MATH 253 Honors
EXAM 1
Sections 201-203
Solutions
Spring 1999
P. Yasskin
Multiple Choice: (5 points each)
Problems 1 – 4: Consider the vectors:
c Ÿ1, 1, 1 .
a Ÿ2, 0, "2 ,
b Ÿ0, "3, 3 and
1. The angle between a and b is
(
a. 30
b. 45 (
c. 90 (
d. 120 (
correctchoice
e. 150 (
cos 2 a b a b
"6 8 18
" 12
2 120 (
b is
2. The vector projection of a along correctchoice
a. Ÿ0, 1, "1 b. Ÿ0, "1, 1 c. Ÿ"2, "3, 5 d. 0, 3 2 , "3 2
e.
a
proj b
0, "3 2 , 3 2
a b2 b "6 Ÿ0, "3, 3 Ÿ0, 1, "1 18
b
a and 3. The area of a triangle with b as two sides is
3
a.
b. 6
correctchoice
c. 3 3
d. 6 3
e. 54
a • b i j
k
2 0
"2
0
"3 3
iŸ"6 " jŸ6 kŸ"6 Ÿ"6, "6, "6 A 1 a•
b 3 3
2
1
4. The volume of the parallelepiped with edges a, b and c is
a. "18
b. "6
c. 3
d. 6
correctchoice
e. 18
a•
b c |Ÿ"6, "6, "6 Ÿ1, 1, 1 | |"18| 18
V Volume cannot be negative!
Problems 5 – 7: The pressure in an ideal gas is given by P k>T where k is a constant, >
is the density and T is the temperature. The pressure, density and temperature are all
functions of position. At the point Q Ÿ1, 2, 3 , the density is >ŸQ 1.5 and its gradient is
>ŸQ Ÿ.2, .3, ".1 . Also at that point, the temperature is TŸQ 24 and its gradient is
TŸQ Ÿ"3, 1, 2 .
5. At the point Q, the pressure is PŸQ 36k. What is the gradient of the pressure?
PŸQ kŸ.7, ".1, 1.1 a. PŸQ kŸ.3, 8.7, .6 correctchoice
b. PŸQ kŸ.3, "8.7, .6 c. PŸQ kŸ"2.8, 1.3, 1.9 d. PŸQ kŸ"2.8, "1.3, 1.9 e. Apply the product rule to P k>T: (Or do each partial derivative separately.)
PŸQ k T >ŸQ k > TŸQ k24Ÿ.2, .3, ".1 k1.5Ÿ"3, 1, 2 kŸ4.8, 7.2, "2.4 kŸ"4.5, 1.5, 3 kŸ.3, 8.7, .6 6. If a fly is located at the point Q, in what direction should the fly travel to cool off as soon
as possible?
a. Ÿ".2, ".3, .1 correctchoice
b. Ÿ3, "1, "2 c. Ÿ"3, 1, 2 d. Ÿ2, "1, 3 e. Ÿ2, 1, 3 T ŸQ The temperature decreases fastest in the direction of "
"Ÿ"3, 1, 2 Ÿ3, "1, "2 2
7. If a fly is located at the point Q and travelling with velocity v Ÿ3, 4, 12 , how fast is the
density changing at the location of the fly?
d>
a.
ŸQ "7.8
dt
d>
b.
ŸQ ".6
dt
d>
.6
c.
ŸQ dt
13
d>
correctchoice
d.
ŸQ .6
dt
d>
e.
ŸQ 7.8
dt
d>
>ŸQ Ÿ3, 4, 12 Ÿ.2, .3, ".1 .6 1.2 " 1.2 .6
v
ŸQ dt
8.
The graph at the right is the
contour plot of which function?
a. y 2
" x2
correctchoice
b. xy
c. x 2 y 2
d. y " x 2
e. x " y 2
y2
" x2 C
is a hyperbola which opens up and down if C 0 and left and right if C 0.
3
9. (24 points) Consider the parametric curve rŸt Ÿt 3 , 3t 2 , 6t .
a. Compute the velocity and acceleration:
v Ÿ3t 2 , 6t, 6 a Ÿ6t, 6, 0 b. Find a parametric equation for the line tangent to the curve at t 1.
