Exponential Models with Differential Equations

In Grade 11 you modelled exponential growth and decay with equations like $A = A_0(1.03)^t$. Calculus explains why those models work: whenever a quantity changes at a rate proportional to its own size, it must be exponential. This page solves that differential equation once and for all, then uses it for populations, medicine, radioactive dating, and cooling.

Key ideas

The equation dy/dt = ky

“The rate of change of $y$ is proportional to $y$” is the differential equation

$$\frac{dy}{dt} = ky$$

Solve it by separation of variables:

$$\frac{1}{y}\,dy = k\,dt \quad\Rightarrow\quad \ln|y| = kt + C \quad\Rightarrow\quad y = Ae^{kt}$$

At $t = 0$, $y = Ae^0 = A$, so $A$ is the initial value $y_0$:

$$\boxed{\ \frac{dy}{dt} = ky \quad\Longleftrightarrow\quad y = y_0e^{kt}\ }$$

You can quote this result on the AP exam without re-deriving it each time, unless the question asks you to solve by separation of variables.

Growth or decay

• $k \gt 0$: exponential growth (populations, bacteria, money with continuous interest).
• $k \lt 0$: exponential decay (radioactive material, medicine leaving the body).

The constant $k$ is the relative growth rate: $k = \dfrac{1}{y}\cdot\dfrac{dy}{dt}$. If $k = 0.04$ per year, the quantity grows at $4\%$ of its current size per year, at every instant.

This connects to the Grade 11 form: $y_0e^{kt} = y_0\left(e^k\right)^t$, so the growth factor per unit of time is $b = e^k$, and $k = \ln b$.

Doubling time and half-life

For growth ($k \gt 0$), the doubling time is the time for $y$ to double. Set $y_0e^{kt} = 2y_0$:

$$e^{kt} = 2 \quad\Rightarrow\quad t_{\text{double}} = \frac{\ln 2}{k}$$

For decay ($k \lt 0$), the half-life is the time for $y$ to halve. Set $y_0e^{kt} = \tfrac{1}{2}y_0$:

$$e^{kt} = \frac{1}{2} \quad\Rightarrow\quad t_{\text{half}} = \frac{\ln(1/2)}{k} = \frac{\ln 2}{|k|}$$

Neither depends on $y_0$: a substance takes just as long to go from $200$ mg to $100$ mg as from $10$ mg to $5$ mg.

With a half-life of $4$ hours, the amount halves every $4$ hours, no matter how much is left.

Finding k from data

If you know two values, substitute them to find $k$:

1. Use the initial value to write $y = y_0e^{kt}$.
2. Substitute the second data point and solve for $k$ with $\ln$.
3. Keep $k$ exact (like $\dfrac{\ln 1.25}{8}$) or store it in your calculator. Rounding $k$ early can change the 3rd decimal place of later answers.

Newton’s law of cooling

An object’s temperature $T$ changes at a rate proportional to the difference between $T$ and the surrounding temperature $T_s$:

$$\frac{dT}{dt} = k(T - T_s)$$

with $k \lt 0$. Separating variables (Example 4 shows the steps) gives

$$T = T_s + (T_0 - T_s)e^{kt}$$

The difference $T - T_s$ decays exponentially, so the temperature levels off at $T_s$, the room temperature.

On the AP exam, problems with messy logarithms are usually calculator-active, but you should be able to set up and solve $\dfrac{dy}{dt} = ky$ by hand.

Worked examples

Example 1: Bacterial growth

A bacteria population $P$ satisfies $\dfrac{dP}{dt} = 0.4P$, where $t$ is in hours, and $P(0) = 500$. Find $P(5)$ and the doubling time.

Solution. The equation has the form $\dfrac{dP}{dt} = kP$ with $k = 0.4$, so

$$P(t) = 500e^{0.4t}$$ $$P(5) = 500e^{2} \approx 3694.528$$

About $3695$ bacteria. The doubling time is

$$t = \frac{\ln 2}{0.4} \approx 1.733 \text{ hours}$$

Example 2: Half-life of a medicine

A patient takes $200$ mg of a medicine. The amount $A$ in the body decreases at a rate proportional to $A$, and the half-life is $4$ hours.

(a) Find $k$ and write $A(t)$. (b) How much remains after $10$ hours? (c) When will $20$ mg remain?

Solution.

