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UNIT-I (2marks questions)
12
1. Find the characteristic equation of the matrix2.
Sol.0
A I 0
The characteristic equatin of A is
1210
021 0
0
1 2 0
02
(1 )(2 ) 0 0 2 2 2 0 2 3 2 0
The required characteristic equation is 2 3 2 0 .
12
2. Obtain the characteristic equation of5.
Sol.4
12
Let A=5
4
The characteristic equation of A is 2 c1 c2 0
c1 sumof the maindiagonal elements
1 4 5
c2A2
1
54
4 10
6
Hencethecharacteristic equationis 2 (5) (6) 0 2 5 6 0
3. Find the sum and product of the eigenvalues of the matrix 1 1 1111 .
111
Sol.
sumof theeigenvalues sum ofthe diagonal elements
(1) (1) (1)
3
1 1 1 product of theeigenvalues 1 1 1 1 1 1
1(1 1) 1( 1 1) 1(11)
1(0) 1( 2) 1(2)
4
1147
4. Two eigen values of the matrix72
5 are 0 and 1,
1046
find the third eigen value.
Sol.
Given 1 0, 2 1, 3 ?
sumof theeigenvalues sum of the main diagonal elements
123 11 ( 2) (6)
0 1 3 3
3 2
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5. Verify the statement that the sum of the elements in the diagonal of a matrix is the sum of the eigenvalues of the matrix 223
21
6
120
sol .sum of theeigenvalues sumof the maindiagonal elements
(2) (1) (0)
1
223
product of theeigenvalues 216
120
2(0 12) 2(0 6) 3(4 1)
24 12 9
45
622
23
6. The product of the eigenvalues of the matrix A 1
213
is16, Find the third eigenvalue. Sol.
let theeigenvalues of the matrix Abe 1, 2 , 3.
Given12 16
we knowthat 12 3 A2
62
231
213
6(9 1) 2( 6 2) 2(2 6)
6(8) 2(4) 2(4)
32
163 32
3 2
7. Two eigenvalues of the matrix
862
674 are 3and 0.what is the product of theeigenvalues of A?
24
3
sol .given1 3, 2 0, 3 ?
w.k .tThe sum of theeigenvalues sumof the main diagonal elements 1 2 3 8 7 3
3 0 3 18
3 15
productofeigenvalues 123 (3)(0)(15) 0
8. Find the sum and product of the eigen values of the matrix
201
020
102
sol .sumof theeigenvalues sum of the main diagonal elements
2 2 2
6
product of theeigenvalues A1
20
02
0
102
2(4 0) 0(0) 1(0 2)
8 2
6
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9.Find the characteristic equation of the matrix
eigenvalues.Sol.Given is a upper triangular matrix. Hence the eigenvalues are 1,2
1 2 and get its 0 2
11.Find the eigenvalues of A given
123
A 027
003
sol.
1 2 3
W.k.t the chacteristic equation of the given matrix is
2 (sumof theeigenvalues ) ( product of theeigenvalues) 0 2 (1 2) (1)(2) 0 2 3 2 0
10.Prove that if is an eigenvalues of a matrix A, then1is the
eigenvalue of A1
proof ;
If X betheeigenvector corresponding to
then AX X
premultiplying bothsides by A 1, weget
A 1 AX A 1 X
IX A 1 X
X A 1 X
1X A 1 X
i.e, A 1 X 1X
027
A
003
clearly given Ais aupper triangular matrix Hencetheeigenvalues are1,2,3 theeigenvalues of the given matrix Aare1,2,3By the property theeigenvalues of the matrix A 3 are13 ,23 , 33.
31
12.If and are cthe eigen values of15form the
matrix whose eigenvalues are 3and 3
1 75
02 9 0
005
Sol. (1 )[(2 )(5 ) 0] 7[0 0] 5[0 0] 0
(1 )(2 )(5 ) 0
1, 2, 5
sumof theeigenvalues 12 22 52
30
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175
13.Sum of square of the eigenvalues of02
9 is..
005
Sol.
1 The characteristic equatin of A isA I 0
75
02 9 0
005
(1 )[(2 )(5 ) 0] 7[0 0] 5[0 0] 0 (1 )(2 )(5 ) 0
1, 2, 5
sumof theeigenvalues 12 22 52
30
466
132
14 .two eigenvalues of A=are equal and they are
1 5 2
double thethird.Find the eigenvalues of A. Sol.Letthethirdeigenvaluebe
Theremainingtwoeigenvaluesare2 ,2 sumftheeigenvalues sumofthemaindiagonalelements
2 2 (4) (3) (2) 5 5
1 theeigenvaluesofAare2,2,1
HencetheeigenvaluesofA2 are 22 ,22 ,12
15.show that the matrix12
2satisfies its own characteristic
equation.1
Sol.12
LetA
21
The cha.equation of the given matrix is
A I 0
2 S S2 0
1
S1 sumof main iagonal elements
1 1 2
S2 A1 2 1 4 5
21
Thecharacteristicis 2 2 5 0
Toprove A2 2A 5I 0
A2 A.A
1212
2121
34
43
A2 2A 5I 3 4 21 21 0
5
43
2 10 1
00
00
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10
16.If A=45express A3 in terms of A and I using Cayley
Hamilton theorem.A I 0
Sol.The cha.equation of the given matrix is
1010
451 0
0
1 0 0
5
(1 )(5 ) 0 0 (1 )(5 ) 0 2 6 5 0
By Cayley Hamilton theorem,
A 2 6A 5I 0, A 2 6A 5I multiply Aon both sides
A3 6A 2 5A 0
A3 6A 2 5A
6(6A 5I ) 5A
36A 30I 5A
31A 30I
17.Write the matrix of the quadratic form
2x 2 8z 2 4xy 10xz 2 yz .
