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(MATH100)[2011](s)midterm~wcpoon^_91095.pdf
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HKUST
MATH100 Introduction to Multivariable Calculus
Midterm Examination Name:
6th April 2011 20:30-21:30 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 10

1. Given that S is a surface whose spherical coordinate equation is = cos . csc2 .. Let
11
C be the curve described by the vector valued function r (t)= .0, 6 t6 , 4 t4. where 0 t< +. Answer the following parts:
(a)
Find the rectangular coordinate equation of the surface S.

(b)
We know that the curve C intersects the surface S at two distinct points P1 and P2. Find these two distinct intersection points P1 and P2.

(c)
Find the arclength of the curve C between P1 and P2.


Remark: The conversion formula from the spherical to the rectangular coordinates: x = sin . cos , y = sin . sin , z = cos ..
2.
Locate all relative maxima, relative minima and saddle points of the function f(x, y)=2xy . x 3 . y 2

3.
Let n be an odd integer which is at least 3. S is the surface de.ned by



nnn
x + y + z =0.
P =(a, b, c) is a point lying on S which is not the origin.
(a)
Write down the equation of the line joining P and the origin and then show that the line lies entirely on S.

(b)
Show that the line joining P and the origin lies entirely on the plane tangent S at P .

(c)
Find the line contained in the plane tangent to S at P which is perpendicular to the line joining P and the origin.


Solution to Q1(a) We may express , sin . and cos . in terms of x, y, z as follows:

2 + y2
xz
= x2 + y2 + z2 , sin . = , cos . = .

Hence the rectangular coordinate equation of S is z = x2 + y2 .
Solution to (b) To .nd the intersection points, we substitute x = 0, y = 16 t6 and z = 14 t4 into the equation z = x2 + y2 .
11 6)2
t4 =02 +( t,
46 9t4 = t12 , t4(9 . t8)=0,

88
9 and t = .points are (0, 0, 0) and (0, 23 , 43 ).
Solution to (c) The arclength of C between P1 and P2 is given by

9 (rejected because 0 t). The two intersection

Hence we have t = 0, t =

8
8

t=
8
9


8


9


9
9
t=
t= t=

5,t3.| dt = t10 + t6 dt = t3 t4 +1 dt,
t=0 t=0
|r (t)| dt = |.0,t
t=0
t=0

u=4
17
u du = let u = t4 +1.
=

u=1 46
Solution to Q2:

fx =2y . 3x 2,fy =2x . 2y fxx = .6x, fyy = .2,fxy =2 D(x, y) = 12x . 4 Set fx = fy = 0, we have critical points (x, y) = (0, 0), (x, y) = (2/3, 2/3) D(0, 0) = .4,D(2/3, 2/3) = 4,fxx(2/3, 2/3) = .4 (0, 0) is a saddle point and (2/3, 2/3) is a local maximum.
Solution to Q3(a) For every point Q lying on the line joining P and the origin,