Preliminary Exam II – Solutions
1
. Find
dy
dx
by implicit differentiation given that
cos(
πy
)

3 sin(
πx
) = 1
.
Solution: Differentiation yields

π
sin(
πy
)
dy
dx

3
π
cos(
πx
) = 0
hence
π
sin(
πy
)
dy
dx
=

3 cos(
πx
)
sin(
πy
)
.
2
. An object is moving along the circle
x
2
+
y
2
= 16. If
dy
dt
= 2 when it reaches
the point with coordinates
x
= 3 and
y
= 1, find
dx
dt
at that point.
Solution: Differentiating with respect to the variable
t
yields:
2
x
dx
dt
+ 2
y
dy
dt
= 0
.
Substituting the given information yields:
3
dx
dt
+ 2 = 0
.
Thus
dx
dt
=

2
3
.
3
. Find the absolute extrema of the function
f
(
x
) =
x
3

3
2
x
2
+ 5
on the closed interval [

1
,
2].
Solution: Differentiation yields
f
0
(
x
) = 3
x
2

3
x
= 3
x
(
x

1)
hence the critical numbers are
x
= 0
,
1. The candidates are

2
,
0
,
1
,
2 and 4
f
(

1) = (

1)
3

3
2
(

1)
2
+ 5 =

1

3
2
+ 5 =
5
2
f
(0) = 5
f
(1) = 1

3
2
+ 5 =
9
2
f
(2) = 2
3

3
2
2
2
+ 5 = 8

6 + 5 = 7
.
Thus (a).
max =
7
min =
5
/
2
1
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2
4
. Find all of the critical numbers of the function
f
(
x
) =
x
2
+ 1
x
.
Solution: Differentiation yields
f
0
(
x
) =
x
2

1
x
2
.
The critical numbers are obtained by setting the numerator equals zero as well as
setting the denominator equals. Thus the critical numbers are
x
= 0
,
1
,

1.
5
. Compute
d
2
y
dx
2
for the curve
x
2

y
2
= 25
.
First we compute
dy/dx
:
2
x

2
y
dy
dx
= 0
hence
dy
dx
=
x
y
.
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 Spring '09
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 Calculus, Derivative, Convex function, 1 g, 3 seconds, 0 2 G, 0 8 1 g

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