Sol.11
coeff of x2
2 coeff of xy2 coeff of xz
11
coeff ofy2
Q=2 coeff of xy2 coeff of yz
1 coeff of xz1 coeff of
yzcoeff of z2
22
225
Q=20
1
518
18.Determine the nature of the following quadratic form
f x1 , x2 , x3 x12 2x22
sol .The matrix of Q.F is
1Q=1
coeff of x2
2 coeff of xy2 coeff of xz
1 coeff of xy1 coeff of
coeff ofy 2yz
22
1 coeff of xz1 coeff of
yzcoeff of z2
22
100
02
=0
000
There for the eigenvalues are0,1,2. so find the eigenvalues one
eigenvalue is Zero another two eigenvalues are positive .so given Q.F is positive semi definite.
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19. State Cayley Hamilton theorem.Every square matrix satisfies its own characteristic equation.
20. Prove that the Q.F x 2 2 y 2 3z 2 2xy 2 yz 2zx .
Sol.The matrix of the Q.F form,
coeff of x211
2 coeff of xy2 coeff of xz
11
coeff ofy2
Q=2 coeff of xy2 coeff of yz
1 coeff of xz1 coeff of
yzcoeff of z2
22
111
121
=
113
D1 a1
D2 a1 a2
a1 D3 a2 a3
1 1(ve)
b1 1 1 (2 1) 1(ve) b2 1 2
b1 c1
b2 c2 1(6 1) 1(3 1) 1(1 2) 2(ve) b3 c3
The Q.F is indefinite.
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UNIT II - SEQUENCES AND SERIESPart A
1. Given an example for (i) convergent series (ii) divergent series (iii) oscillatory series Solution:
(i) The series
+ is convergent
(ii) 1+2+3+.+n+ is divergent
(iii) 1-1+1-1+ is oscillatory
2. State Leibnitzs test for the convergence of an alternating series
Solution:
The series a1-a2+a3-a4+. In which the terms are alternately +ve and ve and all ais are positive, is convergent if
Solution:
(i) The converges or diverges of an infinite series is not affected when each of its terms is multiplied by a finite quantity
(ii) If a series in which all the terms are positive is convergent, the series will remain convergent even when some or all of its terms are made negative
5. Define alternating series Solution:
A series whose terms are alternatively positive and negative is called alternating series
Eg: + is an alternating series
6. Prove that the series is convergent
(i) and
(ii)
3. State the comparison test for convergence of series Solution: Let an and bn be any two series and let a
finite quantity 0, then the two series converges or diverges together
4. State any two properties of an infinite series
Solution:
The nth term of the series is an=
Then an+1 =
now = =
= =0(
Hence by DAlemberts test, an is convergent
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7. When is an infinite series is said to be (i) convergent (ii) divergent (iii) oscillatory?
Solution:
Let an be an infinite series and let Sn be the sum of the first n terms of an infinite series then
(i) If is finite the series is said to be convergent
(ii) If If the series is said to be divergent
(iii) If not tend to a definite limit or , then the series is oscillatory.
8. State true or false
(i) If an is convergent, an2 is also convergent.
(ii) If the nth term of a series does not tend to zero as n, the series is divergent.
The nth term of the series be an=
Then =1/n and =1/n+1
Since , n+1n ,
an is decreasing and = =0
By Leibnitz test, the given series is convergent. Also the series formed by the absolute value of its terms is divergent. Hence the series is conditionally convergent.
10. For what values of p, the series + ++ + will be
(i) convergent (ii) divergent
Solution:
The p-series is convergent if p1 and divergent if
(iii) The convergence or divergence of an infinite seies is not affected by the removal of a finite number of terms from the beginning
(iv) An absolutely convergent series is convergent Solution:
All are true.
9. Prove that the series is conditionally convergent
Solution:
UNIT-IIIDIFFERENTIAL CALCULUS
1) Find the curvature of x 2 y 2 4 x 6 y 1 0
Solution:
f x3
2 f y2 2
f xx f y 2 2 f xy f x f y f yy fx2
f = x2 y2 4x 6 y 1
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f x 2x 4f y 2 y 6
f xx 2f yy 2
fxy 0
223223
22
2x 4 2 y 6 2x 4 2 y 6
2 2 y 6 2 0 2 2x 422 2 y 6 2 2x 4 2
12 2 y 62 (2x4)2
curvature 3 2 2 y 6 2 (2x 4
2 (2 y 6)22
2x 4
21/ 2
2 2 y 6 (2x4)2
12 2 y 6 2 (2x 4)2
2) What is the formula of radius of curvature in Cartesian form and parametric form? Sol:
Cartesian form: (1 y12 )3/ 2 y2
Parametric form: x ' 2 y ' 2 3/ 2 x ' y '' y ' x ''
3 Find the radius of curvature at x=0 on y ex
Solution:
Given y ex
3Radius of curvature 1 y12 2 y2
y ex
y e xy ]x0e01
11
y 2 e xy 2 ]x0 e01
1 y12 3/ 2113/ 2
2 2
y21
4 Find the radius of curvature of the curve xy c 2 at ( c , c)
Sol:
Given xy c 2 at ( c , c)
y c2
x
y c 2yc21
x 2c2
11
y 22c 2y22c22
x 3c 3c
1 y12 3/ 2113/ 2c.23/ 2
y 22 / c2
c2.
