cadena (jc59484) – HW09 – benzvi – (55400)
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001
10.0points
A 15 foot ladder is leaning against a wall.
If the foot of the ladder is sliding away from
the wall at a rate of 6 ft/sec, at what speed is
the top of the ladder falling when the foot of
the ladder is 9 feet away from the base of the
wall?
1.
speed = 4 ft/sec
2.
speed =
9
2
ft/sec
correct
3.
speed =
15
4
ft/sec
4.
speed =
17
4
ft/sec
5.
speed =
19
4
ft/sec
Explanation:
Let
y
be the height of the ladder when the
foot of the ladder is
x
feet from the base of
the wall as shown in figure
x
ft.
15 ft.
We have to express
dy/dt
in terms of
x, y
and
dx/dt
. But by Pythagoras’ theorem,
x
2
+
y
2
= 225
,
so by implicit differentiation,
2
x
dx
dt
+ 2
y
dy
dt
= 0
.
In this case
dy
dt
=

x
y
dx
dt
.
But again by Pythagoras, if
x
= 9, then
y
=
12. Thus, if the foot of the ladder is moving
away from the wall at a speed of
dx
dt
= 6 ft/sec
,
and
x
= 9, then the velocity of the top of the
ladder is given by
dy
dt
=

3
4
dx
dt
.
Consequently, the speed at which the top of
the ladder is falling is
speed =
vextendsingle
vextendsingle
vextendsingle
dy
dt
vextendsingle
vextendsingle
vextendsingle
=
9
2
ft/sec
.
keywords: speed, ladder, related rates
002
10.0points
The radius of a circle is increasing at a
constant rate of 5 ft/sec.
Express the rate at which the area of the
circle is changing in terms of the circumfer
ence,
C
of the circle.
1.
rate = 5
πC
sq. ft./sec
2.
rate =
5
2
C
sq. ft./sec
3.
rate = 10
πC
sq. ft./sec
4.
rate = 10
C
sq. ft./sec
5.
rate = 5
C
sq. ft./sec
correct
6.
rate =
5
2
πC
sq. ft./sec
Explanation:
cadena (jc59484) – HW09 – benzvi – (55400)
2
The area and circumference of a circle of
radius
r
are given by
A
=
πr
2
.
C
= 2
πr
respectively. Thus
dA
dt
= 2
πr
dr
dt
=
C
dr
dt
.
When
dr/dt
= 5 ft/sec, therefore,
dA
dt
= 5
C
sq. ft./sec
.
003
10.0points
Determine the value of
dy/dt
at
x
= 2 when
y
=
x
2

5
x
and
dx/dt
= 3.
1.
dy
dt
vextendsingle
vextendsingle
vextendsingle
x
=2
= 5
2.
dy
dt
vextendsingle
vextendsingle
vextendsingle
x
=2
=

1
3.
dy
dt
vextendsingle
vextendsingle
vextendsingle
x
=2
= 1
4.
dy
dt
vextendsingle
vextendsingle
vextendsingle
x
=2
=

3
correct
5.
dy
dt
vextendsingle
vextendsingle
vextendsingle
x
=2
= 3
Explanation:
Differentiating implicitly with respect to
t
we see that
dy
dt
= (2
x

5)
dx
dt
= 3 (2
x

5)
.
At
x
= 2, therefore,
dy
dt
= 3(

1) =

3
.
004(part1of2)10.0points
A point is moving on the graph of
3
x
3
+ 2
y
3
=
xy .
When the point is at
P
=
parenleftBig
1
5
,
1
5
parenrightBig
,
its
x
coordinate is decreasing at a speed of 4
units per second.
What is the speed of the
y
coordinate at
that time?
1.
speed
y
coord = 15 units/sec
2.
speed
y
coord = 16 units/sec
correct
3.
speed
y
coord =

17 units/sec
4.
speed
y
coord = 17 units/sec
5.
speed
y
coord =

16 units/sec
Explanation:
Differentiating
3
x
3
+ 2
y
3
=
xy
implicitly with respect to
t
we see that
9
x
2
dx
dt
+ 6
y
2
dy
dt
=
y
dx
dt
+
x
dy
dt
.
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 Accounting, Critical Point, Mathematical analysis, dt