white (taw933) – HW08 – benzvi – (55600)
1
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beFore answering.
001
10.0 points
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 12 Ft/sec, at what speed
is the top oF the ladder Falling when the Foot
oF the ladder is 12 Feet away From the base oF
the wall?
1.
speed = 15 Ft/sec
2.
speed = 16 Ft/sec
correct
3.
speed =
47
3
Ft/sec
4.
speed =
44
3
Ft/sec
5.
speed =
46
3
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 fgure
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 di±erentiation,
2
x
dx
dt
+ 2
y
dy
dt
= 0
.
In this case
dy
dt
=
−
x
y
dx
dt
.
But again by Pythagoras, iF
x
= 12, then
y
= 9.
Thus, iF the Foot oF the ladder is
moving away From the wall at a speed oF
dx
dt
= 12 Ft/sec
,
and
x
= 12, then the velocity oF the top oF the
ladder is given by
dy
dt
=
−
4
3
dx
dt
.
Consequently, the speed at which the top oF
the ladder is Falling is
speed =
v
v
v
dy
dt
v
v
v
= 16 Ft/sec
.
keywords: speed, ladder, related rates
002
10.0 points
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
equal to the rate oF change oF
C
with respect
to
t
?
1.
r
= 1
correct
2.
r
= 2
3.
r
= 2
π
4.
r
=
1
2
5.
r
=
π
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2
6.
r
=
π
2
Explanation:
Diferentiating
A
=
πr
2
,
C
= 2
implicitly with respect to
t
we see that
dA
dt
= 2
dr
dt
,
dC
dt
= 2
π
dr
dt
.
Thus the rate oF change,
dA/dt
, oF area is
equal to the rate oF change,
dC/dt
, oF circum
Ference when
dA
dt
=
dC
dt
,
i.e.
, when
2
dr
dt
= 2
π
dr
dt
.
This happens when
r
= 1
.
003
10.0 points
Determine the value oF
dy/dt
at
x
= 3 when
y
=
x
2
−
2
x
and
dx/dt
= 3.
1.
dy
dt
v
v
v
x
=3
= 16
2.
dy
dt
v
v
v
x
=3
= 20
3.
dy
dt
v
v
v
x
=3
= 14
4.
dy
dt
v
v
v
x
=3
= 18
5.
dy
dt
v
v
v
x
=3
= 12
correct
Explanation:
Diferentiating implicitly with respect to
t
we see that
dy
dt
= (2
x
−
2)
dx
dt
= 3 (2
x
−
2)
.
At
x
= 3, thereFore,
dy
dt
= 3(4) = 12
.
004
10.0 points
A point is moving on the graph oF
xy
= 2.
When the point is at (3
,
2
3
), its
x
coordinate
is increasing at a rate oF 6 units per second.
What is the speed oF the
y
coordinate at
that moment and in which direction is it mov
ing?
1.
speed =
−
10
3
units/sec, increasing
y
2.
speed =
4
3
units/sec, decreasing
y
correct
3.
speed =
7
3
units/sec, increasing
y
4.
speed =
−
7
3
units/sec, decreasing
y
5.
speed =
−
4
3
units/sec, decreasing
y
6.
speed =
10
3
units/sec, increasing
y
Explanation:
Provided
x, y
n
= 0, the equation
xy
= 2
can be written as
y
= 2
/x
. Diferentiating
implicitly with respect to
t
we thus see that
dy
dt
=
−
2
x
2
dx
dt
.
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 Spring '08
 schultz
 Derivative, Differential Calculus, Trigraph, dt, Inch

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