What is the curvature of the curve x 2 y2 25 at the
point (4,3) on it. Sol:
Since the given curve is a circle &We know that radius of given circle is 5 unitsradius of curvature of a circle is equal to the radius of the given circle 5
curvature 1 15 .
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6 Find radius of curvature of the curve x a cos , y b sin at any point ' '
Sol:
x a cosy bsin
x ' a siny ' b cos
x '' a cosy '' b sin
x '2 y '2 3/ 2 a 2 sin2 b2 cos2 3/ 2
x ' y '' x '' y 'ab sin2 ab cos2
a 2 sin2 b2 cos2 3/ 2(Qsin2 cos2 1)
ab
Find the radius of curvature at any point on the curve r e .
Sol:3/ 2 r 2 r12 r 2 rr1 2r22
Given r er e&r e
12
e2 e 23/ 2 e 23/ 2
2
e 2 e e 2 e 2 e2 e 2 2 e 2
3/ 233
2 e 22 1 ee
2
e 2
2
2.r
Find the radius of curvature at y=2a on the curve y 2 4ax
Sol:
Given y 2 4ax1
Formula 1 y12 3/ 2 y2
diff 1 w.r .to x, 2 yy1 4ayy1 2a 2
y2a
1y
y1 at y 2a 2a 1
diff 2 w.r .to x,2a
yy2 y y 0 yy2y 2
111
y2 y 2
1
y
y 2 at y 2a 1/ 2a
113/ 223/ 2 2a 23/ 2
1/ 2a
1/ 2a
25/ 2 a
i.e. 25/ 2.a 4a
2
9 Find the radius of curvature at (a,0) on y2 a 3 x3 xSol: Given y2 a 3 x3 x
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y 2 a3 x2
x
2 yya3 2x
x2
1
ya 3x
2x 2 yy
1
at (a ,0) y1
Hence we finddxdy
xy 2 a 3 x3
x. 2y y 2 . dydx 0 3x2 dydx 2xy ( y 2 3x2 ) dydx 0
dx2xy
dy3x 2 y2
at (a,0)dx0
dy
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3x2 y2 2 ydxdx 2 y
d2xdy2x 2xy 6xdy
dy2 3x 2 y2 2
at (a,0)d 2 x 3a 2 0 0 2a 0 6a32
dy 2 3a2 029a 43a
1dx2 3/ 2
dy1 03/ 23
a
d2x22
dy23a
32 a
10 Find the radius of curvature at x 2 on the curve
y 4sin x sin 2x .
Sol:
y 4sin x sin 2x
y1 dydx 4cos x 2cos 2x
y 2 d 2 y 4sin x 4sin 2x dx2at x / 2, y1 4(0) 2cos 2 at x / 2, y2 4(1) 4sin 43/ 21 (2)23/ 2
1 y12
y24
1 43/ 253/ 25.51/ 2 5
5
4444
5
5Q is ve
4
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11 Define the curvature of a plane curve and what is the curvature of a straight lineSol:
The curvature of a plane curve at K dds
The curvature of a straight line is zero.12 Find the radius of curvature at any point (x,y) on the
curve y c logx
sec
c
Sol:
x
y c logsec
c
y1 c.1xx 1 tanx
tan. sec c
xc
secc cc
c
y2 1sec2x
cc
tan2x 3/ 22 x3/ 2
23/ 21sec
cc
1 y1
y21sec2x1sec2x
ccc
c
sec3x
x
c
c. c.sec
xc
sec2
c
13 Find the radius of the curve given by x 3 2cos ,
4 2sin
Sol:
x 3 2cosy 4 2sin
dx2sindy 2cos
d
2cosd
dy cot
dx2sin
d 2 yddyd d cot 1
2
dxdxd2sin
d dx
cos ec21 cos ec3
2sin2
1 y 23/ 23/ 2
1 cot2 3
1cos ec 2
y2 1313
2cos ec 2cos ec
2
14 Write the formula for centre of curvature and equation of circle of curvature.Sol:
Centre of curvature: x x y1 1 y12 y2(1 y 2 )
y y
&1
y2
Circle of curvature: x x 2 y y 2 2
15 Find the centre of curvature of y x2 of the origin.
Sol:
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The centre of curvature is given by
y11 y1 2
X x 1 ( y1 )2 , Y y
y2
y 2
Given y x 2 ; y 2x ;y2 2.
1
at (0,0),y1 0
at (0,0),y2 2
x 02 1 (0)2 x
X
y 1 0 2 y 1
Y
22
at (0,0),X 0
at (0,0), 1
Y
2
0,1
Centreof curvatureis2
16 Write properties of evolutes. Sol:
(i) The normal at any point of a curve touches the evolute at the corresponding Centre of curvature.
