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11.27: A.4.1- Section 4.1 Answers

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Dec 31, 2022
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- 11.26: A.3.3- Section 3.3 Answers
- 11.28: A.4.2- Section 4.2 Answers

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1. $Q=20e^{-(t\ln 2)/3200}$

2. $\frac{2\ln 10}{\ln 2}\text{days}$

3. $\tau = 10\frac{\ln 2}{\ln 4/3}\text{minutes}$

4. $\tau\frac{\ln (p_{0}/p_{1})}{\ln 2}$

5. $\frac{t_{p}}{t_{q}}=\frac{\ln p}{\ln q}$

6. $k=\frac{1}{t_{2}-t_{1}}\ln \frac{Q_{1}}{Q_{2}}$

7. $20\text{ g}$

8. $\frac{50\ln 2}{3}\text{yrs}$

9. $\frac{25}{2}\ln 2%$

10.

$=20\ln 3\text{yr}$
$Q_{0}=100000e^{-5}$

11.

$Q(t)=5000-4750e^{-t/10}$
$5000\text{lbs}$

12. $\frac{1}{25}\text{ yrs}$

13. $V=V_{0}e^{t\ln 10/2} \: 4\text{ hours}$

14. $\frac{1500\ln \frac{4}{3}}{\ln 2}\text{yrs};2^{-4/3}Q_{0}$

15. $W(t)=20-19e^{-t/20};\lim_{t\to\infty }W(t)=20\text{ ounces}$

16. $S(t)=10(1+e^{-t/10};\lim_{t\to\infty}S(t)=10\text{ g}$

17. $10\text{ gallons}$

18. $V (t) = 15000 + 10000e ^{t/20}$

19. $W(t) = 4 × 10^{6} (t + 1)^{2}$ dollars $t$ years from now

20. $p=\frac{100}{25-24e^{-t/2}}$

21.

$P(t)=1000e^{.06t}+50\frac{e^{.06t}-1}{e^{.06/52}-1}$
$5.64\times 10^{-4}$

22.

$P-=rP-12M$
$P=\frac{12M}{r}(1-e^{rt})+P_{0}e^{rt}$
$M\approx\frac{rP_{0}}{12(1-e^{-rN})}$
For (i) approximate $M = $402.25$ , exact $M = $402.80$ for (ii) approximate $M = $1206.05$ , exact $M = $1206.93$ .

23.

$T(\alpha )=-\frac{1}{r}\ln (1-(1-e^{-rN})/\alpha ))\text{ years}$ $S(\alpha )=\frac{P_{0}}{(1-e^{-rN})}[rN+\alpha\ln (1-(1-e^{-rN})/\alpha )]$
$T(1.05) = 13.69 \text{yrs}, S(1.05) = $3579.94 T(1.10) = 12.61 \text{yrs}, S(1.10) = $6476.63 T(1.15) = 11.70 \text{yrs}, S(1.15) = $8874.$

24. $P_{0}=\left\{\begin{array}{cc}{\frac{S_{0}(1-e^{(a-r)T})}{r-a}}&{\text{if }a\neq r}\\[4pt]{S_{0}T}&{\text{if }a=r}\end{array} \right.$

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