Mahon, Kevin – Homework 8 – Due: Oct 20 2005, 3:00 am – Inst: Edward Odell
1
This
printout
should
have
20
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
A rock is thrown into a still pond and causes
a circular ripple. If the radius of the ripple is
increasing at a rate of 4 ft/sec, at what speed
is the area of the ripple increasing when its
radius is 6 feet?
1.
speed = 47
π
sq. ft/sec
2.
speed = 49 sq. ft/sec
3.
speed = 45 sq. ft/sec
4.
speed = 46
π
sq. ft/sec
5.
speed = 45
π
sq. ft/sec
6.
speed = 47 sq. ft/sec
7.
speed = 48 sq. ft/sec
8.
speed = 48
π
sq. ft/sec
correct
Explanation:
The area,
A
, of a circle having radius
r
is
given by
A
=
πr
2
.
Differentiating implicitly
with respect to
t
we thus see that
dA
dt
= 2
πr
dr
dt
.
When
r
= 6
,
dr
dt
= 4
,
therefore, the speed at which the area of the
ripple is increasing is given by
speed = 48
π
sq. ft/sec
.
keywords: Stewart5e,
002
(part 1 of 1) 10 points
A point is moving on the graph of
xy
= 5.
When the point is at (3
,
5
3
), its
x
coordinate
is increasing at a rate of 1 units per second.
What is the speed of the
y
coordinate at that
moment and in which direction is it moving?
1.
speed =

5
9
units/sec, decreasing
y
2.
speed =
14
9
units/sec, increasing
y
3.
speed
=
5
9
units/sec, decreasing
y
correct
4.
speed =

14
9
units/sec, decreasing
y
5.
speed =
23
9
units/sec, increasing
y
Explanation:
Provided
x, y
6
= 0, the equation
xy
= 5
can be written as
y
= 5
/x
.
Differentiating
implicitly with respect to
t
we thus see that
dy
dt
=

5
x
2
dx
dt
.
whenever
x
6
= 0. When
x
= 3
,
dx
dt
= 1
,
therefore, the corresponding rate of change of
the
y
coordinate is given by
dy
dt
fl
fl
fl
x
=3
=

‡
5
x
2
·fl
fl
fl
x
=3
=

5
9
.
Consequently, the speed of the
y
coordintate
is
5
9
units per second and the negative sign
indicates that the point is moving in the di
rection of decreasing
y
.
keywords: Stewart5e,
003
(part 1 of 1) 10 points
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Mahon, Kevin – Homework 8 – Due: Oct 20 2005, 3:00 am – Inst: Edward Odell
2
A street light is on top of a 10 foot pole.
A person who is 5 feet tall walks away from
the pole at a rate of 4 feet per second.
At
what speed is the tip of the person’s shadow
moving when he is 10 feet from the pole?
1.
tip speed
=
42
5
ft/sec
2.
tip speed
=
39
5
ft/sec
3.
tip speed
=
41
5
ft/sec
4.
tip speed
= 8 ft/sec
correct
5.
tip speed
=
38
5
ft/sec
Explanation:
If
x
denotes the distance of the tip of the
person’s shadow from the pole and
y
denotes
the distance of the person from the pole, then
the shadow and the lightpole are related in
the following diagram
(0
,
10)
(10
,
5)
y
x
By similar triangles,
5
x

y
=
10
x
,
so (10

5)
x
= 10
y
.
Thus, after implicit
differentiation with respect to
t
,
(10

5)
dx
dt
= 10
dy
dt
.
When
y
= 10
,
dy
dt
= 4
,
therefore,
tip speed
= 8 ft/sec .
keywords:
Stewart5e,
related rates,
street
light problem, implicit differentiation appli
cation
004
(part 1 of 1) 10 points
At noon, ship
A
is 150 miles due west of
ship
B
.
Ship
A
is sailing south at 10 mph
while ship
B
is sailing north at 30 mph.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 schultz
 Differential Calculus, triangle, dt, dt dt, Mahon, Edward Odell

Click to edit the document details