(ii)The length of an of the evolute is equal to the of curvature at the points on the original curve corresponding to the extremities of the arc
(iii)There is only one evolute, but an infinite number of involutes.
17 Find the envelope of the family of straight lines y mx am2 , m being the parameter.
Sol:
Given y mx am2
Diff. partially w.r.to m, we get,
0 x 2am
2ax
y mx am2
xx2
y 2ax a
2a
x2 ax2
2a4a2
y x 2x 2x2
2a4a4a
x 2 4ay is the required envelope
18 efine envelope of a family of curves. Definition:
A curve which touches each member of a family of curve is called the envelope of that family curves.
19 efine Evolute and Involute.
The locus of the centre of the given curve is called the evolute of the curve.The given curve is called the Involute of its evolute. 20 Find the envelope of the family of lines xt yt 2c ,
t being the parameter. Sol:
Given family of lines can be written as, yt 2 2ct x 0 --------- (1)The envelope of At 2 Bt C 0 is B 2 4AC 0 From (1) we get A = y, B= -2c, C = x
Putting these values in (2) we get,
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(2c ) 2 4 yx 0 4c 2 4 yx 0 c 2 xy 0
(1)2 (2)2 we get,
xy2xy222
cos sinsin cos 1 0
abb
a
xy c2
This is required envelope.
21 Find the Envelope of the family of Straight lines a
x 2cos2 y2sin2
a 2b2
x 2 sin2 y2 cos2 a 2 b2
2xy cos sin 0 2xy cos sin
y mx m , where m is a parameter.Sol:
Given y mx a(1)
m
x 2cos2 sin2
y 2
x 2 y2 1 a 2 b2
y2sin2 cos2 1
b2
ym m 2 x a
m 2 x ym a 0 This is a quadratic in m
So the envelope is B 2 4AC 0
Here A x , B y , c a
y 2 4ax 0
y 2 4ax
22 Find the Envelope of the family of lines ax cos by sin 1, being the parameter
Sol:
Given,xcos ysin 11
ab
' ' we get
diffpartially (1) w.r .to
x sin ycos 0(2)
ab
23 Find the envelope of the straight lines
x cos y sin a sec , where is the parameter.
Sol:
Given x cos y sin a sec 1 Dividing equation (1) by cos we get,
x y tan a cossec a sec2 a(1 tan2 ) a tan2 y tan2 (a x) 0
Which is a quadratic equation in tan Here A=a, B=-y, C = (a-x).
B 2 4AC 0,
y 2 4a ( a x) 0
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24 Find the envelope of y mx a 2 m 2 b2 , where m is a
parameter.Sol:
mx a 2 m 2 b2 ( y mx ) 2 a 2 m 2 b2 2 m 2 x 2 2mxy a 2 m 2 b2
m 2 (x 2 a 2 ) 2mxy y 2 b2 0
Which is a quadratic equation in m. Hence the envelope is B 2 4AC 0
Here A= ( x 2 a2 ), B=-2xy, C = y 2 b2
4x 2 y 2 4(x 2 a 2 )( y 2 b2 ) 0
x 2 (cos2 sin2 ) y 2 (sin2 cos2 ) a2 x 2 y 2 a2
26 Find the envelope of the family given by x my m1 ,
m is parameter. Sol:
The given equation can be written as m 2 y mx 1 0,
Which is quadratic equation in m , Here, A y , B x , c 1
Hencethe envelopeis B 2 4AC 0
x 2 y 2 (x 2 a 2 )( y 2 b2 ) 0
x 2 y 2 x 2 y 2 b 2 x 2 a 2 y 2 a 2 b2 0
i.e, b 2 x 2 a 2 y 2 a 2 b2
x 2y2 1
22
x 2
27 Find the envelope of
parameter.Sol:
4 y 0 x 2 4 yy mx 1 m2 where m is a
ab
25 Find the envelope of x cos y sin a, where is
a parameter.
Sol:Given x cos y sin a(1)
Diff w.r.to
x sin y cos 0(2)
Eliminate between (1) and (2)
1 2 22, we have
(x cos y sin )2 ( x sin y cos )2 a2 02
x 2 cos2 y 2 sin2 2xy sin cos
2222 a2
xsin ycos
2xy sin cos
Given y mx 1 m2 y mx 1 m2
Squaring onboth sides ( y mx ) 2 1 m2 y 2 2mxy m 2 x 2 1 m2m 2 (x 2 1) 2mxy y2 1 0.
Here A x 2 1, B 2xy , C y2 1. B 2 4AC 0(2xy ) 2 4(x 2 1)( y2 1) 0 4x 2 y 2 4(x 2 1)( y2 1) 0
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UNIT IV
FUNCTIONS AND SEVARAL VARIABLE
PART-A
1x yuu1
1. If u cos, P.T . x ycot u .
xy2
x y
Proof:
Given f ( x , y ) cosu x y
xy
As f is hom ogeneous function of deg ree n 1,
2
it is satisfiesthe Euler ' s equation.
xf yfnf
xy
x(cos u ) y(cos u)1 cos u .
x
y2
x ( sin u )u y ( sin u )u1 cos u.
xy
2
xu yu1 cos u .
xy
2 sin u
xu yu1 cot u.
xy
2
As f is a homogeneous function of order n=2, it satisfies the Eulers theorem.
