Mathematics for Engineers 1

1. EE107 Mock Exam with Solutions March 2013
Answer any FIVE (5) questions. Total 100 marks.
1 Given the matrix,
 2 4  5
A   3 1 3 
 
 4 4 7 
 
1
(a) Show that  0  is an eigenvector of A and find the corresponding eigenvalues.
 
  1
 
(8 marks)
(b) Find the other eigenvalues of A and for each find the corresponding eigenvectors.
(12 marks)
2(a) Find the partial derivative f xxyzz for f x, y, z   z 3 y 2 ln( x)
(10 marks)
3 f
(b) Find the partial derivative for f ( x, y)  e xy
yx 2
(6 marks)
dy
(c) Find for x cos(3 y)  x 3 y 5  3x  e xy by using Implicit Differentiation method.
dx
(4 marks)
3(a) Find the equations of normal vector, tangent plane, and normal line to the surface:
2 x 2  3 y 2  4 z 2  16  0 at the point P0 : 2,1,3 .
(12 marks)

2. 
(b) Determine divergence and curl of the vector function F  yz ,3zx, z .
(8 marks)
     
4(a) Evaluate the line integral  F  dr , where
C
F ( x, y, z )  8x 2 yzi  5zj  4 xyk and C is
   
the curve given by r (t )  ti  t 2 j  t 3 k , and 0  t  1 .
(10 marks)
 ( x  xy  y 2 )dA where R is the ellipse given by x 2  xy  y 2  2 and
2
(b) Evaluate
R
2 2
using the transformation x  2u  v and y  2u  v.
3 3
(10 marks)
5(a) By using Green’s theorem, evaluate the line integral  ( y 3 dx  x 3 dy ) , where C is the
C
positively oriented circle of radius 2 centered at the origin.
F2 F1
Green’s theorem is defined by  ( x
R

y
)dxdy   ( F1dx  F2 dy )
C
(10 marks)
     
(b) By using Stokes’s theorem, evaluate  curlF  dS , where F  z 2 i  3xyj  x 3 y 3 k and
S
S is the part of z  5  x 2  y 2 above the plane z  1 . Assume that S is oriented
upwards.
   
Stoke’s theorem is defined by  F (r )  dr   curlF  dS .
C S
(10 marks)
1

3.   
6(a) Evaluate the surface integral,  F  ndA , when F  x 2 ,0,3 y 2 and S is the portion of
S
the plane x  y  z  1 (hint.: Assume new variables for x  u and y  v ).
(10 marks)
 
(b) Use Gauss’s Divergence Theorem to evaluate  F  ndA , the surface integral of vector
S

F  x, y, z , where the surface S is the region bounded by the coordinate planes and
the plane x  y  z  1 .
   
Gauss’s Divergence Theorem is defined by  F  ndA     FdV
S V
(10 marks)
Solution
1(a) A  I X   0
 2   4  5   1  0 
 3 1   3   0   0 
    
 4
 4 7     1 0
   
Then,
2 5  0
33  0
47  0
We can see from the above three equations satisfied by   3 .
1
  0  is an eigenvector of A and   3 is the corresponding eigenvalue.
 
 1
 
(b)
2

4. 2(a)
(b)
(c)
3

5. 3(a) S: 2 x 2  3 y 2  4 z 2  16  0 ; P0 : 2,1,3 ;
The equation of the normal vector:
f f f
f  i j k,
x y z
f  4 xi  6 yj  8 zk
  
f P0  4( 2)i  6( 1) j  8(3)k
  
f P0  8i  6 j  24k
The tangent plane:
     
x  x0 i   y  y0  j  z  z0 k  x  2i   y  1 j  z  3k
 (8i  6 j  24k )x  2i   y  1 j  z  3k 
     
 8x  6 y  24 z  82
The normal line:
  
x  x0 i   y  y0  j  z  z0 k  tf
     
x  2i   y  1 j  z  3k  t (8i  6 j  24k )
 
8  2i  8ti  x  8t  2
 
 y  1 j  6tj  y  6t  1
 
z  3k  24tk  z  24t  3
  
(b) divergence F    F
  F F F
  F  1  2  3  0  0 1  1
x y z
  
curl F    F
  
i j k
          
 F   (0  3x)i  j (0  y )  (3z  z )k  3xi  yj  2 zk
x y z
yz 3zx z
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,
4

6. (b)
5

7. 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 = -x3
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)
6

8. 6(a) R is the projection of S in the xy-plane.
x y  z 1
Let’s z  0  x  y  1
The boundary of R;
0  x  1 y
0  y 1
Let’s x  u and y  v , then z  1  x  y  1  u  v
 r (u, v)  u, v,1  u  v
  
i j k
N  ru xrv  1,0,1 x 0,1,1  1 0  1  1,1,1
0 1 1
Hence,
F (S )  N  u 2 ,0,3v 2  r ,1,1  u 2  3v 2
1 1v
 
 F  ndA   u  3v dudv  
2 2
 u
2

 3v 2 dudv 
1
3
S R 0 0
F1 F2 F3
(b) divF     111  1
x y z
  

S
F  ndA   divFdv  dv
T T
7

9. The boundary,
0  x  1; 0  y  1 x ; 0  z  1 x  y
  1 1 x 1 x  y

S
F  ndA     dzdydx
0 0 0
1 1 x 1
y 2 y 1 x
   (1  x  y )dydx   ( y  xy  ) y 0 dx
0 0 0
2
(1  2 x  x 2
1
1
  ((1  x)  x(1  x)  )dx 
0
2 6
8