102
Midterm HKUST
MATH Examination
Multivariable and Vector Calculus
21 Dec 2005
Answer ALL 8 questions
Time allowed 180 minutes
Directions 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
6
/20
7
/20
8
/20
Total
/160
1
Problem 1 (20 points)
Your Score:
Problem 2 (20 points)
(a) Assume a, b and c are three dimensional vectors and if
Your Score:
(a) Sketch and describe the parametric curve C
(a b) c = a + b + c.
r = t cos t i + t sin t j + (2 t) k,
Use sux notation to nd , and in terms of the vectors a, b and c. Can you say something
about the direction of the vector (a b) c.
0
t
2.
Show the direction of increasing t. Find the project curve C onto the yz -plane.
(b) Find a change of parameter t = g ( ) for the semicircle
(b) Let a be a constant vector and r = (x, y, z ), use sux notation to evaluate
(i)
r,
(ii)
(a r),
(iii)
r(t) = cos t i + sin t j,
(a r).
0
t
such that (i) the semicircle is traced counterclockwise as varies over the interval [0, 1],
(ii) the semicircle is traced clockwise as varies over the interval [0, 0.5].
Solution:
(a)
[(a b) c]i =
ijk (a
axb is normal to the plane P
c
b)j ck
=
ijk j pq
ap bq ck
=
j ki j pq
ap bq ck
Plane P
(a)
b
= (kp iq kq ip )ap bq ck
= a k bi ck a i bk ck
i.e. (a b) c = (a c)b (b c)a,
Solution:
a
(ax b)x c
x = t cos t
y = t sin t
z = 2 t
x2 + y 2 = t2
t=
x2 + y 2
since
t
0.
i.e. z = 2 x2 + y 2 .
r is a conical helix wound around the cone z =
i.e. = a c, = b c and = 0.
2
The resulting vector of (a b) c is a linear combination of the vectors a and b, hence it lies
x2 + y 2 starting at the vertex (0, 0, 2 ) and com-5
pleting one revolution to end up at (2, 0, 0).
on the plane containing the vectors a and b.
x
0
The projection curve can be obtained by considering
5
(b)
(i)
r = i ri = 3.
6
y = t sin t
(ii)
z
(a r) = i (a r)i
= i ijk aj rk
4
2
z = 2 t
0
-5
0
r(t) = t sin t j + (2 t) k
y
= ijk aj i rk
5
y
= ijk aj ik
(b)
= iji aj = 0
(i) r( ) = cos i + sin j,
0
1
(iii)
1
[
x
(a r)]i = ijk j (a r)k
= ijk j kpq ap rq
= kij kpq ap jq
(ii) r( ) = cos (1 2 ) i + sin (1 2 ) j,
= kij kpj ap
= (ip jj ij jp ) ap
= 3ai ai = 2ai
0
0.5
0
= cos(2 ) i + sin(2 ) j,
0.5.
y
2
(1 2 )
1
(a r) = 2a.
2
3
x
Problem 3 (20 points)
Your Score:
Problem 3 (20 points)
Your Score:
Since the parametric representation of the curve is
If a wheel with radius a rolls along a at surface without slipping, a point P on the rim of the
wheel traces a curve C , nd the parametric equation of the point P . Suppose that the point P on
the wheel is initially at the origin. Find also the arc length of the curve C if the wheel makes one
r( ) = a( sin ) i + a(1 cos ) j
r ( ) = a(1 cos ) i + a sin j
2
r ( )
complete turn (no need to carry out the integration).
= a2 (2 2 cos )
Therefore, the arc length is given as
Solution:
2
s=
y
r ( ) d
0
2
=a
0
1
(2 2 cos ) 2 d
= 8a.
a
x
2a
P
y
P
y
3/2
P
3
2
x
A
j
A
a
O
a
x
OA = a i + a j
AP = a cos
3
2
i + a sin
= a sin i a cos j
OP = OA + AP
= (a i + a j) + (a sin i a cos j)
= a( sin ) i + a(1 cos ) j.
y
2
1.5
1
0.5
2
4
6
4
8
10
12
x
5
Problem 4 (20 points)
Your Score:
Problem 4 (20 points)
Your Score:
(b) In spherical coord.
(a) Verify the formula for the arc length element is cylindrical coordinates,
2
dr
dt
ds =
+ (r (t))2
d
dt
2
+
dz
dt
x = sin cos ,
dt.
dx
d
d
d
=
sin cos + cos cos
sin sin
dt
dt
dt
dt
d
d
d
dy
=
sin sin + cos sin
+ sin cos
dt
dt
dt
dt
(c) Use part (b) or otherwise, nd the arc length of the curve in spherical coordinates:
t
dz
d
d
=
cos sin
dt
dt
dt
dx 2
dy 2
dz 2
2
+
+
r (t) =
dt
dt
dt
2
d 2
d
2 d
=
+
+ 2 sin2
dt
dt
dt
5.
