −
)!
1
(
)
( 1
−
− −
n
a
b n
)
(
!
)
(
c
f
n
a
b n
n
−
!
2
)
( 2
a
x −
)!
1
(
)
( 1
−
− −
n
a
x n
n
a
x n
∠
− )
(
!
)
(
n
a
x n
−
!
n
xn
)
(
)]
(
)
(
[ x
R
Lim
x
S
x
f
Lim n
n
n
n ∞
→
∞
→
=
−
)
(
)
(
0
)
( x
S
Lim
x
thenf
x
R
Lim n
n
n
n ∞
→
∞
→
=
)
(x
S
Lim n
n ∞
→
!
2
)
( 2
a
x −

33. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
33
2011
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
)
( 2
x
!
2
)
( 2
x
!
)
(
n
x n
2
)
( 2
∠
− a
x
2
)
1
( 2
−
x
2
)
1
( 2
−
x
Sinx
Cosx
!
2
)
( 2
a
x −
!
3
)
( 3
a
x −
!
2
)
3
( 2
−
x
!
3
)
3
( 3
−
x
!
2
)
3
( 2
−
x
!
3
)
3
( 3
−
x

34. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
34
2011
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.
⎟
⎠
⎞
⎜
⎝
⎛
−
2
π
x
⎥
⎦
⎤
⎢
⎣
⎡
−
−
−
−
−
−
−
−
−
+
+
+
+
!
3
!
2
1
3
2
h
h
h
{ }
⎭
⎬
⎫
⎩
⎨
⎧
+
=
+
= −
−
x
dx
d
x
dx
d
x
f n
n
n
n
n
1
1
)
1
log(
)
( 1
1
n
n
n
n
)
1
(
)!
1
(
)
1
( 1
+
−
− −
!
2
2
x
!
n
xn
!
2
2
x
!
3
3
x
!
4
4
x
2
2
x
3
3
x
4
4
x

35. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
35
2011
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
2
1
1
x
+
2
.
1
)
1
( −
n
n
!
2
2
x
!
3
3
x
3
3
x
5
5
x
!
2
2
x
!
4
5 4
x
6
3
x
40
3 5
x
3
3
x

36. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
36
2011
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
)
( x
x
e
x
f x
−
= ⎥
⎦
⎤
⎢
⎣
⎡
4
5
,
4
π
π
2
/
)
2
(
)
( x
e
x
x
x
f −
= [ ]
2
,
0
2
sin 2
( )
x
x
f x
e
= ⎥
⎦
⎤
⎢
⎣
⎡
4
5
,
4
π
π
( ) ( 1)( 2)
f x x x x
= − −
1
0,
2
⎡ ⎤
⎢ ⎥
⎣ ⎦
1
( )
f x Tan x
−
= [ ]
0,1
( )
f x x
=
1
( )
g x
x
=
1
,1
4
⎡ ⎤
⎢ ⎥
⎣ ⎦
2
1
( )
f x
x
=
1
( )
g x
x
= [ ]
,
a b
( )
f x Sin x
= ( )
g x Cos x
= [ ]
,
a b

37. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
37
2011
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
)
(x
f
Lim
a
x→
)
(x
g
Lim
a
x→
)
(
)
(
)
(
)
(
x
g
Lim
x
f
Lim
x
g
x
f
Lim
a
x
a
x
a
x
→
→
→
=
)
(x
f
Lim
a
x→
)
(x
g
Lim
a
x→
0
0
)
(
)
(
=
→ x
g
x
f
Lim
a
x
∞
−
∞
∞
×
∞
∞
,
0
,
0
0
∞
∞
)
(x
f
Lim
a
x→
)
(x
g
Lim
a
x→
)
(
)
(
)
(
)
(
1
1
x
g
Lim
x
f
Lim
x
g
x
f
Lim
a
x
a
x
a
x
→
→
→
=
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
→
→
)
(
1
1
)
(
)
(
x
f
x
g
Lim
x
g
x
f
Lim
a
x
a
x 0
0

38. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
38
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
)
(
)
(
)
(
)
(
)]
(
[
)
(
)]
(
[
)
(
)
(
)
(
⎥
⎦
⎤
⎢
⎣
⎡
=
−
−
=
→
→
→ x
g
x
f
x
f
x
g
Lim
x
f
x
f
x
g
x
g
Lim
x
g
x
f
Lim
a
x
a
x
a
x
2
1
1
)
(
)
(
)
(
)
(
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
=
→
→ x
g
x
f
Lim
x
f
x
g
Lim
a
x
a
x
∞
≠
≠
→
and
x
g
x
f
Lim
a
x
0
)
(
)
(
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
=
→
→ )
(
)
(
)
(
)
(
1
1
x
g
x
f
Lim
x
f
x
g
Lim
a
x
a
x
)
(
)
(
)
(
)
(
1
1
x
g
x
f
Lim
x
g
x
f
Lim
a
x
a
x →
→
=
)
(
)
(
x
g
x
f
Lim
a
x→
0
0
=
→ x
Sinx
Lim
a
x
1
1
1
1
=
=
→
θ
Cos
Cosx
Lim
a
x
1
=
∴
→ x
Sinx
Lim
a
x
Cotx
Sinx
Lim
a
x
log
→
Cotx
Sinx
Lim
a
x
log
→ ∞
∞
−
=
∞
=
0
log
0
0
log
Cot
Sin
xCotx
xCo
Co
x
Cosce
Lim
a
x sec
sec
2
2
−
−
→

39. Engineering Mathematics – I Dr. V. Lokesha
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
0
2
1
=
=
→ Cotx
Lim
a
x
0
log
=
∴
→ Cotx
Sinx
Lim
a
x
x
x
Lim
x
tan
0
→
x
x
x
Lim
1
0
)
1
( +
→
x
a
Lim
x
x
1
−
∞
→
a
x
a
x
Lim
n
n
x −
−
→0
)
(x
f
Lim
a
x→
∞
=
→
)
(x
g
Lim
a
x
)
(x
f
Lim
a
x→ 0
0
∞
∞
[ ]
0
0
)
(
1
)
(
)
(
).
( =
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
→
→
x
g
x
f
Lim
x
g
x
f
Lim
a
x
a
x
[ ]
∞
∞
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
→
→
)
(
1
)
(
)
(
).
(
x
f
x
g
Lim
x
g
x
f
Lim
a
x
a
x
)
(x
f
Lim
a
x→
∞
=
→
)
(x
g
Lim
a
x
[ ] ∞
−
∞
=
→
)
(
).
( x
g
x
f
Lim
a
x
0
0
∞
∞

40. Engineering Mathematics – I Dr. V. Lokesha
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
⎭
⎬
⎫
⎩
⎨
⎧ +
−
→ 2
0
)
1
log(
1
x
x
x
Lim
x
⎭
⎬
⎫
⎩
⎨
⎧ +
−
→ 2
0
)
1
log(
1
x
x
x
Lim
x
0
0
1
log(
2
0
=
⎭
⎬
⎫
⎩
⎨
⎧ +
−
→ x
x
x
Lim
x
x
x
Lim
x 2
1
1
1
0
+
−
→
2
1
)
1
(
2
1
2
1
0
0
=
+
=
+
=
→
→ x
Lim
x
x
x
Lim
x
x
⎭
⎬
⎫
⎩
⎨
⎧
−
→
Cotx
x
Lim
x
1
0
⎭
⎬
⎫
⎩
⎨
⎧
−
→ Sinx
Cosx
x
Lim
x
1
0
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎠
⎞
⎜
⎝
⎛ −
→ 0
0
0 xSinx
xCosx
Sinx
Lim
x
Sinx
xCosx
xSinx
Cosx
Cosx
Lim
x +
+
−
→0
⎟
⎠
⎞
⎜
⎝
⎛
=
+
→ 0
0
0 Sinx
xCosx
xSinx
Lim
x

41. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
41
2011
=
=
7. Evaluate tan x log x
Solution: Given limit is (0 × - ∞) form at x = 0
∴ Required limit = form
Apply L’ Hospitals rule
=
= form
Apply L Hospitals rule
=
8. Evaluate
Solution: Given limit is (∞ × 0) form at x = 1
∴ Required limit = form
Apply L’ Hospitals rule
Cosx
xSinx
Cosx
Sinx
xCosx
Lim
x +
−
+
→0
0
2
0
0
2
0
=
=
−
0
→
x
Lim
⎥
⎦
⎤
⎢
⎣
⎡
∞
∞
−
=
→ Cotx
x
Lim
x
log
0
x
Co
x
Lim
x 2
0 sec
1
−
→
⎟
⎠
⎞
⎜
⎝
⎛
−
→ 0
0
2
0 x
x
Sin
Lim
x
0
1
2
0
=
−
→
SinxCosx
Lim
x
x
x
Sec
Lim
x
log
.
2
1
⎟
⎠
⎞
⎜
⎝
⎛
→
π
0
0
2
log
1 x
Cos
x
Lim
x π
→
π
π
π
2
2
.
2
1
1
−
=
−
=
→ x
Sin
x
Lim
x

