1 Problem (20 points)
HKUST
Your Score:
MATH 102
Identify the following surfaces
Midterm One Examination
(a) r u = 0.
Multivariable and Vector Calculus
(b) (r a) (r b) = k .
30 Oct 2007
(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 ).
Answer ALL 5 questions
Time allowed 120 minutes
Solution:
Directions This is a closed book examination. No talking or whispering are allowed. Work must
(a) Let u = (u1 , u2 , u3 ), then u1 x + u2 y + u3 z = 0, this is a plane with its normal in the direction
of u and passing through the origin.
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
(b)
or unstaple the packet.
(r a) (r b) = k
r r (a + b) r + a b = k
r
Student Name:
a+b
2
2
=kab+
=k+
Student Number:
ab
4
2
It is a sphere center at (a + b)/2 with radius k + a b 2 /4
Tutorial Session:
2
a+b
4
1/2
.
u
(c) Note that p = (r u)u is the vector compo-
nent of r onto the vector u and r (r u)u
is the vector component of r orthogonal to
Question No.
Marks
1
/20
2
/20
3
/20
4
k . Therefore, we can conclude that it is a
/20
5
u and the norm of this vector is a constant
/20
Bonus
p
parallel to u.
r
0
/5
Total
circular cylinder of radius k with its axis
/100
1
Problem 2 (20 points)
Your Score:
Problem 2 (20 points)
Your Score:
(b) On the rst quadrant part of the circle x 2 + y 2 = a2 , we have
(a) Find the velocity, speed and acceleration at time t of the particle whose position is r(t).
r = r( ) = a sin i + a cos j,
Describe the path of the particle.
(0
/2)
The arc length s = a, hence
r = at cos t i + at sin t j + b ln t k
r(s) = a sin
y
s
s
i + a cos j,
a
a
(b) Find the required parametrization of the rst quadrant part of the circular arc x 2 + y 2 = a2
in terms of arc length measured from (0, a), oriented clockwise.
(c) Let C be the curve x2/3 + y 2/3 = a2/3 on the xy -plane, nd the parametric equation of the
curve C . Hence nd the tangent line to the curve C at (a, 0).
o
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/t 2 ) k
Speed: v = a2 (1 + 2 t2 ) + b2 /t2
1/2
Let
x = at cos t
(1)
y = at sin t
z = b ln t
(2)
t = ez/b
(3)
(c) Let x = a cos3 , y = a sin3 , then
From (1) and (2)
r( ) = a cos3 i + a sin3 j
x2
y2
+ 2 2 =1
a2 t2
at
then
r ( ) = 3a cos2 sin i + 3a sin2 cos j.
x2 + y 2 = a2 t2 = a2 e2z/b
At (a, 0), = 0 and r (0) = 0.
Path: a spiral on the surface (acone - see gure below)
The curve is not dierentiable at (a, 0), hence, there is no tangent line.
x2 + y 2 = a2 e2z/b .
x
10
0
-10
2
2
0
0
0
z
z
z -2
-2
-2
-10
-4
0
10
-10
0
y
-4
-10
0
x
-10
0
10
Taking a = 2 and b = 2
-10
10
10
y
2
2
0
y
2 2z/b
The surface x + y = a e
2
-4
x
10
The surface and the curve
3
a
x
Problem 3 (20 points)
Your Score:
Problem 3 (20 points)
Your Score:
(b) The vector equation of the line is
(a) Assume a, b, c, x and y are three dimensional vectors and if
(a b) (b c) (c a) = [x y]2 .
Use sux notation to nd x and y in terms of the vectors a, b and c.
r(t) = (1, 3, 0) + (1, 2, 1)t = r1 + vt
The given line is parallel to the plane if v is perpendicular
0
1
0
-1
-0.5
-1
to the normal vector n of the plane, i.e.
z -2
v n = (1, 2, 1) (2, 2, 2) = 2 4 + 2 = 0
(b) Show that the line x = 1 + t, y = 3 + 2t, z = t and the plane 2x 2y 2z + 3 = 0 are
parallel, and nd the distance between them.
x
0.5
-3
-4
3
3.5
v is perpendicular to n.
4
4.5
y
5
Solution:
Now the problem becomes to nd the distance between any point on the line and the given
(a) Note that from question (1a), we have
a (b c) = ijk ai bj ck .
L.H.S. = (a b) (b c) (c a)
plane, i.e.
r1
d = r 1 r0
n
|cos |
= (a b)i [(b c) (c a)]i
= ijk aj bk ipq (b c)p (c a)q
d
= r1 r0 n |cos |
= |(r1 r0 ) n|
= ijk ipq aj bk pmn bm cn qrs cr as
r0
= (jp kq jq kp ) pmn qrs aj bk bm cn cr as
= (jmn krs kmn jrs ) aj bk bm cn cr as
= (jmn aj bm cn ) (krs bk cr as ) (kmn bk bm cn ) (jrs aj cr as )
= [a (b c)] [a (b c)] [b (b c)] [a (c a)]
= [a c)]2
x=a
To (b nd a point r0 on the plane, let x = y = 0, then z = 3/2, so
d = ((1, 3, 0) (0, 0, 3/2)) (2, 2, 2)
= |(1, 3, 3/2) (2, 2, 2)|
12
12
= |2 6 + 3|
= 5 12
and y = b c.
