Prelim 3
November 27, 2001
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Your reasoning is important.
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You are allowed to
use a 5 by 8 index card. The number of points each problem is worth is indicated in parantheses. Keep in
mind there may be simpler methods than direct applications of calculus. Good luck.
1.
(16 points) Find the directional derivative of
f
(
x, y
) = 2
x
2
+ 3
xy
−
y
2
at the point (2
,
3) in the same
direction as
v
=
i
+ 3
j
.
SOLUTION:
Step 1
Find the direction.
The direction of
v
is
1
√
10
i
+
3
√
10
j
.
Step 2
Find the gradient of
f
.
∇
f
(
x, y
) =
4
x
+ 3
y,
3
x
−
2
y
∇
f
(2
,
3) =
17
,
0
Step 3
Find the directional derivative.
D
v
f
(2
,
3) =
∇
f
(2
,
3)
·
1
√
10
,
3
√
10
=
17
√
10
2.
(16 points) Find the centroid (ie. center of mass with uniform density) of the region bounded by
y
= 1
−
x
2
and the
x
−
axis.
SOLUTION:
Step 1
Find the area of the region.
The formula for the area of a parabola is Area =
2
3
·
base
·
height. Our parabola has a base of 2 and a height
of 1 so its area is
4
3
.
Step 2
Find the centroid.
Since our region is symmetric about the
y
axis we must have that the centroid lies on the
y
axis. So ¯
x
= 0.
Now we need to find ¯
y
.
M
x
=
1

1
1

x
2
0
y dydx
=
1

1
(1
−
x
2
)
2
2
dx
=
1
0
(1
−
x
2
)
2
dx
=
1
0
1
−
2
x
2
+
x
4
dx
= 1
−
2
3
+
1
5
=
8
15
So ¯
y
=
M
x
M
=
8
15
·
3
4
=
2
5
.
So the centroid is (¯
x,
¯
y
) =
0
,
2
5
.
3.
Let
g
(
x, y
) =
x
2
+ 2
xy
+ 4
y
2
, where
x
= 2 cos
θ
and
y
= sin
θ
.
a.
(8 points) Find
dg
dθ
.
b.
(8 points) Find the maximum and minimum values of
g
as a function of
θ
.
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SOLUTION:
a.
If
x
= 2 cos
θ
and
y
= sin
θ
then we have
g
(
x, y
) =
g
(
θ
) = 4 cos
2
θ
+ 2(2 cos
θ
)(sin
θ
) + 4 sin
2
θ
= 4 sin
θ
cos
θ
+ 4 = 2 sin(2
θ
) + 4
Which gives us that
dg
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