Answers
1. A = 2
π
r +
π
, r
≥
0
2. (a) A(1)
≈
30, A(4)
≈
40, (b) A(x) =40 for
x
≈
3
and
x
≈
4
.
A(x) = 60 has no solution with 1
≤
x
≤
6.
(c)
A
(2)
≈
26, [
A
(2)]
2
≈
26
2
=
676
A(2
2
)= A(4)
≈
40 ,
[A(2)]
2
is greater.
3. If A is a function, then A(2) is its value at 2.
If A is a constant, then A(2) is A multiplied by 2.
4. p(1) = 18,
p(2) = –6,
p(3) =
2
5. Possible answers: (a) She is stopped from about t = 40 to t = 55, which is about 15 seconds.
(b)
At
about t = 110
(c) From about t = 70 to t = 115
(d)
From about t = 7 to t = 27,
She goes about 1/6
mile.
6. (a) Least inrease in number of PAC's = 111 (from 1986 to 1988), Greatest increase = 898 (from about
1978 to 1980)
(b)
Least increase in contributions = $10,045,000 (from 1974 to 1976),
Greatest
increase = $31,403,000 (from 1980 to 1982)
(c)
$20,604 per PAC in 1974 and $34,653 per PAC in
1988(rounded to the nearest dollar)
7. (a) 10Q(10) = 210
(b) Q(6 + 3 + 1) = 21
(c) Q(100/20) = 7 (d)
Q(10)/Q(5) = 3
(e) Q(Q(5) + 3)
=21
8. (a) Change = 4300 – 3500 = 800 cigarettes per person, Relative change = 800/3500 = 0.23
(b) Change
= 3300 – 4300 = 1000 cigarettes per person,
Relative change = 1000/4300 = 0.23
(c)
C(1953) =
3700 cigarettes per person,
Error = 3700  3500 = 200 cigarettes per person,
Relative error =
200/3500 = 0.057
(d) The first graph does not show relative change well because the Caxis is not
shown for
0
≤
C
≤
3300.
9. P(100) = 3P(50) = 3(2P(1)) = 6P(1),
Relative change =
P(100) – P(1)
P(1)
= 5
10. A(1960) = 100(1610)/(1610 + 18581) = 7.97,
A(1965) = 14.04,
A(1970) = 30.24,
A(1975) =
38.44,
A(1980) = 45.84,
A(t) is the percentage in year t of the km traveled by Japanese in private
cars, trains, and busses that were traveled in private cars.
11. Possible answer: Set F(0)=1.
Then F(1) = 4F(0) = 4, F(2) = 4F(1) = 16, F(3) = 4F(2) = 64, F(4) =
4F(3) = 256, and F(5) = 4F(4) = 1024.
12. (a)
4 –
π
4
X100 = 21.5%
(b)
π
– 2
π
X 100 = 36.3%
(c) Circumscribed area = 2(inscribed area)
13. y = 36/x
2
 25,
y = 20 – 3x
2
, y = – 24/x,
y = x
3
– 5.
14.
x
1
dominates x
2
for large positive and large negative.
x
dominates x
1
for small positive and
large negative x.
15.
(a) x
3
is the dominant term in the numerator and 7x
2
is the dominant term in the denominator
when x is a large positive or negative number.
(b) –5 is the dominant term in the numerator and 3x
is the dominant term in the denominator when x is close to zero.
16.
(a) figure 12
(b) figure 11, b is positive
(c) figure 13, a is positive
17.
Figure 15, b is negative since
y
→−∞
as x
→
0
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View Full Document18.
Figure 17, Both a and b are positive since y = a when x = 0 and
y
→
+
∞
as x
→
+
∞
19.
Figure 14, a is positive since
y
→
a as x
→
±
∞
,
b is negative because
y
→−∞
as x
→
0
+
20.
200 miles + (t hours)(60
miles
hours
) = 200 + 60t miles
21.
1
0.05 gallons/mile
= 20
gallon
22.
V = 20 – (s miles0 ( 0.05
gallons
mile
)=20 – 0.05s gallons
The tank is empty when 20 – 0.05s = 0,
which is at 400 miles, so the domain of V as a function of s is the interval 0
≤
s
≤
400.
23.
s = 60t miles.
Since s = 400 at t = 6 2/3, the domain of s a function of t is 0
≤
t
≤
6 2/3.
24.
(a) V = 20 – 3t for 0
≤
t
≤
6 2/3
(b) 3 gallons per hour
(c) The volume of gas in your tank
decreases 3 gallons every hour.
25.
6
26.
f(x) = 10 + 50(x – 3)
27.
1/3
28.
The average rate of change of Z(x) with respect to x from x = 0 to x = 5 is
200 – 100
5 – 0
= 20.
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 Fall '11
 JamesLang
 Calculus, Imperial units, dt, Dh

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