6.6 Exponential growth and decay in modelling — Practise

Work through these on your own. Write down the model with its units, decide the sign of the rate constant, and keep the rounding until the final answer.

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

£1000 is invested at $5\%$ per year, compounded continuously, so $A = 1000e^{0.05t}$ . Find the amount after $10$ years, to the nearest penny.
An investment grows as $A = A_0e^{0.07t}$ . Find the exact time for the money to double.
A radioactive isotope has a half-life of $5$ years. Find the decay constant $k$ in the model $m = m_0e^{-kt}$ .
A radioactive sample has a half-life of $8$ days. What fraction of the sample remains after $24$ days?
The amount of a drug in the blood is $A = 50e^{-0.3t}$ mg after $t$ hours. Find the amount after $4$ hours, to three significant figures.
A drug is modelled by $A = 50e^{-0.3t}$ . Find the exact time for the amount to fall to $5$ mg.
A population is modelled by $P = 1500e^{0.02t}$ , where $t$ is in years. Find $P$ after $25$ years, to the nearest whole number.
A car worth £ $20000$ loses value as $V = 20000e^{-0.15t}$ , where $t$ is in years. Find its value after $3$ years, to the nearest penny.
A quantity grows as $Q = Q_0e^{0.07t}$ . Write the doubling time to three significant figures.
Show that $y = Ae^{kt}$ satisfies the equation $\dfrac{dy}{dt} = ky$ .
A model predicts $1000e^{0.5}$ for a measurement. The measured value is $1600$ . Find the difference, model minus measurement, to two decimal places.
A concentration is modelled by $C = 120e^{-0.25t}$ . Find the exact time for $C$ to fall to $30$ .

Answers and concise solutions

1. The amount

Answer: Substitute $t = 10$ : $A = 1000e^{0.5} = 1648.7209\ldots$ , which is £ $1648.72$ . Continuous compounding is the limiting case of more and more frequent compounding.

2. The time

Answer: Doubling means $A = 2A_0$ , so $e^{0.07t} = 2$ . Take logarithms: $0.07t = \ln 2$ , and $t = \dfrac{\ln 2}{0.07} = \dfrac{100\ln 2}{7} \approx 9.90$ years. The starting amount cancels, so the doubling time does not depend on it.

3. The value of $k$

Answer: Half-life means $m = \dfrac{m_0}{2}$ when $t = 5$ , so $e^{-5k} = \dfrac{1}{2}$ . Take logarithms: $-5k = -\ln 2$ , giving $k = \dfrac{\ln 2}{5} \approx 0.139$ per year.

4. The fraction remaining

Answer: $24$ days is three half-lives, because $24 \div 8 = 3$ . Each half-life leaves half of what was there, so the fraction left is $\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$ .

5. The amount

Answer: Substitute $t = 4$ : $A = 50e^{-1.2} = 15.059\ldots$ , which is $15.1$ mg to three significant figures. The negative exponent means the amount falls.

6. The time

Answer: Divide by $50$ : $e^{-0.3t} = 0.1 = \dfrac{1}{10}$ . Take logarithms: $-0.3t = -\ln 10$ , so $t = \dfrac{\ln 10}{0.3} \approx 7.68$ hours. The amount has fallen by a factor of ten.

7. The population

Answer: Substitute $t = 25$ : $P = 1500e^{0.5} = 2473.08\ldots$ , which is $2473$ to the nearest whole number. The population grows, so the model has a positive exponent.

8. The value

Answer: Substitute $t = 3$ : $V = 20000e^{-0.45} = 12752.563\ldots$ , which is £ $12752.56$ . The value falls, so the exponent is negative.

9. The doubling time

Answer: The doubling time is $\dfrac{\ln 2}{0.07} = 9.9021\ldots$ , which is $9.90$ to three significant figures. A larger growth constant means a shorter doubling time.

10. The derivative

Answer: Differentiate: $\dfrac{dy}{dt} = kAe^{kt}$ . Since $y = Ae^{kt}$ , this is exactly $ky$ . The rate of change is proportional to the current amount, which is why the exponential model describes so many processes.

11. The difference

Answer: The model predicts $1000e^{0.5} = 1648.72\ldots$ , so the difference is $1000e^{0.5} - 1600 = 48.72$ to two decimal places. A positive difference means the model overestimates.

12. The time

Answer: Divide by $120$ : $e^{-0.25t} = 0.25 = \dfrac{1}{4}$ . Take logarithms: $-0.25t = -\ln 4$ , so $t = \dfrac{\ln 4}{0.25} = 4\ln 4 \approx 5.55$ . The time for a quarter to remain is two half-lives.