(a) $A = 200e^{kt}$. After $4$ hours, half remains:

$$100 = 200e^{4k} \quad\Rightarrow\quad e^{4k} = \frac{1}{2} \quad\Rightarrow\quad k = \frac{\ln(1/2)}{4} = -\frac{\ln 2}{4} \approx -0.173$$

So $A(t) = 200e^{-(\ln 2/4)t}$. (This is the same as $200\left(\tfrac{1}{2}\right)^{t/4}$, the Grade 11 form.)

(b) $A(10) = 200\left(\tfrac{1}{2}\right)^{10/4} = 200(0.5)^{2.5} \approx 35.355$ mg.

(c) Solve $200e^{kt} = 20$:

$$e^{kt} = 0.1 \quad\Rightarrow\quad t = \frac{\ln 0.1}{k} = \frac{4\ln 10}{\ln 2} \approx 13.288 \text{ hours}$$

Example 3: Finding k from two data points

A town’s population grows at a rate proportional to its size. It was $12\,000$ in 2015 and $15\,000$ in 2023. Predict the population in 2030, and find how fast it was growing in 2023.

Solution. Let $t$ be years since 2015, so $P = 12\,000e^{kt}$. In 2023, $t = 8$:

$$15\,000 = 12\,000e^{8k} \quad\Rightarrow\quad e^{8k} = 1.25 \quad\Rightarrow\quad k = \frac{\ln 1.25}{8} \approx 0.028$$

In 2030, $t = 15$:

$$P(15) = 12\,000e^{15k} = 12\,000(1.25)^{15/8} \approx 18\,234$$

The growth rate in 2023 comes straight from the differential equation:

$$\frac{dP}{dt} = kP = \frac{\ln 1.25}{8}(15\,000) \approx 418.394$$

About $418$ people per year.

Example 4: Newton’s law of cooling

A bowl of soup at $90^\circ\text{C}$ is left in a $20^\circ\text{C}$ room. Its temperature satisfies $\dfrac{dT}{dt} = k(T - 20)$, with $t$ in minutes. After $10$ minutes it is $60^\circ\text{C}$. When will it be $40^\circ\text{C}$?

Solution. Separate variables (the soup stays hotter than the room, so $T - 20 \gt 0$):

$$\frac{1}{T - 20}\,dT = k\,dt \quad\Rightarrow\quad \ln(T - 20) = kt + C \quad\Rightarrow\quad T - 20 = Ae^{kt}$$

At $t = 0$: $90 - 20 = A$, so $A = 70$ and $T = 20 + 70e^{kt}$.

At $t = 10$: $60 = 20 + 70e^{10k}$, so $e^{10k} = \dfrac{4}{7}$ and $k = \dfrac{\ln(4/7)}{10} \approx -0.056$.

Now solve $40 = 20 + 70e^{kt}$:

$$e^{kt} = \frac{2}{7} \quad\Rightarrow\quad t = \frac{\ln(2/7)}{k} = \frac{10\ln(2/7)}{\ln(4/7)} \approx 22.386 \text{ minutes}$$

Common mistakes

Using $y = e^{kt} + y_0$. The initial value multiplies: $y = y_0e^{kt}$. Check at $t = 0$: $y_0e^0 = y_0$ ✓, but $e^0 + y_0 = 1 + y_0$ ✗.

Getting the sign of $k$ wrong. For decay, $k$ must be negative. If you find a positive $k$ for a cooling or decaying quantity, you have probably flipped a fraction, for example writing $e^{4k} = 2$ instead of $e^{4k} = \tfrac{1}{2}$.

Rounding $k$ too early. Rounding $k$ to $0.03$ in Example 3 would give $12\,000e^{0.45} \approx 18\,820$ instead of $18\,234$. Keep $k$ exact or stored in the calculator.

Forgetting the surrounding temperature in Newton’s law. It’s the difference $T - T_s$ that decays, so $T = T_s + (T_0 - T_s)e^{kt}$, not $T = T_0e^{kt}$. A $T_0e^{kt}$ model would cool all the way to $0^\circ\text{C}$, which is not what happens in a $20^\circ\text{C}$ room.

Mixing up the rate and the amount. $\dfrac{dP}{dt}$ is how fast the population is changing (people per year); $P$ is the population (people). Read the question carefully, and include units.

Practice

1. (Warm-up) Solve $\dfrac{dy}{dt} = -0.05y$ with $y(0) = 80$. Is this growth or decay?

Solution

$$y = 80e^{-0.05t}$$

Since $k = -0.05 \lt 0$, this is decay.

