Exponential growth and decay

When a quantity changes at a rate proportional to itself, it is an exponential, and the constant in the exponent is found from one measurement. That single idea covers radioactive decay, population growth, compound interest and Newton’s law of cooling, where the quantity that decays is the temperature difference rather than the temperature itself.

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In this topic

Taught: Before the final, flavours A and B
Questions: 11
Total marks: 30
Suggested time: 52 min
Difficulty: Introductory to Capstone
Show the 3 hardest

Key results

Exponential growth and decay: $\frac{dy}{dt} = ky \iff y(t) = y(0)\,e^{kt}$
For a constant $k$ : growth when $k > 0$ , decay when $k < 0$ .
Doubling time and half-life: $t_{\text{double}} = \frac{\log 2}{k}, \qquad t_{\text{half}} = -\frac{\log 2}{k}$
For $y(t) = y(0)\,e^{kt}$ : doubling time when $k > 0$ , half-life when $k < 0$ , neither depending on $y(0)$ .
Newton’s law of cooling: $\frac{dT}{dt} = k\,(T - A) \iff T(t) = A + \bigl(T(0) - A\bigr)e^{kt}$
For a constant surrounding temperature $A$ and a constant $k < 0$ .
Constant inflow or outflow: $\frac{dy}{dt} = a - by \iff y(t) = \frac{a}{b} + \Bigl(y(0) - \frac{a}{b}\Bigr)e^{-bt}$
For constants $a$ and $b \neq 0$ ; when $b > 0$ , $y(t) \to a/b$ as $t \to \infty$ .

Read the theoryUBC Math 100 §3.5 Derivatives of exponential functionsUBC Math 100 §11.3 Radioactive decayUBC Math 100 §11.4 Equations of the form y′ = a − byCLP-1 §3.3 Exponential growth and decay

Worked example

Standard3 marks · 5 min

A radioactive isotope decays so that 30 % of any sample is gone after 10 days. How long does it take for a 200 g sample to decay to 50 g?

Give the answer in days.

Strategy

Two measurements, two logarithms: the first fixes the decay constant from the 30 % fact, the second finds the time for the sample to reach a quarter of its mass.

Working

$m(t) = 200e^{kt}, \qquad m(10) = 200e^{10k} = 140 \quad\Longrightarrow\quad k = \frac{\log 0.7}{10}$
Losing 30 % leaves 70 %. Since 0.7 < 1, log 0.7 is negative, so k < 0 as decay requires.
$200e^{kt} = 50 \quad\Longrightarrow\quad e^{kt} = \frac{1}{4} \quad\Longrightarrow\quad t = \frac{\log(1/4)}{k} = \frac{10\log 4}{\log(10/7)}$
log(1/4) = −log 4 and log 0.7 = −log(10/7); the signs cancel.
$t = \frac{10\log 4}{\log(10/7)} = \frac{20\log 2}{\log(10/7)} \approx 38.9 \text{ days}$
The exact form is the answer; the decimal is only a sense of scale.

Answer

t = \frac{10\log 4}{\log(10/7)} = \frac{20\log 2}{\log(10/7)} \approx 38.9 \text{ days}

Check

After 40 days the sample is 200 · 0.7⁴ ≈ 48 g, just under 50 g, so reaching 50 g takes a little under 40 days.

Where this goes wrong

Reasoning linearly: “30 % per 10 days, so 75 % takes 25 days”. Decay is proportional to what remains, so each 10 days removes 30 % of a smaller amount and the process slows down.

On the paper

Write the model, find its constant from the data, then solve with logarithms, keeping exact values until the last line and giving units.

Leans on

Solving an exponential equation with logarithms, from Functions, graphs and inverses

Your turn

Standard2 marks · 4 min

A yeast culture grows at a rate proportional to its size, and increases by 50 % every 4 hours. What is its doubling time?

Give the answer in hours.

