AMS 151 (Fall, 2009)
Joe Mitchell
Applied Calculus I
Practice Final Exam – Solution Notes
1. Solve for
x
:
10
·
3
2
x
= 3
·
(1
.
12)
x
Taking the natural log of both sides, we get
ln 10 + 2
x
ln 3 = ln 3 +
x
ln 1
.
12
x
(2 ln 3

ln 1
.
12) = ln 3

ln 10
x
=
ln 3

ln 10
2 ln 3

ln 1
.
12
=
ln(3
/
10)
ln(9
/
1
.
12)
2. Determine which function has a larger value as
x
→ ∞
:
(a).
f
(
x
) = 0
.
0005
·
x
3
or
g
(
x
) = 52
,
000
·
2
x
f
(
x
) grows like
x
3
(we can ignore the constant factor).
g
(
x
) grows like 2
x
, and exponential function (again, we can ignore the constant factor).
Thus,
g
(
x
) has a larger value as
x
→ ∞
.
Another way to say this is that
lim
x
→∞
f
(
x
)
g
(
x
)
= 0
,
so that
g
dominates
f
as
x
→ ∞
, which we compute using l’Hopital’s rule (since it has form
∞
∞
):
lim
x
→∞
0
.
0005
·
x
3
52
,
000
·
2
x
= lim
x
→∞
0
.
0005
·
3
x
2
52
,
000
·
2
x
ln 2
= lim
x
→∞
0
.
0005
·
6
x
52
,
000
·
2
x
ln 2 ln 2
= lim
x
→∞
0
.
0005
·
6
52
,
000
·
2
x
ln 2 ln 2 ln 2
= 0
.
(b).
f
(
x
) = log
x
7
or
g
(
x
) = 6
·
log
2
x
2
f
(
x
) grows like log
x
(since log
x
7
= 7 log
x
), while
g
(
x
) grows like log
2
x
(since log
2
x
2
= 2 log
2
x
), so
g
(
x
) dominates
f
(
x
) as
x
→ ∞
.
3. Find the equation of the line that goes through the point (8,3) and is perpendicular to the plot of the equation
2
x
+1
y
−
1
= 2.
We rewrite
2
x
+1
y
−
1
= 2: 2
y

2 = 2
x
+ 1, so that it is the equation of the line
y
=
x
+
3
2
. The slope of this line is 1;
thus, the slope of a line perpendicular to it is
−
1
1
=

1.
Thus, the equation of the line,
L
, we desire is
y
=

x
+
b
.
We find
b
by requiring that (8,3) lies on line
L
:
3 =

8 +
b
, so
b
= 11. Thus, line
L
has equation
y
=

x
+ 11.
4. The quantity of moisture in a loaf of bread decreases with time as it sits on the table. Suppose that the moisture,
M
(
t
), at time
t
minutes after being placed on the table decreases according to the function
M
(
t
) =
Qe
−
kt
. If 10%
of the moisture is gone at the end of 1 hour,
(a). What percentage of the original moisture is present after 30 minutes?
We are told that
M
(60) = (
.
9)
M
(0), which implies that
Qe
−
60
k
= (
.
9)
Q

60
k
= ln(
.
9)
k
=

ln(
.
9)
60
Now, we want to find
M
(30)
M
(0)
·
100 =
Qe
−
30
k
Q
·
100 = 100
e
−
30

ln(
.
9)
60
= 100
e
1
2
ln(
.
9)
= 100(
.
9)
1
/
2
(b). How long will it take until the moisture is reduced to 60% of its original quantity?
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We want to find
t
so that
M
(
t
) = 0
.
6
M
(0) = 0
.
6
Q
. Thus, we want
t
so that
Qe
−
kt
= 0
.
6
Q
, i.e., so that
e
−
kt
= 0
.
6.
Thus, we want
t
so that

kt
= ln(0
.
6), i.e.,
t
=
ln(0
.
6)

k
=
ln(0
.
6)
ln(0
.
9)
60
= 60
ln(0
.
6)
ln(0
.
9)
5. Evaluate the following expressions:
(a). cos(sin
−
1
(
5
11
)) =
Draw a right triangle and let
θ
be one of the two angles that is not
π/
2. Then,
θ
= sin
−
1
(
5
11
) if the side opposite
to
θ
is of length 5 and the hypotenuse is of length 11. Then, the side adjacent to
θ
is of length
√
11
2

5
2
=
√
96.
Thus, cos(sin
−
1
(
5
11
)) = cos
θ
=
√
96
11
.
(b). cos(
32
π
3
) = Note that
32
π
3
= 10
π
+
2
π
3
(i.e., it is 10 revolutions, plus an additional angle 2
π/
3 (120 degrees).
Thus, cos(
32
π
3
) = cos(2
π/
3) =

1
2
. (Draw a picture! The angle refers to a point in the second quadrant.)
6. On February 10, 1990, high tide in Boston was at midnight. The water level at high tide was 9.9 feet; later, at
low tide, it was 0.1 feet. Assuming the next high tide is at exactly 12 noon and that the height of the water is given
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