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(math102)[2007](f)midterm1~ma_yxf^_10423.pdf
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HKUST MATH 102
Midterm One Examination
Multivariable and Vector Calculus
30 Oct 2007
Answer ALL 5 questions
Time allowed C 120 minutes
Directions C This is a closed book examination. No talking or whispering are allowed. Work must be shown to receive points. An answer alone is not enough. Please write neatly. Answers which are illegible for the grader cannot be given credit.
Note that you can work on both sides of the paper and do not detach pages from this exam packet or unstaple the packet.
Student Name:
Student Number:
Tutorial Session:
Question No. Marks
1 /20
2 /20
3 /20
4 /20
5 /20
Bonus /5
Total /100
Problem 1 (20 points) Your Score:
Identify the following surfaces
(a)
r u.= 0.
(b)
(r . a) (r . b)= k.
(c)
r . (r u.)u. = k. [Hint: What are the vectors (r u.)u.and r . (r u.)u.?] Here k is .xed scalar, a, b are .xed 3D vectors and u.is a .xed 3D unit vector and r =(x, y, z).
Solution:
(a)
Let u.=(u1,u2,u3), then u1x + u2y + u3z = 0, this is a plane with its normal in the direction of u and passing through the origin.
(b)
(r . a) (r . b)= k r r . (a + b) r + a b = k
a + b a + b2
2
r . = k . a b +
24
a . b2
= k + 4
. It is a sphere center at (a + b)/2 with radius .k + a . b2/4.1/2 .
(c) Note that p =(ru)u.is the vector compo-
u
nent of r onto the vector u and r . (r u.)u.
is the vector component of r orthogonal to
u and the norm of this vector is a constant
k. Therefore, we can conclude that it is a
circular cylinder of radius k with its axis
parallel to u.
(a) Find the velocity, speed and acceleration at time t of the particle whose position is r(t). Describe the path of the particle.
r = at cos .t i + at sin .t j + b ln t k
22 2
(b) Find the required parametrization of the .rst quadrant part of the circular arc x+ y= ain terms of arc length measured from (0,a), oriented clockwise.
2/32/32/3
(c)
Let C be the curve x+ y= aon the xy-plane, .nd the parametric equation of the curve C. Hence .nd the tangent line to the curve C at (a, 0).
Solution:
(a)
Position: r(t)= at cos .t i + at sin .t j + b ln t k Velocity: v = r.(t)= a(cos .t . .t sin .t) i + a(sin .t + .t cos .t) j +(b/t) k Acceleration: a = r..(t)= .a.(2 sin .t + .t cos .t) i + a.(2 cos .t . .t sin .t) j . (b/t2) k
Speed: v = .a2(1 + .2t2)+ b2/t2.1/2
Let
x = at cos .t (1) y = at sin .t (2)
z/b
z = b ln t . t = e(3)
From (1) and (2)
22
xy
+ =1
2t22t2 22 22 22z/b
aa
x + y = at= ae
. Path: a spiral on the surface (acone -see .gure below)
22 22z/b
x + y = ae .
x
0
-10
10
2
2
00
00
00
z
z
z -2-2
-2-2
-2-2
-4
-10
-4
-4
-10
0
xx 0
xx
10
-10
-10
-10
0
10
0
0
10
10
10
yy y
22 22z/b
Taking a =2. and b = 2 The surface x+ y= aeThe surface and the curve