Applied Maths i u II Solution

7/31/2019 Applied Maths i u II Solution

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APPLIED MATHS I

Sol u t i on :CSVTU Ex a m i n a t i on

Papers

Depa r tm en t of Ma th em a ticsDIMAT

DI FFERENTI AL CALCULUS


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UNIT IV( I Semester)

Depar tm ent o f M athemat ics, DIM AT Page 2

APPLIED MATHS ITime Allowed : Three hours

Maximum Marks : 80

Minimum Pass Marks : 28

Note : Solve any two parts from each question. All questions carry equal marks.

UNIT II

DI FFERENTI AL CALCULUS

SOLUTION (Nov-Dec-2005)

(a) If 22log 1y x x prove that 2 22 1(1 ) (2 1) 0n n nx y n xy n y andhence show that

1 2 1 2

2 0( ) ( 1) .2 [( 1)!]k k

ky k

where k is a positive integer.

Ans: 2

2log 1y x x

2

22

12log 1 . 1

11

dy xx x

dx xx x

2

1 22

1 1

2 . 11

x x

y y xx x

1 2

12 .

1y y

x

2

11 2x y y ------------- (1)

2 211 4x y y Differentiating with respect to x we get

2 21 2 1 11 2 2 4x y y xy y

2

2 11 2x y xy ------------- (2)

By using Leibnitz rule we get

20 2 1 1 2 0 1 1( 1) 2 2 0 ... 0 ... 0n n n n nn n n n nC x y C xy C y C xy C y 2

0 2 1 1 2 0 1 1(1 ) 2 2 0n n n n n

n n n n nC x y C xy C y C xy C y 2

2 1 1(1 ) 2 ( 1) 0n n n n nx y n xy n n y xy ny

2 2

2 1(1 ) (2 1) 0n n nx y n xy n y (Proved)--------- (3)


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(0) 0y , 1(0) 0y , at 0x ,

(2) becomes2

2y

(3) becomes 22(0) (0)n ny n y -------- (4)

Putting all values in (4) we get3 5 2 1(0) (0) ........... (0) 0

ky y y .2 2

4 2

(0) 2 (0) 2 2y y 2 2 2

6 4(0) 4 (0) 2 2 4y y

2 2 2 2

8 4(0) 6 (0) 2 2 4 6y y

Similarly,2 1 2 2 2 2

2 2 2(0) (2 2) (0) ( 1) 2 2 4 6 ........... (2 2)

k

k ky k y k

21

2(0) ( 1) 2 2 4 6 ........... (2 2)k

ky k

21 2 2

2(0) ( 1) 2 2 1 2 3 ........... ( 1)k k

ky k

21 2 1

2(0) ( 1) 2 ( 1)!k k

ky k (Proved).

(b)Use Taylors theorem to prove that1 1 2 3sin sin 2 sin 3tan ( ) tan ( sin ) ( sin ) ( sin ) .....

1 2 3

z z zx h x h z h z h z

where 1cotz x .

Ans: Given that 1cotz x cotx z

21 cos

dzec z

dx

2sindz

zdx

Let 1 1( ) tan ( ) ( ) tanf x h x h f x x

2

2 2 2

1 1 1'( ) sin

1 1 cot cosf x z

x z ec z

2''( ) 2sin .cos . sin 2 .( sin )

dzf x z z z z

dx

2'''( ) (2cos 2 .sin 2sin .cos .sin 2 )

dzf x z z z z z

dx

22sin (cos 2 .sin .cos .sin 2 )( sin )z z z z z z 3

2sin .sin 3z z

So,2 3

1tan ( ) ( ) ( ) '( ) ''( ) '''( ) ......2! 3!h hx h f x h f x hf x f x f x

2 31 2 2 3tan sin ( sin .sin 2 ) 2sin .sin 3 ......

2! 3!

h hx h z z z z z

2 31 ( sin ) ( sin ) ( sin )tan sin sin 2 sin 3 ......

1 2 3

h z h z h zx z z z

(Ans).


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(c)Trace the curve 2 2( ) (3 )y a x x a x .Ans: 2 2( ) (3 )y a x x a x

22 (3 )

( )

x a xy

a x

2 2 2 33 0ay xy ax x ------------------ (1)

(i) The curve is symmetrical about x-axis.(ii)The curve is passes through the origin.(iii) The tangent at origin is 2 23 0ay ax 3y x

(iv)The curve meets the x axis at (0, 0), (3a, 0) and y axis at (0, 0).(v)When 3x a , 2 0y . So no curve for 3x a .(vi)Asymptote to the curve is x a .So, the curve is looks as follows:-

SOLUTION (Apr-May-2006)

(a)If 1/ 1/ 2m my y x , prove that 2 2 22 1( 1) (2 1) ( ) 0n n nx y n xy n m y .Ans: 1/ 1/ 2m my y x

2/ 1/ 1 2m my xy 2/ 1/ 2 1 0m my xy

21/ 2 4 4

2

m x xy

1/ 2 1my x x

2 1m

y x x


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Taking logarithm both side we get,

2log log 1y m x x

By differentiating with respect to x we get,

12 2

1 11

1 1

xy m

yx x x

2

12 2

1 1 1

1 1

x xy m

y x x x

1 2

1

1

my

y x

2 2 2 2

1( 1)x y m y

Again differentiating with respect to x we get,2 2 2

1 2 1 12( 1) 2 2x y y xy m yy

2 2

2 1( 1)x y xy m y

2 2

2 1( 1) 0x y xy m y


2 20 2 1 1 2 0 1 1( 1) 2 2 0 ... 0 ... 0n n n n nn n n n n nC x y C xy C y C xy C y m y 2 2

0 2 1 1 2 0 1 1( 1) 2 2 0n n n n n

n n n n n nC x y C xy C y C xy C y m y

2 2

2 1 1( 1) 2 ( 1) 0

n n n n n nx y n xy n n y xy ny m y

2 2 2

2 1( 1) (2 1) ( ) 0n n nx y n xy n m y (Proved)

(b)Find the first three terms in the expansion of log(1 tan )x by Maclaurinstheorem.

