6.6 Exponential growth and decay in modelling — Practise with help

Work each question closed book first. Write down the model, decide the sign of $k$ , then decide whether the question asks for a value or for a time.

Warm-up: remember

Solve $e^x = 9$ , giving $x$ exactly.
Write $\ln 2 + \ln 5$ as a single logarithm.
Work out $e^0$ .

Answers

$x = \ln 9$ .
$\ln 2 + \ln 5 = \ln 10$ .
$e^0 = 1$ .

If question 1 was hard, revise solving exponential equations in Section 6.4 before you continue.

Practise with help

1. £ $1500$ is invested with continuous compounding at $4\%$ per year, so $A = 1500e^{0.04t}$ . Find the amount after $6$ years, to the nearest penny.

Hint

Substitute $t = 6$ into the model. The rate is $0.04$ as a decimal.

Next step

Work out $A = 1500e^{0.24}$ , then round to two decimal places.

Full working

A = 1500e^{0.24} = 1906.87\ldots,

so the amount is £ $1906.87$ to the nearest penny.

2. A radioactive isotope has a half-life of $10$ years. Find the decay constant $k$ in $m = m_0e^{-kt}$ , to three significant figures.

Hint

Half-life means the mass is $\dfrac{m_0}{2}$ after $10$ years. Divide by $m_0$ first.

Next step

Set $e^{-10k} = \dfrac{1}{2}$ , take logarithms, and make $k$ the subject.

Full working

e^{-10k} = \frac{1}{2}, \qquad -10k = -\ln 2, \qquad k = \frac{\ln 2}{10} \approx 0.0693.

The decay constant is about $0.0693$ per year.

3. A mass is modelled by $m = 200e^{-0.05t}$ . Find the time for the mass to fall to $50$ , giving the answer exactly.

Hint

Divide by $200$ first: $\dfrac{50}{200} = 0.25$ . Then take logarithms.

Next step

Solve $e^{-0.05t} = 0.25$ , so $-0.05t = \ln 0.25$ . Make $t$ the subject and simplify the logarithm.

Full working

e^{-0.05t} = 0.25, \qquad -0.05t = -\ln 4, \qquad t = \frac{\ln 4}{0.05} = 20\ln 4.

The mass takes $20\ln 4 \approx 27.7$ time units to fall to a quarter of its value.

4. A population is modelled by $P = 300e^{0.02t}$ , where $t$ is in years. Find the doubling time.

Hint

Doubling means $P = 2 \times 300$ . Divide by $300$ and take logarithms.

Next step

Solve $e^{0.02t} = 2$ , so $0.02t = \ln 2$ .

Full working

t = \frac{\ln 2}{0.02} = 50\ln 2 \approx 34.7.

The population doubles about every $34.7$ years. The starting value $300$ does not affect the doubling time.

5. A savings account pays $2\%$ per year. Explain the difference between compounding once a year and compounding continuously.

Hint

With yearly compounding the amount after $t$ years is $A = A_0\left(1.02\right)^t$ . Continuous compounding uses $A = A_0e^{0.02t}$ .

Next step

Compare the two formulas after $1$ year and after a long time. Continuous compounding adds interest on interest at every instant.

Full working

Yearly compounding multiplies the amount by $1.02$ once each year. Continuous compounding multiplies by $e^{0.02}$ once each year, which is about $1.0202$ . The continuous model is slightly larger, because interest is added all the time rather than once at the end of the year. The difference grows as the years pass, but it stays small for a rate of $2\%$ .

6. A drug is modelled by $A = 60e^{-0.5t}$ mg after $t$ hours. Find the amount after $6$ hours, and the time for the amount to fall to $6$ mg.

Hint

For the amount, substitute $t = 6$ . For the time, divide by $60$ and solve $e^{-0.5t} = 0.1$ .

Next step

After $6$ hours, $A = 60e^{-3}$ . For the time, take logarithms of $e^{-0.5t} = 0.1$ to get $-0.5t = \ln 0.1$ .

Full working

After $6$ hours,

A = 60e^{-3} = 2.988\ldots \approx 2.99\text{ mg}.

For a fall to $6$ mg,

e^{-0.5t} = 0.1, \qquad t = \frac{\ln 0.1}{-0.5} = 2\ln 10 \approx 4.61\text{ hours}.

The amount has fallen by a factor of ten.

7. A model predicts a population of $500e^{0.15t}$ after $t$ years. At $t = 200$ the model gives an enormous number. Explain why the prediction is not believable, and suggest a refinement.

Hint

Think about what a population needs to keep growing at the same percentage rate for two hundred years.

Next step

List the resources a population needs, and decide whether they can grow without limit.

Full working

The exponential model assumes that the growth rate stays constant. After a long time the population would need unlimited food, water and space, which a real environment cannot supply. The model is only reliable near the data used to fit it. A refinement is to include a maximum population, so the growth slows as the population approaches the limit, or to quote the prediction only for a short time ahead.