wei (jw35975) – HW10 – milburn – (54685)
1
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001
10.0points
A circle of radius
r
has area
A
and circum
ference
C
are given respectively by
A
=
πr
2
,
C
= 2
πr .
If
r
varies with time
t
, for what value of
r
is
the rate of change of
A
with respect to
t
half
the rate of change of
C
with respect to
t
?
1.
r
= 1
2.
r
=
1
2
correct
3.
r
=
π
4.
r
=
π
2
5.
r
= 2
6.
r
= 2
π
Explanation:
Differentiating
A
=
πr
2
,
C
= 2
πr
implicitly with respect to
t
we see that
dA
dt
= 2
πr
dr
dt
,
dC
dt
= 2
π
dr
dt
.
Thus the rate of change,
dA/dt
, of area is half
the rate of change,
dC/dt
, of circumference
when
dA
dt
=
1
2
dC
dt
,
i.e.
, when
2
πr
dr
dt
=
1
2
parenleftBig
2
π
dr
dt
parenrightBig
.
This happens when
r
=
1
2
.
002
10.0points
A point is moving on the graph of
3
x
3
+ 5
y
3
=
xy.
When the point is at
P
=
parenleftBig
1
8
,
1
8
parenrightBig
,
its
y
coordinate is increasing at a speed of 7
units per second.
What is the speed of the
x
coordinate at
that time and in which direction is the
x

coordinate moving?
1.
speed = 48 units/sec
,
decreasing
x
2.
speed = 51 units/sec
,
increasing
x
3.
speed = 48 units/sec
,
increasing
x
4.
speed
=
49 units/sec
,
decreasing
x
correct
5.
speed = 50 units/sec
,
increasing
x
6.
speed = 51 units/sec
,
decreasing
x
7.
speed = 49 units/sec
,
increasing
x
8.
speed = 50 units/sec
,
decreasing
x
Explanation:
Differentiating
3
x
3
+ 5
y
3
=
xy
implicitly with respect to
t
we see that
9
x
2
dx
dt
+ 15
y
2
dy
dt
=
y
dx
dt
+
x
dy
dt
.
Thus
dx
dt
=
parenleftBig
x

15
y
2
9
x
2

y
parenrightBig
dy
dt
.
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wei (jw35975) – HW10 – milburn – (54685)
2
Now at
P
,
x

15
y
2
=

7
64
,
while
9
x
2

y
=
1
64
.
Hence, at
P
,
dx
dt
=

7
dy
dt
.
When the
y
coordinate at
P
is increasing at a
rate of 7 units per second, therefore,
dx
dt
=

49
.
Consequently, the
x
coordinate is moving at
speed
= 49 units/sec
,
and the negative sign indicates that it is mov
ing in the direction of decreasing
x
.
003
10.0points
If the radius of a melting snowball decreases
at a rate of 2 ins/min, find the rate at which
the volume is decreasing when the snowball
has diameter 2 inches.
1.
rate = 10
π
cu.ins/min
2.
rate = 7
π
cu.ins/min
3.
rate = 8
π
cu.ins/min
correct
4.
rate = 9
π
cu.ins/min
5.
rate = 11
π
cu.ins/min
Explanation:
The volume,
V
, of a sphere of radius
r
is
given by
V
=
4
3
πr
3
.
Thus by implicit differentiation,
dV
dt
= 4
πr
2
dr
dt
=

8
πr
2
,
since
dr/dt
=

2 ins/min.
When the snow
ball has diameter 2 inches, therefore, its ra
dius
r
= 1 and
dV
dt
=

8 (1)
π .
Consequently, when the snowball has diam
eter 2 inches, the volume of the snowball is
decreasing at a
rate = 8
π
cu.ins/min
.
004
10.0points
A street light is on top of a 12 foot pole.
Joe, who is 3 feet tall, walks away from the
pole at a rate of 3 feet per second.
At what
speed is the tip of Joe’s shadow moving from
the base of the pole when he is 12 feet from
the pole?
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 Fall '11
 MILBURN
 Calculus, Derivative, dt, Inch, Wei

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