xf yf nf
y
x
x(tan u ) y(tan u) 2 tan u.
xy
x (sec2 u )u y (sec2 u )u 2 tan u.
xy
xu yu 2 sin u1.
xysec2 u
cos u
2 sin ucos2 u.
cos u
2sin u cos u.
xu yu sin 2u.
xy
3. If u logx 3 y 3u yu 2
, P.T. x
x yxy
2. If u tan1x 3 y 3u yu sin 2u .
, P.T . x
xy
x y
Solution:
Givenf ( x , y ) tan u x 3 y3
x y
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3322222
x yx y.1 x2x
fyx
Solution:Given u log
x y
2 x 2 y 2 2 x 2 y2 2
x
Similarly,2 fx 2 y2
x 2 y2 2
y2
Letf eux 3 y3
x y
Asf is hom ogeneous functionof deg ree n 2,
it is satisfies the Euler ' s equation.
xf yfnf
xy
2 f2 f 0
x 2y2
5. If u sin1x tan1yxu yu 0
show that
yxxy
Solution: Here u is a homogeneous function of degree n = 0.
x(e u ) y(eu ) 2eu
xy
x ( e u )u y ( e u )u 2eu .
y
x
xu yu 2.
y
x
Hence the proof.
4. If f ( x , y ) log x 2 y2 , show that
Solution: Given f ( x , y ) log x 2 y2 f 12 log x 2 y2
f1 2xx
x2x 2 y 2x 2 y2
2 f2 f 0 .
x 2y2
Using Eulers theorem,xu yu 0
xy
6. If u xyzshow that xu yu zu 0 .
yzxz
xy
Solution: Given u xyz
yz
x
u1z
xyx2
xuxz.........(1)
xyx
ux1
yy2z
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yuxy..........(2)
yyz
z21z2
R.H .S
rr2
uy 1
zz 2
x
zuyz..........(3)
zzx
Add eqn. (1),(2) & (3), we get
zz
cos sin
xy
z22z
cos
xy
2zz
( sin ) (cos )
xy
2zz
sin2 2sin cos
x
y
xu yu zu 0.
xyz
z
x
22z
sin
y
2zz
cos2 2sin cos
x
y
7. If z f ( x , y )
z2z
wherex r cos , y r sin .Show that
2z21z 2
2
z2
x
2zy
xy
Solution:Wkt,
z
1 z r
rr
zzxz y
rxryr
zcos zsin
xy
zxzy
xy
z( r sin ) z(r cos )
x
y
zsin zcos
y
x
Thus, R.H.S = L.H.S
8.If z f x , y , x e u cos v ,y e u sin v show that
xz yz e2uz
.
vuy
Solution: Givenz f x , y , x e u cos v , y e u sin v
zzxzy
uxuyu
ze u cos v ze u sin v
xy
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yzzye u cos v zye u sin v
uxy
e 2 u sin v cos vz e 2 u sin 2 vz....(1)
xy
zzxzy
vxvyv
zu sin v zucos v
ee
xy
xzzxe u sin v zxe ucos v
v
xy
e 2 u sin v cos vz e 2 ucos 2 vz....(2)
xy
(1) (2)
zz2 u sin22z
x yev cosv
vuy
e2u z y Hence proved. 9. If u x log( xy) wherex 3 y 3 3 xy 1 find du .
dx
Solution:
Given , u x log ( xy ) & x 3 y 3 3 xy 1
duuudy....(1)
dxxy
dx
u x1( y ) log ( xy)
xxy
u 1 log (xy)
x
u x1xx
y
xyy
consider , x 3 y 3 3xy 1
Diff. w.r.to x,
3x 2 3y 2 dy 3y 3x dy 0
dxdx
3x 2 3y 3y 2 3x dydx 0
dy 3 x 2 3y x 2 y
dx3 y 2 3xy 2 x
dux x 2 y
(1)dx1 log( xy) y y 2 x
10. Finddywhenx 3 y 3 3axy
dx
Solution:
Letf ( x , y ) x 3 y 3 3axy
f 3x 2 3ay ;f 3y 2 3ax
xy
dyf x3x 2 3ay x 2 ay
dxf3y 2 3axy 2 ax
y
11. Finddywheny sin x x cos y
dx
Solution:
Giveny sin x x cos y
y sin x x cos y0
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Let f ( x , y) x cos y y sin x
f cos y y cos x &f x sin y sin x
x
y
dyfx cos y y cos x
dxfy x sin y sin x
dy cos y y cos x dx x sin y sin x
12.If u x2 y2 z2andx et , y et sin t , z et cos t finddu
dt
withactual substitution.
Solution: Given u x2 y2 z2 , x et , y et sin t, z etcos t
duudxudyudz
dtxdtydtz
dt
2x et 2 y (et sin t et cos t) 2z (et cos t et sin t)
2 et x y (sin t cos t) z (cos t sin t)
2 et et etsin2 t et sin t cos t et cos2 t et sin t cos t
2etet etsin2 t cos2 t
2et2et
13. Find duif u sin (x / y) , where x et , y t2 .
dt
Solution:
duu. dx u. dy
dttdtydt
x1
cos.et cosxx2t
2
yy
yy
duetet2et
dt cost22t3
t
uuu
14. If u = f( y z , z x, x y ) find.
xyz
Solution: Given u f y z, z x, x y
Let r y z, s z x and t x y
uurus ut
xrxs
xtx
u(1) u(1).....(1)
st
uurusut
yrysyty
u(1) u(1).....(2)
r
t
uurusut
zrzsztz
u(1) u(1).....(3)
rs
(1) (2) (3)uuu0
xyz
15.Find the minimum value of F = x2+y2 subject to the
Constraint x=1. Solution: Given F = x2+y2
= square of the distance from the origin The minimum of F is 1.
