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Exercise B1
2.
a.
The general antiderivative of
f
(
x
)=12
x
2
– 24
x
+1
is
F
(
x
1
}
1
3
}
x
3
2
– 24
1
}
1
2
}
x
2
2
+
x
+
C
=4
x
3
– 12
x
2
+
x
+
C.
Since
F
(1) = –2, we have:
F
(1)=4
–
12+1+
C
= –2.
Thus, –7 +
C
= –2
C
=5
.
The specific antiderivative is
F
(
x
)=4
x
3
– 12
x
2
+
x
+5.
b.
The general antiderivative of
f
(
x
)=3
Ï
x
w
– sin
x
is
F
(
x
1
}
2
3
}
x
}
3
2
}
2
+ cos
x
+
C
=2
x
}
3
2
}
+ cos
x
+
C.
Since
F
(0) = 0, we have:
F
(0)=0+1+
C
=0
.
Thus,
C
= –1 and the specific antiderivative is
F
(
x
)=2
x
}
3
2
}
+ cos
x
– 1.
d.
The general antiderivative of
f
(
x
)=
e
3
x
–
}
2
1
x
}
is
F
(
x
}
1
3
}
e
3
x
–
}
1
2
}
ln
x
+
C.
Since
F
(1) =
e
3
, we have:
F
(1) =
}
1
3
}
e
3
–
}
1
2
}
ln 1 +
C
=
}
1
3
}
e
3
– 0 +
C
=
e
3
.
Thus,
C
=
}
2
3
}
e
3
and the specific antiderivative is
F
(
x
}
1
3
}
e
3
x
–
}
1
2
}
ln
x
+
}
2
3
}
e
3
.
e.
The general antiderivative of
f
(
x
}
Ï
x
3
x
+
2
1
w
}
is
F
(
x
)= (2)
1
}
1
3
}
2
(
x
3
+1)
}
1
2
}
+
C.
=
}
2
3
}
(
x
3
}
1
2
}
+
C
Since
F
(0) = 4, we have:
F
(0) =
}
2
3
}
(0+1)
}
1
2
}
+
C
=
}
2
3
}
+
C
.
Thus,
C
=
}
1
3
0
}
and the specific antiderivative is
F
(
x
}
2
3
}
Ï
x
3
w
+
}
1
3
0
}
.
f.
The general antiderivative of
f
(
x
)= cos
x
sin
4
x
is
F
(
x
}
1
5
}
(sin
x
)
5
+
C.
Since
F
(0) = –1, we have:
F
(0) =
}
1
5
}
(sin 0)
5
+
C
=
C
= –1.
The specific antiderivative is
F
(
x
)= sin
5
x
– 1.
3.
We wish to determine a function
P
(
t
) that gives the
population at any time
t
. We are given that the rate of
change of the population is 3 + 4
t
}
1
3
}
.
Using
P'
(
t
)=3+4
t
}
1
3
}
, we can determine
P
(
t
), the
general antiderivative:
P
(
t
t
+4
1
}
3
4
}
t
}
4
3
}
2
+
C
=3
t
+3
t
}
4
3
}
+
C.
In order to determine the specific population function,
we use the fact that the current population is 10 000,
i.e.,
P
(0) = 10 000:
P
(0) = 10 000
0+0+
C
= 10 000
C
= 10 000.
Thus, the population at any time
t
is given by
P
(
t
t
t
}
4
3
}
+ 10 000. Six months from now
the population will be
P
(6) = 3(6)+3(6)
}
4
3
}
+ 10 000
= 10 081.
4.
We first need to determine a function
V
(
t
) that gives
the volume of water in the tank at any time
t
. Since
the water is leaking from the tank at the rate of
}
5
t
0
}
L/min, we have
V'
(
t
}
5
t
0
}
.
Thus,
V
(
t
)=–
}
5
1
0
}
1
}
1
2
}
t
2
2
+
C
=–
}
1
1
00
}
t
2
+
C.
Since
V
= 400 L at time
t
,
C
= 400.
The volume of water in the tank at any time
t
is
V
(
t
}
1
1
00
}
t
2
+ 400.
Appendix B
271
Appendix B
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View Full DocumentTo determine when the tank will be empty, we solve
V
(
t
)=0:
–
}
1
1
00
}
t
2
+ 400 = 0
t
2
= 40 000
t
= 200,
t
≥
0.
The tank will be empty 200 min or 3 h 20 min from
the time at which there were 400 L of water in it.
5.
Let the measure of the inner radius of a water pipe
at any time
t
be
r
(
t
). We are given that
r'
(
t
)= –0.02
e
–0.002
t
cm/year.
a.
Thus,
r
(
t
1
}
–0.
1
002
}
e
–0.002
t
2
+
C
= 10
e
–0.002
t
+
C.
Since
r
=1
when
t
=0:
1= 10
e
°+
C
=10+
C
C
= –9.
The inner radius of a pipe at any time
t
is
r
(
t
)=10
e
–0.002
t
– 9.
b.
When
t
=3
years,
r
(3) = 10
e
(–0.002)(3)
– 9
= 0.94
After three years, the inner radius of a pipe will
be 0.94 cm.
c.
The pipe will be completely blocked when
r
=0
.
To determine when this occurs, we solve
r
(
t
10
e
–0.002
t
– 9 = 0
e
–0.002
t
=
}
1
9
0
}
–0.002
t
= ln(0.9), by definition
t
=
}
–
ln
0
(
.
0
0
.
0
9
2
)
}
= 52.68.
The pipe will be completely blocked in
approximately 52.7 years from the time when
its inner radius is 1 cm.
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 Spring '10
 Mr.M
 Calculus, Derivative

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