Introduction to Related Rates — AP® Calculus AB

Part 1 · Hook

Why this matters

Drop a stone in a pond and a ring of ripples spreads. The radius grows, so the area inside grows too, and their rates are linked: knowing one tells you the other. Whenever two changing quantities are tied by an equation, their rates of change are tied by the derivative of that equation.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $y$ depends on $x$ , what is $\frac{d}{dx}\left(y^3\right)$ ?

$3y^2\frac{dy}{dx}$
$3y^2$
$3x^2$
$y^3\frac{dy}{dx}$

Show the answer

The chain rule, as in implicit differentiation.

Correct: $3y^2\frac{dy}{dx}$ :
$3y^2$ :
$3x^2$ :
$y^3\frac{dy}{dx}$ :

2. If $g$ is a differentiable function of $t$ , what is $\frac{d}{dt}[g(t)]^2$ ?

$2g(t)g'(t)$
$2g(t)$
$2g'(t)$
$[g'(t)]^2$

Show the answer

The chain rule: outside derivative $2u$ at $u = g(t)$ , times $g'(t)$ .

Correct: $2g(t)g'(t)$ :
$2g(t)$ :
$2g'(t)$ :
$[g'(t)]^2$ :

Part 4 · See it

See it first

An expanding ripple:

A = \pi r^2

with both

A

and

r

depending on time, so

\frac{dA}{dt} = 2\pi r\frac{dr}{dt}

. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Two or more quantities change with time and are linked by an equationeach one is a function of $t$ , even if the equation shows no $t$
Differentiate both sides of the equation with respect to $t$ the chain rule attaches a rate to each variable: $\frac{d}{dt}r^2 = 2r\frac{dr}{dt}$
The result is an equation among the ratesknown rates and values give the unknown rate
Substitute the values of one instant only after differentiatingthe rates are not lost to derivatives of constants

Part 6 · Key ideas

Key ideas

Related rates: differentiate an equation linking quantities with respect to time $t$ .
Each variable that changes gets its own rate, $\frac{dx}{dt}$ , $\frac{dr}{dt}$ , from the chain rule. Constants get none.
A decreasing quantity has a negative rate: shrinking at 2 cm/s means $\frac{dr}{dt} = -2$ .
Products need the product rule: $\frac{d}{dt}(xy)$ $= \frac{dx}{dt}y + x\frac{dy}{dt}$ .

Part 7 · Misconception

A common mistake

The wrong idea: For $V = \frac{4}{3}\pi r^3$ , $\frac{dV}{dt} = 4\pi r^2$ .

What actually happens: $4\pi r^2$ is $\frac{dV}{dr}$ . With respect to time, the chain rule adds the factor $\frac{dr}{dt}$ : $\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$ .

Part 8 · Check yourself

Check yourself

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

Table

A spherical balloon of radius r centimeters and volume V cubic centimeters; both change with time t in seconds
Quantity	Formula or value
Volume	$V = \frac{4}{3}\pi r^3$
Radius	$r$ cm, a function of $t$
Time	$t$ seconds

1. Air is pumped into the balloon so that $\frac{dr}{dt} = 2$ cm per second. Differentiate $V = \frac{4}{3}\pi r^3$ with respect to $t$ and write $\frac{dV}{dt}$ in terms of $r$ .

Type an expression for $\frac{dV}{dt} =$ .

Show the answer

$\frac{dV}{dt}$ $= 4\pi r^2 \cdot \frac{dr}{dt}$ $= 4\pi r^2 \cdot 2$ $= 8\pi r^2$ . The factor $\frac{dr}{dt}$ comes from the chain rule, because $r$ depends on $t$ .

Answer: $\frac{dV}{dt} =$ $8\,\pi\,r^{2}$

2. With $\frac{dr}{dt} = 2$ cm per second, how fast is the volume increasing when $r = 3$ cm? Include units.

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

Show the answer

$\frac{dV}{dt}$ $= 4\pi r^2\frac{dr}{dt}$ $= 4\pi(9)(2)$ $= 72\pi$ cubic centimeters per second, about 226 cm³ per second.

Answer: $72\,\pi$ ≈ 226.195 cubic centimeters per second

3. Now air is pumped in at $\frac{dV}{dt} = 100$ cubic centimeters per second. How fast is the radius increasing when $r = 5$ cm? Include units.

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

Show the answer

$100$ $= 4\pi(5)^2\frac{dr}{dt}$ $= 100\pi\frac{dr}{dt}$ , so $\frac{dr}{dt} = \frac{1}{\pi} \approx 0.318$ centimeters per second.

Answer: $\frac{1}{\pi}$ ≈ 0.318 centimeters per second

4. Later, air leaks out and the radius shrinks by 0.5 cm every second. Which value goes into the related-rates equation?

$\frac{dr}{dt} = -0.5$
$\frac{dr}{dt} = 0.5$
$\frac{dV}{dt} = -0.5$
$r = -0.5$

Show the answer

A decreasing quantity has a negative derivative: $\frac{dr}{dt} = -0.5$ cm per second.

Correct: $\frac{dr}{dt} = -0.5$ : The radius is decreasing, so its rate of change is negative.
$\frac{dr}{dt} = 0.5$ : A shrinking radius has a negative rate of change; the sign is part of the value.
$\frac{dV}{dt} = -0.5$ : 0.5 cm per second is a rate for the radius, not for the volume.
$r = -0.5$ : A radius is a length; $-0.5$ is how fast it changes.

5. A circle's radius grows at $\frac{dr}{dt} = 3$ cm per second. Write $\frac{dA}{dt}$ in terms of $r$ , where $A = \pi r^2$ .

Type an expression for $\frac{dA}{dt} =$ .

Show the answer

$\frac{dA}{dt}$ $= 2\pi r\frac{dr}{dt}$ $= 2\pi r \cdot 3$ $= 6\pi r$ .

Answer: $\frac{dA}{dt} =$ $6\,\pi\,r$

6. The area of a circular stain grows at a constant rate, $\frac{dA}{dt} = 5$ cm² per second. As the stain gets larger, what happens to $\frac{dr}{dt}$ ?

It increases
It stays the same
It becomes negative
It decreases

Show the answer

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$ , so $\frac{dr}{dt} = \frac{5}{2\pi r}$ , which shrinks as $r$ grows.

It increases: Adding the same area each second to a larger circle widens it less.
It stays the same: A constant $\frac{dA}{dt}$ does not mean a constant $\frac{dr}{dt}$ : they are linked by $2\pi r$ .
It becomes negative: The radius still grows, so $\frac{dr}{dt} > 0$ .
Correct: It decreases: $\frac{dr}{dt} = \frac{5}{2\pi r}$ : a larger $r$ gives a smaller rate.

Part 9 · Summary

Summary

When quantities linked by an equation change with time, differentiate the equation with respect to

t

. The chain rule gives every changing variable its rate, producing an equation among the rates. Substitute the values of one instant afterward, with signs that match increasing or decreasing.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

A spherical balloon

Summary

What comes next

Connections