4
5
12
Problem 4 (20 points)
Your Score:
Problem 4 (20 points)
Your Score:
(b) We can see that the limit does not exist by looking at
(a) Find the parametric equation of the curve of intersection C between the plane z = 2y + 3
and the surface z = x2 + y 2 . Find also the equation of the projection curve of the curve of
the limit along the paths x = 0 or y = 0 and x = y . On
the one hand
intersection C onto the xz -plane.
|y |
lim
(x,y )(0,0)
|x| + |y |
=
lim
(b) Evaluate
lim
|x| + |y |
|x|
= = 1,
x2
(0,y )(0,0)
|x| + |y |
x2 + y 2
.
x2 + y 2
y2
= 1,
or
(x,0)(0,0)
Solution:
x2 + y 2
-1-1
-0.5
x
while
(a) For intersection, we have
lim
y =x, x0
x2 + y 2 = 2y + 3
0.5
1
Two dierent limits along two dierent paths, hence the
x2 + y 2 2y = 3
2
2 | x|
=
= 2.
x2 + y 2
2x2
|x| + |y |
2
x + (y 1) = 4
limit does not exist.
(This is a circle.)
Let
x( ) = 2 cos
y ( ) = 2 sin + 1,
0
< 2
0
< 2 .
z ( ) = 4 sin + 5
or r( ) = 2 cos i + (2 sin + 1) j + (4 sin + 5) k,
The projection curve has to be independent of y , therefore from
x = 2 cos
z = 4 sin + 5
we have
(z 5)2
x2
+
= 1.
4
25
It is an ellipse.
-2
-2
x
x
0
2
10
10
z
0
2
z
5
5
0
0
-2
-2
0
y
0
2
y
6
2
7
0
-0.5
y
0
0.5
1
1.4
1.3
1.2 z
1.1
1
Problem 5 (20 points)
Your Score:
Problem 5 (20 points)
Your Score:
(c) Note that the equaation for sphere: x 2 + y 2 + z 2 = a2 , and for cylinder: x2 + y 2 = b2 .
(a) Let f (x, y ) =
xy
, (x, y ) = (0, 0)
x2 + y 2
0,
(x, y ) = (0, 0).
(i) In cylindrical coordinates:
r 2 + z 2 = a2
(i) Use the denition of partial derivative to show that f x (0, 0) and fy (0, 0) exist.
z = a2 r 2
(sphere)
r=b
(ii) Is the function f continuous at (0, 0)?
(b) Determine (sketch) the graph of the spherical-coordinate equation = 2 cos .
z
(c) A sphere of radius a is centered at the origin. A hole of radius b
is drilled through the sphere, with the axis of the hole lying on the
(r, , z ) b
r
a,
0
(cylinder)
2,
z=
a2 r 2 .
(ii) In spherical coordinates: (x = sin cos , y = sin sin , z = cos )
z -axis. Describe the solid region that remains (see Figure) in a (i)
y
a
b
=a
x
(, , )
b
sin
Solution:
Note that sin =
(a) (i)
fx (0, 0) = lim
x0
fy (0, 0) = lim
f ( x, 0) f (0, 0)
00
= lim
=0
x0
x
x
f (0,
y 0
y ) f (0, 0)
00
= lim
= 0.
y 0
y
y
b
a
a,
0
= sin1
(sphere)
sin = b
cylindrical coordinates; (ii) spherical coordinates.
(cylinder)
2,
b
.
a
(blue)
-0.5
y
-0.25
0
0.25
(red)
0.5
-0.5
-0.25
0
(ii) Take the path along y = 0, x = 0 (blue), then
x
0.25
0.5
lim
(x,y )(0,0)
xy
0
=
lim
= 0.
x2 + y 2 (x,y)(0,0) x2
0.25
z0
-0.25
-0.5
0.5
Take the path along y = x (green), then
lim
(x,y )(0,0)
xy
x2
1
= lim
=.
x2 + y 2 x0 x2 + x2
2
So dierent limits along dierent paths, hence
lim
(x,y )(0,0)
f (x, y ) is not continuous at (0, 0).
f (x, y ) does not exist.
z
(b) Multiplying by gives
2 = 2 cos ;
(0,0,1)
then substituting 2 = x2 + y 2 + z 2 and z = cos yields
x2 + y 2 + z 2 = 2z
1
y
x
as the rectangular-coordinate equation of the graph. Completing the square in z now gives
x2 + y 2 + (z 1)2 = 1,
so the graph is a sphere with center (0, 0, 1) and radius 1. It is tangent to the xy -plane at the
8
origin (as shown in the gure).
9
sin1
b
a
sin1
b
a
.
Bonus question (5 points)
Your Score:
Find the maximum and minimum distances between the point (1, 1, 1) and a point on the curve of
x2 + y 2 and the sphere x2 + y 2 + z 2 = z . (Note that the curve of
intersection of the cone z =
intersection is not the origin).
Solution:
First, nd the curve of intersection.
z2 + z2 = z
z
2z 2 z = 0
If z =
12
, x + y2 =
2
z=0
1
2
or
z=
1
1
.
2
x2 + y 2 + z 2 = z
r( )
2
1/2
.
z=
x2 + y 2
(Note that z = 0.)
/4
y
x
The parametric form of the curve of intersection is
r( ) =
1
1
1
cos i + sin j + k,
2
2
2
0
< 2.
Let = 0 be the required point on the curve, then
D = d2 =
2
1
cos 1
2
1
sin 1
2
+
2
+
1
1
2
2
1
1
1
cos2 cos + 1 + sin2 sin + 1 +
4
4
4
5
= cos sin
2
=
dD
= sin cos
d
d2 D
= cos + sin .
d 2
For critical points,
dD
= 0, i.e. tan 0 = 1
d
0 =
dmin =
5
cos sin =
2
4
4
dmax =
5
5
5
+ cos
sin
=
2
4
4
10
5
or
.
4
4
5
2
2
5
+2
2
(D > 0)
(D < 0)