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(math100)[2009](f)midterm1~ylmaaa^_10411.pdf
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Math100 Introduction to Multivariable Calculus
Fall 2009
C Midterm Examination C
Name:
Student ID:
Tutorial Section:
.
There are FIVE 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
Total
1. Answer the following questions:
(a)
Given a surface S1 with cylindrical coordinate equation r =csc cot . Find the rectangular coordinate equation of S1. [5 points]
(b)
Given a sphere with rectangular coordinate equation x2 + y2 + z2 . 4z = 0. Find the spherical coordinate equation of the sphere. [5 points]
(c)
Determine if the points A(1, .1, 2), B(2, 0, 1), C(3, 2, 0), D(5, 4, .2) lie on the same plane. Justify your answer for full credit. [10 points]
Solution
yx
(a) Notice that r = x2 + y2,sin = and cos = . Hence we have,
rr
r =csc cot , (r sin )(sin )=cos , y (y )= x,
rr
2
y = x.
(b) Notice that 2 = x2 + y2 + z2 and z = cos . Hence we have,
x 2 + y 2 + z 2 . 4z =0, 2 . 4 cos =0, ( . 4cos )=0.
Since the solution set of equation . 4cos = 0 includes a point for which = 0, nothing is lost by discarding the factor . So the desired spherical coordinate equation is .4cos =0.
(c) It is enough to show that the vectors AB = . 1, 1, .1 ., AC = . 2, 3, .2 . and AD = . 4, 5, .4 .
all lie on the same plane. Their triple scalar product is given by
11 .1
. 23 .2 . =(1) (.2) . (1) (0) + (.1) (.2) = 0.
. 45 .4 .
Thus, the volume of the parallelepiped formed by vectors AB, AC and AD is zero. So all
these vectors lie on the same plane.
2. Let S1 be the plane with equation x+ y=0. Let S2 be the surface with equation
3/2
2 |x| + |y| z
z=
32
y
x
(a)
Find a vector-valued function .r(t) that describes the intersecting curve of S1 and S2.[5points]
(b)
Find the arc-length of the intersecting curve of S1 and S2 from the origin to the point (1,.1, ).
2
3
[15 points]
Solution
(a) Let x= t.Then express yand zin terms of tonly by using equations x+ y=0, z=
3/2 3/2
2 |x|+|y| 2 |t|+|.t| 2 |3/2
and x= t. Hence y= .tand z== |t. Hence the desired
32 323
vector-valued function is given by 2
.r(t)= . t, .t, |t|3/2 ..
3
(b) Notice that when t=0,we have .r(0) = . 0 , 0 , 0 . and when t=1,we have .r(1) =
2 23/2
. 1 , .1 , .. Moreover for each 0 t 1, .r(t)= . t, .t, t.. Thus the desired
33
arc-length is given by
t=1 1
. .r .(t) . dt =12 +(.1)2 +( t)2 dt,
t=0 0
1
= 2+ tdt,
0
2 t=1 )3/2
= (2+ t,3t=0
2
33/2 . 23/2
= ,
3 = 1.57848.
3. Given y implicitly de.ned as a function of x through the equation
2
xey +2xy +cos y =10.
dy
(a) Use