42. Engineering Mathematics – I Dr. V. Lokesha
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
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
→ x
x
x
Lim
x log
1
1
1
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
−
→ a
x
Cot
x
a
Lim
x 0
⎟
⎠
⎞
⎜
⎝
⎛
−
−
→ Sinx
Secx
Lim
x 1
1
2
π
x
a
Lim x
x
⎟
⎠
⎞
⎜
⎝
⎛ −
∞
→
1
1
0
0 ∞
1 0
∞ ∞
0
[ ] )
(
)
(
x
g
x
f
0
0
a
x
Lim
→ a
x
Lim
→
∞
1
a
x
Lim
→ a
x
Lim
→
0
∞
a
x
Lim
→ a
x
Lim
→
∞
0
a
x
Lim
→ a
x
Lim
→
a
x
Lim
→
[ ] )
(
)
(
x
g
x
f
a
x
Lim
→
[ ] )
(
)
(
x
g
x
f
a
x
Lim
→
∞
∞
or
0
0

43. Engineering Mathematics – I Dr. V. Lokesha
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
Apply L’ Hospitals rule
=
0
→
x
Lim
0
→
x
Lim
0
→
x
Lim
⎟
⎠
⎞
⎜
⎝
⎛
∞
∞
=
=
→
→ Coscex
x
Lim
Sinx
x
Lim
LogL
x
x
log
1
log
0
0
( )
0
0
tan
.
1
0
0 x
x
Sinx
Lim
CoscexCotx
x
Lim
x
x
−
=
− →
→
−∞
=
−
→ 1
tan
sin 2
0
x
Cosx
x
xSec
Lim
x
0
1
1
=
∞
=
=
=
⇒
−∞
=
∴ ∞
∞
−
e
e
L
LogL
x
x
x
Lim −
→
1
1
1
)
(
x
x
x
Lim
−
→
1
1
1
)
(
⎟
⎠
⎞
⎜
⎝
⎛
≡
⎟
⎠
⎞
⎜
⎝
⎛
−
=
∴
→ 0
0
log
1
1
1
x
x
Lim
LogL
x
1
1
1
1
1
1
−
=
−
=
− →
→ x
Lim
x
Lim
x
x

44. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
44
2011
∴ Log L = -1
⇒ L =
12. Evaluate
Solution: let L = form
form
form
Apply L’ Hospitals rule
=
form
Apply L’ Hospital rule, we get
Log L =
e
e
1
1
=
−
2
1
0
tan x
x x
x
Lim ⎟
⎠
⎞
⎜
⎝
⎛
→
∞
→
≡
⎟
⎠
⎞
⎜
⎝
⎛
1
tan 2
1
0
x
x x
x
Lim
)
0
(
tan
log
1
2
0
×
∞
≡
⎭
⎬
⎫
⎩
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
=
∴
→ x
x
x
Lim
LogL
x
( )
⎟
⎠
⎞
⎜
⎝
⎛
≡
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
∴
→ 0
0
tan
log
2
0 x
x
x
Lim
LogL
x
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧ −
→ x
x
x
x
xSec
x
x
Lim
x 2
tan
tan
1 2
2
0
⎟
⎠
⎞
⎜
⎝
⎛
=
⎥
⎦
⎤
⎢
⎣
⎡ −
=
→ 0
0
tan
2
1
3
2
0 x
x
x
xSec
Lim
LogL
x
2
2
2
2
0 3
tan
2
2
1
x
x
Sec
x
x
xSec
x
Sec
Lim
x
−
+
=
→
⎟
⎠
⎞
⎜
⎝
⎛
=
→ x
x
x
Sec
Lim
x
tan
)
(
3
1 2
0
3
1
3
1
e
L =
∴

45. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
45
2011
Exercise: 3
Evaluate the following limits.
a) b)
c) d)
Evaluate the following limits.
Cotx
x
Secx
Lim )
(
0
→
⎟
⎠
⎞
⎜
⎝
⎛
→
⎟
⎠
⎞
⎜
⎝
⎛
−
a
x
a
x a
x
Lim
2
tan
2
π
x
e
x
Lim
x
x
−
+
→
1
0
)
1
( 2
)
(
0
x
x
Cosax
Lim
∞
→
3
3
0 0
2log(1 ) log(1 )
( ) lim ( ) lim
sin sin
x x
x x
e e x x
i ii
x x x
−
→ →
− − + +
2
0 2
2
1 sin cos log(1 ) logsin
( ) lim ( ) lim
tan ( )
2
x x
x x x x
iii iv
x x x
π π
→ →
+ − + −
−
1
2 2
0
2
cosh log(1 ) 1 sin sin
( ) lim ( ) lim
x x
x x x x x
v vi
x x
π
−
→ →
+ − − +
2 2
0
(1 )
( ) lim
log(1 )
x
x
e x
vii
x x
→
− +
+
( )
0
( ) lim 2 cot( ) ( ) lim cos cot
x a x
x
i x a ii ecx x
a
→ →
⎛ ⎞
− − −
⎜ ⎟
⎝ ⎠
2
2
0
2
1
( ) lim tan sec ( ) lim cot
2 x
x
iii x x x iv x
x
π
π
→
→
⎡ ⎤ ⎛ ⎞
− −
⎜ ⎟
⎢ ⎥
⎣ ⎦ ⎝ ⎠
[ ]
2
0
2
1 1
( ) lim ( ) lim 2 tan sec
tan
x x
v vi x x x
x x x π
π
→ →
⎡ ⎤
− −
⎢ ⎥
⎣ ⎦
2
2
1
0 0
1 cos
( ) lim(cos ) ( ) lim
2
b
x
x
x x
x
i ax ii
→ →
⎛ ⎞
+
⎜ ⎟
⎝ ⎠

46. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
46
2011
Evaluate the following limits.
2
1
1
2 log(1 )
1 0
sin
( ) lim(1 ) ( ) lim
x
x
x x
x
iii x iv
x
−
→ →
⎛ ⎞
− ⎜ ⎟
⎝ ⎠
( )
cot
tan
0 0
( ) lim(sin ) ( ) lim 1 sin
x
x
x x
v x iv x
→ →
+
( )
2 tan 2
cos
0
4
( ) lim(cos ) ( ) lim tan
x
ec x
x x
vii x viii x
π
→ →
1
0
1
( ) lim( ) ( ) lim
1 3
x x x x
x
x x
ax a b c
ix x
ax
→∞ →
⎛ ⎞
+ + +
⎜ ⎟
− ⎝ ⎠
2
2
1
0 0
1 cos
( ) lim(cos ) ( ) lim
2
b
x
x
x x
x
i ax ii
→ →
⎛ ⎞
+
⎜ ⎟
⎝ ⎠
2
1
1
2 log(1 )
1 0
sin
( ) lim(1 ) ( ) lim
x
x
x x
x
iii x iv
x
−
→ →
⎛ ⎞
− ⎜ ⎟
⎝ ⎠
( )
cot
tan
0 0
( ) lim(sin ) ( ) lim 1 sin
x
x
x x
v x iv x
→ →
+
( )
2 tan 2
cos
0
4
( ) lim(cos ) ( ) lim tan
x
ec x
x x
vii x viii x
π
→ →
1
0
1
( ) lim( ) ( ) lim
1 3
x x x x
x
x x
ax a b c
ix x
ax
→∞ →
⎛ ⎞
+ + +
⎜ ⎟
− ⎝ ⎠

47. Engineering Mathematics – I Dr. V. Lokesha
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
S
PQ δ
=
∩
S
AP =
∩
( )
2 2 1
, tan (1)
y
r x y
x
θ −
−−−−−−
= + =

48. Engineering Mathematics – I Dr. V. Lokesha
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)
2
2
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
=
⎟
⎠
⎞
⎜
⎝
⎛
x
y
x
s
δ
δ
δ
δ
2
0
2
0
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
=
⎟
⎠
⎞
⎜
⎝
⎛
∴
→
→ x
y
Lim
x
s
Lim
x
x δ
δ
δ
δ
δ
δ
2
2
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
=
⎟
⎠
⎞
⎜
⎝
⎛
dx
dy
dx
ds
2
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
=
dx
dy
dx
ds
2
1 ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
dy
dx
dy
ds

49. Engineering Mathematics – I Dr. V. Lokesha
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,
dx
ds
dy
ds
dx
dy
dx
ds
ψ
2
tan
1+
ψ
ψ sec
2
=
= Sec
dx
ds
ds
dx
dy
ds
ψ
2
1 Cot
+ ψ
2
sec
Co
=
dy
ds
ds
dy