Ans:3 52

log(1 tan ) log 1 .......3 15

x xx x

23 5 3 5

3 43 5 3 5

2 1 2....... .......

3 15 2 3 15

1 2 1 2....... ....... ..........

3 3 15 4 3 15

x x x xx x

x x x xx x

3 5 42 3 42 1 2 1 1

....... ...... ....... ....... ....3 15 2 3 3 4

x x x

x x x x

23 41 1 4 3 .......

2 3 12

xx x x

23 42 7 .......

2 3 12

xx x x (Ans)


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(c)Trace the curve 2 2 2 2 2( )a y x a x .Ans: 2 2 2 2 2( )a y x a x

(i) The curve is symmetrical about x-axis and y axis.(ii)The curve is passes through the origin.(iii) The tangent at origin is

2 2 2 2 2 2

a y x a y x y x (iv)The curve meets the x axis at (0, 0), (a, 0), (-a, 0) and y axis at (0, 0).(v)When ,x a x a , 2 0y . So no curve for ,x a x a .(vi)There is no asymptote to the curve.So, the curve is looks as follows:-


(a)If 1sina xy e prove that 2 2 22 1

(1 ) (2 1) ( ) 0n n nx y n xy n a y and find the

value of (0)n

y .

Ans:1sina x

y e

1sin

1 2 2

1.

1 1

a x ayy e ax x

2

11 x y ay -------------- (1)

2 2 2 211 x y a y On differentiating with respect to x we get

2 2 21 2 1 11 2 2 2x y y xy a yy

2 22 11 x y xy a y

2 22 11 0x y xy a y --------------- (2)By using Leibnitz rule we get

2 20 2 1 1 2 0 1 1(1 ) 2 2 0 ... 0 ... 0n n n n nn n n n n nC x y C xy C y C xy C y a y


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2 2

0 2 1 1 2 0 1 1(1 ) 2 2 0n n n n n

n n n n n nC x y C xy C y C xy C y a y 2 2

2 1 1(1 ) 2 ( 1) 0

n n n n n nx y n xy n n y xy ny a y 2 2 2

2 1(1 ) (2 1) ( ) 0

n n nx y n xy n a y -----------------(3)

At 0, (0) 1x y -------------(4)

Then equation (1) becomes1

(0) (0)y ay a --------- (5)

From (2) 2 22(0) (0)y a y a ------------- (6)

From (3) 2 22(0) ( ) (0)

n ny n a y

2 2 2 2 2 2 2

3 1 4 2

2 2 2 2

5 3 6 4

2 2 2 2 2 2 2 2

(0) (1 ) (0) (1 ) (0) (2 ) (0) (2 )

(0) (3 ) (0) (0) (4 ) (0)

(1 )(3 ) (2 )(4 )

y a y a a y a y a a

y a y y a y

a a a a a a

Similarly2 2 2 2 2

2 2 2 2 2 2 2

(1 )(3 ).....(( 2) ) if n is odd(0)

(2 )(4 )....(( 2) ) if n is evenn

a a a n ay

a a a n a

(Ans)

(b)Use Taylors theorem to prove that1 1 2 3sin sin 2 sin 3

tan ( ) tan ( sin ) ( sin ) ( sin ) .....1 2 3


where1

cotz x .


21 cosdz

ec zdx

2sin

dzz

dx


2

2 2 2

1 1 1'( ) sin

1 1 cot cosf x z

x z ec z

2''( ) 2sin .cos . sin 2 .( sin )

dzf x z z z z

dx

2'''( ) (2cos 2 .sin 2sin .cos .sin 2 )dz

f x z z z z zdx


2sin .sin 3z z

So,2 3

1tan ( ) ( ) ( ) '( ) ''( ) '''( ) ......2! 3!h hx h f x h f x hf x f x f x


2! 3!

h hx h z z z z z


1 2 3

h z h z h zx z z z

(Ans).


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(c)Trace the curve (1 cos )r a .Ans: Given that (1 cos )r a .

(i) The curve is symmetrical about the initial line since its equation remainsunchanged, when changes to .

(ii)We have sindr

ad

Now2

2

2

2 2

2cos(1 cos )tan cot

/ sin 2sin cos

r a

dr d a

2 2 2 2tan tan( )

Now 32 2 2 2

Since is the angle between the tangent and initial line, therefore

At2

0, , So, the tangent is perpendicular to the initial line.

For , 2 . So, the tangent is the initial line.

(iii) Table having some values of ,r .2

6 3 2 30

2 1.86 1.5 0.5 0r a a a a a

Hence the curve is as follows:-

SOLUTION (May-June-2007)

(a)If 1sin( sin )y m x , prove that 2 22 1(1 ) 0x y xy m y and deduce that2 2 2

2 1(1 ) (2 1) ( ) 0

n n nx y n xy n m y .

Ans: 1sin( sin )y m x

1

1 2cos( sin )

1

my m xx

------------- (1)

2 1

11 cos( sin )x y m m x

Differentiating with respect to x we get

2 12 12 21 sin( sin )1 1

x mx y y m m x

x x


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2 22 11 x y xy m y

2 22 11 0x y xy m y -------------- (2)Taking nth derivative of (2) with respect to x by using Leibnitz rule we get

2 2

0 2 1 1 2 0 1 1(1 ) 2 2 0 ... 0 ... 0n n n n n

n n n n n nC x y C xy C y C xy C y m y

2 2

0 2 1 1 2 0 1 1(1 ) 2 2 0n n n n n

n n n n n nC x y C xy C y C xy C y m y 2 2

2 1 1(1 ) 2 ( 1) 0n n n n n nx y n xy n n y xy ny m y

2 2 2

2 1(1 ) (2 1) ( ) 0n n nx y n xy n m y (Proved)

(b)Use Taylors theorem to prove that1 1 2 3sin sin 2 sin 3tan ( ) tan ( sin ) ( sin ) ( sin ) .....

1 2 3


Where1

cotz x .