X P t v
X Ÿx, y, z P rŸ1 Ÿ1, 3, 6 Ÿx, y, z Ÿ1, 3, 6 t Ÿ3, 6, 6 Ÿ1 3t, 3 6t, 6 6t vŸ1 Ÿ3, 6, 6 c. Find a non-parametric (symmetric) equation for the line tangent to the curve at
t 1.
x 1 3t
y 3 6t
z 6 6t
3
y
"
6
z"
t x"1 6
3
6
d. Find a parametric equation for the plane instantaneously containing the curve at
t 1.
X P t v s a
aŸ1 Ÿ6, 6, 0 vŸ1 Ÿ3, 6, 6 X Ÿx, y, z P rŸ1 Ÿ1, 3, 6 Ÿx, y, z Ÿ1, 3, 6 t Ÿ3, 6, 6 sŸ6, 6, 0 Ÿ1 3t 6s, 3 6t 6s, 6 6t e. Find a non-parametric equation for the plane instantaneously containing the curve
at t 1.
v • a N
i j k
3 6 6
iŸ"36 " jŸ"36 kŸ18 " 36 Ÿ"36, 36, "18 6 6 0
X N
P
N
" 36x 36y " 18z "36 1 36 3 " 18 6 "36
2x " 2y z 2
f. Find the arclength of the curve between t 0 and t 2.
|v| |Ÿ3t 2 , 6t, 6 | 2
2
0
0
9t 4 36t 2 36 L ; |v|dt ; 3t 2 6 dt t 3 6t
9Ÿt 2 2 2 3Ÿt 2 2 3t 2 6
2
0
8 12 20
4
10. (10 points) Does each limit exist? Why or why not? Find the value of the one that
exists.
(Up to 4 points extra credit for a good explanation.)
a.
lim
Ÿx,y vŸ0,0 2xy
x 2 2y 2
Look at the path y mx :
2xy
2m
lim
lim 2 2xmx2 2 lim 2m 2 ymx
2
2
xv0 x 2m x
xv0 1 2m
1 2m 2
xv0 x 2y
This depends on direction. So the limit does not exist.
b.
lim
Ÿx,y vŸ0,0 2xy
x 2y 2
2
Look at the path y mx :
2xy
2xmx
lim
lim
lim 2xm
0
ymx
xv0
xv0
2
2
2
2
2
x v0
1 2m 2
x 2m x
x 2y
This is independent of direction. So we expect the limit exists.
To prove it, we look at the path Ÿx, y Ÿr cos 2Ÿr , r sin 2Ÿr lim
Ÿx,y vŸ0,0 2xy
x 2 2y 2
lim
rv0
2r 2 cos 2 sin 2
cos 2 sin 2
lim 2r
rv0
2
2 cos 2
2
cos 2 2 2 sin 2 2
2 2r sin 2
r
lim 2r cos 2 sin 2 0
rv0
1 sin 2 2
because cos 2 sin 2 is bounded since
1 sin 2 2
i. the numerator is bounded:
|cos 2 sin 2 | 1
and
2
1 sin 2 1
ii. the denominator cannot go to zero:
5
11. (13 points) Find the equation of the plane tangent to the graph of the function
fŸx, y 3x sin y " 2y cos x at the point Ÿx, y 0, = .
2
f 0, = 0 " 2 = cos 0 "=
2
2
f x 0, = 3 sin = 0 3
2
2
f y 0, = 0 " 2 cos 0 "2
2
fŸx, y 3x sin y " 2y cos x
f x Ÿx, y 3 sin y 2y sin x
f y Ÿx, y 3x cos y " 2 cos x
z f 0, =
2
f x 0, =
2
Ÿx
" 0 f y 0,
"= 3Ÿx " 2 y " =
2
z " = 3Ÿ x " 2 y " =
2
=
2
y" =
2
"= 3x " 2y = 3x " 2y
or
z 3x " 2y
12. (13 points) Find the equation of the plane tangent to the surface
FŸx, y, z x 2 y y 3 z z 4 x 29 at the point P Ÿx, y, z Ÿ3, 2, 1 .
F Ÿ2xy z 4 , x 2 3y 2 z, y 3 4z 3 x F
N
Ÿ3,2,1 X N
P
N
Ÿ2 3 2 1 4 , 3 2 3 2 2 1, 2 3 4 1 3 3 Ÿ13, 21, 20 Ÿ13, 21, 20 Ÿx, y, z Ÿ13, 21, 20 Ÿ3, 2, 1 39 42 20 101
13x 21y 20z 101
6