Math 16A, Fall 2007
Professor Stankova
Félice Lê
Homework 4 Solutions – Part 2
Section 1.8
1.8#4
a)
The average rate of change is given by
5
2
7
1
2
12
2
)
2
(
3
3
12
2
)
3
(
3
2
3
)
2
(
)
3
(
=

=

+


+
=


f
f
.
b)
The instantaneous rate of change when
2
=
t
is given by
)
2
(
f
′
.
We use the Sum Rule to differentiate each term of
)
(
t
f
separately.

+
+
=
′
t
dt
d
dt
d
t
dt
d
t
f
12
2
3
)
(
Then we apply the Power Rule to each term.
2
1
12
3
12
0
3
12
2
3
)
(


+
=

+
=

+
+
=
′
t
t
dt
d
t
dt
d
dt
d
t
dt
d
t
f
.
Then
6
4
12
3
)
2
(
12
3
)
2
(
2
=
+
=
+
=
′

f
.
1.8#6
Note that the
t
value in the graph represents the number of years since January 1, 1950.
Therefore,
t
=10 represents January 1, 1910,
t
=20 represents January 1, 1920, and so forth.
a)
We are looking at the points (20, 150) and (60, 300) on the curve.
The average rate of change is
4
15
20
60
150
300
20
60
)
20
(
)
60
(
=


=


f
f
.
b)
The speed at which the mean farm size was rising on January 1, 1950, is equivalent to the
instantaneous rate of change at
t
= 50, or
)
50
(
f
′
.
This in turn is equivalent to the slope of the
tangent line to the graph at
t
= 50.
This tangent line is shown on the graph.
We can choose any two
1
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points on the line to calculate its slope; using the points (40, 150) and (80, 400),
)
50
(
4
25
40
80
150
400
f
x
y
m
′
=
=


=
∆
∆
=
.
c)
We can see that at
t
= 60, the graph is steeper than at
t
= 80. In other words, the slope of the
graph (and of the tangent line) is greater at
t
= 60. Therefore, we conclude that the mean farm size
was rising faster on January 1, 1960, than on January 1, 1980.
1.8#10 The rate at which the liquid is flowing into the vat is given by
(
29
t
t
dt
d

5
. In order to
calculate this rate at
t
= 2, we need to calculate
(
29
t
t
dt
d

5
│
2
=
t
.
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 Spring '07
 Stankova
 Math, Derivative, Rate Of Change, dt

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