2. (Warm-up) The function $y = 300e^{0.02t}$ is the solution of a differential equation with an initial condition. Write them both.

Solution

$$\frac{dy}{dt} = 0.02y, \qquad y(0) = 300$$

Check: $\dfrac{dy}{dt} = 300(0.02)e^{0.02t} = 0.02y$ ✓.

3. (Warm-up) A substance decays according to $\dfrac{dA}{dt} = -0.1A$, with $t$ in years. Find its half-life.

Solution

$$t_{\text{half}} = \frac{\ln 2}{0.1} = 10\ln 2 \approx 6.931 \text{ years}$$

4. (Core) A culture of yeast starts with $1000$ cells and doubles every $6$ hours. Its rate of growth is proportional to its size.

• (a) Find $k$.
• (b) How many cells are there after $15$ hours?
• (c) How fast is the population growing at $t = 15$?

Solution

(a) $2 = e^{6k}$, so $k = \dfrac{\ln 2}{6} \approx 0.116$ per hour.

(b) $P(15) = 1000e^{15k} = 1000 \cdot 2^{15/6} = 1000 \cdot 2^{2.5} \approx 5656.854$, about $5657$ cells.

(c) $\dfrac{dP}{dt} = kP = \dfrac{\ln 2}{6}(1000 \cdot 2^{2.5}) \approx 653.505$ cells per hour.

5. (Core) Carbon-14 has a half-life of about $5730$ years. A piece of bone has $30\%$ of the carbon-14 it had originally. About how old is it?

Solution

$A = A_0e^{kt}$ with $k = -\dfrac{\ln 2}{5730}$. Set $A = 0.3A_0$:

$$e^{kt} = 0.3 \quad\Rightarrow\quad t = \frac{\ln 0.3}{k} = \frac{5730\ln 0.3}{\ln 0.5} \approx 9952.813$$

The bone is about $9950$ years old (the half-life is only known to the nearest few years, so a rounded answer makes sense).

6. (Core) A drink comes out of a $4^\circ\text{C}$ fridge into a $24^\circ\text{C}$ room. It warms according to $\dfrac{dT}{dt} = k(T - 24)$, with $t$ in minutes. After $15$ minutes it is $10^\circ\text{C}$. When will it reach $18^\circ\text{C}$?

Solution

$T = 24 + (4 - 24)e^{kt} = 24 - 20e^{kt}$.

At $t = 15$: $10 = 24 - 20e^{15k}$, so $e^{15k} = 0.7$ and $k = \dfrac{\ln 0.7}{15} \approx -0.024$.

Now solve $18 = 24 - 20e^{kt}$: $e^{kt} = 0.3$, so

$$t = \frac{\ln 0.3}{k} = \frac{15\ln 0.3}{\ln 0.7} \approx 50.633 \text{ minutes}$$

7. (Core) Money in an account grows according to $\dfrac{dA}{dt} = 0.04A$, with $t$ in years, starting from $2000. How long until it reaches $3000?

Solution

$A = 2000e^{0.04t}$. Solve $2000e^{0.04t} = 3000$:

$$e^{0.04t} = 1.5 \quad\Rightarrow\quad t = \frac{\ln 1.5}{0.04} \approx 10.137 \text{ years}$$

8. (Challenge) A radioactive sample decays at a rate proportional to its mass. When $30$ g remain, the mass is decreasing at $0.9$ g per day. Find $k$ and the half-life.

Solution

From $\dfrac{dm}{dt} = km$ with $m = 30$ and $\dfrac{dm}{dt} = -0.9$:

$$-0.9 = k(30) \quad\Rightarrow\quad k = -0.03 \text{ per day}$$$$t_{\text{half}} = \frac{\ln 2}{0.03} \approx 23.105 \text{ days}$$

9. (Challenge) A cake comes out of a $175^\circ\text{C}$ oven into a $20^\circ\text{C}$ kitchen. At that moment its temperature is dropping at $9.3^\circ\text{C}$ per minute. Assuming Newton’s law of cooling, $\dfrac{dT}{dt} = k(T - 20)$, find $k$ and the cake’s temperature after $30$ minutes.

Solution

At $t = 0$, $T = 175$ and $\dfrac{dT}{dt} = -9.3$:

$$-9.3 = k(175 - 20) = 155k \quad\Rightarrow\quad k = -0.06$$

So $T = 20 + 155e^{-0.06t}$, and

$$T(30) = 20 + 155e^{-1.8} \approx 45.621^\circ\text{C}$$