21 cosdz

ec zdx

2sindz

zdx


2

2 2 2

1 1 1'( ) sin

1 1 cot cosf x z

x z ec z

2''( ) 2sin .cos . sin 2 .( sin )

dzf x z z z z

dx

2

'''( ) (2cos 2 .sin 2sin .cos .sin 2 )dz

f x z z z z z dx 22sin (cos 2 .sin .cos .sin 2 )( sin )z z z z z z

32sin .sin 3z z

So,2 3

1tan ( ) ( ) ( ) '( ) ''( ) '''( ) ......2! 3!

h hx h f x h f x hf x f x f x


2! 3!

h hx h z z z z z


1 2 3

h z h z h zx z z z

(Ans).

(c)Find the first three terms in the expansion of log(1 tan )x by Maclaurinstheorem.

Ans:3 52

log(1 tan ) log 1 .......3 15

x xx x


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23 5 3 5

3 43 5 3 5

2 1 2....... .......

3 15 2 3 15

1 2 1 2....... ....... ..........

3 3 15 4 3 15

x x x xx x

x x x xx x

3 5 42 3 42 1 2 1 1....... ...... ....... ....... ....

3 15 2 3 3 4x x xx x x x

23 41 1 4 3 .......

2 3 12

xx x x

23 42 7 .......

2 3 12

xx x x (Ans)

(d)Trace the curve 2 2 3(2 )a y x a x .Ans: 2 2 3(2 )a y x a x

(i) The curve is symmetrical about x axis.(ii)The curve passes through origin.(iii) Tangent at origin is 2 2 0 0a y y , x-axis is the tangent.

(iv)The curve meets x axis at (0, 0) and (2a, 0) and y axis at (0, 0)(v)At 20, 2 0x x a y . So no curve when 0, 2x x a The curve is as follows:-

SOLUTION (Nov-Dec-2007)(a)State Taylors theorem and Maclaurins theorem.Ans: Taylors Theorem: - If ( )f x h can be expanded an infinite series, then

2 3

( ) ( ) '( ) ''( ) '''( ) ..........2! 3!

h hf x h f x hf x f x f x

Maclaurins Theorem: If ( )f x h can be expanded an infinite series, then


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2 3

( ) (0) '(0) ''(0) '''(0) ..........2! 3!

x xf x f xf f f .

(b)If 1sin( sin )

y m x , prove that 2 22 1

(1 ) 0x y xy m y and find(0)n

y .

Ans: 1sin( sin )y m x

1

1 2cos( sin )

1

my m x

x

------------- (1)

2 1



2 12 12 21 sin( sin )1 1

x mx y y m m x

x x

2 22 11 x y xy m y

2 22 11 0x y xy m y -------------- (2)Taking nth derivative of (2) with respect to x by using Leibnitz rule we get

2 20 2 1 1 2 0 1 1(1 ) 2 2 0 ... 0 ... 0n n n n nn n n n n nC x y C xy C y C xy C y m y 2 2

0 2 1 1 2 0 1 1(1 ) 2 2 0n n n n n


2 1 1(1 ) 2 ( 1) 0

n n n n n nx y n xy n n y xy ny m y 2 2 2

2 1(1 ) (2 1) ( ) 0

n n nx y n xy n m y -----------------(3)

At 0, (0) 0x y -------------(4)

Then equation (1) becomes1(0)y m --------- (5)

From (2) 22(0) (0) 0y m y ------------- (6)

From (3) 2 22(0) ( ) (0)n ny n m y

2 2

3 1

42 2

5 3

62 2 2

(0) (1 ) (0) (1 )(0) 0

(0) (3 ) (0)(0) 0

(1 )(3 )

y m y m my

y m yy

m m m

So,2 2 2 2 2

0 if n is even(0)

(1 )(3 )........((2 1) ) if n is oddny

m m m n m

(c)Obtain the Maclaurins expansion oftan( / 4 )

x and hence find the value

of 0tan 46 30' , to four decimal places.

Ans:2 3

( ) ( ) '( ) ''( ) '''( ) ...........2! 3!


Let ( ) tanf x x

4( ) tan , 1f x x f

2 4'( ) sec , ' 2f x x f


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2 3 4''( ) 2 tan sec 2 tan 2 tan , '' 4f x x x x x f

2 2 2 4'''( ) 2sec 6 tan sec , '' 16f x x x x f , .

So,2 3

4 4 4 4 4tan( ) ( ) '( ) ''( ) '''( ) ...........

2! 3!

h hh f hf f f

2 3

4tan( ) 1 2 4 16 ...........2! 3!

h hh h

2 3

4tan( ) 1 2 4 16 ...........

2! 3!

x xx x

Now, by putting 03 3 3 22/ 7 33

1 30 ' deg 0.02622 2 180 2 180 1260

x ree

2 30 (0.0262) (0.0262)tan(46 30 ') 1 2 (0.0262) 4 16 ...........

2! 3!

0 16tan(46 30 ') 1 0.0524 2 0.00068644 0.00001798 ...........

6

0tan(46 30') 1.0524 0.00137288 0.00004792 ........... 0tan(46 30') 1.0538208 1.0538 (Ans)

(d)Expand 1tan x in the power of ( / 4)x upto three terms.Ans: From Taylor Series

2 3( ) ( )( ) ( ) ( ) '( ) ''( ) '''( ) ...........

2! 3!

x a x af x f a x a f a f x f a

Let 1( ) tanf x x 1

4( ) tan , ( ) 1f x x f

2

42 2 2

1 4'( ) , '( )

1 4f x f

x

3

42 22 2 2

2 2 4''( ) , '( )

1 4

xf x f

x

22 2

42

1 2 2 2(1 ) 2''( )

1

x x x xf x

x

2 2 2 4 2 2

43 3 32 2 2 2

1 4 2 2 1 3 2 4 4 3''( ) , ''( )

1 1 4

x x xf x f

x x

So,2 34 4

4 4 4 4 4( ) ( )( ) ( ) ( ) '( ) ''( ) '''( ) ...........

2! 3!x xf x f x f f f

4 2 22 32 3

4 44 2 32 2 2 2 2 2

2 4 4 3( ) ( )4 2 4( ) 1 ( ) ....