50. Engineering Mathematics – I Dr. V. Lokesha
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
2
2
2
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎠
⎞
⎜
⎝
⎛
t
y
t
x
t
s
δ
δ
δ
δ
δ
δ
2
0
2
0
2
0
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎠
⎞
⎜
⎝
⎛
→
→
→ t
y
Lim
t
x
Lim
t
s
Lim
t
t
t δ
δ
δ
δ
δ
δ
δ
δ
δ
2
2
2
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎠
⎞
⎜
⎝
⎛
∴
dt
dy
dt
dx
dt
ds
2
2
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
=
dt
dy
dt
dx
dt
ds

51. Engineering Mathematics – I Dr. V. Lokesha
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
δθ
Sin
OP
PN
=
r
PN
OP
ON
r
ON
2
2
2
⎟
⎠
⎞
⎜
⎝
⎛
+
=
⎟
⎠
⎞
⎜
⎝
⎛
δθ
δ
δθ
δ r
r
S

52. Engineering Mathematics – I Dr. V. Lokesha
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
2
2
0
⎟
⎠
⎞
⎜
⎝
⎛
+
=
⎟
⎠
⎞
⎜
⎝
⎛
→
→ δθ
δ
δθ
δ
δθ
δθ
r
Lim
r
S
Lim
2
2
2
⎟
⎠
⎞
⎜
⎝
⎛
+
=
⎟
⎠
⎞
⎜
⎝
⎛
θ
θ d
dr
r
d
ds
2
2
⎟
⎠
⎞
⎜
⎝
⎛
+
=
θ
θ d
dr
r
d
ds
2
0
2
2
0
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎠
⎞
⎜
⎝
⎛
→
→ r
r
Lim
r
S
Lim
r
r δ
δθ
δ
δ
δ
δ
2
2
2
⎟
⎠
⎞
⎜
⎝
⎛
=
⎟
⎠
⎞
⎜
⎝
⎛
dr
d
r
dr
ds θ
1
2
2
+
⎟
⎠
⎞
⎜
⎝
⎛
=
dr
d
r
dr
ds θ
dr
dθ
=
φ
φ
Cos
Sin
dr
dθ
ds
dθ
dr
ds
=
φ
φ
Cos
Sin
ds
dr
ds
d
r θ
ds
dθ
ds
dr

53. Engineering Mathematics – I Dr. V. Lokesha
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
dx
ds
dy
ds
⎟
⎠
⎞
⎜
⎝
⎛
c
x
⎟
⎠
⎞
⎜
⎝
⎛
c
x
dx
dy
⎟
⎠
⎞
⎜
⎝
⎛
c
x
2
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
=
∴
dx
dy
dx
ds
( ) ( )
c
x
Cosh
c
x 2
2
sinh
1 =
+
≡
dx
ds
⎟
⎠
⎞
⎜
⎝
⎛
c
x
⎟
⎠
⎞
⎜
⎝
⎛
c
x
C
1
dy
dx
dy
dx
⎟
⎠
⎞
⎜
⎝
⎛
c
x
( )
c
x
h
Co
dy
ds 2
sec
1 +
=
∴

54. Engineering Mathematics – I Dr. V. Lokesha
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)
dy
dx
dx
dy
dy
dx
2
3
2
x
ay
dx
dy
ay
x
2
3 2
dx
ds
2
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
dx
dy
2
2
2
3
1 ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
ay
x
2
2
2
2
2
3
4
9
1
4
9
1
y
a
x
ay
y
a
x
x
+
=
+
dx
ds
a
x
4
9
1+
2
1 ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
dy
dx
dy
ds
2
2
3
2
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
x
ay
2
1
2
1
4
2
2
9
4
1
9
4
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
x
a
x
y
a
dy
ds
dx
dy
Cosx
1
dx
dy
Cosx
1
dy
dx

55. Engineering Mathematics – I Dr. V. Lokesha
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
dy
dx
dy
dx
dx
ds
2
1 ⎟
⎠
⎞
⎜
⎝
⎛
+
dx
dy
2
1 ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
dy
dx
dy
ds
x
dx
ds 2
tan
1+
=
∴
dy
ds
x
Cot2
1+
x
Sec2
dx
ds 2
1
2
)
1
( x
Cot
dy
ds
+
=
dt
ds
( )
2
tan
log t
Cost +
dt
dx
dt
dx
dt
dy
dt
dy
2
2
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
=
∴
dt
dy
dt
dx
dt
ds

56. Engineering Mathematics – I Dr. V. Lokesha
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
t
Sin
t
a
t
Cos
t
a 2
2
2
2
2
2
+
dt
ds
dt
dx
dt
dy
2
2
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
=
∴
dt
dy
dt
dx
dt
ds
2
1
2
1
2
1
dt
ds 2
1
2
t
dt
dx
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
+
−
2
1
.
2
2
tan
1
int 2 t
Sec
t
S
dt
dy
2
2
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
=
∴
dt
dy
dt
dx
dt
ds
2
1
2
2
2
2
2
2
tan
2
2
int
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
+
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
− t
Cos
a
t
t
Sec
S
a
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
+
−
2
tan
4
2
2
tan
2
int
2
4
2
2
2
2
t
t
Sec
a
t
t
Sec
S
a
a

57. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
57
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 =
=
θ
d
ds
θ
d
dr
θ
d
dr
θ
d
ds 2
1
2
2
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
θ
d
dr
r
2
1
2
1
2
1
2
1
2
1
2
θ 2
1
θ
d
ds
2
θ
dr
ds 2
1
2
2
1
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
dr
d
r
θ
2
1
2
2
2
2
)
1
(
1
⎭
⎬
⎫
⎩
⎨
⎧ −
+
θ
θ
Sin
a
Cos
a

58. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
58
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
=
=
=
=
=
2
1
2
2
2
)
1
(
⎭
⎬
⎫
⎩
⎨
⎧ −
+
θ
θ
θ
Sin
Cos
Sin
{ }
2
2
2
2
2
2
2
)
1
(
2 2
1
θ
θ
θ
θ
θ
θ
θ
Cos
Sin
Sin
Sin
Sin
Sin
Cos
=
⎟
⎠
⎞
⎜
⎝
⎛
=
−
dr
ds
2
1
θ
Cos
θ
d
dr
θ
d
dr
θ
d
dr
r
Sin
a βθ
2
2
θ
d
ds 2
1
2
2
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
θ
d
dr
r
2
1
2
2
2 2
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+
r
Sin
a
r
θ
2
1
2
2
4
2 2
⎭
⎬
⎫
⎩
⎨
⎧
+
r
Sin
a
r
θ
2
1
2
2
4
2
2
2
⎭
⎬
⎫
⎩
⎨
⎧
+
θ
θ
θ
Cos
a
Sin
a
Cos
a
2
1
2
2
4
2
4
2
2
2
⎭
⎬
⎫
⎩
⎨
⎧ +
θ
θ
θ
Cos
a
Sin
a
Cos
a

59. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
59
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}
θ
θ 2
2
2
1
2
4
Cos
a
Cos
a
a
=
⎭
⎬
⎫
⎩
⎨
⎧
θ
d
ds
r
a
r
a 2
2
1
2
4
=
⎭
⎬
⎫
⎩
⎨
⎧
dr
ds
2
1
2
2
1
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
dr
d
r
θ
2
1
2
2
2
2
1
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−
+
θ
Sin
a
r
r
2
1
2
4
4
2
1
⎭
⎬
⎫
⎩
⎨
⎧
+
θ
Sin
a
r
2
1
2
4
2
4
2
4
2
sin
2
2
⎭
⎬
⎫
⎩
⎨
⎧ +
θ
θ
θ
a
Cos
a
Sin
a
2
1
2
4
4
2
sin ⎭
⎬
⎫
⎩
⎨
⎧
θ
a
a
θ
2
1
Sin
θ
d
dr
θ
d
ds 2
1
2
2
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
θ
d
dr
r
2
1
2
1
2
1

60. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
60
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.
=
θ
d
ds
dr
ds
2
1
2
2
1
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
dr
d
r
θ
2
1
2
2
2
2
2
1
1
⎭
⎬
⎫
⎩
⎨
⎧
+ α
α
θ
α
θ
Cot
Cot
Cot
e
a
e
a
2
1
2
1
dr
ds
θ
d
ds
ds
dθ
ds
dr
ds
dr 2
1
2
1
2
2
1 ⎥
⎦
⎤
⎢
⎣
⎡
−
r
p
ds
dr
r
p
r 2
2
−

61. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
61
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 ⎟
⎠
⎞
⎜
⎝
⎛
=
dr
d
r
θ
φ
tan
Let P (r, θ) be any point on the curve r = f (θ)
r
OP
and
P
O
X =
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
dr
ds
∴
2
2
p
r
r
−
r
p
OP
OR
=