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16. Define Jacobian.
If u and v are functions of the two independent variables
uu
x and y, then thedeterminantxyis called the Jacobian
vv
yy
18. If u 2xy , v x2 y2and x r cos , y r sin ,
evaluate(u, v)
(r, )
Solution:
u, vu, vx, y
r,x, yr,
of u ,v with respect to x,y.
(x, y)17. Find the Jacobian (r, )
It is denoted by x, y
.
u, v
if x r cos , y r sin .
uuxx
xy
r
vvyy
xyr
Given u 2xyv x2 y2
Solution: Given x r cos
x cos
r
y r sin
r
x, yxx
r
r, y
y
r
x, y r, r sin
y sin
r
y r cos
cosr sin
sinr cos
r cos2 r sin2
r cos2 sin2
r
u 2 yv2x
xx
u 2xv 2 y
yy
Given x r cosy r sin
x cosy sin
rr
y r siny r cos
r
u, v2 y2xcosr cos
r, 2x2 ysinr cos
4 y2 4x2 r cos2 r sin2
4x2 y2 r cos2 sin2
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4 r 2 r
u , v 4r3
r,
19. If u y 2, v x 2thenfind(u , v).
xy(x , y)
Solution:
Givenu y 2v x2
xy
uy 2v2x
xx 2xy
u2 yvx2
yxyy2
u , vuuy 22 y
xyx
x 2
x , yvv2xx2
xyyy2
y 2x 22 y2x
x2y2xy
1 4 3
u , v 3
x , y
20. If x u (1 v ), y uv compute J & J , and prove J . J 1.
Solution: Given x u 1 v and y uv
x 1 v y v
uu
x uyu
v
v
x , yxx
J uv
u , v
yy
uv
1 v
v u
u (1 v ) (uv)
u uv uvu x , y ' u , v1
J u& J u
u , v x , y
To prove: J .J = 1
' x , y u , v1
J Ju
u , v x ,y
u
J J '1
21. If x r sin cos , y r sin sin ,z r cos .Find J.
Solution:
Given x r sin cos , y r sin sin , z r cos
xxx
x , y , z r
J yyy
r , , r
yyy
r
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sin cosr cos cosr sin sin
sin sinr cos sinr sin cos
cosr sin0
cos (r 2 cos sin cos2 r2 cos sin sin2 ) r sin (r sin2 cos2 r sin2 sin2 )
r 2 sin cos2 sin2 cos2 r2 sin3 (sin2 cos2 )
r2 sin sin2 cos2
J r2 sin
22. Expand f ( x , y ) exy in Taylors series at (1, 1) up to
second degree.
f x , y f a , b 1f x a , b x a f y a , b y b
1
1f xx a , b x a 2 2 f xy a , b y b x a
...
22
f yy a , b y b
1 x 1 y 1
exy1
ex 12 4x 1y 1y 12
2
23. Find the Taylors series expansion of ex sin y near the
up to the first degree terms.
point 1,4
Solution:
Solution:
Givenf x , y e xy and the po int a 1,b 1
fx , y e xyf1,1e
fxx , y e xy yfx1,1e
f y x , y e xy xf y 1,1e
f x , y exsin y1,
f4
f x x , y exsin y1,
f x4
f y x , y excos y1,
f y4
e1 sin
4 e1 sin
4 e1 cos
4
e12e12e12
f xx x , y e xy y 2f xx 1,1 e
f xy x , y e xy (1) y e xy (x )f xy 1,1 e e 2e
f yy x , y e xy x 2f yy 1,1e
The Taylors series is
The required expansion is
f x , y f a , b 1f x a , b x a f y a , b y b
1
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fx , y f 1,x 1ff
x 1,yy1,
4444
exsin y11x 1 y
4
e2
24. Write condition for finding maxima and minima. Necessary Conditions:
The necessary conditions for f(x, y) to have a maximum
ff
or minimum at (a, b) are that 0 and 0 at (a,b)
xy
Sufficient Conditions:
Let r fxx a,b ; s fxy a,b and t f yy a,b
(i ) If rt s 2 0 and r 0at (a, b) , then f is maximum and
f (a, b) is maximum value
(ii) If rt s2 0 and r 0 at (a,b) , then f is minimum and f(a, b) is minimum value.
(iii) If rt s2 0 , then f is neither maximum nor minimumat (a, b).
(iv) If rt s2 = 0 , in this case further investigation are
required.
25. Find the stationary points of
f (x, y) x3 y3 3x 12 y 20 .
Solution: Given f (x, y) x3 y3 3x 12 y 20
fx 3x2 3 fy 3y2 12 For stationary points fx 0, f y 03x2 3 0 x2 1 x 1 3y2 12 0 y2 1 y 2
The stationary points are (1,2), (1,-2),(-1,2) & (-1,-2).