4 2! 3!4 4

x xf x x


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4 2 22 32 3

4 4

4 2 32 2 2 2 2 2

2 4 4 3( ) ( )4 2 4( ) 1 ( ) ....

4 2! 3!4 4

x xf x x

(Ans).

SOLUTION (May-June-2008)

(a)Define point of inflexion and write Test to check point ofinflexion.Ans: If the two portions of a curve lie on different sides of the tangent at a pointC, then this point C is said to be a point of inflexion.

Test: Let y = f(x) be any curve. If at a point2 3

2 30, 0

d y d y

dx dx then the curve has a

point of inflexion at that point.

(b)If 1cosa xy e , prove that 2 2 22 1(1 ) (2 1) ( ) 0n n nx y n xy n a y . Also findthe nth differential coefficient of

1cosa xy e

at 0x .

Ans:1cosa x

y e

1cos

1 2 2

1.

1 1

a x ayy e a

x x

2

11 x y ay -------------- (1)


2 2 21 2 1 11 2 2 2x y y xy a yy

2 22 11 x y xy a y


2 20 2 1 1 2 0 1 1(1 ) 2 2 0 ... 0 ... 0n n n n nn n n n n nC x y C xy C y C xy C y a y 2 2

0 2 1 1 2 0 1 1(1 ) 2 2 0

n n n n n


2 1 1(1 ) 2 ( 1) 0n n n n n nx y n xy n n y xy ny a y

2 2 22 1(1 ) (2 1) ( ) 0n n nx y n xy n a y -----------------(3)

At 20, (0)a

x y e

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

Then equation (1) becomes 21(0) (0)a

y ay ae

--------- (5)

From (2) 2 2 22(0) (0)

a

y a y a e

------------- (6)


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From (3) 2 22(0) ( ) (0)

n ny n a y

2 2 2 2 2 2 22 23 1 4 2

2 2 2 2

5 3 6 4

2 2 2 2 2 2 2 22 2

(0) (1 ) (0) (1 ) (0) (2 ) (0) (2 )

(0) (3 ) (0) (0) (4 ) (0)

(1 )(3 ) (2 )(4 )

a a

a a

y a y a a e y a y a a e

y a y y a y

a a a e a a a e

Similarly

2 2 2 2 2 2

2 2 2 2 2 2 2 2

(1 )(3 ).....(( 2) ) if n is odd(0)

(2 )(4 )....(( 2) ) if n is even

a

n a

a a a n a ey

a a a n a e

(Ans)

(c)Expand 1sin( sin )y a x by Maclaurins theorem as for as 5x . Hence expandsin m .

Ans: 1sin( sin )y a x

1

1

2

cos( sin )

1

ay a x

x

------------- (1)

2 1

11 cos( sin )x y a a x


2 12 12 21 sin( sin )1 1

x ax y y a a x

x x

2 22 11 x y xy a y

2 22 11 0x y xy a y -------------- (2)Taking nth derivative of (2) with respect to x by using Leibnitz rule we get


0 2 1 1 2 0 1 1(1 ) 2 2 0n n n n n


2 1 1(1 ) 2 ( 1) 0

n n n n n nx y n xy n n y xy ny a y 2 2 2

2 1(1 ) (2 1) ( ) 0

n n nx y n xy n a y -----------------(3)

At 0, (0) 0x y -------------(4)

Then equation (1) becomes1(0)y a --------- (5)

From (2) 22(0) (0) 0y a y ------------- (6)

From (3) 2 22(0) ( ) (0)

n ny n a y

2 23 1

42 2

5 3

62 2 2

(0) (1 ) (0) (1 )(0) 0

(0) (3 ) (0)(0) 0

(1 )(3 )

y a y a ay

y a yy

a a a

Now, 1sin( sin )y a x 2 3 4 5

1 2 3 4 5(0) (0) (0) (0) (0) (0) ......2! 3! 4! 5!

x x x xy y xy y y y y


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3 52 2 2 20 . 0 (1 ) 0 (1 )(3 ) ......

3! 5!

x xy x a a a a a a

3 51 2 2 2 2sin( sin ) (1 ) (1 )(3 ) ......

3! 5!

x xa x ax a a a a a

By putting 1sin sinx x , then we get

3 52 2 2 2sin sinsin( ) sin (1 ) (1 )(3 ) ......3! 5!

a a a a a a a (Ans)

(d)Trace the curve 2 2( ) ( )y a x x a x .Ans: 2 2( ) ( )y a x x a x

(i) It is symmetrical about y axis.(ii) It passes through the origin.(iii) Tangent at origin is 2 2 2 2ay ax y x y x .(iv) Asymptote parallel to y axis is x a (v) At x axis it passes through (a, 0) and at y axis it passes through (0, 0).(vi) At , , 0x a x a y . So no curve when ,x a x a .The curve is as follows: -

SOLUTION (Dec-Jan-2008-2009)


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(a)The nth differential of: cos .cos2 .cos3x x x .Ans: cos .cos2 .cos3x x x

11 cos6 cos 4 cos 2

4x x x

Its nth derivative1

6 cos 6 4 cos 4 2 cos 24 2 2 2

n n nn n nx x x

(b)If 1siny x , find (0)ny .Ans: 1siny x

1 2

1

1y

x

2

1 2

1

1y

x

2 2

1(1 ) 1x y

Again differentiating with respect to x we get2 2

1 2 1(1 )2 2 0x y y xy

2

2 1(1 ) 0x y xy ----------- (1)

Now by applying Leibnitz rule differentiating n times.

20 2 1 1 2 0 1 1(1 ) 2 2 0 ... 0 ... 0n n n n nn n n n nC x y C xy C y C xy C y 2

0 2 1 1 2 0 1 1(1 ) 2 2 0n n n n n

n n n n nC x y C xy C y C xy C y 2

2 1 1(1 ) 2 ( 1) 0

n n n n nx y n xy n n y xy ny 2 2

2 1(1 ) (2 1) 0

n n nx y n xy n y

Now by putting n = n-2 we get2 2

1 2(1 ) (2 1) ( 2) 0

n n nx y n xy n y -----------(2)

At x = 0. equation (2) becomes 2 2( 2) 0n ny n y 2

2( 2)n ny n y ----------(3)

Given 1siny x , so (0) 0y

Now 22 0

(0) 0y y .

