roberts (eer474) – Quest HW 7 – seckin – (56425)
1
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17
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before answering.
001
10.0 points
A storm system is headed towards Austin.
The Weather Service Office forecasts that
t
hours after midday the storm will be at a
distance of
s
(
t
) = 4 + 3
t

t
2
miles from Austin.
At what speed (in miles
per hour) will the storm be moving when it
reaches Austin?
1.
storm speed = 5 mph
correct
2.
storm speed =

7 mph
3.
storm speed =

6 mph
4.
storm speed = 6 mph
5.
storm speed = 7 mph
6.
storm speed =

5 mph
Explanation:
The storm will reach Austin when
s
(
t
) = (4

t
)(1 +
t
) = 0
,
i.e.
, when
t
= 4.
Now the velocity of the
storm is given by
s
′
(
t
) = 3

2
t,
the negative sign indicating that the storm is
moving towards Austin. Thus at
t
= 4,
s
′
parenleftBig
4
parenrightBig
= 3

8 =

5
.
Consequently,
since
speed
is
always
non
negative,
storm speed = 5 mph
.
002
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 10 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 =
37
3
ft/sec
2.
speed = 13 ft/sec
3.
speed = 12 ft/sec
4.
speed =
40
3
ft/sec
correct
5.
speed =
38
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 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
.
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roberts (eer474) – Quest HW 7 – seckin – (56425)
2
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
= 10 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 =
vextendsingle
vextendsingle
vextendsingle
dy
dt
vextendsingle
vextendsingle
vextendsingle
=
40
3
ft/sec
.
keywords: speed, ladder, related rates
003
(part 1 of 3) 10.0 points
A Calculus student leaves the RLM build
ing and walks in a straight line to the PCL Li
brary. His distance (in multiples of 40 yards)
from RLM after
t
minutes is given by the
graph
1
0
1
2
3
4
5
6
7
8
9
10
11
2
4
6
8
10
2
4
6
8
10
t
distance
(i) What is his speed after 3 minutes, and in
what direction is he heading at that time?
1.
towards RLM at 40 yds/min
2.
away from RLM at 60 yds/min
3.
away from RLM at 80 yds/min
correct
4.
away from RLM at 40 yds/min
5.
towards RLM at 80 yds/min
Explanation:
The graph is linear and has positive slope
on [2
,
4], so the speed of the student at time
t
= 3 coincides with the slope of the line on
[2
,
4]. Hence
speed = 40
·
8

4
4

2
=
80 yds/min
.
As the distance from RLM is increasing on
[2
,
4] the student is thus moving away from
the RLM.
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 Spring '09
 Gualdani
 Derivative, Differential Calculus, Trigraph, Imperial units, Inch

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