26. Find the stationary points of z x2 xy y2 2x y .
Solution: Given z x2 xy y2 2x y
zx 2x y 2 , zy x 2 y 1 For stationary points fx 0, f y 0
2x y 2 and x 2 y 1Solving x =1, y =0
The stationary point is (1,0)
27. Find the maximum and minimum values of x2 xy y2 2x y
Solution: Given f (x, y) x2 xy y2 2x y
fx 2x y 2f y x 2 y 1
fxx 2f yy 2
fxy 1
At maximum and minimum point: fx = fy = 0 (1,0) may be maximum point or minimum point. At (1,0): fxx . fyy ( fxy)2 = 4-1 = 3 > 0 & fxx =2 > 0
(1,0) is a minimum point
Minimum value = f(1,0) = -1
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28. A flat circular plate is heated so that the temperature at any point (x,y) is u(x,y) = x2+2y2-x. Find the coldest
point on the plate.
Solution: Givenu x 2 2 y 2 x
u x 2 x 1u y 4 y
u xx 2uyy 4
uxy 0
For stationary points u x 02x 1 0x1
2
u y 0 4 y 0 y 0
The point is1,0
2
At1,0uu uxy 280&u2 0
xxyyxx
2
The point1,0
2is the minimum point.
1,0
Hence the point2is the coldest point.
29. Find the shortest distance from the origin to the curve
x 2 8xy 7 y2 225 .
Solution:
Let f x 2 y 2 & x 2 8xy 7 y2 225
f x 2 y 2 x 2 8xy 7 y2 225
f x x 0 1 x 4 y 0(1)
fyy 04 x 1 7y 0(2)
Solving (1) & (2) = 1, = 19
If 1 x 2 y & 5y2 225 (no real valueof y)If 1y 2x x
95, y 20
1.y3x1 x2
x3
(12. (i) Find the Jacobian u , v , w, if
x , y , z
x y z u , y z u v , z u v w
(ii) If u x 2 y 2 , v 2xy . f ( x , y ) (u , v) show that
2 f2 f2222
4(x y)
x2y2u22
v
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UNIT-VMULTIPLE INTEGRALSPART-A
1. Evaluate 1dx x e yx dy
00
Sol:= 1x e y x dydx
Let I
00
=1e y x xaxdx eax
0dydxQ e
1 x 0a
1
xe y x xdx
0
0
1 (xe x x xe 0 )dx
0
1 (xe x ) dx
0
1 x ( e 1) dx
0
(e 1)1 x dx
0
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(e 1) x2 1 2 0
(e 1) 1 0 2
e 1
2
2. Evaluate b a dxdy
1 1xy
Sol:
b adxdydx
Let I Q log x
1 1xyx
adx bdy
y
1x 1
log x1a log y1b
log a log1 log b log1
(log a 0)(log b 0) (Qlog1 0)
log a log b
3. Evaluate a a 2 x2 dydx
0
Sol:
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Let I a a 2 x2 dydx
0
a y0 a 2 x2 dx
0
a 0dx
a2 x2
0
a
a2 x2 dx
0
x a
x a2 x2a21
sin
22
a
0
2 a222
a a a sin10 a sin1 (0)
(1)
222
sin1(0)0,
Qsin 0 0
sin1sin1(1)
22
a2 0 0
0 22
a2
4
4. Evaluate 1
x xy(x y) dxdy
00
Sol:
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Let I 1 x xy(x y) dxdy
0 0
1 x x2 y xy2 dxdy
0 0
1
x (x2 y xy2 ) dydx (correct form)
00
xy3
1x2 y2x
dx
23
00
x2 2x 3
1xx
dx
23
0
1
x3x5 2Q x3 25 2
dxx x
23
0
1x3dx 1 x5 2dx
0203
1x11x7 21
4
23
4 07 20
11 012 0
2431
7
18 212
2116
168
16837
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2ydxdy
5. Evaluate 0 0 x2 y2
Sol:y
2dxdy
Let I 0 0 x2 y2
2y
0 0 x2 dxy2dy
21tan1x
1yy
211y
1tan
yy
21tan1 (1) dy
1y
21 dy
1y 4
21dy
41y
log 2 log1
4
2 a cos
6. Evaluater2 drd
00
Sol:
ydy0 tan1 (0) dy
4 log 2 (Qlog1 0)
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2 a cos
Let I r2drd
00
2r3 a cos
d
03 0
2a3 cos3 d
03
a3 2
cos3 d
30
a33 1
33
2a3
9
sin
7. Evaluate2
r d dr
00
Sol:
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Let I 2 sin r d dr
0
2 sin r dr d correct form
0
sin
2r2
n 1 n 3d
2......1,if nis odd020
n n 2
Qcosn
n 1 n 32 sin 2
0......,if nis even
0d
n n 222
0
n 1 n 3
12 2......1,if nis odd
0 sin20 sinnn n 2
2 dQn 1 n 3
......,if niseven
n n 22
1.1 .