So2 4 2

(0) (0) (0) ........ (0) .... 0ky y y y

Again1 12

1(0) 1

1y y

x

So, 23 1

(3 2) 1 1 1y y 2 2 2 2

5 3(5 2) 3 1 3 1y y 2 2 2 27 5

(7 2) 5 3 1y y 2 2 2 2 2

9 7(9 2) 7 5 3 1y y

Similarly,2 2 2 2

0 when n is even(0)

( 2) ( 4) ( 6) ..........1 when n is oddny

n n n

(c)Expand by Maclaurins theorem the function log(1 sin )x .


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Ans:3 5

log(1 sin ) log 1 .......6 120

x xx x

23 5 3 5

3 43 5 3 5

1....... .......

6 120 2 6 120

1 1....... ....... ..........

3 6 120 4 6 120

x x x xx x

x x x xx x

3 5 4

2 3 41 1 1....... ...... ....... ....... ....6 120 2 3 3 4

x x xx x x x

23 41 2 2 3 .......

2 6 12

xx x x

2 3 4

.......2 6 12

x x xx (Ans)

(d)Trace the curve 2 2cos2r a .Ans: 2 2cos2r a

2 2 2 2cos sinr a 2 2 2 2 2cos sinr r a

By changing into Cartesian form we get 2 2 2x y a

Which represents a hyperbola.

The curve is as follows:-

Solution (Apr-May-2009)

(a)Select the correct answer.


18/32



For the curve 2 2(1 ) (1 )y x x x , the origin is a

(i) Cusp.(ii) Node.(iii) Point of Inflexion.(iv) None of the above.

Ans: (ii) Node (Ans).

(b)If 1sina xy e , Prove that: 2 2 22 1

(1 ) (2 1) ( ) 0n n nx y n xy n a y . Hence find

(0)n

y .

Ans:1sina x

y e

1sin

1 2 2

1.

1 1

a x ayy e ax x

2

11 x y ay -------------- (1)


2 2 21 2 1 11 2 2 2x y y xy a yy

2 22 11 x y xy a y



0 2 1 1 2 0 1 1(1 ) 2 2 0n n n n n

n n n n n nC x y C xy C y C xy C y a y

2 2

2 1 1(1 ) 2 ( 1) 0

n n n n n nx y n xy n n y xy ny a y

2 2 22 1(1 ) (2 1) ( ) 0n n nx y n xy n a y -----------------(3)

At 0, (0) 1x y -------------(4)

Then equation (1) becomes1(0) (0)y ay a --------- (5)

From (2) 2 22(0) (0)y a y a ------------- (6)

From (3) 2 22(0) ( ) (0)n ny n a y

2 2 2 2 2 2 2

3 1 4 2

2 2 2 2

5 3 6 4

2 2 2 2 2 2 2 2

(0) (1 ) (0) (1 ) (0) (2 ) (0) (2 )

(0) (3 ) (0) (0) (4 ) (0)

(1 )(3 ) (2 )(4 )

y a y a a y a y a a

y a y y a y

a a a a a a

Similarly2 2 2 2 2

2 2 2 2 2 2 2

(1 )(3 ).....(( 2) ) if n is odd(0)

(2 )(4 )....(( 2) ) if n is evenn

a a a n ay

a a a n a

(Ans)

(c)Find the Taylors series expansion for log(cos )x about the point3

.


19/32



Ans: Given that ( ) log(cos ), log(0.5)3

f x x f

'( ) , ' 33

f x tanx f

2

''( ) sec , '' 43f x x f

2'''( ) 2sec tan , ''' 8 33

f x x x f

So,

2 3

3 3log(cos ) ( ) ' '' ''' ......

3 3 3 2! 3 3! 3

x x

x f x f x f f f

2 3

3 3log(cos ) log(0.5) 3 ( 4) ( 8 3) ......3 2! 3!

x x

x x

2 34 3

log(cos ) log(0.5) 3 2 ......3 3 3 3

x x x x

(Ans).

(d)Trace the curve 2 2 2 2 2 2( ) ( )y a x x a x .Ans: 2 2 2 2 2 2( ) ( )y a x x a x

(i) It is symmetrical about x axis and y axis.(ii) It passes through (0, 0).(iii) No Asymptote parallel to x axis or y axis.(iv) When ,x a x a curve does not exist.(v) At x = 0,y = 0 and at y = 0, x = -a, a. So Curve passes through (0, 0),(-a, 0)

and (a, 0).

(vi) Tangent at origin are 2 2 2 2 2 2y a x a y x y x (vii) Hence the curve is as follows:



20/32



(a) State Leibnitzs theorem for the nth derivative of the product of twofunctions.

Ans: - Leibnitz theorem for nth derivative of product of two functions is

nn

n

nn

n

n

n

n

n

n

n

nuvCvuCvuCvuCvuCuv 1112221110 ..........................)( .

(b) If 1tany x , prove that 0)1()1(2)1(12

2 nnn

ynnxynyx . Hence

find (0)n

y .

