MATH 253 Honors
EXAM 1
Fall 1999
Sections 201-202
Solutions
P. Yasskin
Multiple Choice: (8 points each)
1. Find the volume of the parallelepiped with edges a
c Ÿ0, 1, "2 .
a. "14
b. "12
c. 2
d. 12
e. 14
correctchoice
2
a
b • c
4
0
"1 0
3
0
"14
V
a
Ÿ2, 4, 0 ,
|"14|
b • c
b
Ÿ"1, 0, 3 and
14
1 "2
x 4 y " 1 z 2 intersects the plane
4
3
2
Ÿ4 " 2s t, 3 2s, 6 " t we have 2x y z correctchoice
2. At the point where the line
Ÿx, y, z a. 15
b.
c.
d.
e.
13
11
9
7
Substitute the plane x 4 " 2s t, y 3 2s, z 6 " t into the line:
8 " 2s t 1 s 2 " t
4 " 2s t 4 3 2s " 1 6 " t 2
®
4
3
3
2
4
From the second and third parts:
s 1" t
4
t
t
8"2 1"
t
4
2"
From the first and third parts:
3
4
4 8 " 2 1 " t t 3Ÿ8 " t Solve for t:
Multiply by 12:
4
32 " 2Ÿ4 " t 4t 24 " 3t
®
24 6t 24 " 3t
®
9t 0
®
Substitute back:
s 1,
x 4 " 2 2,
y 3 2 5,
z 6,
2x y z 15
t
1
0
3. Find the equation of the plane which is perpendicular to the line through P
and Q Ÿ3, 7, 1 and is equidistant from P and Q.
a. x 2y " 2z 12
correctchoice
b. x 2y " 2z 6
c. x 2y " 2z "6
d. x 2y " 2z "3
e. 2x 4y " 4z 24
The plane passes through the midpoint R
PQ
2
PQ Q " P Ÿ2, 4, "4 .
Its normal is N
So its equation is N
X N
R or 2x 4y " 4z
x 2y " 2z 6
Ÿ1, 3, 5 Ÿ2, 5, 3 .
2245"43
12 or
T 4P
Problems 4 – 6: The temperature in an ideal gas is given by
where 4 is a
>
constant, P is the pressure and > is the density. The pressure, density and temperature are
all functions of position. At a certain point
Q Ÿ1, 2, 3 ,
we have
PŸQ 4
PŸQ Ÿ"3, 2, 1 >ŸQ 2
>ŸQ Ÿ3, "1, 2 4. Use the linear approximation to estimate the pressure at the point R
a. 2.9
b. 3.1
correctchoice
c. 3.3
d. 4.7
e. 7.3
Ÿ1.2, 1.9, 3.1 .
The linear approximation says:
PŸx, y, z X PŸa, b, c P Ÿx " a P Ÿy " b P Ÿz " c y
z
x
Here
Ÿx, y, z R Ÿ1.2, 1.9, 3.1 ,
Ÿa, b, c Q Ÿ1, 2, 3 ,
P , P , P PŸQ Ÿ"3, 2, 1 and
x y z
So
PŸR X 4 " 3Ÿ1.2 " 1 2Ÿ1.9 " 2 1Ÿ3.1 " 3 4 " 3Ÿ.2 2Ÿ".1 1Ÿ.1 3.3
2
5. At the point Q, the temperature is
a. 4Ÿ"4.5, 0, 2.5 b. 4Ÿ1.5, 0, 2.5 c. 4Ÿ1.5, 2, "4.5 d. 4Ÿ"4.5, 2, "1.5 T ŸQ 24
and its gradient is
T ŸQ correctchoice
e. 4Ÿ"1.5, 2, 2.5 Apply chain rule with T 4 P
> and P PŸx, y, z and > >Ÿx, y, z :
T T P T > 4 P " 4 P >
and similarly for T and T
> x
y
y
P x
> x
x
> 2 x
These fit together as the vector gradient:
4
> 4 Ÿ"3, 2, 1 " 442 Ÿ3, "1, 2 4 " 3 " 3, 1 1, 1 " 2
P " 4P2 T >
2
2
2
2
>
4Ÿ"4.5, 2, "1.5 6. If a fly is located at the point Q and travelling with velocity v
density changing at the location of the fly?
d>
a.
ŸQ 37
dt
d>
correctchoice
b.
ŸQ 29
dt
d>
c.
ŸQ 21
dt
d>
d.
ŸQ 2.1
dt
d>
e.
ŸQ .21
dt
Chain rule:
d>
> dx
dt
x dt
7.
