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Math 112 Solutions for 2.1, p.100

1.
(a) 8; 14.4; 15.84. (b) about 16.

(c) -32 ft/sec, -24 ft/sec, -17.6 ft/sec, -16.8 ft/sec

(d) about -16 ft/sec (e) The negative sign indicates motion downward.

2.
(a)19.67; 19.74; 19.79. (b) about 19.8.

4.
$96.58 per year, $94.67 per year, $93.18 per year. At $ t=2$, the rate of change is about $92.81 per year, which is about 7.92% of $ A(2)$. $ A(t)=1000(1.02)^{4t}=1000(1+\frac{.08}{4})^{4t}$, so this is compounding 4 times per year at 8%.

5.
(a) 25.66, 36.8; 3.47, 4.98; 0.47, 0.67. (b) about 40; about 6; about 1. [Actual: 40.6, 5.49, 0.74.] (c) The velocity will approach 150 and acceleration will approach 0.

8.
The slope at $ a$ is about -1. No tangent line exists at 1 or at -1.

9.
The slope at $ a$ is about -2. No tangent line exists at -1 or at 2.

12.
(a) 8.1, 25.5, and 16.8 people per year. The last is indeed the average of the first two.

(c) about 16.8 people per year (using the average growth rate over [1960, 1980]).

(d) 1980: Use the average for [1970, 1990], which is 31.6 people per year.

1990: Use the average for [1980, 2000], which is 49.25 people per year.

13.
If $ b=a+h$, then $ b\to a$ if and only if $ h\to 0$. Hence the result, upon substitution.

14.
If $ b=a+h$, then $ b-a=h$ and $ b\to a$ if and only if $ h\to 0$. Hence, by substitution, $ m_{\text{tan}}=\lim_{b\to
a}\dfrac{f(b)-f(a)}{b-a}=\lim_{h\to 0} \dfrac{f(a+h)-f(a)}{h}$.

16.
(a) The slope of $ Q(t)=b+mt$ is $ m$, which is the rate of change.

(b) $ Q'(t)=\lim_{a\to t}\dfrac{Q(a)-Q(t)}{a-t}= \lim_{a\to
t}\dfrac{(b+ma)-(b+mt)}{a-t} =\lim_{a\to t}m=m$.




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Jason Grout 2003-01-24