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Section 6.12 Exponential Growth and Decay

These answers correspond to the problems in Section 3.7.

  1. (a)\(\ds \frac{dA}{dt}=pA\text{,}\) \(A(0)=A_0\text{;}\) \(A=A_0e^{pt}\text{.}\) (b) Solve \(15,000=10,000\cdot e^{4p}\)

  2. (a) \(\ds \frac{dA}{dt}=k(M-A(t))\text{.}\) (b) \(A(t)=M-ce^{kt}\text{.}\) (c) It is given that \(M=100\text{,}\) \(A(0)=0\text{,}\) and \(A(100)=50\text{.}\) Hence \(\ds A(t)=100(1-e^{-\frac{t\ln 2}{100}})\text{.}\) The question is to solve \(\ds 75=100(1-e^{-\frac{t\ln 2}{100}})\) for \(t\text{.}\) It follows that the student needs to study another 100 hours.

  3. (a) The model is \(\ds C(t)=C_0e^{-\frac{t}{2.5}}\) where \(C_0=1\) and the question is to solve \(\ds 0.5=e^{-\frac{t}{2.5}}\) for \(t\text{.}\) Hence \(t=2.5\ln2\approx 1.75\) hours. (b) \(\ds C'(0)=-\frac{1}{2.5}=-\frac{2}{7}\) hours.

  4. The percentage of alcohol in the blood at time \(t\) can be modelled as \(\ds c(t)=c_0 e^{-0.4t}\text{.}\) If \(c(t)=1\) and \(c_0=2\text{,}\) it follows that \(\ds t=\frac{5\ln 2}{2}\approx 1.73\) hours.

  5. (b) \(\ds f(t)=e^{-\frac{t\ln 2}{5700}}\text{.}\) (c) The question is to solve \(\ds 0.15=e^{-\frac{t\ln 2}{5700}}\) for \(t\text{.}\) Hence the age of the skull is \(\ds t=-\frac{5700\ln 0.15}{\ln 2}\approx 15600\) years.

  6. The proportion of radioactive sample remaining after time \(t\) can be modelled as \(\ds m(t)=m_0 e^{-5t}\text{.}\) If \(m(t)=0.5\text{,}\) \(m_0=1\text{,}\) it follows that the half-life of the particle is \(t=\frac{\ln 2}{5}\approx 0.138\) units of time.

  7. (a) The model is \(\ds m(t)=10e^{-kt}\) where \(t\) is in years, \(m(t)\) is in kilograms, and \(k\) is a constant that should be determined from the fact that \(m(24110)=5\text{.}\) Hence \(\ds k=-\frac{\ln 2}{24110}\) and \(m(t)=10e^{-\frac{t\ln 2}{24110}}\text{.}\) (b) \(m(1000)=10e^{-\frac{\ln 2}{24.11}}\approx 9.716\) kilograms. (c) We solve \(\ds 1=10e^{-\frac{t\ln 2}{24110}}\) to get \(\ds t=24110\frac{\ln 10}{\ln 2}\approx 80091.68\) years.

  8. The model is \(\ds A=A(0)e^{-kt}\text{.}\) It is given that \(A(0)=1\) kg and \(A(6)=0.027\) kg. Hence \(\ds A(t)= e^{\frac{t\ln 0.0027}{6}}\text{.}\) It follows that at 3:00 there are \(A(2)=e^{\frac{\ln 0.0027}{3}} \approx 0.1392476650\) kg of substance \(X\text{.}\)

  9. The model is \(\ds P=P(t)=P_0e^{kt}\) where \(k\) is a constant, \(P_0\) is the initial population and \(t\) is the time elapsed. It is given that \(\ds 10P_0=P_0e^{10k}\) which implies that \(\ds k=\frac{\ln 10}{10}\text{.}\) The question is to solve \(\ds 2=e^{\frac{t\ln 10}{10}}\) for \(t\text{.}\) Hence \(\ds t=\frac{10\ln 2}{\ln 10}\approx 3.01\) hours.

  10. (a) The model is \(\ds P=500e^{kt}\text{.}\) From \(\ds 8000=500e^{3k}\) it follows that \(\ds k=\frac{4\ln 2}{3}\text{.}\) Thus the model is \(\ds P=500e^{\frac{4t\ln 2}{3}}\text{.}\) (b) \(128000\) bacteria. (c) Solve \(\ds 10^6=500e^{\frac{4t\ln 2}{3}}\) for \(t\text{.}\) It follows that \(\ds t=\frac{3(4\ln 2+3\ln 5)}{4\ln 2}\approx 4.7414\) hours.

  11. \(\displaystyle 20158\)

  12. (a) The model is \(\ds P(t)=10e^{kt}\) where \(t\) is time in hours. From \(40=10e^{\frac{k}{4}}\) it follows that \(k=4\ln 4\text{.}\) Hence \(P(3)=10e^{12\ln 4}=167,772,160\) bacteria. (b) \(3.32\) hours.

  13. (a) \(\displaystyle \frac{dP}{dt}=kP\text{;}\) (b) \(\displaystyle P=160e^{t\ln \frac{3}{2}}\text{;}\) (c) \(P(0)=160\) bacteria; (d) \(\displaystyle \frac{dP}{dt}(0)=160\ln \frac{3}{2}\)

  14. The model is \(\ds P(t)=500e^{kt}\) where \(t\) is time in years. From \(P(2)=750\) it follows that \(\ds k=\frac{\ln 3-\ln 2}{2}\text{.}\) Thus \(\ds P(500)=500e^{250(\ln 3-\ln 2)}=5.27\cdot 10^{46}\text{.}\)

  15. Use Newton's Law of Cooling. \(t\approx 20.2\) minutes.

  16. (a) \(\ds \left. \frac{dT}{dt}\right| _{T=80^0}=-0.09\cdot(80-20)=-5.4\ ^0\)C/min. (b) Note that 6 seconds should be used as 0.1 minutes. From \(T\approx 80- 5.4\Delta T=80-5.4\cdot 0.1\) it follows that the change of temperature will be \(\ds T-80\approx -0.54^0\)C. (c) \(\ds t= -\frac{100}{9}\ln \frac{9}{16}\approx 6.4\) minutes. Find the function \(T=T(t)\) that is the solution of the initial value problem \(\ds \frac{dT}{dt}=-0.09(T-20)\text{,}\) \(T(0)=90\text{,}\) and then solve the equation \(T(t)=65\) for \(t\text{.}\)

  17. The model is \(\ds \frac{dT}{dt}=k(T-32)\) where \(T=T(t)\) is the temperature after \(t\) minutes and \(k\) is a constant. Hence \(\ds T=32+(T_0-32)e^kt\) where \(T_0\) is the initial temperature of the drink. From \(14=32+(T_0-32)e^{25k}\) and \(20=32+(T_0-32)e^{50k}\) it follows \(\ds \frac{3}{2}=e^{-25k}\) and \(\ds k=-\frac{1}{25}\ln \frac{3}{2}\text{.}\) Answer: \(T_0=5^0\)C.

  18. \(t\approx 4.5\) minutes.

  19. \(t\approx 2.63\) hours.