8.8 Interpreting differential-equation solutions in context — Practise

Work through these on your own. For each context, write the differential equation, solve it, then answer the question about the quantity or the time. Give one limitation of the model where a question asks for it.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A liquid cools so that $\dfrac{dT}{dt} = -0.1(T - 20)$ , with $T = 90$ °C when $t = 0$ and $t$ in minutes. Find $T$ in terms of $t$ .
A drink cools so that $\dfrac{dT}{dt} = -k(T - 20)$ , with $T = 90$ °C when $t = 0$ . After $20$ minutes the temperature is $55$ °C. Show that $k = \dfrac{\ln 2}{20}$ and find $T$ in terms of $t$ .
A population $P$ grows so that $\dfrac{dP}{dt} = 0.02P$ , with $P = 5000$ when $t = 0$ and $t$ in years. Find $P$ in terms of $t$ .
For the population $P = 5000e^{0.02t}$ , find the time taken for the population to double, giving an exact answer in years.
A radioactive mass $m$ decays so that $\dfrac{dm}{dt} = -0.05m$ , with $m = 200$ g when $t = 0$ and $t$ in years. Find $m$ in terms of $t$ .
For the mass $m = 200e^{-0.05t}$ , find the half-life, giving an exact answer in years.
A ball falls through a fluid so that $\dfrac{dv}{dt} = 9.8 - 2v$ , with $v = 0$ when $t = 0$ and $t$ in seconds. Find $v$ in terms of $t$ .
For the velocity $v = 4.9\left(1 - e^{-2t}\right)$ , find the terminal velocity, that is the limit of $v$ as $t \to \infty$ .
A body cools so that $\dfrac{dT}{dt} = -k(T - 20)$ , where $k > 0$ . Find the general solution for $T$ .
A cooling model gives $T = 20 + 70e^{-0.1t}$ . Find the initial temperature and the limiting temperature.
For the mass $m = 50e^{-0.02t}$ , find the mass after $10$ years, giving an exact answer in grams.
A model for a population is $P = 5000e^{0.02t}$ . Find the rate of change of the population at $t = 0$ , and give one limitation of the model.

Answers and concise solutions

1. The temperature

Answer: Separating gives $\ln|T - 20| = -0.1t + C$ , so $T = 20 + Ae^{-0.1t}$ . The condition $T(0) = 90$ gives $A = 70$ , so $T = 20 + 70e^{-0.1t}$ . The temperature approaches the room temperature of $20$ °C.

2. The temperature

Answer: The solution is $T = 20 + 70e^{-kt}$ . At $t = 20$ , $55 = 20 + 70e^{-20k}$ , so $e^{-20k} = \dfrac{1}{2}$ and $k = \dfrac{\ln 2}{20}$ . Hence $T = 20 + 70e^{-(\ln 2)t/20}$ . The excess above room temperature halves every $20$ minutes.

3. The population

Answer: Separating gives $\ln P = 0.02t + C$ , so $P = Ae^{0.02t}$ . The condition $P(0) = 5000$ gives $P = 5000e^{0.02t}$ . The model predicts unlimited growth at $2\%$ per year.

4. The doubling time

Answer: Set $P = 10000$ : $2 = e^{0.02t}$ , so $0.02t = \ln 2$ and $t = \dfrac{\ln 2}{0.02} = 50\ln 2 \approx 34.7$ years.

5. The mass

Answer: Separating gives $\ln m = -0.05t + C$ , so $m = Ae^{-0.05t}$ . The condition $m(0) = 200$ gives $m = 200e^{-0.05t}$ . The mass falls towards zero but never reaches it in the model.

6. The half-life

Answer: Set $m = 100$ : $\dfrac{1}{2} = e^{-0.05t}$ , so $-0.05t = -\ln 2$ and $t = \dfrac{\ln 2}{0.05} = 20\ln 2 \approx 13.9$ years.

7. The velocity

Answer: Separating gives $-\dfrac{1}{2}\ln|9.8 - 2v| = t + C$ , so $9.8 - 2v = Ae^{-2t}$ . The condition $v(0) = 0$ gives $A = 9.8$ , so $v = 4.9\left(1 - e^{-2t}\right)$ .

8. The terminal velocity

Answer: As $t \to \infty$ , $e^{-2t} \to 0$ , so $v \to 4.9$ . The terminal velocity is $4.9$ m/s, where the drag balances gravity.

9. The general solution

Answer: Separating gives $\ln|T - 20| = -kt + C$ , so $T = 20 + Ae^{-kt}$ . The value $20$ is the limiting temperature, and $A$ is set by the initial temperature.

10. The initial and limiting temperatures

Answer: At $t = 0$ , $T = 20 + 70 = 90$ °C. As $t \to \infty$ the exponential term tends to $0$ , so the limiting temperature is $20$ °C.

11. The mass

Answer: Substitute $t = 10$ : $m = 50e^{-0.2} \approx 40.9$ g to three significant figures.

12. The rate of change at $t = 0$

Answer: The rate of change is $\dfrac{dP}{dt} = 100e^{0.02t}$ , which is $100$ per year at $t = 0$ . One limitation is that the model predicts unlimited growth: it ignores limits on food, space and disease, so it is only reliable while those resources are plentiful.