Ans: - 1tany x

1 2

1

1y

x

------------- (1)

2

1(1 ) 1x y

Differentiating with respect to x we get2

2 1(1 ) 2 0x y xy -------------- (2)

Taking nth derivative of (2) with respect to x by using Leibnitz rule we get

2

0 2 1 1 2 0 1 1

(1 ) 2 2 0 ... 2 0 ... 0n n n n n

n n n n n

C x y C xy C y C xy C y

2 2 1 1(1 ) 2 ( 1) 2 0n n n n nx y nxy n n y xy ny 2

2 1 1(1 ) 2 ( 1) 2 2 0

n n n n nx y nxy n n y xy ny

2 2

2 1(1 ) 2( 1) ( ) 0n n nx y n xy n n y

2

2 1(1 ) 2( 1) ( 1) 0n n nx y n xy n n y -----------------(3)

At 0, (0) 0x y -------------(4)

Then equation (1) becomes 1(0) 1y --------- (5)

From (2)2(0) 0y ------------- (6)

From (3)2(0) ( 1) (0)

n ny n n y 3 1

23 1

45 1

25 3 6

7 182

7 3

(0) 1 2 (0) 1 2 2! ( 1) .2!(0) 0

(0) 3 4 (0) 4! ( 1) .4! (0) 0

(0) 0(0) 5 6 (0) 6! ( 1) .6!

y yy

y y y

yy y

So, 12

0 if n is even(0)

( 1) . ! if n is oddnny

n

(c) Expand xae1sin

in ascending powers of x, by Maclaurins theorem.Ans: -

1sina x

y e

1sin

1 2 2

1.

1 1

a x ayy e a

x x

2

11 x y ay -------------- (1)

2 2 2 211 x y a y


21/32



On differentiating with respect to x we get

2 2 21 2 1 11 2 2 2x y y xy a yy

2 22 11 x y xy a y

2 22 11 0x y xy a y --------------- (2)



0 2 1 1 2 0 1 1(1 ) 2 2 0

n n n n n


2 1 1(1 ) 2 ( 1) 0n n n n n nx y n xy n n y xy ny a y

2 2 2

2 1(1 ) (2 1) ( ) 0n n nx y n xy n a y -----------------(3)

At 0, (0) 1x y -------------(4)

Then equation (1) becomes 1(0) (0)y ay a --------- (5)

From (2) 2 22(0) (0)y a y a ------------- (6)

From (3) 2 22(0) ( ) (0)

n ny n a y

2 2 2 2 2 2 2

3 1 4 2

2 2 2 2

5 3 6 4

2 2 2 2 2 2 2 2

(0) (1 ) (0) (1 ) (0) (2 ) (0) (2 )

(0) (3 ) (0) (0) (4 ) (0)

(1 )(3 ) (2 )(4 )

y a y a a y a y a a

y a y y a y

a a a a a a

So, ..............)0(!5

)0(!4

)0(!3

)0(!2

)0(.)0( 5

5

4

4

3

3

2

2

1 yx

yx

yx

yx

yxyy

.......)3)(1(.!5

)2(!4

)1(.!3!2

.1 2225

2224

23

22

aaax

aax

aax

ax

axy

.......!5

)3)(1(!4

)2(!3

)1(!2

15

2224

2223

222

xaaaxaaxaaxaaxy

(Ans).

(d) Expand 423 23 xxx in powers of 2x .Ans: - 423)( 23 xxxxf 1442824)2( f

149)( 2 xxxf 291836)2( f

418)( xxf 32436418)2( xf

18)( xf 18)2( f

0.............)()( xfxf viv 0.............)2()2( viv ff

So,

.....)2(!4

)2()2(

!3

)2()2(

!2

)2()2().2()2()(

432

ivfx

fx

fx

fxfxf

.....024

)2(18

6

)2(32

2

)2()2(2914)(

432

xxx

xxf

32 )2(3)2(16)2(2914)( xxxxf 3223 )2(3)2(16)2(2914423 xxxxxx (Ans).


22/32



Solution (May-June-2010)

(a) If xy 3sin , find ny .Ans: - xxxy 3sinsin3

4

1sin 3

x

nx

nxy

n

n 32

sin32

sin34

1sin 3

(Ans).

(b) Determine )0(n

y wherexm

ey1cos .

Ans: -xm

ey1cos ------------------- (1)

on Differentiating222

1

2

1

2

2

cos

1 )1(11

11ymyxmyyx

xmey

xm

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

Again on differentiating

1

22

121

2 222)1( yymxyyyx

0)1( 212

2 ymxyyx ---------------- (3)

On differentiating n times by using Leibnitz theorem we get

022)1( 2112112

2 nnn

nn

n

n

n

nymyCxyyCxyCyx

0)1(2)1( 2112

2 nnnnnn ymnyxyynnnxyyx

0)12()1(22

12

2 nnn ymnxynyx ------------------- (4)

From (1) 2/)0( mey

From (2)2/

1 )0(mmey

From (3) 2/22 )0(memy

From (4) 0)0()0( 222 nn ymny )0()0( 222 nn ymny

So, 2/212

3)1()1()0( memmymy

2/222

3

22

5)3)(1()3()0( memmmymy

2/22222

5

22

7)5)(3)(1()5()0( memmmmymy

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

Silmilary, 2/22 )0(memy

2/222

2

22

4)2()0()2()0( memmymy

2/22222

4

22

6)4)(2()0()4()0( memmmymy

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


23/32



So,

evennmmmmme

oddnmmmmmey

m

m

n)8)(6)(4)(2(

)7)(5)(3)(1()0(

2222222222/

22222222/

(c) Expand xelog in powers )1( x and hence evaluate )1.1(loge correct to 4decimal places.

Ans: - xxf elog)( , 01log)1( ef

xxf

1)( , 1

1

1)1( f

2

1)(

xxf , 1)1( f

3

2)(

xxf , !22)1( f

4

6

)( xxf

iv , !36)1(

iv

f Now, by Taylors series

........)1(!4

)1()1(

!3

)1()1(

!2

)1()1()1()1()(

432

ivfx

fx

fx

fxfxf

........)!3(!4

)1()!2(

!3

)1()1(

!2

)1()1)(1(0)(

432

xxx

xxf

........4

)1(

3

)1(

2

)1()1(log

432

xxx

xxe (Ans)

Now

........4

)11.1(

3

)11.1(

2

)11.1()11.1()1.1(

432

f

........4

0001.0

3

001.0

2

01.01.0)1.1(log

e

........00002.00003.0005.01.0)1.1(log e

0953.0)1.1(log e upto 4 decimal places

(d) Trace the curve 2cos22 ar .Ans: -

i. The curve is symmetrical about the pole.ii. The curve lies wholly within the circle ar . No portion of the curve lies

between 4/ and 4/3 .iii. 2

22

2tan2cottan

dr

dr . So, 0 at

4

and2

at 0 . So tangent at origin O is

4

and tangent at A is

perpendicular to initial line is


24/32



iv. When varies from 0 to 4/ , r varies from a to 0 and when varies from4/3 to , r varies from 0 to a.