Solution:
(a) In cylindrical coord.
x = r cos
r (t)
2
=
z=z
dx
dt
2
dr
dt
+ sin2
dr
dt
2
2
dy
dt
+
= cos2
=
dx = cos dr r sin d
dt
dt
dt
dr
d
dy
= sin
+ r cos
dt
dt
dt
dz
dz
=
dt
dt
y = r sin
2
dz
dt
2
d
dt
+
dz
dt
Hence
d
dt
ds =
2
+ ((t))2
d
dt
2
+ ((t) sin (t))2
d
dt
(c)
d
dr d
+ r 2 sin2
dt dt
dt
+ 2r cos sin
2
2
2
2r cos sin
dr
dt
+ r2
+
z = cos
therefore,
(b) Find a similar formula as in (a) for the arc length element in spherical coordinates.
= 2t, = ln t, = /6; 1
y = sin sin ,
2
2
2
dr d
d
+ r 2 cos2
dt dt
dt
d
dt
2
+
dz
dt
2
2
+ 2 sin2
11
+0=5
4 t2
5
5 dt = 4 5.
= 4 + 4t2
L=
1
Hence the answer.
6
d
dt
7
2
+ 2
d
dt
2
2
dt.
Problem 5 (20 points)
Your Score:
Problem 5 (20 points)
Your Score:
For the value at (0, 0), we use
2
2
2xy (x y )
Let f (x, y ) =
x2 + y 2
0
if (x, y ) = (0, 0),
fx (0, 0) = lim
x0
if (x, y ) = (0, 0).
f (x, 0) f (0, 0)
= 0 = fy (0, 0)
x
fxy (0, 0) = lim
(a) Is the function continuous at (0, 0)?
fx (0, y ) fx (0, 0)
2y (y )4
= 2
= lim
y 0 y (y )4
y
fyx (0, 0) = lim
fy (x, 0) fy (0, 0)
2x(x)4
= 2.
= lim
x0 x(x)4
x
y 0
(b) Calculate fx (x, y ), fy (x, y ), fxy (x, y ) and fyx (x, y ) at point (x, y ) = (0, 0). Also calculate these
derivatives (0, at 0).
x0
(c) Is fyx (x, y ) continuous at (0, 0)?
(d) Explain why fyx (0, 0) = fxy (0, 0).
(c) Along then line y = x, i.e.
Solution:
lim f (x, x) =
xo
(a) Examine
lim
(x,y )(0,0)
0
= 0.
x6
Along y = 0, x = 0, then
f (x, y )
lim f (x, 0) = 2.
Along x = 0, y = 0
or along x = 0, y = 0
x0
0
lim f (0, y ) = 2 = 0
y 0
y
0
lim f (x, 0) = 2 = 0.
x0
x
Let x = r cos (r ),
Dierent paths with dierent limits, hence, the limit does not exist.
(d) Observe that fyx (0, 0) = 2 and fxy (0, 0) = 2. This is because the partials f xy and fyx are
y = r sin (r ), then
not continuous at (0, 0).
f (r cos (r ), r sin (r )) =
2r 2 cos sin r 2 (cos2 sin2 )
r2
2
= 2r 2 cos sin (cos2 sin )
As (x, y ) (0, 0)
(For instance, fxy (x, x) = 0 while fxy (x, 0) = 2 for x = 0).
2r 2
r 0+ , i.e. f (x, y ) 0.
lim
(x,y )to(0,0)
f (x, y ) = 0 = f (0, 0)
f (x, y ) is continuous at (0, 0).
(b) Let f (x, y ) = (x2 y 2 )
fx =
4x2 y
2y (y 2 x2 )2
2
+y
(x2 + y 2 )2
x2
2
fy =
fxy =
2xy
, then
x2 + y 2
2
-1
-0.5
22
4xy
2x(x y )
+
x2 + y 2
(x2 + y 2 )2
x
0
0.5
2(x6 + 9x4 y 2 9x2 y 4 y 6 )
= fyx .
(x2 + y 2 )3
1
0.5
0.25
z
0
-0.25
-0.5
-1
-0.5
0
y
0.5
1
Surface of f (x, y )
8
9
Problem 6 (20 points)
Your Score:
Problem 6 (20 points)
Your Score:
(b) Let (x, y, z ) be the point on the given plane closest to 0, so the problem becomes: minimize
Find the distance from the origin to the plane x + 2y + 2z = 3,
s(x, y, z ) = x2 + y 2 + z 2 .
(a) using a geometric argument (no calculus),
Since x + 2y + 2z = 3, we have x = 3 2y 2z
(b) by reducing the problem to an unconstrained problem in two variables, and
s = s(y, z ) = (3 2y 2z )2 + y 2 + z 2
(c) using the method of Lagrange multipliers.
For critical points, sy = sz = 0
sy = 12 + 10y + 8z = 0
sz = 12 + 8y + 10z = 0
y=z=
2
,
3
x=
1
3
Solution:
The distance is 1 unit as in part (a).
