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(Math100)[2011](s)final~4211^_12641.pdf
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HKUST
MATH100 Introduction to Multivariable Calculus
Final Examination Name:
12:30-14:30pm, May 30, 2011 Student I.D.:
Tutorial Section:

Directions:
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DO NOT open the exam until instructed to do so.

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Please have your student ID ready for checking.

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You may not use a calculator during the exam.

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You may write on both sides of the examination papers.

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You must show the steps in order to receive full credits.


Question No. Points Out of
1 10
2 10
3 12
4 13
5 15
Total 60

1. (10pts) Evaluate the double integral


4xy 3dA;
R

over the rectangular region R : .1 x 1 and .3 y 3.
Solution:

3 1 3 4xy 3dA =4xy 3dxdy =2y 3 x 2|.11dy =0
R .3 .1 .3
2. (10pts) De.ne a vector .eld in the three dimensional space F by


( x2 + y2 + z2 . 1)

F (x, y, z)= .
(xi + yj + zk) for (x, y, z) = (0, 0, 0).
x2 + y2 + z2

Show that F is not conservative, or write down a potential of F to reveal that it is conservative.
solution:
De.ne


.(x, y, z)= 1( x2 + y2 + z2 . 1)2 for all x, y, z.
2Then,
..(x, y, z)
.. ..
=(.. ,, )

.x .x .x
xyy
= (( x2 + y2 + z2 . 1) , ( x2 + y2 + z2 . 1) , ( x2 + y2 + z2 . 1) )
222
x2+y2+zx2+y2+zx2+y2+z

(x,y,z)
=( x2 + y2 + z2 . 1)
2
x2+y2+z

= F (x, y, z). Therefore, F is conservative with potential ..
3. (12pts) Use Greens Theorem to evaluate the integral
xy
(e + y 2)dx +(e+ x 2)dy,
C

where C is the boundary curve of region R oriented in counterclockwise direction, where R has .nite area enclosed by the curves of y = x2 and y =2x.
Solution:

xy
(e + y 2)dx +(e+ x 2)dy
C


yx
= . (e+ x 2) . . (e + y 2)dxdy
R .x.y


= (2x . 2y)dxdy
R 2 2x

= (2x . 2y)dydx
2
0 x 2

2)|2x
= (2xy . y x2 dx
0
8
= .
5

4. (13pts) A bird is .ying in a region over which the temperature is varying. At a certain moment, the components of the velocity of this .ying bird due north and east are u, v (in ms.1) respectively. At the same moment, the bird is located at a point so that the directional derivatives of the temperature along the directions due north and east are a and b (in C.m.1) respectively. Evaluate the rate of change of temperature experienced by this bird at the speci.c moment in terms of u, v, a, b.
Solution:

Let the bird be located at the origin at the speci.c moment. The positive x, y axis are due east, north respectively. T (x, y) is the temperature (in C.) at the point (x, y). Then, it is given that
.T

.x (0, 0) = (.T (0, 0)) (1, 0) = D(1,0)T (0, 0) = b,
.T

.y (0, 0) = (.T (0, 0)) (0, 1) = D(0,1)T (0, 0) = a,
and if (x(t),y(t)) is the position of the bird at time t, then
x (0) = v and y (0) = u.
By chain rule,

d .T .T
T (x(t),y(t)) = x (t)(x(t),y(t)) + y (t)(x(t),y(t)).
dt .x .x
Evaluating at 0 yields
d T (x(t),y(t))|t=0 = au + bv, dt