So, the curve is looks like: -


1.(a)

Express )(xf in ascending process of )( ax .Ans from

Taylors Theorem: - If ( )f x h can be expanded an infinite series, then2 3

( ) ( ) '( ) ''( ) '''( ) ..........2! 3!


)(xf in ascending process of )( ax . Will be

(b) If 1sin( sin )y m x , prove that 2 22 1

(1 ) 0x y xy m y and find (0)ny .

Ans:1sin( sin )y m x

1

1 2cos( sin )

1

my m x

x

-------------(1)

2 1



2 12 12 21 sin( sin )1 1

x mx y y m m x

x x

2 22 11 x y xy m y 2 22 11 0x y xy m y --------------(2)

Taking nth derivative of (2) with respect to x by using Leibnitz rule we get

2 20 2 1 1 2 0 1 1(1 ) 2 2 0 ... 0 ... 0n n n n nn n n n n nC x y C xy C y C xy C y m y 2 2

0 2 1 1 2 0 1 1(1 ) 2 2 0

n n n n n


2 1 1(1 ) 2 ( 1) 0

n n n n n nx y n xy n n y xy ny m y 2 2 2

2 1(1 ) (2 1) ( ) 0n n nx y n xy n m y -----------------(3)


25/32



At 0, (0) 0x y -------------(4)Then equation (1) becomes

1(0)y m ---------(5)

From (2) 22(0) (0) 0y m y -------------(6)

From (3)2 2

2(0) ( ) (0)n ny n m y 2 2

3 1

42 2

5 3

62 2 2

(0) (1 ) (0) (1 ) (0) 0(0) (3 ) (0)

(0) 0(1 )(3 )

y m y m m yy m y

ym m m

So,2 2 2 2 2

0 if n is even(0)

(1 )(3 )........((2 1) ) if n is oddny

m m m n m

(c) Use Taylors theorem to prove that

.......3

3sinsin

2

2sinsin

1

sinsintan)(tan

3211 z

zhz

zhz

zhxhx .

Where xz 1cot .


21 cosdz

ec zdx

2sin

dzz

dx


2

2 2 2

1 1 1'( ) sin

1 1 cot cosf x z

x z ec z

2

''( ) 2sin .cos . sin 2 .( sin )

dz

f x z z z zdx

2'''( ) (2cos 2 .sin 2sin .cos .sin 2 )

dzf x z z z z z

dx


2sin .sin 3z z

So,2 3

1tan ( ) ( ) ( ) '( ) ''( ) '''( ) ......2! 3!



2! 3!

h hx h z z z z z

2 31 ( sin ) ( sin ) ( sin )tan sin sin 2 sin 3 ......1 2 3

h z h z h zx z z z (Ans).

(d) Trace the curve 22 )3(9 axxay .

SOLUTION (Apr-May-2011)


26/32



a) Write the Leibnnitzs theorem for the derivative of the product of two function.Ans: - Leibnitz theorem for nth derivative of product of two functions is

nn

n

nn

n

n

n

n

n

n

n

n uvCvuCvuCvuCvuCuv 1112221110 ..........................)(

.b) If 1sina xy e prove that 2 2 22 1(1 ) (2 1) ( ) 0n n nx y n xy n a y and find the

value of (0)n

y .

Ans: -

1sina xy e

1

sin

1 2 2

1.

1 1

a x ayy e a

x x

2

11 x y ay --------------(1)


2 2 21 2 1 11 2 2 2x y y xy a yy 2 22 11 x y xy a y 2 22 11 0x y xy a y ---------------(2)


2 20 2 1 1 2 0 1 1(1 ) 2 2 0 ... 0 ... 0n n n n nn n n n n nC x y C xy C y C xy C y a y

2 2

0 2 1 1 2 0 1 1(1 ) 2 2 0n n n n nn n n n n nC x y C xy C y C xy C y a y

2 22 1 1

(1 ) 2 ( 1) 0n n n n n nx y n xy n n y xy ny a y 2 2 2

2 1(1 ) (2 1) ( ) 0

n n nx y n xy n a y -----------------(3)

At 0, (0) 1x y -------------(4)Then equation (1) becomes

1(0) (0)y ay a ---------(5)

From (2) 2 22(0) (0)y a y a -------------(6)

From (3)2 2

2(0) ( ) (0)

n ny n a y 2 2 2 2 2 2 2

3 1 4 2

2 2 2 2

5 3 6 4

2 2 2 2 2 2 2 2

(0) (1 ) (0) (1 ) (0) (2 ) (0) (2 )

(0) (3 ) (0) (0) (4 ) (0)(1 )(3 ) (2 )(4 )

y a y a a y a y a a

y a y y a ya a a a a a

So,

..............)0(!5

)0(!4

)0(!3

)0(!2

)0(.)0( 5

5

4

4

3

3

2

2

1 yx

yx

yx

yx

yxyy


27/32



.......)3)(1(.!5

)2(!4

)1(.!3!2

.1 2225

2224

23

22

aaax

aax

aax

ax

axy

.......!5

)3)(1(!4

)2(!3

)1(!2

15

2224

2223

222

x

aaax

aax

aaxa

axy

(Ans).c) Expand tan( / 4 )x as the term and evaluate 0tan 46 30' , to four decimal places.

Ans:

2 3

( ) ( ) ( ) '( ) ''( ) '''( ) ...........2! 3!

h hf x h f a h f x hf x f x f x

Let ( ) tanf x x

4( ) tan , 1f x x f

2 4'( ) sec , ' 2f x x f

2 3 4''( ) 2 tan sec 2 tan 2 tan , '' 4f x x x x x f

2 2 2

4

'''( ) 2sec 6 tan sec , '' 16f x x x x f , .

