8.8 Interpreting differential-equation solutions in context — Practise with help

Work each question closed book first. Write down the initial value, the direction of change and the expected limiting value before you solve the equation.

Warm-up: remember

Solve $\dfrac{dy}{dx} = ky$ , giving the general solution.
Solve $e^{0.05t} = 3$ for $t$ , giving an exact answer.
Find $\displaystyle\lim_{t \to \infty} \left(20 + 70e^{-0.1t}\right)$ .

Answers

$y = Ae^{kx}$ .
$t = \dfrac{\ln 3}{0.05} = 20\ln 3$ .
The exponential term tends to $0$ , so the limit is $20$ .

If question 1 was hard, revise separable equations in Section 8.7.

Practise with help

1. A population $P$ grows so that $\dfrac{dP}{dt} = 0.03P$ , with $P = 2000$ when $t = 0$ . Find $P$ in terms of $t$ , then find the population after $20$ years.

Hint

The rate is positive, so the population grows. The general solution is $P = Ae^{0.03t}$ .

Next step

Substitute $t = 0$ , $P = 2000$ to find $A$ , then evaluate at $t = 20$ .

Full working

P = 2000e^{0.03t}.

After $20$ years,

P = 2000e^{0.6} \approx 3644 \text{ to the nearest whole number}.

The population grows without bound in this model. The model does not include limits on food or space.

2. A radioactive mass $m$ decays so that $\dfrac{dm}{dt} = -0.02m$ , with $m = 50$ g when $t = 0$ and $t$ in years. Find $m$ in terms of $t$ , and find the half-life.

Hint

The rate is negative, so the mass falls. The general solution is $m = Ae^{-0.02t}$ . The half-life is the time for $m$ to reach $25$ g.

Next step

With $m = 50e^{-0.02t}$ , set $m = 25$ and solve $25 = 50e^{-0.02t}$ .

Full working

m = 50e^{-0.02t}.

For the half-life:

\frac{1}{2} = e^{-0.02t}, \qquad -0.02t = \ln\frac{1}{2} = -\ln 2, \qquad t = \frac{\ln 2}{0.02} \approx 34.7 \text{ years}.

The half-life does not depend on the starting mass.

3. A drink cools so that $\dfrac{dT}{dt} = -k(T - 15)$ , with $T = 95$ °C when $t = 0$ . After $10$ minutes the temperature is $70$ °C. Find $k$ , then find the temperature after $30$ minutes.

Hint

The general solution is $T = 15 + Ae^{-kt}$ . Substitute $t = 0$ to find $A$ , then use the temperature at $10$ minutes to find $k$ .

Next step

$T = 15 + 80e^{-kt}$ . At $t = 10$ , $70 = 15 + 80e^{-10k}$ , so $e^{-10k} = \dfrac{11}{16}$ .

Full working

From $e^{-10k} = \dfrac{11}{16}$ ,

e^{-30k} = \left(\frac{11}{16}\right)^3.

Then

T = 15 + 80\left(\frac{11}{16}\right)^3 \approx 41.0 \text{°C}.

In the long run the temperature approaches $15$ °C, the surrounding temperature.

4. A falling object has velocity $v$ satisfying $\dfrac{dv}{dt} = 10 - 0.5v$ , with $v = 0$ when $t = 0$ and $t$ in seconds. Find $v$ in terms of $t$ and state the terminal velocity.

Hint

Separate to $\displaystyle\int \frac{1}{10 - 0.5v}dv = \int 1dt$ . The left side integrates to $-2\ln|10 - 0.5v|$ .

Next step

Rearrange to $10 - 0.5v = Ae^{-0.5t}$ . Substitute $t = 0$ , $v = 0$ to find $A$ .

Full working

At $t = 0$ : $10 = A$ . Then

10 - 0.5v = 10e^{-0.5t}, \qquad v = 20\left(1 - e^{-0.5t}\right).

As $t \to \infty$ , $v \to 20$ m/s. This is the terminal velocity, where the acceleration $10 - 0.5v$ is zero.

5. A model for a population gives $P = 5000e^{0.02t}$ , with $t$ in years. Describe the long-term behaviour and give one limitation of the model.

Hint

There is no limiting value here: the exponential grows without bound. Think about what a real population needs in order to keep growing.

Next step

State the growth rate and the starting value, then name one resource that the model ignores.

Full working

The population grows at $2\%$ per year from a starting value of $5000$ , and the model predicts unlimited growth. In the long run the population becomes arbitrarily large.

One limitation is that the model ignores limits on food, space and disease. A real population cannot grow without bound, so the model is only useful while those resources are plentiful. Another limitation is that the rate constant $0.02$ is assumed constant, but it can change as conditions change.

6. A cooling model gives $T = 18 + 60e^{-0.05t}$ , with $T$ in °C and $t$ in minutes. Find the initial temperature, the limiting temperature, and the time when the temperature reaches $30$ °C.

Hint

Substitute $t = 0$ for the initial temperature. The limiting value is the constant when the exponential term vanishes. For the time, set $T = 30$ .

Next step

$30 = 18 + 60e^{-0.05t}$ , so $e^{-0.05t} = \dfrac{12}{60} = 0.2$ . Take logarithms.

Full working

At $t = 0$ , $T = 18 + 60 = 78$ °C. As $t \to \infty$ , the exponential term tends to $0$ , so the limiting temperature is $18$ °C.

For $T = 30$ :

0.2 = e^{-0.05t}, \qquad -0.05t = \ln 0.2 = -\ln 5, \qquad t = \frac{\ln 5}{0.05} = 20\ln 5 \approx 32.2 \text{ minutes}.

The model approaches $18$ °C but never reaches it in finite time.