Exponential Models — AP® Calculus AB Notes

One differential equation appears in more places than any other: $\frac{dy}{dt} = ky$ . It says the rate of change is proportional to the amount present. Bacteria with room to grow, an account earning continuous interest, a medicine cleared by the kidneys, a radioactive sample: each changes faster when there is more of it. This page solves the equation once, so you can quote the result, and shows how to read and fit the model.

The solution, once and for all

Separate: $\frac{dy}{y} = k\,dt$ . Integrate: $\ln|y| = kt + C$ . Exponentiate: $y = Ae^{kt}$ . At $t = 0$ , $y = Ae^0 = A$ , so the constant $A$ is the starting amount $y_0 = y(0)$ :

$\frac{dy}{dt} = ky, \quad y(0) = y_0 \qquad \text{gives} \qquad y = y_0e^{kt}.$

This is the exponential model. You may quote it on the exam when a problem says the rate of change is proportional to the amount.

Left: P = 50e^(0.2t) for t from 0 to 10, starting at 50 and doubling to 100 at t ≈ 3.47 and to 200 at t ≈ 6.93. Right: A = 80e^(−0.3t), starting at 80 and halving to 40 at t ≈ 2.31 and to 20 at t ≈ 4.62. — **Figure 1.** Growth with $k = 0.2$ and decay with $k = -0.3$ . Each doubling (or halving) takes the same time, whatever the starting amount. LevlPrep original diagram.

What $k$ means

Rearrange the equation: $k = \frac{dy/dt}{y}$ . So $k$ is the rate of change as a fraction of the current amount. If $k = 0.05$ per year, the quantity grows at 5% of its current size per year, at every moment. If $k = -0.12$ per hour, it shrinks at 12% of its current size per hour. That is why $k$ is called the growth constant ( $k > 0$ ) or the decay constant ( $k < 0$ ).

Units follow from the fraction: $\frac{dy}{dt}$ is in "amount per hour" and $y$ in "amount", so $k$ is in "per hour". The product $kt$ has no units, as an exponent must.

	Growth, $k > 0$	Decay, $k < 0$
Graph	rises ever faster	falls ever more slowly toward 0, never reaching it
Rate of change	$\frac{dy}{dt} = ky$ grows with $y$	$\frac{dy}{dt} = ky$ shrinks in size with $y$
Fixed time to	double: $\frac{\ln 2}{k}$	halve (half-life): $\frac{\ln 2}{\|k\|}$

Finding $k$ from data

Worked example 1. A culture grows at a rate proportional to its size. There are 400 bacteria at $t = 0$ and 1200 at $t = 3$ hours. Find $k$ , the model, and the population at $t = 6$ .

The model is $P = 400e^{kt}$ . Use the second count: $1200 = 400e^{3k}$ .

Divide first: $e^{3k} = 3$ . Take natural logs: $3k = \ln 3$ , so $k = \frac{\ln 3}{3} \approx 0.366$ per hour.

Model: $P(t) = 400e^{\frac{\ln 3}{3}t}$ . Then $P(6)$ $= 400e^{2\ln 3}$ $= 400 \cdot 9$ $= 3600$ bacteria.

Notice the pattern: 3 hours multiplies the population by 3, so 6 hours multiplies it by 9. A linear guess (add 800 every 3 hours, giving 2000) is the trap.

Worked example 2 (calculator). A medicine leaves the blood by $\frac{dA}{dt} = -0.12A$ , with $A(0) = 250$ mg and $t$ in hours. Find the half-life and the time when 50 mg remain.

The model is $A = 250e^{-0.12t}$ . Half-life: $125 = 250e^{-0.12t}$ , so $e^{-0.12t} = \frac{1}{2}$ and $t = \frac{\ln 2}{0.12} \approx 5.776$ hours.

50 mg: $50 = 250e^{-0.12t}$ , so $e^{-0.12t} = 0.2$ and $t = \frac{\ln 5}{0.12} \approx 13.412$ hours.

Keep full precision in the calculator and round to three decimal places only at the end.

Cooling and warming

Newton's law of cooling, $\frac{dT}{dt} = k(T - T_r)$ with room temperature $T_r$ , is an exponential model for the difference $T - T_r$ , not for $T$ . So $T - T_r$ $= (T_0 - T_r)e^{kt}$ , or $T$ $= T_r + (T_0 - T_r)e^{kt}$ . A pie at 90 °F in a 20 °F freezer has $T - 20 = 70e^{kt}$ ; if it is at 55 °F after 10 minutes, then $35 = 70e^{10k}$ and $k = -\frac{\ln 2}{10}$ per minute.

Common slips

Thinking linearly: an exponential multiplies by the same factor in equal times; it does not add the same amount.
Taking the log before dividing: from $1200 = 400e^{3k}$ , divide by 400 first.
Losing the sign of $k$ for decay, or using the rate at $t = 0$ when a later rate is asked: $\frac{dA}{dt} = kA(t)$ changes with time.
Applying the model to $T$ instead of $T - T_r$ in a cooling problem.

The solution, once and for all

What kk means

Finding kk from data

Cooling and warming

Common slips

What $k$ means

Finding $k$ from data