> dy
y dt
> dz
z dt
Ÿ3, 4, 12 ,
> Ÿ3, 4, 12 Ÿ3, "1, 2 v
how fast is the
9 " 4 24
29
The graph at the right is the
contour plot of which function?
a. y 2 " x 2
b. xy
c. x 2
y2
d. y " x 2
correctchoice
e. x " y 2
The contour plot is the level curves y " x 2
parabolas opening up.
C for various values of C. These are
3
8. (12 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
v
Ÿx,y Ÿ0,0 2x 2 y
x4 y2
First check the straight lines y mx:
2x 2 y
2x 2 mx lim 2mx
lim
lim
4
2
4
xv0 x m 2 x 2
xv0 x 2 m 2
Ÿx,y vŸ0,0 x y
ymx
0
So all straight lines give 0. Now check the parabola y
2x 2 y
2x 2 x 2 lim 2 1
lim
lim
4
2
x
xv0 2
v0 x 4 x 4
Ÿx,y vŸ0,0 x y
2
x2:
yx
This is different. So the limit does not exist.
b.
lim
v
Ÿx,y Ÿ0,0 2x 2 y
x2 y2
First check the straight lines y mx:
2x 2 y
2x 2 mx lim 2mx
lim
lim
2
2
2
xv0 x m 2 x 2
xv0 1 m 2
Ÿx,y vŸ0,0 x y
ymx
Now check the parabola y mx 2
2x 2 y
2x 2 mx 2 lim 2mx 2
lim
lim
2
2
xv0 x 2 m 2 x 4
xv0 1 m 2 x 2
Ÿx,y vŸ0,0 x y
0
0
ymx 2
x r cos 2
So try polar:
y r sin 2
3
2
2x 2 y
2r cos 2 sin 2 lim 2r cos 2 2 sin 2
lim
lim
2
2
rv0
rv0
Ÿx,y vŸ0,0 x y
r2
2
since cos 2 sin 2 is bounded and r v 0.
0
4
9. (16 points) Consider the parametric curve rŸt Ÿt, 4e
t/2
, 2e t .
a. Compute the velocity and acceleration:
v Ÿ1, 2e t/2 , 2e t a Ÿ0, e t/2 , 2e t b. Find a parametric equation for the line tangent to the curve at t
rŸ0 Ÿ0, 4, 2 0.
vŸ0 Ÿ1, 2, 2 X Ÿx, y, z rŸ0 tvŸ0 Ÿ0, 4, 2 tŸ1, 2, 2 Ÿt, 4 2t, 2 2t or
x t, y 4 2t, z 2 2t
c. Find a parametric equation for the plane instantaneously containing the curve at
t 0. This is the plane containing the velocity and the acceleration.
rŸ0 Ÿ0, 4, 2 aŸ0 Ÿ0, 1, 2 vŸ0 Ÿ1, 2, 2 Ÿ0 Ÿ0, 4, 2 tŸ1, 2, 2 sŸ0, 1, 2 Ÿx, y, z rŸ0 tvŸ0 sa
X
Ÿt, 4 or
x
2t s, 2 2t 2s t, y 4 2t s, z
2 2t 2s
d. Find a non-parametric equation for the plane instantaneously containing the curve
at t
0.
The normal to the plane is
î j k
N
vŸ0 • aŸ0 Ÿ1, 2, 2 • Ÿ0, 1, 2 iŸ4
1 2 2
" 2 " jŸ2 kŸ1 Ÿ2, "2, 1 0 1 2
The point is
rŸ0 Ÿ0, 4, 2 .
So the plane is
N
X
2x " 2y z 2 0 " 2 4 2
or
2x " 2y z "6.
e. Find the arclength of the curve between t
0 and t
2.
2e 2 ¢ " ¡0 2 ¢
N
r Ÿ0 or
HINT: Factor inside the square root.
|v|
1 4e t
2
L
; |v|dt
0
4e 2t
Ÿ1 2e t 2
2
; 1 2e t dt
t 2e t
0
1 2e t
2
0
¡2 2e 2
10. (16 points) Find the equation of the plane tangent to the graph of the function
fŸx, y 3xe y " 2x 2 e 2y at the point Ÿx, y Ÿ1, ln 2 .
fŸx, y f x Ÿx, y f y Ÿx, y z
z
"2
fŸ1, ln 2 3e y " 4xe 2y
f x Ÿ1, ln 2 32"44
"10
3xe " 4x e
f y Ÿ1, ln 2 32"44
"10
y
2 2y
" 1 f y Ÿ1, ln 2 Ÿy " ln 2 "10x " 10y 8 10 ln 2
fŸ1, ln 2 f x Ÿ1, ln 2 Ÿx
32"24
3xe y " 2x 2 e 2y
"2 " 10Ÿx " 1 " 10Ÿy " ln 2 5