Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

Section 4.1 Answers

  • Page ID
    28901
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    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.

    1. \(=20\ln 3\text{yr}\)
    2. \(Q_{0}=100000e^{-5}\)

    11. 

    1. \(Q(t)=5000-4750e^{-t/10}\)
    2. \(5000\text{lbs}\)

    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. 

    1. \(P(t)=1000e^{.06t}+50\frac{e^{.06t}-1}{e^{.06/52}-1}\)
    2. \(5.64\times 10^{-4}\)

    22.

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

    23.

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

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