So,

2 3

4 4 4 4 4tan( ) ( ) '( ) ''( ) '''( ) ...........

2! 3!

h hh f hf f f

2 3

4tan( ) 1 2 4 16 ...........

2! 3!

h hh h

2 3

4tan( ) 1 2 4 16 ...........

2! 3!

x xx x

Now, by putting

0 3 3 3 22 / 7 331 30 ' deg 0.0262

2 2 180 2 180 1260x ree

2 3

0 (0.0262) (0.0262)tan(46 30 ') 1 2 (0.0262) 4 16 ...........2! 3!

0 16tan(46 30 ') 1 0.0524 2 0.00068644 0.00001798 ...........

6

0tan(46 30') 1.0524 0.00137288 0.00004792 ........... 0tan(46 30') 1.0538208 1.0538 (An

d) Trace the curve 2 2( ) ( )y a x x a x .Ans: 2 2( ) ( )y a x x a x

a.It is symmetrical about y axis.b.It passes through the origin.c.Tangent at origin is 2 2 2 2ay ax y x y x .d.Asymptote parallel to y axis is x a e.At x axis it passes through (a, 0) and at y axis it passes through (0, 0).f.At , , 0x a x a y . So no curve when ,x a x a .

The curve is as follows: -


28/32




1. Express )(xf in ascending process of )( ax .Ans:-( ) = [ + ( ) ] =

(

) + (

)

(

) +

( )

! (

) +

( )

! (

) + .

2. If 1sin( sin )y m x , prove that 2 22 1(1 ) 0x y xy m y and find (0)ny Ans:- Given

1sin( sin )y m x

= ( ) . = ( ) . . ( ) ( !) . . ,

+

= ( )

( ) = ( ) ( ) = ( ) + = (2)Again di f ferent iat ing eq(2 ) w.r . t to x ,n t ime s by Leibni tz rule ,we get( ) + + [ + ] + = ( ) ( + ) + ( ) = If x=0 ,then equation(1) becomes

) =mIf x=0 ,then equation (2) becomes( y) =0s

Therefore ( ) = ( ) ( ) ( ) . ( ( )

3. Use Taylors theorem to prove that


29/32



.......3

3sinsin

2

2sinsin

1

sinsintan)(tan

3211 z

zhz

zhz

zhxhxwhere xz

1cot

.


21 cos

dzec z

dx

2sin

dzz

dx

Let1 1( ) tan ( ) ( ) tanf x h x h f x x

2

2 2 2

1 1 1'( ) sin

1 1 cot cosf x z

x z ec z

2

''( ) 2sin .cos . sin 2 .( sin )dz

f x z z z zdx

2'''( ) (2cos 2 .sin 2sin .cos .sin 2 )dz

f x z z z z zdx


2sin .sin 3z z

So,

2 31tan ( ) ( ) ( ) '( ) ''( ) '''( ) ......

2! 3!



2! 3!

h hx h z z z z z


1 2 3

h z h z h zx z z z

(Ans).

4. Trace the curve 22 )3(9 axxay .Ans:- (1) . The curve is symmetrical about x-axis .

(2) The curve passes through the origin .the tangent at the origin is x=0 .(3) The curve cuts x-axis also at (3a,0) .shifting the origin to (3a,0) ,the

new equation of the curve is

9=( + 3) (4). No asymptote .

(5) Solving the given equation for y, we have

= ( ) When x is positive ,

When 0 < < 3, = + .


30/32



(ii) When > 3 , = + .(2). When x is negative , = ..

Therefore curve doesnot lie in the region a


31/32



() 0 = 0 (

1)

0=0

( + 2 ) 0 = 2 ( ) 0 Case 1:- When n is a odd number , putting n=1,3,5,-------------n-2 in equation (5) ,we obtain

Putting n=1 in equation(

3)

0= 0

( ) = 0 ( ) = 0 ( ) = 0

Case2;- When n is a even number , putting n=2,4,6,-------------n-2 in equation (5) ,we obtain

( ) = 22 ( ) = 2 2

( ) = 42 2 ( ) ) = ( 2 ) ) 42 2

(c) Expand xelog in powers )1( x and hence evaluate )1.1(loge correct to 4decimal places.

Ans: -

xxf elog)( ,01log)1( ef

xxf

1)(

,1

1

1)1( f

2

1)(

xxf

, 1)1( f

32)( xxf

, !22)1( f

4

6)(

xxf

iv , !36)1( ivf

Now, by Taylors series

........)1(!4

)1()1(

!3

)1()1(

!2

)1()1()1()1()(

432

ivfx

fx

fx

fxfxf

........)!3(!4

)1()!2(

!3

)1()1(

!2

)1()1)(1(0)(

432

xxx

xxf

........

4

)1(

3

)1(

2

)1()1(log

432

xxx

xxe

Now

........4

)11.1(

3

)11.1(

2

)11.1()11.1()1.1(

432

f


32/32


........4

0001.0

3

001.0

2

01.01.0)1.1(log

e

........00002.00003.0005.01.0)1.1(log e

0953.0)1.1(log e upto 4 decimal places

(d) Trace the curve 3 3 3x y axy .Sol: (1) Interchanging x and y, the given equation remains unchanged, hence there is symmetry

about the line y=x.(2) The curve passes through the origin and the tangents at the origin are given by 3axy=0, i.e.,

x=0, y=0 i.e., the coordinate axes. Hence (0,0) is a node.(3) The curve cuts the axes at (0,0) only.

(4) Asymptote. From the given equation, we have

( ) ( + ) = .The only asymptote is + + = 0 .

(5) From the given equation it is clear that x and y both cannot be negative since in this case lefthand side will become negative and right hand side will become positive, which is impossible.Hence the curve will not exist in the 3

rdquadrant.

(6) Intersection with the line (y=x): Putting y=x in the given equation, we get

2 = 3.. , = 3 = .Hence the curve meets the line y=x in the point (3 a/2, 3 a/2). Also from the given equation

=

3 33 3 = 1 3

2

,32

.Thus the tangent at the point is inclined at an angle 135

0

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Applied Maths i u II Solution