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(MATH100)[2010](s)midterm~2680^_10416.pdf
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Math100 Introduction to Multivariable Calculus
Spring 2010
C Mid-Term Examination C

Name:
Student ID:

Lecture and Tutorial Section:

.
There are SIX questions in this midterm examination.

.
Answer all questions.


.
You may write on both sides of the paper if necessary.

.
You may use a HKEA approved calculator. Calculators with symbolic calculus functions are not allowed.

.
The full mark is 100.



Question Points
Q1
Q2
Q3
Q4
Q5
Q6
Total




1. [15 pts.] Let F = .0,A. where A> 0 is a .xed number, e = .cos , sin . where 0


.

Answer the following two parts:




(a) Find the vector component of F along e .




(b) Find the vector component of F orthogonal e .



F . e
Solution to (a)

= A sin .


|
e |
Solution to (b)



F . ee


= .0,A...A sin cos , A sin2 .
F .


,
| e || e |

= A2 sin2 cos2 + A2 (1 . sin2 )2 ,
= A sin2 cos2 + cos4 ,
= A cos .




2. [15 pts.] Let r (t) = cos t i + sin t j + k . Evaluate the following integral:
2
1




(t) dt,
r (t) r
2 0


[ r (t)].

d
(t)=
where

r

dt

(t)= . sin ti + cos tj.

Hence we have

Solution We notice that

r

i jk


r (t)

r

(t)=

. cos t sin t 1 . = . cos t i . sin t j + k.
. . sin t cos t 0 .
Hence
2 2
1

(t) dt =1
r (t)

r


. cos t i . sin t j + k dt = k.

2

2

0
0

3. [15 pts.] Answer the following two parts:
(a)
Find the shortest distance from the point S(1, 1, 5) to the line:

L : x =1+ t, y =3 . t, z =2t.

(b)
Find the shortest distance from the point P (1, 1, 3) to the plane which passes through the


1
following points: (2, 0, 0), (0, 3, 0) and (1, 0, ).
2

Solution to (a) We see that the line L passes through the point P (1, 3, 0) and is parallel to the
.


vector v = .1, .1, 2.. Hence PS = .1 . 1, 1 . 3, 5 . 0. = .0, .2, 5. and
. P S v = . . . . . . . i 0 1 j .2 .1 k 5 2 . . . . . . . = i + 5j + 2k.
Hence the desired distance is given by

.


d == = =5.
|PS v | 1+25+4 30


| v | 1+1+4 6
Solution to (b) The equation of the plane passing through the three given points is 3x +2y +6z . 6=0. Then the desired distance is given by |3(1) + 2(1) + 6(3) . 6| 17
d = = .
32 +22 +62 7
4. [15 pts.] Find the parametric equations of the tangent line to the curve C at the point (0, 1, 3), where the curve C is the intersection curve of two surfaces S1 and S2 whose equations are given by
2 22
S1 : x 2 +2y 2 +2z = 20,S2 : x + y + z =4.
22

Solution Let F (x, y, z)= x2 +2y2 +2z2 . 20 and G(x, y, z)= x+ y+ z . 4. Then we have
.F (x, y, z)=2xi +4yj +4zk, .G(x, y, z)=2xi +2yj + k, .F (0, 1, 3) = 0i +4j + 12k, .G(0, 1, 3) = 0i +2j + k.
The cross-product of these two