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Math 112 Solutions for 4.3, p. 340

3.
$ Q_{0}\approx 89$.
5.
about 10.58 days.
6.
$ \dfrac{dV}{dt}=-kV\Rightarrow V=V_{0}e^{-kt},k=\dfrac{\ln 2}{h}$. $ V=0.1V_{0}\Rightarrow t\approx 19.93$ weeks.
7.
$ Q=Q_{0}e^{-kt},k=\dfrac{\ln 2}{h}; Q=0.52Q_{0}\Rightarrow
t\approx 4.245\cdot 10^{9}$ years.
8.
(a) $ Q_{0}=0.00362$ moles. (b) about 218.9 million years.
11.
$ t=\dfrac{9\ln\frac{17}{28}}{\ln\frac{13}{14}}\approx 61$ minutes.
13.
(b) $ \frac{1}{2}L$. (c) The graph of $ Q$ is increasing where $ \frac{dQ}{dt}$ is positive, concave upward where $ \frac{dQ}{dt}$ is increasing, and concave downward where $ \frac{dQ}{dt}$ is decreasing. Thus the graph of $ Q$ is always increasing, concave upward between $ Q_{0}$ and $ \frac{1}{2}L$, and concave downward between $ \frac{1}{2}L$ and $ L$.

15.
Use $ L=1$, $ Q_{0}=0.1$, and the condition $ Q=0.4$ when $ t=20$. When $ t=120$, $ Q=$ 99.98%.

16.
$ t=\dfrac{\ln\frac{1/2}{199}}{-3000 k}\approx 3.67$ hours.
17.
The rate of increase begins to decrease between 1983 and 1984, so half capacity is about 1900, making the carrying capacity about 3800.
18.
(a) $ k=\dfrac{\ln 2}{h}\approx 0.0347$. (b) $ \dfrac{dQ}{dt}=U-kQ$. (c) $ Q=\dfrac{U}{k}+Ae^{-kt}$. (d) $ t\to\infty\Rightarrow Q\to\dfrac{U}{k}$. (e) ... an essay summarizing (a) - (d)...
19.
(a) $ t=\frac{1}{2}\Rightarrow T\approx 131.8^{\circ}$F. (b) Using the same $ k$, $ T\approx 122.33^{\circ}$F. (c) about 35.12 minutes.
20.
(a) Show that $ H'(x)=0$, so that $ H(x)$ is constant; find $ H(0)$.

(b) Show that $ K'(x)=0$, so that $ K(x)$ is constant; then show that $ K(0)=0$. The sum of squares is zero only if each square is zero.




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Jason Grout 2003-03-19