62. Engineering Mathematics – I Dr. V. Lokesha
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
θ
φ
φ
+
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
the
of
slope
dx
dy
tan =
=
ψ
ie.,
d
dx
d
dy
tan
θ
θ
=
ψ since x and y are function of θ
ie.,
( )
( ) θ
=
′
θ
′
+
θ
θ
′
+
θ
=
θ
θ
θ
θ
=
ψ
d
dr
r
where
cos
r
sin
r
-
sin
r
cos
r
cos
r
d
d
sin
r
d
d
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
r
sin
r
cos
r
sin
r
cos
r
cos
r
tan
Or
θ
′
θ
+
′
=
ψ
tan
.
r
r
-
1
an
t
r
r
tan …(2)
Comparing equations (1) and (2) we get
⎟
⎠
⎞
⎜
⎝
⎛ θ
=
φ
⎟
⎠
⎞
⎜
⎝
⎛
θ
=
′
=
φ
dr
d
r
or tan
d
dr
r
r
r
tan

63. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
63
2011
Prove with usual notations
2
4
2
2
d
dr
r
1
r
1
p
1
⎟
⎠
⎞
⎜
⎝
⎛
θ
+
= or
r
1
u
where
d
du
u
1
2
2
2
=
⎟
⎠
⎞
⎜
⎝
⎛
+
=
θ
p
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
=
∧
P
O
L
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
=
=
∧
∧
P
O
L
90
P
N
O
Now from the right angled triangle ONP
OP
ON
sin =
φ
ie., φ
=
=
φ sin
r
p
or
r
P
sin
we have p = r sin φ …(1)
and
θ
=
φ
d
dr
r
1
cot …(2)
Squaring equation (1) and taking the reciprocal we get,

64. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
64
2011
φ
=
φ
= cosec
r
1
p
1
ie.,
sin
1
.
r
1
p
1 2
2
2
2
2
2
Or ( )
φ
+
= cot
1
r
1
p
1 2
2
2
Now using (2) we get,
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
θ
+
=
2
2
2
2
d
dr
r
1
1
r
1
p
1
Or
2
4
2
2
d
dr
r
1
r
1
p
1
⎟
⎠
⎞
⎜
⎝
⎛
θ
+
= …(3)
Further, let u
r
1
=
Differentiating w.r.t. θ we get,
2
2
4
2
d
du
d
dr
r
1
d
du
d
dr
r
1
⎟
⎠
⎞
⎜
⎝
⎛
θ
=
⎟
⎠
⎞
⎜
⎝
⎛
θ
⇒
θ
=
⎟
⎠
⎞
⎜
⎝
⎛
θ
− , by squaring
Thus (3) now becomes
2
2
2
d
du
u
p
1
⎟
⎠
⎞
⎜
⎝
⎛
θ
+
= …(4)
1. Find the angle of intersection of the curves:
( ) ( )
θ
=
θ
+
= cos
-
1
b
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
cos
1
sin
-
0
d
dr
r
1
θ
+
θ
+
=
θ
:
cos
1
sin
0
d
dr
r
1
θ
−
θ
+
=
θ
( ) ( )
( )
/2
cos
2
/2
cos
/2
sin
2
cot 2
1
θ
θ
θ
−
=
φ :
( ) ( )
( )
/2
sin
2
/2
cos
/2
sin
2
cot 2
2
θ
θ
θ
=
φ
ie., ( ) ( )
/2
/2
cot
/2
tan
-
cot 1 θ
+
π
=
θ
=
φ : ( )
/2
cot
cot 2 θ
=
φ
/2
/2
1 θ
+
π
=
φ
⇒ : /2
2 θ
=
φ

65. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
65
2011
∴ angle of intersection = /2
/2
-
/2
/2
- 2
1 π
θ
θ
π
φ
φ =
+
=
Hence the curves intersect orthogonally.
2. S.T. the curves ( ) ( )
θ
=
θ
+
= sin
-
1
a
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
d
dr
r
1
:
θ
−
θ
=
θ sin
1
cos
-
d
dr
r
1
ie.,
sin
1
cos
cot 1
θ
+
θ
=
φ :
sin
1
cos
-
cot 2
θ
−
θ
=
φ
We have
θ
θ
=
φ
θ
θ
+
=
φ
cos
-
sin
-
1
tan
and
cos
sin
1
tan 2
1
1
-
cos
-
cos
cos
sin
-
1
tan
.
tan 2
2
2
2
2
1 =
θ
θ
=
θ
−
θ
=
φ
φ
∴
Hence the curves intersect orthogonally.
3. Find the angle of intersection of the curves:
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
+
θ
θ
θ
=
θ cos
sin
sin
-
cos
d
dr
r
1
:
sin
cos
d
dr
r
1
θ
θ
=
θ
ie.,
( )
( )
θ
+
θ
θ
θ
=
φ
tan
1
cos
tan
-
1
cos
cot 1 : θ
=
φ
⇒
θ
=
φ cot
cot 2
2
ie., ( ) θ
+
π
=
φ
⇒
θ
+
π
=
φ /4
/4
cot
cot 1
1
/4
-
/4
- 2
1 π
=
θ
θ
+
π
=
φ
φ
∴
The angle of intersection is 4
/
π

66. Engineering Mathematics – I Dr. V. Lokesha
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,
θ
θ
=
θ .
log
1
d
dr
r
1
:
θ
θ
−
=
θ .
log
1
d
dr
r
1
ie.,
log
1
cot 1
θ
θ
=
φ :
log
1
-
cot 2
θ
θ
=
φ
Note : We can not find 2
1 and φ
φ explicitly.
θ
θ
=
φ
∴ log
tan 1 : θ
θ
=
φ log
tan 2
Now consider, θ
θ
=
φ log
tan 1 : θ
θ
=
φ log
-
tan 2
Now consider, ( )
2
1
2
1
tan
tan
1
log
2
-
tan
φ
φ
θ
θ
φ
φ
+
=
( )2
log
1
log
2
θ
θ
−
θ
θ
= ......…(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
log
or
1
1
log 2
=
θ
⇒
θ
=
=
θ
Substituting ( ) get
we
1
in
e
=
θ
( ) 2
2
1
e
-
1
2e
tan =
φ
−
φ ( )
1
e
log =
∵
∴ angle of intersection e
tan
2
e
-
1
2e
tan
- 1
-
2
1
-
2
1 =
⎟
⎠
⎞
⎜
⎝
⎛
=
φ
φ
5. Find the angle of intersection of the curves:
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,

67. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
67
2011
cos
-
1
sin
d
dr
r
1
θ
θ
=
θ
:
cos
sin
-
d
dr
r
1
θ
θ
=
θ
ie.,
( ) ( )
( )
θ
=
φ
θ
θ
θ
=
φ tan
-
cot
:
/2
sin
2
/2
cos
/2
sin
2
cot 2
2
1
ie., ( )
/2
cot
cot 1 θ
=
φ : ( )
θ
+
π
=
φ /2
cot
cot 2
/2
1 θ
=
φ
⇒ : θ
+
π
=
φ /2
2
2
2
/
-
/2
-
/2
2
1 θ
π
θ
π
θ
φ
φ +
=
=
−
∴ …(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
= a
r and θ
= /
a
r
Solution : θ
= a
r : θ
= /
a
r
θ
+
=
⇒ log
a
log
r
log : θ
= log
-
a
log
r
log
Differentiating these w.r.t θ, we get,
θ
=
θ
1
d
dr
r
1
:
θ
=
θ
1
-
d
dr
r
1
ie.,
θ
=
φ
1
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
⇒
=
θ
θ
=
θ
When and
1
-
tan
1,
tan
1, 2
1 =
φ
=
φ
=
θ
When 1
tan
1,
-
tan
1,
- 2
1 =
φ
=
φ
=
θ .
/2
-
1
-
tan
.
tan 2
1
2
1 π
=
φ
φ
⇒
=
φ
φ
∴
The curves intersect at right angles.

68. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
68
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
θ
θ
=
θ
=
θ
θ
+
θ cos
-
1
sin
-
d
dr
r
1
or
0
cos
-
1
sin
d
dr
r
1
( ) ( )
( )
( )
/2
cot
-
/2
sin
2
/2
cos
/2
sin
2
-
cot 2
θ
=
θ
θ
θ
=
φ
∴
ie., ( ) ( )
/2
-
/2
-
cot
cot θ
=
φ
⇒
θ
=
φ
∴
Consider φ
= sin
r
p
( ) ( )
/2
sin
r
-
p
or
/2
-
sin
r
p θ
=
θ
=
∴
Now we have, ( ) 2a
cos
-
1
r =
θ …(1)
( )
/2
sin
r
-
p θ
= …(2)
We have to eliminate θ from (1) and (2)
(1) can be put in form ( ) 2a
2
/
sin
2
. 2
=
θ
r
ie., ( ) a
/2
sin
r 2
=
θ
But p/-r = sin (θ/2), from (2)
ar
p
or
a
r
p
r 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
=
θ
θ
θ
θ
=
θ
2
tan
d
dr
r
1
ie.,
2
sec
2
tan
2
sec
2
d
dr
r
2
ie., ( ) θ
π
=
φ
⇒
θ
π
=
φ 2
-
/2
2
-
/2
cot
cot

69. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
69
2011
Consider ( ) θ
=
θ
π
=
∴
φ
= 2
cos
r
p
ie.,
2
-
/2
sin
r
p
sin
r
p
Now we have, θ
= 2
sec
a
r 2
2
…(1)
θ
= 2
cos
r
p …(2)
From (2) p/r = cos 2θ or r/p = sec 2 θ
Substituting in (1) we get, ( )
r/p
a
r 2
2
= 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
log
n
r
log
n
Differentiating w.r.t q we get
θ
=
θ
θ
θ
=
θ
n
tan
-
d
dr
r
1
ie.,
n
cos
n
sin
n
-
d
dr
r
n
( ) θ
+
π
=
φ
⇒
θ
+
π
=
φ
∴ n
/2
n
/2
cot
cot
Consider φ
= sin
r
p
( ) θ
=
θ
=
π
=
∴ n
cos
r
p
ie.,
n
/2
sin
r
p
Now we have, θ
= n
cos
a
r n
n
…(1)
θ
= n
cos
r
p …(2)
( )
p/r
a
r
is
(2)
of
e
consequenc
a
as
(1) n
n
=
∴
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
+
θ
θ
+
θ
=
θ m
sin
m
cos
m
cos
m
m
sin
m
-
d
dr
r
m
ie.,
( )
( )
θ
+
θ
θ
θ
=
θ
+
θ
θ
θ
=
θ m
tan
1
m
cos
m
tan
-
1
m
cos
m
sin
m
cos
m
sin
-
m
cos
d
dr
r
1
( ) θ
+
π
=
φ
⇒
θ
+
π
=
φ
∴ m
/4
m
/4
cot
cot

70. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
70
2011
Consider φ
= sin
r
p
( )
θ
+
π
=
∴ m
/4
sin
r
p
ie., ( ) ( )
[ ]
θ
π
+
θ
π
= m
sin
/4
cos
m
cos
/4
sin
r
p
ie., ( )
θ
+
θ
= m
sin
m
cos
2
r
p
(we have used the formula of sin (A + B) and also the values ( ) ( ) )
2
1/
/4
cos
/4
sin =
π
=
π
Now we have, ( )
θ
+
θ
= m
sin
m
cos
a
r m
m
…(1)
( )
θ
+
θ
= m
sin
m
cos
2
r
p …(2)
Using (2) in (1) we get,
p
a
2
r
or
r
2
p
.
a
r m
1
m
m
m
=
= +
Thus p
a
2
r m
1
m
=
+
is the required pedal equation.
11. Establish the pedal equation of the curve:
θ
+
θ
= n
cos
b
n
sin
a
r n
n
n
in the form ( ) 2
n
2
n
2
2n
2
r
b
a
p +
=
+
Solution : We have θ
+
θ
= n
cos
b
n
sin
a
r n
n
n
( )
θ
+
θ
=
⇒ n
cos
b
n
sin
a
log
r
log
n n
n
Differentiating w.r.t θ we get
θ
+
θ
θ
θ
=
θ n
cos
b
n
sin
a
n
sin
nb
-
n
cos
na
d
dr
r
n
n
n
n
n
Dividing by n,
θ
+
θ
θ
θ
=
φ
n
cos
b
n
sin
a
n
sin
b
-
n
cos
a
cot n
n
n
n
Consider φ
= sin
r
p
Since φ cannot be found, squaring and taking the reciprocal we get,
( )
φ
+
=
φ
= cot
1
r
1
p
1
or
cosec
r
1
p
1 2
2
2
2
2
2
( )
( ) ⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
θ
+
θ
θ
θ
+
=
∴ 2
n
n
2
n
n
2
2
n
cos
b
n
sin
a
n
sin
b
-
n
cos
a
1
r
1
p
1

71. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
71
2011
( ) ( )
( ) ⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
θ
+
θ
θ
θ
+
θ
+
θ
= 2
n
n
2
n
n
2
n
n
2
2
n
cos
b
n
sin
a
n
sin
b
-
n
cos
a
n
cos
b
n
sin
a
r
1
p
1
e.,
i
( ) ( )
( ) ⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
θ
+
θ
θ
+
θ
+
θ
+
θ
= 2
n
n
2
2
n
2
2
2
n
2
2
2
n
cos
b
n
sin
a
n
sin
n
cos
b
n
cos
n
sin
a
r
1
p
1
e.,
i
(product terms cancels out in the numerator)
( )2
n
n
n
2
n
2
2
2
n
cos
b
n
sin
a
b
a
.
r
1
p
1
.,
ie
θ
+
θ
+
=
( )
,
r
b
a
.
r
1
p
1
or 2
n
n
2
n
2
2
2
+
= by using the given equation.
( ) 2
n
2
n
2
n
2
2
r
b
a
p +
=
+
∴ 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.
ds
dψ
ψ
ψ
ρ
d
ds
ds
d
=
=
∴
1

72. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
72
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)
ψ
ρ
d
ds
=
∴
ds
dψ
ρ
=
1
K
1
=
ρ
ds
dψ
dx
dy
dx
dψ
2
2
dx
y
d
2
2
dx
y
d
dx
dψ
dx
dψ
dx
dψ
dx
ds
ρ
1 2
1
2
1
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
dx
dy
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
2
1
dx
dy
ρ
1 2
1
2
1
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
dx
dy

73. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
73
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.
ρ
1 2
3
2
1
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
dx
dy
2
3
2
2
2
1
dx
y
d
dx
dy
⎥
⎥
⎦
⎤
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
+
dx
dy
2
2
dx
y
d
ds
dψ
ds
dψ
r
1
r
1
O r

74. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
74
2011
15. Derive an expression for radius of curvature in parametric form.
Solution: We have ρ = (1 + y 1
2 )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
dt
dx
dt
dy
dx
dy
2
2
dx
y
d
dx
d
⎟
⎠
⎞
⎜
⎝
⎛
dx
dy
dx
d
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
•
•
X
Y
dt
d
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
•
•
X
Y
dx
dt
dt
d
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
•
•
X
Y
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
dt
dx
1
dt
d
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
•
•
X
Y
•
X
1
( )2
x
x
y
y
x −
•
)
(
1
X
3
⎟
⎠
⎞
⎜
⎝
⎛
−
•
•
x
x
y
y
X
( )
2
2
3
2
1
1
y
Y
+
( )
( )
( )3
2
3
2
1
x
x
y
y
x
x
Y
−
⎥
⎦
⎤
⎢
⎣
⎡ +
( ) ( )
[ ]
x
y
y
x
y
x
−
+
=
2
3
2
2
ρ

75. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
75
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
dr
dθ
θ
d
dr
1
( )
θ
d
dr
r
θ
φ
d
d
2
2
2
.
⎟
⎠
⎞
⎜
⎝
⎛
−
θ
θ
θ
θ
d
dr
d
r
d
r
d
dr
d
dr
( )
( ) ⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
∴ 2
2
2
2
2
1
θ
θ
θ
φ
θ
φ
d
dr
d
r
d
r
d
dr
Sec
d
d
( )
( ) ⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧ −
+ 2
2
2
2
2
tan
1
1
θ
θ
θ
φ
d
dr
d
r
d
r
d
dr
⎪
⎪
⎭
⎪
⎪
⎬
⎫
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
−
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
+
2
2
2
2
2
2
1
1
θ
θ
θ
θ
d
dr
d
r
d
r
d
dr
d
dr
r
θ
φ
d
d
2
2
2
2
2
r
d
dr
d
r
d
r
d
dr
+
⎟
⎠
⎞
⎜
⎝
⎛
−
⎟
⎠
⎞
⎜
⎝
⎛
θ
θ
θ

76. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
76
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
θ
ψ
d
d
θ
φ
d
d
θ
ψ
d
d
2
2
2
2
2
r
d
dr
d
r
d
r
d
dr
+
⎟
⎠
⎞
⎜
⎝
⎛
−
⎟
⎠
⎞
⎜
⎝
⎛
θ
θ
θ
θ
ψ
d
d
2
2
2
2
2
2
2
r
d
dr
d
r
d
r
d
dr
r
+
⎟
⎠
⎞
⎜
⎝
⎛
−
⎟
⎠
⎞
⎜
⎝
⎛
+
θ
θ
θ
θ
d
ds 2
1
2
2
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
θ
d
dr
r
ψ
d
ds
θ
d
ds
ψ
θ
d
d
2
1
2
2
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎟
⎠
⎞
⎜
⎝
⎛
+
θ
d
dr
r
2
2
2
2
2
2
2
θ
θ
θ
d
r
d
r
d
dr
r
r
d
dr
−
⎟
⎠
⎞
⎜
⎝
⎛
+
+
⎟
⎠
⎞
⎜
⎝
⎛
( )
2
2
1
2
2
3
2
1
2
2 rr
r
r
r
r
−
+
+
=
θ
d
dr
2
2
θ
d
r
d
dr
dp
dr
dφ

77. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
77
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,
ds
dθ
ds
dr
dr
dp
ds
dθ
ds
dr
dr
dφ
ds
dθ
dr
dφ
ds
dr
ds
dθ
ds
dφ
ds
d
ds
dψ
dr
dp
ds
dψ
dr
dp
ρ
1
ds
dψ
ρ
1
dp
dr
( )
2
2
3
2
1
1
y
y
+
=
ρ

78. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
78
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 =
a
x
a
x
a
x
ay
x
y
dx
dy
x
dx
dy
2
3
2
3
2
3
3
2
1
3
2
2
1
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
=
=
⇒
=
a
x
y
2
3
1 =
∴
x
a
x
a
dx
y
d
y
4
3
2
1
2
3
2
2
2 =
=
=
( )
a
x
a
x
6
9
4 2
3
+
=
ρ
⎟
⎠
⎞
⎜
⎝
⎛
c
x
⎟
⎠
⎞
⎜
⎝
⎛
c
x
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
⎟
⎠
⎞
⎜
⎝
⎛
=
c
x
c
x
Sec
c
x
Sec
c
y tan
.
1
.
1 ⎟
⎠
⎞
⎜
⎝
⎛
c
x
c
1
⎟
⎠
⎞
⎜
⎝
⎛
c
x
⎟
⎠
⎞
⎜
⎝
⎛
=
c
x
Sec
y 2
2
c
1
c
1
⎟
⎠
⎞
⎜
⎝
⎛
c
x
Sec2
( )
2
2
3
2
1
1
y
y
+
=
δ
( )
[ ]
( )
( )
[ ]
( )
c
x
Sec
c
c
x
Sec
c
x
Sec
c
c
x
2
2
3
2
2
2
3
2
1
1
tan
1
=
+
=
δ
( )
c
x
cSec

79. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
79
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
( ) ( )
( )
x
y
y
x
y
x
−
+
=
2
3
2
2
ρ
dt
dx
x = y
dt
dy
x y
x y y x
( )
int)
int(
)
1
(
int
)
1
( 2
3
2
2
2
aS
aS
aCost
Cost
a
S
a
Cost
a
−
−
+
+
+
=
ρ
{ }
{ }
t
Sin
t
Cos
Cost
a
t
Sin
t
Cos
Cost
a
2
2
2
2
3
2
2
3
2
1
+
+
+
+
+
{ } { }
2
2
2
.
8
2
2
2
2
.
2
)
1
(
)
1
(
2
2
3
2
2
3
2
2
3
t
Cos
t
Cos
a
t
Cos
t
Cos
a
Cost
Cost
a
=
=
+
+
2
4 t
aCos
=
ρ
( )
{ }
2
tan
log t
Cost
a +
( )
{ }
2
tan
log t
Cost
a +
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
+
−
=
2
1
.
2
.
2
tan
1
int 2 t
Sec
t
S
a
dt
dx
2
/
cos
2
/
sin
2
1
t
t
2
/
sin
2
1
t
dt
dx

80. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
80
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)
aCost
dx
dy
and =
dt
dx
dt
dy
dx
dy
/
/
=
∴
t
t
t
a
aCost
sin
/
cos
cos
t
dx
dy
tan
=
∴
int
1
.
1
2
2
2
2
2
2
S
t
aCos
t
Sec
dt
dx
t
Sec
dx
dt
t
Sec
dx
y
d
=
=
=
int
1 4
2
2
tS
Sec
a
dx
y
d
=
=
dx
dy
2
2
dx
y
d ( )
2
2
3
2
1
1
y
y
+
=
ρ
( ) aCott
S
Cost
a
tS
Sec
t
aSec
tS
Sec
a
t
=
=
=
+
=
int
int
int
1
tan
1
4
3
4
2
3
2
ρ
a
y
x =
+ ⎟
⎠
⎞
⎜
⎝
⎛
4
,
4
a
a
a
y
x =
+
0
2
1
2
1
=
+
dx
dy
y
x
( )
x
x
a
x
y
dx
dy
y
−
−
=
−
=
=
1
x
a
y −
=1
1

81. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
81
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
=
=
2
3
2
3
2
2
2
2
2
1
x
a
x
a
dx
y
d
y =
⎟
⎠
⎞
⎜
⎝
⎛
−
−
=
=
−
⎟
⎠
⎞
⎜
⎝
⎛
4
,
4
a
a
2
1
1
a
a
y −
=
( ) a
a
a
y
4
4
2
1
2
3
2 =
−
=
( )
2
2
3
2
1
1
y
y
+
=
ρ
( )
2
4
2
2
4
2
4
)
1
(
1 2
3
2
3
2
a
a
a
a
=
=
=
−
+
2
a
=
θ
d
dr
2
4
2
2
1
1
1
⎟
⎠
⎞
⎜
⎝
⎛
+
=
θ
d
dr
r
r
P
4
2
1
1
r
r
+
4
2
2
2
2
4
2
2
2
)
1
(
r
Sin
a
Cos
a
r
Sin
a
r θ
θ
θ +
+
=
+
4
2
)
1
(
2
r
Cos
a θ
+

82. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
82
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
⎟
⎠
⎞
⎜
⎝
⎛
=
∴
a
r
r
a
P 4
2
2
2
1
θ
Cos
a
r
+
=
∴ 1
3
2
2
1
r
a
P
=
a
r
P
2
3
2
=
∴
dp
dr
r
a
2
3
2
1
2
3
4
r
ap
dp
dr
=
r
ap
dp
dr
r
3
4
=
=
ρ
9
8
2
.
9
16
9
16
1 3
3
2
2
2
2
2
2
a
a
r
r
a
r
p
a
r
r
=
=
⎥
⎦
⎤
⎢
⎣
⎡
=
ρ
2
2
2
2
2
2
1
1
1
b
a
r
b
a
P
−
+
=
2
2
2
2
2
2
1
1
1
b
a
r
b
a
P
−
+
=
dp
dr
r
b
a
P
2
1
2
2
2
3
−
=
−
r
p
b
a
dp
dr
3
2
2
=
∴
3
2
2
3
2
2
.
.
p
b
a
r
p
b
a
r
dp
dr
r =
=
=
∴ ρ

83. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
83
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
θ
a
θ
a
θ
θ
θ
θ
θ
r
a
a
d
dr −
=
−
=
−
=
1
.
2
θ
θ
r
d
dr −
=
2
4
2
2
1
1
1
⎟
⎠
⎞
⎜
⎝
⎛
+
=
θ
d
dr
r
r
P
2
2
2
2
2
4
2
1
1
1
1
θ
θ r
r
r
r
r
+
=
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
=
⎟
⎠
⎞
⎜
⎝
⎛
+
= 2
2
2
2
2
2
1
1
1
1
1
1
θ
θ
θ r
r
P
1
2
+
=
∴
θ
θ
r
P
2
2
2
2
2
.
1
.
1 r
a
r
a
r
a
r
a
r
r
P
+
=
+
=
+
=
θ
θ
( )
2
2
2
2
2
2
2
2
1
.
.
1
r
a
dp
dr
r
r
a
a
r
dp
dr
a
r
a
+
/
+
/
−
+
=
dp
dr
r
a
a
r
a
r
a
r
a ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
+
=
+
2
2
2
2
2
2
2
.
dp
dr
r
a
a
r
a
r
a
r
a
2
2
2
2
2
2
2 ).
(
+
−
+
=
+

84. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
84
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
( ) dp
dr
a
r
a
r
a 3
2
2
2
2
=
+
+
( )
dp
dr
a
r
a
=
+
3
2
3
2
2
( )
3
2
3
2
2
.
.
a
r
a
r
dp
dr
r
+
=
=
ρ
( ) 2
3
2
2
3
r
a
a
r
+
=
∴ ρ
⎟
⎠
⎞
⎜
⎝
⎛
+
2
4
ψ
π
9
16 2
2
2
2
1
a
=
+ ρ
ρ

85. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
85
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.

86. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
86
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.

87. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
87
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 /1
T x y x y
= + +
dT
dx
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 y
T x x y
x
T
x
−
−
−
= + + = + +
∂ = − + +
∂
∂ −
= − + + =
∂
20
3
−

88. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
88
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 y
T y x y
x
T
x
−
−
−
= + + = + +
∂ = − + +
∂
∂ −
= − + + =
∂
10
3
−

89. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
89
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
1
I R
E
−
∂ =
∂
1
15 0.0667
I
E
−
∂ = ≈
∂
1
I ER
E
−
∂ =
∂
1
(120,15) 120(15) 0.5333
I
E
−
∂ = − ≈ −
∂
0
( , ) ( , )
lim
x
x
z f x x y f x y
z
x x
δ
δ
δ
→
∂ + −
= =
∂
0
( , ) ( , )
lim
y
y
z f x y y f x y
z
y y
δ
δ
δ
→
∂ + −
= =
∂

90. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
90
2011
1. If ( )
by
ax
sin
e
u by
-
ax
+
= show that u
ab
2
a
- =
∂
∂
∂
∂
y
u
x
u
b
Solution : ( )
by
ax
sin
e
u by
-
ax
+
=
( ) ( )
by
ax
sin
a.e
a
.
by
ax
cos
e
x
u by
-
ax
by
-
ax
+
+
+
=
∂
∂
∴
( ) u
a
by
ax
cos
e
a
x
u
e.,
i by
-
ax
+
+
=
∂
∂
…(1)
Also ( ) ( ) ( )
by
ax
sin
e
b
-
b
.
by
ax
cos
e by
-
ax
by
-
ax
+
+
+
=
∂
∂
x
u
( ) bu
by
ax
cos
e
b
y
u
.,
ie by
-
ax
−
+
=
∂
∂
…(2)
Now
y
u
a
x
u
b
∂
∂
=
∂
∂
by using (1) and (2) becomes
( ) ( ) abu
by
ax
cos
abe
-
abu
by
ax
cos
abe by
-
ax
by
ax
+
+
+
+
= −
= 2 abu
Thus abu
2
y
u
a
-
x
u
b =
∂
∂
∂
∂
2. If ( ) that
prove
,
by
-
ax
f
e
u by
ax+
=
abu
2
y
u
a
x
u
b =
∂
∂
=
∂
∂
Solution : ( )
by
-
ax
f
e
u by
ax+
= , by data
( ) ( )
by
-
ax
f
ae
a
by
-
ax
f
.
e
x
u by
ax
by
ax +
+
+
′
=
∂
∂
Or ( ) u
a
by
-
ax
f
.
e
a by
ax
+
′
=
∂
∂ +
x
u
…(1)
Next, ( ) ( ) ( )
by
-
ax
f
e
b
b
-
.
by
-
ax
f
e
y
u by
ax
by
ax +
+
+
′
=
∂
∂
Or ( ) ba
by
-
ax
f
e
b
y
u by
ax
+
′
−
=
∂
∂ +
…(2)
Now consider L.H.S
y
u
a
x
u
b
∂
∂
+
∂
∂
=

91. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
91
2011
( )
{ } ( )
{ }
bu
by
-
ax
f
be
-
a
au
by
-
ax
f
ae
b by
ax
by
ax
+
′
+
+
′
= +
+
( ) ( ) abu
by
-
ax
f
e
ab
-
abu
by
-
ax
f
e
ab by
ax
by
ax
+
′
+
′
= +
+
= 2abu = R.H.S
Thus u
b
a
2
y
u
a
x
u
b =
∂
∂
+
∂
∂
3. If 2
2
2
z
y
x
log
u +
+
= , show that ( ) 1
z
u
y
u
x
u
z
y
x 2
2
2
2
2
2
2
2
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
+
+
Solution : By data ( )
2
2
2
2
2
2
z
y
x
log
2
1
z
y
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)
2
2
2
2
2
2
y
2
.
1
.
2
1
z
x
x
x
z
y
x
x
u
+
+
=
+
+
=
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
∂
∂
=
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
∂
∂
=
∂
∂
2
2
2
2
z
y
u
u
x
x
x
x
x
x2
ie.,
( )
( ) ( )2
2
2
2
2
2
2
2
2
2
2
2
2
2
z
y
z
y
z
y
2
.
-
1
z
y
+
+
−
+
=
+
+
+
+
x
x
x
x
x
x
( )2
2
2
2
2
2
2
2
2
z
y
z
y
u
+
+
−
+
=
∂
∂
∴
x
x
x
…(1)
Similarly
( )2
2
2
2
2
2
2
2
2
z
y
x
z
u
+
+
−
+
=
∂
∂
∴
x
y
y
…(2)
( )2
2
2
2
2
2
2
2
2
z
y
y
x
u
+
+
−
+
=
∂
∂
x
z
z
…(3)
Adding (1), (2) and (3) we get,
( ) 2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
z
y
1
z
y
z
y
u
u
u
+
+
=
+
+
+
+
=
∂
∂
+
∂
∂
+
∂
∂
x
x
x
z
y
x
Thus ( ) 1
z
u
y
u
u
z
y 2
2
2
2
2
2
2
2
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
+
+
x
x

92. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
92
2011
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
u
2
+
+
=
x
x
x
( )
z
tan
y
tan
tan
.
sec
cos
sin
2
u
2
sin
2
+
+
=
x
x
x
x
x x
Or
z
tan
y
tan
tan
tan
2
u
2
sin
+
+
=
x
x
x x …(1)
Similarly
z
tan
y
tan
tan
tan
2
u
2
sin
+
+
=
x
y
y y …(2)
z
tan
y
tan
tan
tan
2
u
2
sin
+
+
=
x
z
z y …(3)
Adding (1), (2) and (3) we get,
( )
( )
2
z
tan
y
tan
x
tan
z
tan
y
tan
tan x
2
u
2z
sin
u
2y
sin
u
2
sin z
y =
+
+
+
+
=
+
+
x
x
Thus 2
u
2z
sin
u
2y
sin
u
2x
sin z
y
x =
+
+
5. If u = log (x3 + y3 + z3 – 3xyz) then prove that
z
y
x
3
z
u
y
u
x
u
+
+
=
∂
∂
+
∂
∂
+
∂
∂
and hence show
that
( )2
2
z
y
9
-
u
+
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
x
z
y
x
Solution : u = log (x3 + y3 + z3 – 3xyz) is a symmetric function
yz
3
-
z
y
3yz
-
3
u
3
3
3
2
x
x
x
x +
+
=
∂
∂
…(1)
yz
3
-
z
y
3zx
-
3
u
3
3
3
2
x
x
y
y +
+
=
∂
∂
…(2)
yz
3
-
z
y
3xy
-
3
u
3
3
3
2
x
x
z
z +
+
=
∂
∂
…(3)
Adding (1), (2) and (3) we get,

93. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
93
2011
( )
( )
yz
3
z
y
z
yz
y
z
y
3
z
u
u
u
3
3
3
2
2
2
x
x
x
x
x
y
x −
+
+
−
−
−
+
+
=
∂
∂
+
∂
∂
+
∂
∂
Recalling a standard elementary result,
( )( )
ca
bc
ab
c
b
a
c
b
a
abc
3
c
b
a 2
2
2
3
3
3
−
−
−
+
+
+
+
=
−
+
+
We have,
( )
( )( )
x
x
x
x
x
x
x
x z
yz
y
z
y
z
y
z
yz
y
z
y
3
z
u
y
u
u
2
2
2
2
2
2
−
−
−
+
+
+
+
−
−
−
+
+
=
∂
∂
+
∂
∂
+
∂
∂
Thus
z
y
3
z
u
y
u
u
+
+
=
∂
∂
+
∂
∂
+
∂
∂
x
x
Further u
z
y
2
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
x
u
z
y
z
y ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
=
x
x
z
u
y
u
u
z
y ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
=
x
x
z
y
3
z
y ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
=
x
x
, by using the earlier result.
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
∂
∂
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
∂
∂
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
∂
∂
=
z
y
3
z
z
y
3
y
z
y
x
3
x x
x
( ) ( ) ( ) ( )2
2
2
2
z
y
9
z
y
3
z
y
3
z
y
3
+
+
−
=
+
+
−
+
+
+
−
+
+
+
−
=
x
x
x
x
Thus
( )2
2
z
y
9
-
u
z
y
x +
+
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
x

94. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
94
2011
6. If ( ) θ
=
θ
=
= sin
r
y
,
cos
r
and
r
f
u x ,
prove that ( ) ( )
r
r
1
r
2
2
2
2
f
f
y
u
x
u
′
+
′
′
=
∂
∂
+
∂
∂
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.
( ) 2
2
y
x
r
where
r
f
u
have
we +
=
=
∴
( ) ( ) ( )
and
.
r
r
-
r
r
r
u 2
2
2
2
3
2
2
x
f
x
f
x
′
′
+
′
=
∂
∂
( ) ( ) ( )
.
r
r
-
r
r
r
u 2
2
2
2
3
2
2
y
f
y
f
y
′
′
+
′
=
∂
∂
Adding these results we get,
( ) ( )
{ } ( ) ( )
2
2
2
2
2
2
3
2
2
2
2
y
r
r
f
y
-
2
r
r
f
u
u
+
′
′
+
+
′
=
∂
∂
+
∂
∂
x
x
x
y
x
( ) ( ) ( ) ( )
r
r
r
1
r
.
r
r
r
.
r
r 2
2
2
3
f
f
f
f
′
′
+
′
=
′
′
+
′
=
Thus ( ) ( )
r
r
1
r
2
2
2
2
f
f
y
u
x
u
′
+
′
′
=
∂
∂
+
∂
∂
7. Prove that nu
y
u
y
x
u
x =
∂
∂
+
∂
∂
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
( ) ( )
x
x
x
x
x y/
g
n
y
-
.
y/
g
.
x
u 1
-
n
2
n
+
⎟
⎠
⎞
⎜
⎝
⎛
′
=
∂
∂
∴
ie., ( ) ( )
x
x
x
x
x
y /
g
n
y/
g
y
u 1
-
n
2
-
n
+
′
=
∂
∂
…(2)
Also ( ) ⎟
⎠
⎞
⎜
⎝
⎛
′
=
∂
∂
x
x
x
y
1
.
y/
g
.
u n

95. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
95
2011
ie., ( )
y/
g
.
u 1
-
n
x
x
y
′
=
∂
∂
…(3)
Now consider
y
x
x
∂
∂
+
∂
∂ u
y
u
as a consequence of (2) and (3)
( ) ( )
[ ] ( )
[ ]
x
x
x
x
x
x
x y /
g
y
y/
g
n
y/
g
y
- 1
-
n
1
-
n
2
-
n
′
+
+
′
=
( ) ( ) ( )
x
x
x
x
x
x y/
g
y
y/
g
n
y/
g
y 1
-
n
n
1
n
′
+
+
′
−
= −
( )
y/
g
x
.
n n
x
=
= n u, by using (1)
Thus we have proved Euler’s theorem
u
n
yu
u
;
u
n
y
u
y
u
y =
+
=
∂
∂
+
∂
∂
x
x
x
x
8. Prove that ( )u
1
-
n
n
y
2
2
2
=
∂
∂
∂
+
∂
∂
y
x
u
x
x
u
x 2
2
Proof : Since u = f (x, y) is a homogeneous function of degree n, we have Euler’s theorem
u
n
y
u
y
u
=
∂
∂
+
∂
∂
x
x …(1)
Differentiating (1) partially w.r.t. x and also w.r.t y we get,
x
x
x
x
x
∂
∂
=
∂
∂
∂
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂ u
n
y
u
y
u
1.
u 2
2
2
…(2)
Also,
y
u
n
y
u
.
1
y
u
y
y
u
2
2
2
∂
∂
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
∂
x
x …(3)
We shall now multiply (2) by x and (3) by y.
and
u
n
y
u
y
u
x
u 2
2
x
x
x
x
x
x
x 2
2
∂
∂
=
∂
∂
∂
+
∂
∂
+
∂
∂
y
u
ny
y
u
y
y
u
y
y
u
y 2
2
2
2
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
∂
x
x
Adding these using the fact that
x
∂
∂
∂
y
u
2
=
x
∂
∂
∂
y
u
2
we get,

96. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
96
2011
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
∂
+
∂
∂
y
u
y
u
n
y
u
y
u
y
u
y
y
u
y
2
u
2
2
2
2
2
2
2
x
x
x
x
x
x
x
x
ie., ( ) ( )
1
using
by
,
u
n
n
u
n
y
u
y
y
u
y
2
u
2
2
2
2
2
2
2
=
+
∂
∂
+
∂
∂
∂
+
∂
∂
x
x
x
x
or ( ) ( )u
1
-
n
n
nu
-
nu
n
y
u
y
y
u
y
2
u
2
2
2
2
2
2
2
=
+
∂
∂
+
∂
∂
∂
+
∂
∂
x
x
x
x
Thus ( )u
1
-
n
n
y
u
y
y
u
y
2
u
2
2
2
2
2
2
2
+
∂
∂
+
∂
∂
∂
+
∂
∂
x
x
x
x
ie., ( )u
1
-
n
n
u
y
u
y
2
u yy
2
y
2
=
+
+ x
xx x
x
9. If
y
z
z
y
z
y +
+
+
+
+
=
x
x
x
u show that 0
z
u
z
y
u
y
x
u
x =
∂
∂
+
∂
∂
+
∂
∂
Solution : (Observe that the degree is 0 in every term)
y
z
z
y
z
y
u
+
+
+
+
+
=
x
x
x
We shall divide both numerator and denominator of every term by x.
( )
{ }
x
x
x
x
x
x
x
x
z/
,
y/
g
/
y
1
z
1
/
z
/
y
z/
y/
1
u =
+
+
+
+
+
=
⇒ u is homogeneous of degree 0. ∴ n = 0
We have Euler’s theorem, u
n
z
u
z
y
u
y
u
=
∂
∂
+
∂
∂
+
∂
∂
x
x
Putting n = 0 we get, 0
z
u
z
y
u
y
u
=
∂
∂
+
∂
∂
+
∂
∂
x
x
10. If 3
y
that
show
y
log
4
4
=
∂
∂
+
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
+
=
y
u
x
u
x
y
x
x
u
Solution : we cannot put the given u in the form xn g (y/x)
( )
( )
( )
( )⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
+
+
=
+
+
=
+
+
=
∴
x
x
x
x
x
x
x
x
x
/
y
1
y/
1
y/
1
/
y
1
y
y
e
4
3
4
4
4
4
4
u
ie., eu = x3 g (y/x) ⇒ eu is homogeneous of degree 3 ∴ n = 3
Now applying Euler’s theorem for the homogeneous function eu

97. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
97
2011
We have
( ) ( ) u
u
u
e
n
y
e
y
e
=
∂
∂
+
∂
∂
x
x
ie., u
u
u
3e
y
u
e
y
u
e =
∂
∂
+
∂
∂
x
x
Dividing by eu we get 3
y
u
y
u
=
∂
∂
+
∂
∂
x
x
11. If ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
+
=
y
x
y
x
u
3
3
1
-
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) data
by
y
-
y
tan
u
3
3
1
-
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
=
x
x
( )
( )
( )
( )⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
−
+
=
+
=
+
=
⇒
x
/
y
1
y/
1
y/
-
1
/
y
1
y
-
y
u
tan
3
2
3
3
3
3
3
x
x
x
x
x
x
x
x
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
u
tan
y
u
tan
=
=
∂
∂
+
∂
∂
x
x
ie., u
tan
2
y
u
u
sec
y
u
u
sec 2
2
=
∂
∂
+
∂
∂
x
x
or u
2
sin
u
sin
u
cos
2
u
cos
u
sin
u
cos
2
u
sec
u
tan
2
y
u
y
u 2
2
=
=
=
=
∂
∂
+
∂
∂
x
x
2u
sin
yu
u y =
+
∴ x
x
(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
x
x
x
xx
x u
.
2u
cos
2
yu
u
.
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,
x
x
x
xx x
x
x
x u
u.
2
cos
2
u
y
u
u y
2
=
+
+

98. Engineering Mathematics – I Dr. V. Lokesha
10 MAT11
98
2011
y
y
yy
yx y
y
x u
u.
2
cos
2
u
y
u
u
y 2
=
+
+
Adding these by using the fact that uyx = uxy, we get
( ) ( )
y
y
yy
2
y
2
u
y
u
2u
cos
2
yu
y
u
y
u
y
2
u +
=
+
+
+
+ x
x
x
xx x
x
x
x
By using (1) we have,
2u
sin
-
u
2
sin
2u
cos
2
u
y
u
y
2
u yy
2
y
2
=
+
+ x
xx x
x
(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 ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
x
z
,
z
y
,
y
x
f
u Prove that 0
z
u
z
y
u
y
x
u
x =
∂
∂
+
∂
∂
+
∂
∂
>> here we need to convert the given function u into a composite function.
Let ( )
x
x z
r
,
z
y
q
,
y
p
where
r
q,
p,
f
u =
=
=
=
ie., ( ) ( )
{ } z
y,
x,
u
z
y,
x,
r
q,
p,
u →
⇒
→
→
x
x
x
x ∂
∂
∂
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∂
∂
∴
r
r
u
q
q
u
p
p
u
u
ie., ⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
2
z
-
.
r
u
0
.
q
u
y
1
.
p
u
u
x
x
r
u
z
-
p
u
y
u
∂
∂
∂
∂
=
∂
∂
∴
x
x
x
x …(1)
Similarly by symmetry we can write,
p
u
x
-
q
u
z
u
y
∂
∂
∂
∂
=
∂
∂
y
y
y
…(2)
q
u
y
-
r
u
x
u
z
∂
∂
∂
∂
=
∂
∂
z
z
z
…(3)
Adding (1), (2) and (3) we get 0
z
u
z
y
u
y
u
x =
∂
∂
+
∂
∂
+
∂
∂
x