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Practice questions

5 of the 11, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory1 mark · 2 min
Which of the following functions satisfies the differential equation $\dfrac{dy}{dt} = 3y$ ?
Strategy
The equation says the rate is three times the amount. Test each candidate by differentiating; only a constant multiple of e^3t works.
Working
$y = 5e^{3t}:\quad y' = 15e^{3t} = 3\bigl(5e^{3t}\bigr) = 3y \quad\checkmark$
The chain rule brings down the 3.
$y = e^{3t} + 5:\ y' = 3e^{3t} \neq 3y;\qquad y = 3e^{t}:\ y' = 3e^{t} = y \neq 3y;\qquad y = 3t + 5:\ y' = 3 \neq 3y$
Each of the others fails.
Answer
$y = 5e^{3t}$
The answer is y equals 5 e to the 3t.
Check
At t = 0, y = 5e^3t has value 5 and slope 15, three times the value, while e^3t + 5 has value 6 but slope only 3.
Where this goes wrong
Choosing e^3t + 5 because it “has e^3t in it”. Adding a constant breaks the equation: the derivative ignores the 5 but 3y does not.
Leans on
The derivative of e^kt, from The chain rule and logarithms
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A quantity $y(t)$ satisfies $y'(t) = 3y(t)$ for all $t$ , and $y(0) = 4$ . Find $y(1)$ .
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Introductory2 marks · 3 min
A quantity $y(t)$ satisfies $y'(t) = 3y(t)$ for all $t$ , and $y(0) = 4$ . Find $y(1)$ .
Strategy
The general solution of y′ = ky is an exponential with rate constant k; the initial value fixes the constant in front.
Working
$y(t) = Ce^{3t}$
General solution of y′ = 3y.
$y(0) = C = 4 \quad\Longrightarrow\quad y(t) = 4e^{3t}$
Initial condition.
$y(1) = 4e^{3} \approx 80.3$
Evaluate.
Answer
$y(1) = 4e^{3}$
The answer is 4 e cubed, about 80.3.
Check
y = 4e^3t starts at 4 and its derivative 12e^3t is 3y, so it is the solution, and 4e³ ≈ 80.3.
Where this goes wrong
Writing y = 4 + 3t, treating the rate as constant. The rate is proportional to y, so it grows as y grows: the solution is exponential, not linear.
Leans on
Exponential functions, from Functions, graphs and inverses
Try a similar question
Which of the following functions satisfies the differential equation $\dfrac{dy}{dt} = 3y$ ?
Answer this question
Standard3 marks · 5 min
A mug of tea at 90 °C is left in a room held at 20 °C. After 5 minutes it has cooled to 60 °C. Assuming Newton’s law of cooling, when will the tea reach 30 °C?
Give the time, in minutes after the tea was poured.
Strategy
The quantity that decays exponentially is the difference between the tea and the room, not the temperature itself. Fit the constant from the 5-minute reading, then solve for the target difference.
Working
$T(t) = 20 + 70e^{kt}$
Initial difference 70 °C.
$T(5) = 60:\quad 70e^{5k} = 40 \quad\Longrightarrow\quad k = \frac{1}{5}\log\frac{4}{7}$
Since 4/7 < 1, k is negative: the gap to room temperature shrinks.
$T(t) = 30:\quad 70e^{kt} = 10 \quad\Longrightarrow\quad t = \frac{\log(1/7)}{k} = \frac{5\log 7}{\log(7/4)} \approx 17.39 \text{ min}$
log(1/7) = −log 7 and log(4/7) = −log(7/4), so the signs cancel. The exact form is the answer.
Answer
$t = \frac{5\log 7}{\log(7/4)} \approx 17.39 \text{ minutes}$
The answer is 5 log 7 over log of 7/4 minutes, about 17.39 minutes.
Check
Every 5 minutes the gap to the room shrinks by the factor 4/7: 70, 40, about 22.9, 13.1, 7.5, so it passes 10 between 15 and 20 minutes.
Where this goes wrong
Modelling the temperature itself as 90e^kt, which decays towards 0 °C rather than towards room temperature. The model must be written for T − 20.
Try a similar question
A kettle of water at 100 °C is left to cool in a room of unknown constant temperature. After 10 minutes the water is at 80 °C, and after 20 minutes it is at 65 °C. Assuming Newton’s law of cooling, find the room temperature.
Answer this question
Challenging4 marks · 7 min
A kettle of water at 100 °C is left to cool in a room of unknown constant temperature. After 10 minutes the water is at 80 °C, and after 20 minutes it is at 65 °C. Assuming Newton’s law of cooling, find the room temperature.
