Rates of Change in Other Contexts — AP® Calculus AB Notes

Motion is one story the derivative tells. It tells many others: how fast a medicine leaves the blood, how fast a town grows, how much one more kilogram of coffee adds to a roaster's cost. The calculus is identical in every case. What you must get right is the language: which quantity, increasing or decreasing, how fast, and in what units. This page also handles the second derivative in context, the rate at which a rate changes.

One pattern, many settings

Quantity	Its derivative means	Units of the derivative
temperature $H(t)$ , °C, $t$ in minutes	how fast it warms or cools	°C per minute
population $P(t)$ , people, $t$ in years	how fast the town grows	people per year
cost $C(x)$ , dollars, $x$ in kilograms	cost of one more kilogram (approximately)	dollars per kilogram
concentration $C(t)$ , mg/L, $t$ in hours	how fast the medicine level changes	mg/L per hour

The sentence is always built the same way: "At [input value], [the quantity] is increasing (or decreasing) at [the size of the derivative] [units]."

Worked example 1. A town's population is $P(t) = 500e^{0.02t}$ people, $t$ years after 2020. Find and interpret $P'(10)$ .

By the chain rule, $P'(t)$ $= 500 \cdot 0.02\,e^{0.02t}$ $= 10e^{0.02t}$ , so $P'(10) = 10e^{0.2} \approx 12.2$ .

In 2030 the population is increasing at about 12.2 people per year.

Worked example 2. The concentration of a medicine is $C(t) = \frac{8t}{t^2 + 4}$ mg/L, $t$ hours after a dose. Find and interpret $C'(1)$ and $C'(3)$ .

Quotient rule: $C'(t)$ $= \frac{8(t^2 + 4) - 8t \cdot 2t}{(t^2 + 4)^2}$ $= \frac{32 - 8t^2}{(t^2 + 4)^2}$ .

$C'(1) = \frac{24}{25} = 0.96$ : one hour after the dose, the concentration is increasing at 0.96 mg/L per hour.

$C'(3)$ $= \frac{32 - 72}{169}$ $= -\frac{40}{169}$ $\approx -0.24$ : after three hours it is decreasing at about 0.24 mg/L per hour.

Rates of rates

A derivative is a function, so it has its own rate of change, the second derivative. In context, its units carry the input unit twice: if $H$ is in °C and $t$ in minutes, then $H''(t)$ is in °C per minute per minute.

Top: temperature H(t) = 70 − 50e^(−0.1t) of a drink in °F for t from 0 to 40 minutes, about 51.6 °F at t = 10. Bottom: its rate H′(t) = 5e^(−0.1t) in °F per minute, about 1.84 at t = 10 and shrinking. — **Figure 1.** Top: a warming drink, $H(t)$ in °F. Bottom: its rate of change $H'(t)$ , measured in °F per minute, positive but shrinking, so $H''(t) < 0$ . LevlPrep original diagram.

Read the two signs separately. The sign of $H'$ says whether the temperature rises or falls. The sign of $H''$ says whether that rate is growing or shrinking:

$Q'$	$Q''$	In words
$+$	$+$	increasing, faster and faster
$+$	$-$	increasing, more and more slowly
$-$	$-$	decreasing, faster and faster
$-$	$+$	decreasing, more and more slowly

Worked example 3. Soup cools by $H(t) = 20 + 70e^{-0.04t}$ °C. Describe the cooling at $t = 10$ .

$H'(t) = -2.8e^{-0.04t}$ , so $H'(10) \approx -1.877$ °C per minute: the soup is cooling.

$H''(t) = 0.112e^{-0.04t}$ , so $H''(10) \approx 0.075$ °C per minute per minute: positive, so the rate of cooling is easing off.

Together: at 10 minutes the soup is cooling at about 1.88 °C per minute, and the cooling is slowing down.

Watch for a function that is already a rate. If water leaks out of a tank at $L(t)$ liters per hour, then $L'(3) = -2$ means the leak rate is decreasing at 2 liters per hour per hour. It says nothing directly about the amount of water left.

Chained quantities

Sometimes one quantity depends on another that depends on time. A balloon's temperature reading depends on its altitude, and its altitude depends on time. The chain rule multiplies the rates, and the units cancel the same way:

$\frac{d}{dt}T(h(t)) = T'(h(t)) \cdot h'(t), \qquad \frac{\text{°F}}{\text{foot}} \cdot \frac{\text{feet}}{\text{minute}} = \frac{\text{°F}}{\text{minute}}$

Checking that the units cancel is a fast way to catch a setup error.

Slips that cost points

Reporting the amount ( $H(10)$ ) when the question asks for its rate ( $H'(10)$ ).
Giving a second derivative the units of the first ("°C per minute" for $H''$ ).
Reading the sign of $Q''$ as whether $Q$ itself increases.
Dropping the chain rule factor in models like $e^{-0.04t}$ , which also flips the sign.
On a calculator question, rounding before the end, or giving fewer than three decimal places.