A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Section 6.12 Exponential Growth and Decay

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.