2228
cos
8. Evaluater dr d
00
Sol:
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cos
Let I r dr d
00r2 r cos
d
2
0r 0 0 d
cos 2
2
0
cos2 d
02
12
2cos d
0
1 1 cos 2 d
2 02
1sin 2
42
0
1sin 20
4202
1 0
44
e ydxdy is difficult to solve
0 x y
But by changing the order we get, y e y dxdy 0 0 y
eyy x0y dy
0
e y( y 0) dy
0 y
e y dy
0
9. Why do we change the order of integration in multiple integrals? Justify your answer with an example?
Sol :
Some of the problems connected with double integrals,which seen to be complicated,can be made easy to handle by a change in the order of integration.
Example:
e y
10
e y 0
(e e0 ) (0 1) 1
10. Express a a2 x22)dxdyin polar co-ordinates
0 y (x y
Sol:
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The region of integration is bounded by y 0, y a , x y , x a.Let us transform this integral in polar co-ordinates by taking
x r cos , y r sin , dxdy rdrd .
Consider the limitsx y , x a , y 0 .
Ify 0r sin0r0,sin 0
r0, 0
If x yr cosr sincos1
sin
tan1
a4
If x ar cosar
cos
ra sec
a aasec
x 24(r cos )2 rdrd
2 y22 r sin 2 3 2
0 y x00 r cos
4 asecr 3 cos2 drd
r2(cos2 sin23 2
00 )
4 asec
cos2 drd
00
11 .Find dxdy over the region bounded by x 0, y 0, x y 1
Sol:
Given x 0, y 0 & x y 1
The region of integration is the triangle.
Here x varies from x 0 to x 1 y
y varies from y 0 to y 1
I dxdy
R
1 1 y
dxdy
0 0
1 x 10 y dy
0
1 (1 y ) dy
0
y y2 1 2 0
1 12 12
12. Find the area of a circle of radius a by double integration in polar Co-ordinates Sol: The equation of circle whose radius is a is given by
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r 2a cos
The limits for
r : r 0 to r 2a cos
: 0to 2
Area 2 upper area
22 a cos
2rdrd
0 2r22 a cos
202d
0
2
4a 2 cos2 d
0
2
4a 2 cos2 d
0
4a22 1
22
4a 2 1 a2 2 2
13. Define Area in polar Co-ordinates Sol: Area= rdrd
R
14. Express the Volume bounded by x 0, y 0, z 0 and x y z 1
in triple integration.
Sol:For the given region
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z var ies from 0 to 1 x 2 y2
y var iesfrom 0 to1 x2
x var iesfrom 0 to 1
1 x 2 y2
I 112
xdzdydx
000
15. Evaluate2 3 2xy 2 z dzdydx
011
Sol:
2 32
Let I xy2 zdxdydz
01 1
232dy2
xdxyzdz
011
x2 2y33 z 22
23
021
1
4 027141
2332
2
(2) 26 3 26 3 2
16. Find the volume of the region bounded by the surface y x2 , y x2 and the planes z 0, z 3
Sol:
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y 2 x (1) x 2 y (2)
Substituting (2) in (1) we get x 4 xx 4 x 0
xx3 10
x0,1
13
x
Re quired volume dzdydx
0x20
1
x z 30dydx
0x2
1
x 3dydx
0 x2
1
3 y x2xdx
0
31 x 2 dx
x
0
x 3 2x31
3
3 23
0
2x 3 21
x3
33
3 0
3 2(1) 1 3
2 1 1
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17. Sketch roughly the region of integration for 1 x f ( x , y ) dy dx.
0 0
Sol:
The region of integration is bounded by x 0, x 1, y 0, y x
Here x var ies from x 0 to x 1 y var ies from y 0 to y x
a a 2 x2
18. Sketch the region of integrationdydx.
0ax x2
Sol:
Givenx varies from x = 0 to x = a
y varies from y
a 2 x 2to y ax x2
i.e., y 2 x 2 a which is a circle with centre (0,0)
andradius a.
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x2 y2 axa 2a2 y20
x24
a 22a2
i.e., x 2 y4
This is a circle with centre (a/2,0) and radius a/2.
19. Change the order of integration in a x f ( x , y ) dydx
0 0
Sol:
Given a x f ( x , y ) dydx
0 0
The region of integration is bounded by x 0, x a , y 0, y x
i.e., x var iesfrom x 0 to x a represents Vertical path
y var iesfrom y 0 to y x represents Vertical strip
Now changing the order of integration we get
x var ies from x y to x a represents Horizontal strip y var ies from y 0 to y a represents Horizontal path
a x f ( x , y ) dydx a a f ( x , y ) dx dy
0 00 y
20. Sketch roughly the region of integration for the following double integral a a 2 x2 f ( x , y ) dxdy 0
Sol:Given that x var ies from x 0 to x a
y var ies from y 0 to y a 2 x2 i.e., y 2 x 2 a2
x 2 y 2 a2
Which is a circle with centre (0,0) and radius a
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11 y
21.Change the order of integration inf ( x , y ) dxdy
00
Sol:
Given x var ies from x 0 to x 1 y i .e., x y 1 represents Horizontal y var ies from y 0 to y 1 represents Horizontal path
The region of integration is bounded by y 0, y 1, x 0, x y 1
x var ies from x 0 to x 1 represents Vertical path
y var ies from y 0 to y 1 x represents Vertical strip
After changing the order of integration limits of x and y becomes x 0, x 1, y 0 and y 1 x .11 y1 1 x
i.e.,f ( x , y ) dxdy f ( x , y ) dydx
0000
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