(a) The point r0 = (3, 0, 0) is on the given plane
r1
n
d = r r0 |cos | n
(c) Same as in part (b), but now the problem becomes:
d
= |(r r0 ) n|
1
= (3, 0, 0) (1, 2, 2)
3
= 1.
Minimize s = x2 + y 2 + z 2 subject to x + 2y + 2z = 3 = g (x, y, z )
Using Lagrangian multipliers, to nd the critical points, we have
s= g
g (x, y, z ) = 3
r0
Alternatively, let (x, y, z ) be the point on the given plane closest to (0, 0, 0). The vector (1,2,2)
is normal to the plane, so must be parallel to the vector (x, y, z ) from 0 to (x, y, z ). Thus
(x, y, z ) = (1, 2, 2)
for some scalar .
2x =
2y = 2
2z = 2
x + 2y + 2z = 3
Since the point (x, y, z ) is on the given plane, i.e.
So the critical point is once again
t + 4t + 4t = 3
t=
1
3
122
,,
, whose distance from the origin is 1 unit.
333
1
(1, 2, 2)
3
1
1 + 4 + 4 = 1.
d=
3
(x, y, z ) =
10
y = z =
x=
2
= 2
3
11
Problem 7 (20 points)
Your Score:
Problem 8 (20 points)
(a) Find the equation of the tangent plane at the point (1, 1, 0) to the surface
(a) What condition must the constants a, b, and c satisfy to guarantee that
lim
(x,y )(0,0)
Your Score:
x2 2y 2 + z 3 = ez .
xy
ax2 + bxy + cy 2
(b) The temperature at a point (x, y ) on a metal plate in xy -plane is T (x, y ) = x 2 + y 3 degrees
exists. Prove your answer.
Celsius.
2
f (y 2 , xy, x2 ) in terms of partial derivatives of the function f .
(b) Find
yx
(i) Find the rate of change of temperature at (1, 1) in the direction of a = 2 i + j.
(ii) An ant at (1, 1) wants to walk in the direction in which the temperature decreases most
rapidly. Find a unit vector in that direction.
Solution:
(c) Let C be the curve x2/3 + y 2/3 = a2/3 on the xy -plane, nd the parametric equation of the
(a) Suppose (x, y ) (0, 0) along the ray y = kx, then
curve C . Hence nd the tangent line to the curve C at (a, 0).
k
xy
=
.
f (x, y ) = 2
ax + bxy + cy 2
a + bk + ck 2
Solution:
(a) Let F (x, y, z ) = x2 2y 2 + z 3 + ez = 0. This is a level surface in 3D and
Thus f (x, y ) has dierent constant values along dierent rays from the origin unless a = c = 0
1
f (x, y ) = does exist. If this condition
and b = 0. If this condition is satised, then
lim
b
(x,y )(0,0)
is NOT satised, then
lim
(x,y )(0,0)
F = (2x, 4y, 3z 2 ez ).
At (1, 1, 0),
f (x, y ) does not exist.
(1, 1, 0).
Alternatively by using polar, let x = r cos (r ) and x = r sin (r ), then
F = (2, 4, 1). This vector is normal to the level surface at the point
Tangent required plane is
cos sin
f (x, y ) =
.
a cos2 + b cos sin + c sin2
(x + 1, y 1, z 0) (2, 4, 1) = 0
2(x + 1) 4(y 1) z = 0
2x + 4y + z = 2.
Then similar arguement as above and we should end up with the some conclusion.
(b) Let u = u(y ) = y 2 , v = v (x, y ) = xy , w = w(x) = x2 ; then f = f (u, v, w)
f
f v
f dw
=
+
x
v x w dx
(b)
T (x, y ) = (2x, 3y 2 )
(i)
T (1, 1) = (2, 3),
f , fv and fw
= fv y + fw (2x)
= yfv 2xfw
u
(ii)
w
v
2f
=
(yfv 2xfw )
yx
y
y
= fv + y
= fv + y
fv du fv v
fw du fw v
+
+
2x
u dy
v y
u dy
v y
2x
= fv + y [fvu (2y ) + fvv x] 2x[fwu (2y ) + fwv x]
2
2
= fv + 2y fvu + xyfvv 4xyfwu 2x fwv ,
2
2
where all partials are evaluated at (y , xy, x ).
12
x
y
x
1
a = (2 i + j).
5
7
1
Da T = a T = (2 i + j) (2 i + 3 j) =
5
5
1
u = (2 i + 3 j),
13
becasue min(Du T ) = min( u
fw
y
fv
y
and
opposite to
T cos ) =
T
T (1, 1).
when = .
(c) Let x = a cos3 , y = a sin3 , then
r( ) = a cos3 i + a sin3 j
then
r ( ) = 3a cos2 sin i + 3a sin2 cos j.
At (a, 0), = 0 and r (0) = 0.
The curve is not dierentiable at (a, 0), hence, there is no tangent line.
13