Give the answer in degrees Celsius.
Strategy
With two unknowns you need two readings, but the decay constant can be eliminated: over equal time steps the temperature difference is multiplied by the same factor. Setting the two factors equal gives an equation in the room temperature alone.
Working
$T(t) - A = (100 - A)e^{kt}$
Newton’s law with unknown surroundings A.
$\frac{80 - A}{100 - A} = e^{10k} = \frac{65 - A}{80 - A}$
Each 10-minute step multiplies the difference by e^10k.
$(80 - A)^{2} = (100 - A)(65 - A) \quad\Longrightarrow\quad 6400 - 160A + A^{2} = 6500 - 165A + A^{2}$
Cross-multiply and expand; the A² terms cancel.
$5A = 100 \quad\Longrightarrow\quad A = 20$
The differences 80, 60, 45 each fall by a factor of 3/4: consistent.
Answer
$A = 20^{\circ}\text{C}$
The answer is 20 degrees Celsius.
Check
With a 20 °C room the gaps 80, 60 and 45 fall by the same factor 3/4 every 10 minutes, as exponential decay requires.
Where this goes wrong
Assuming a room temperature, such as 0 °C, and using only one reading. The problem gives two readings precisely because the room temperature is unknown and has to be solved for.
Try a similar question
A mug of tea at 90 °C is left in a room held at 20 °C. After 5 minutes it has cooled to 60 °C. Assuming Newton’s law of cooling, when will the tea reach 30 °C?
Answer this question
Capstone5 marks · 9 min
A retiree has $100 000 in an account paying 5 % per year, compounded continuously, and withdraws money continuously at $6000 per year, so the balance satisfies $A'(t) = 0.05A - 6000$ . Solve for $A(t)$ , and find how many years it takes for the money to run out.
Give the time in years.
Strategy
Shift by the equilibrium balance 6000/0.05 = 120 000, at which interest exactly covers the withdrawals. Below that level the shortfall grows exponentially; solve for the time when the balance reaches zero.
Working
$A' = 0.05\,(A - 120000)$
Factor out 0.05.
$A(t) - 120000 = \bigl(100000 - 120000\bigr)e^{0.05t} \quad\Longrightarrow\quad A(t) = 120000 - 20000\,e^{0.05t}$
Initial shortfall of 20 000 grows at 5 %.
$A(t) = 0 \iff e^{0.05t} = 6 \iff t = \frac{\log 6}{0.05} = 20\log 6 \approx 35.8 \text{ years}$
The account is empty after about 35.8 years.
Answer
$A(t) = 120000 - 20000e^{0.05t};\quad t = 20\log 6 \approx 35.8 \text{ years}$
The answer is A of t equals 120,000 minus 20,000 e to the 0.05 t, and the money runs out after 20 natural log 6, about 35.8, years.
Check
A(0) = 120,000 − 20,000 = 100,000, and with no interest the money would last only 100,000/6000 ≈ 16.7 years, so 35.8 is plausible.
Where this goes wrong
Dividing 100 000 by 6000 to get about 16.7 years, ignoring interest, or concluding the money never runs out because the balance is “growing at 5 %”. Interest on 100 000 is only 5000, less than the 6000 withdrawn, so the balance shrinks, and faster every year.
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A savings account pays 4 % per year, compounded continuously, and its owner deposits money continuously at $3000 per year. The balance $A(t)$ therefore satisfies $A'(t) = 0.04A + 3000$ . If the account opens with $20 000, what is the balance after 10 years?
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Browse all 11 questions

Needs
Exponential and trigonometric functions
Derivatives of eˣ, aˣ, sine, cosine and tangent, and the limits behind them.
22 questions
Needs
The chain rule and logarithms
Compositions, nested rules, the natural logarithm and logarithmic differentiation.
30 questions
Leads to
Slope fields and Euler’s method
Checking that a function solves a differential equation, reading a slope field, and stepping a solution forward with Euler’s method.
14 questions

Key results

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Related topics

Exponential and trigonometric functions

The chain rule and logarithms

Slope fields and Euler’s method