EE107 Mock Exam with Solutions March 2013

Answer any FIVE (5) questions. Total 100 marks.

1 Given the matrix,

744

313

542

A

(a) Show that

1

0

1

is an eigenvector of A and find the corresponding eigenvalues.

(8 marks)

(b) Find the other eigenvalues of A and for each find the corresponding eigenvectors.

(12 marks)

2(a) Find the partial derivative xxyzzf for )ln(,, 23 xyzzyxf

(10 marks)

(b) Find the partial derivative 2

3

xy

f

for xyeyxf ),(

(6 marks)

(c) Find dx

dyfor xyexyxyx 3)3cos( 53 by using Implicit Differentiation method.

(4 marks)

3(a) Find the equations of normal vector, tangent plane, and normal line to the surface:

016432 222 zyx at the point 3,1,2:0 P .

(12 marks)

1

(b) Determine divergence and curl of the vector function zzxyzF ,3,

.

(8 marks)

4(a) Evaluate the line integral C

rdF

, where kxyjziyzxzyxF

458),,( 2 and C is

the curve given by ktjtittr 32)( , and 10 t .

(10 marks)

(b) Evaluate R

dAyxyx )( 22 where R is the ellipse given by 222 yxyx and

using the transformation vux3

22 and vuy

3

22 .

(10 marks)

5(a) By using Green’s theorem, evaluate the line integral C

dyxdxy )( 33 , where C is the

positively oriented circle of radius 2 centered at the origin.

Green’s theorem is defined by

CR

dyFdxFdxdyy

F

x

F)()( 21

12

(10 marks)

(b) By using Stokes’s theorem, evaluate S

SdFcurl

, where kyxjxyizF

332 3 and

S is the part of 225 yxz above the plane 1z . Assume that S is oriented

upwards.

Stoke’s theorem is defined by SdFcurlrdrFSC

)( .

(10 marks)

2

6(a) Evaluate the surface integral, S

dAnF

, when 22 3,0, yxF

and S is the portion of

the plane 1 zyx (hint.: Assume new variables for ux and vy ).

(10 marks)

(b) Use Gauss’s Divergence Theorem to evaluate S

dAnF

, the surface integral of vector

zyxF ,,

, where the surface S is the region bounded by the coordinate planes and

the plane 1 zyx .

Gauss’s Divergence Theorem is defined by VS

dVFdAnF

(10 marks)

Solution

1(a) 0 XIA

0

0

0

1

0

1

744

313

542

Then,

052

033

074

We can see from the above three equations satisfied by 3 .

1

0

1

is an eigenvector of A and 3 is the corresponding eigenvalue.

(b)

3

2(a)

(b)

(c)

4

3(a) S: 016432 222 zyx ; 3,1,2:0 P ;

The equation of the normal vector:

kz

fj

y

fi

x

ff

,

kjif

kjif

zkyjxif

P

P

2468

)3(8)1(6)2(4

864

0

0

The tangent plane:

kzjyixkzzjyyixx

312000

kzjyixkji

312)2468(

822468 zyx

The normal line:

ftkzzjyyixx

000

)2468(312 kjitkzjyix

28828 txiti

1661 tyjtjy

324243 tzktkz

(b) divergence F

F

1100321

z

F

y

F

x

FF

curl F

F

kzzyjix

zzxyz

zyx

kji

F

)3()0()30(

3

kzjyix

23

4(a) Firstly, we need the vector field evaluated along the curve:

The derivative of the parameterization.

Finally, get the dot product of.

The line integral is then,

5

(b)

6

5(a) A circle will satisfy the conditions of Green’s Theorem since it is closed and simple

and so there really isn’t a reason to sketch it.

Let’s first identify P and Q from the line integral.

P = y3 Q = -x

3

Be careful with the minus sign on Q!

Now, using Green’s theorem on the line integral gives

where D is a disk of radius 2 centered at the origin.

Since D is a disk it seems like the best way to do this integral is to use polar

coordinates. Here is the evaluation of the integral.

(b)

7

6(a) R is the projection of S in the xy-plane.

1 zyx

Let’s 10 yxz

The boundary of R;

yx 10

10 y

Let’s ux and vy , then vuyxz 11

vuvuvur 1,,),(

1,1,1

110

1011,1,01,0,1

kji

xxrrN vu

Hence,

2222 31,1,3,0,)( vurvuNSF

S

v

R

dudvvududvvundAF

1

0

1

0

2222

3

133

(b) 1111321

z

F

y

F

x

FdivF

TTS

dvdvFdivdAnF

8

The boundary,

10 x ; xy 10 ; yxz 10

1

0

1

0

1

0

x yx

S

dzdydxdAnF

6

1)

2

21()1()1((

)2

()1(

1

0

2

1

0

1

0

21

0

1

0

dxxx

xxx

dxy

xyydydxyx xy

y

x