The Derivative in Context — AP® Calculus AB

Part 1 · Hook

Why this matters

A weather report says the snow is getting deeper at 2 centimeters per hour. A store says sales are dropping by 30 dollars a day. Both are derivatives, read in words. This topic teaches you to say exactly what a derivative means in a real situation, with the right units, which is worth points on every free-response question in context.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The derivative $f'(a)$ is

the slope of the line tangent to the graph of $f$ at $x = a$
the value of $f$ at $x = a$
the slope of a secant line over $[0, a]$
the area under $f$ up to $a$

Show the answer

The derivative is the instantaneous rate of change, the slope of the tangent line.

Correct: the slope of the line tangent to the graph of $f$ at $x = a$ :
the value of $f$ at $x = a$ :
the slope of a secant line over $[0, a]$ :
the area under $f$ up to $a$ :

2. From a table, $f(2) = 10$ and $f(5) = 19$ . What is the average rate of change of $f$ on $[2, 5]$ ?

$3$
$9$
$\frac{19}{5}$
$\frac{29}{7}$

Show the answer

$\frac{19 - 10}{5 - 2} = 3$ .

Correct: $3$ :
$9$ :
$\frac{19}{5}$ :
$\frac{29}{7}$ :

Part 4 · See it

See it first

The temperature T(t) = 20 + 60e^(−t/20) of a cooling drink, in °C, for t from 0 to 60 minutes, with the tangent line at (10, 56.4). A slope triangle runs 10 minutes and drops 18.2 °C, so T′(10) ≈ −1.82 °C per minute. — The slope of the tangent line carries units: a drop of 18.2 °C over a run of 10 minutes gives $T'(10) \approx -1.82$ °C per minute. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$f'(a)$ is a change in output divided by a change in input, as the input change shrinksits units are output units per input unit
The sign of the derivativesays whether the quantity is increasing (positive) or decreasing (negative) at that instant
Its sizesays how fast, in those units
Put together in one sentenceat this instant, this quantity is increasing (or decreasing) at this many units of output per unit of input

Part 6 · Key ideas

Key ideas

Units of $f'$ : units of $f$ divided by units of the input (liters per minute, dollars per item).
Interpret with four parts: the instant, the quantity by name, increasing or decreasing, and the rate with units.
$f(a)$ is an amount; $f'(a)$ is a rate. Do not mix them in a sentence.
From a table, estimate $f'(a)$ with the two closest inputs on either side of $a$ , using the table's actual widths.

Part 7 · Misconception

A common mistake

The wrong idea: $D'(3) = 1.6$ means the snow grew 1.6 centimeters in 3 hours.

What actually happens: $D'(3)$ is a rate at the single instant $t = 3$ : the depth is increasing at 1.6 centimeters per hour then. A change over an interval is a different quantity, in centimeters.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Depth of snow on a deck

Depth D(t) of snow on a deck, in centimeters, at selected times t, in hours after the snow began
$t$ (hours)	$0$	$2$	$5$	$6$	$10$
$D(t)$ (centimeters)	$12.0$	$15.2$	$20.0$	$21.1$	$24.0$

1. $D$ is differentiable. Use the data in the table to estimate $D'(5.5)$ . Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

Use the two times that straddle 5.5: $D'(5.5)$ $\approx \frac{D(6) - D(5)}{6 - 5}$ $= \frac{21.1 - 20.0}{1}$ $= 1.1$ centimeters per hour.

Answer: 1.1 centimeters per hour

2. What are the units of $D'(t)$ ?

centimeters per hour
centimeters
hours per centimeter
centimeters times hours

Show the answer

A derivative is a change in output over a change in input: $\frac{\text{cm}}{\text{hr}}$ , centimeters per hour.

Correct: centimeters per hour: Output units (centimeters) per input unit (hours).
centimeters: Those are the units of $D(t)$ , the depth itself.
hours per centimeter: The ratio is upside down: the output change goes on top.
centimeters times hours: A rate divides the output units by the input units.

3. Estimate $D'(8)$ using the table. Include units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$D'(8)$ $\approx \frac{D(10) - D(6)}{10 - 6}$ $= \frac{24.0 - 21.1}{4}$ $= \frac{2.9}{4}$ $= 0.725$ centimeters per hour.

Answer: 0.725 centimeters per hour

4. From the table, $D'(3) \approx 1.6$ . Which is the best interpretation?

In the first 3 hours, the depth of the snow increased by about 1.6 centimeters
At 3 hours, the depth of the snow is increasing at about 1.6 centimeters per hour
At 3 hours, the snow on the deck is about 1.6 centimeters deep
Over the first 3 hours, the snow fell at an average of 1.6 centimeters per hour

Show the answer

An interpretation names the instant, the quantity, whether it increases or decreases, and the units: at $t = 3$ hours the depth is increasing at about 1.6 cm per hour.

In the first 3 hours, the depth of the snow increased by about 1.6 centimeters: That describes a change over an interval, in centimeters; $D'(3)$ is a rate at $t = 3$ .
Correct: At 3 hours, the depth of the snow is increasing at about 1.6 centimeters per hour: A derivative is a rate at one instant, with units of depth per time.
At 3 hours, the snow on the deck is about 1.6 centimeters deep: That would be $D(3)$ , a depth; the table puts the depth between 15.2 and 20 centimeters then.
Over the first 3 hours, the snow fell at an average of 1.6 centimeters per hour: $D'(3)$ is the rate at the instant $t = 3$ , not an average from 0 to 3.

5. $W(t)$ is the number of liters of water in a tank $t$ minutes after noon. What does $W'(20) = -4$ mean?

At 12:20 there are 4 liters of water in the tank
In the first 20 minutes the tank lost 4 liters
At 12:20 the volume is decreasing at 4 liters per minute
At 12:20 the amount of water is decreasing at 4 minutes per liter

Show the answer

$W'(20) = -4$ liters per minute: at $t = 20$ minutes the volume is decreasing at 4 liters per minute.

At 12:20 there are 4 liters of water in the tank: That would be $W(20) = 4$ .
In the first 20 minutes the tank lost 4 liters: That is a change over an interval; $W'(20)$ is a rate at one instant.
Correct: At 12:20 the volume is decreasing at 4 liters per minute: A negative rate means the amount is going down.
At 12:20 the amount of water is decreasing at 4 minutes per liter: The units are upside down: liters per minute.

6. A square has side $s$ centimeters and area $A(s) = s^2$ square centimeters. Find $A'(5)$ with units.

No calculator. Type the exact value (a fraction, π, e or a root is fine), then its unit.

Show the answer

$A'(s) = 2s$ , so $A'(5) = 10$ square centimeters per centimeter: near $s = 5$ , the area grows about 10 cm² for each extra centimeter of side.

Answer: 10 square centimeters per centimeter

Part 9 · Summary

Summary

In context, a derivative is a rate with units of output per unit of input. Interpret

f'(a)

by naming the instant, the quantity, whether it is increasing or decreasing, and how fast, with units. Keep amounts,

f(a)

, and rates,

f'(a)

, apart.

Part 10 · Up next

What comes next

Part 11 · Connections