Solving Related Rates Problems — AP® Calculus AB

Part 1 · Hook

Why this matters

Firefighters raise ladders, tanks fill and drain, shadows stretch as you walk away from a streetlight. In each case one rate is easy to measure and another is the one you care about. A related rates problem connects them with geometry and the chain rule, and the same five steps solve every one.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Differentiating $x^2 + y^2 = 100$ with respect to $t$ gives

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
$2x + 2y = 0$
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 100$
$2x + 2y\frac{dy}{dx} = 0$

Show the answer

Each variable depends on $t$ ; the constant's derivative is 0.

Correct: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ :
$2x + 2y = 0$ :
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 100$ :
$2x + 2y\frac{dy}{dx} = 0$ :

2. Similar triangles have legs 3 and 6 and $r$ and $h$ . Which is true?

$\frac{r}{h} = \frac{3}{6}$
$r + h = 9$
$r \cdot h = 18$
$r - h = 3$

Show the answer

Corresponding sides are proportional.

Correct: $\frac{r}{h} = \frac{3}{6}$ :
$r + h = 9$ :
$r \cdot h = 18$ :
$r - h = 3$ :

Part 4 · See it

See it first

A 10-foot ladder (crimson) against a wall with its foot x = 6 ft out and top y = 8 ft up, the foot sliding away at dx/dt = 2 ft/s and the top's rate dy/dt unknown; a gray ghost shows a later position. Beside it: x² + y² = 100, differentiated to 2x dx/dt + 2y dy/dt = 0, giving dy/dt = −3/2 ft/s at this instant. — The sliding ladder: write $x^2 + y^2 = 100$ , differentiate with respect to $t$ , and only then substitute $x = 6$ , $y = 8$ and $\frac{dx}{dt} = 2$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Draw the situation and name every changing quantityeach named quantity is a function of time $t$ ; fixed lengths stay numbers
Find an equation linking the quantities at every instant (Pythagoras, similar triangles, a volume formula)eliminate any extra variable with a second relation first, such as r = h/2
Differentiate both sides with respect to $t$ the chain rule gives an equation among the rates
Substitute the values of the instant, with signs for increasing or decreasingsolve for the unknown rate and state it with units

Part 6 · Key ideas

Key ideas

The five steps: draw and label, relate, differentiate with respect to $t$ , substitute, answer with units.
Constants of the problem (a 10-foot ladder, a 6-foot person) stay numbers; quantities that change get letters.
Use similar triangles to remove an extra variable before differentiating, so only one unknown rate remains.
Shrinking or approaching means a negative rate. Report the answer in words too: "the top slides down at 1.5 ft/s".

Part 7 · Misconception

A common mistake

The wrong idea: Substitute the given values first, to make the equation simpler, then differentiate.

What actually happens: The values hold at one instant only. Substituted early they become constants with derivative 0, and the rates disappear. Differentiate the general relation first.

Part 8 · Check yourself

Check yourself

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

Diagram

A 10-foot ladder (crimson) leaning against a wall, its foot x = 6 ft from the wall and its top y = 8 ft up. The foot slides away at dx/dt = 2 ft/s; the top's rate dy/dt is unknown. The angle θ between ladder and floor is marked; a gray ghost shows a later position. — LevlPrep original diagram.

1. The foot of the 10-foot ladder slides away from the wall at 2 feet per second. How fast is the top moving when the foot is 6 feet from the wall? Include units.

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

Show the answer

$x^2 + y^2 = 100$ gives $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ . At $x = 6$ , $y = 8$ : $2(6)(2) + 2(8)\frac{dy}{dt} = 0$ , so $\frac{dy}{dt} = -\frac{3}{2}$ feet per second. The top slides down at 1.5 ft/s.

Answer: $\frac{-3}{2}$ ≈ -1.500 feet per second

2. How fast is the top of the ladder moving when the foot is 8 feet from the wall? Include units.

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

Show the answer

At $x = 8$ , $y = \sqrt{100 - 64} = 6$ . Then $2(8)(2) + 2(6)\frac{dy}{dt} = 0$ , so $\frac{dy}{dt} = -\frac{32}{12} = -\frac{8}{3}$ feet per second.

Answer: $\frac{-8}{3}$ ≈ -2.667 feet per second

3. A student writes $6^2 + y^2 = 100$ first and then differentiates. Why does that lose the answer?

$6^2$ is a constant, so the term with $\frac{dx}{dt} = 2$ disappears
$y$ is a constant, so $\frac{dy}{dt} = 0$
The equation $6^2 + y^2 = 100$ is false at this instant
The derivative of 100 should be 10

Show the answer

The relation $x^2 + y^2 = 100$ holds at every instant; $x = 6$ holds at one. Differentiate the general relation, then substitute the instant.

Correct: $6^2$ is a constant, so the term with $\frac{dx}{dt} = 2$ disappears: The foot's motion is what drives the top; substituting $x = 6$ first throws it away.
$y$ is a constant, so $\frac{dy}{dt} = 0$ : $y$ still changes; it is $x$ that was frozen.
The equation $6^2 + y^2 = 100$ is false at this instant: It is true at this instant ( $y = 8$ ); it is just not true at other instants.
The derivative of 100 should be 10: 100 is a constant, so its derivative is 0.

Diagram

An inverted cone tank, top radius 3 ft and depth 6 ft, with water poured in at 4 ft³/min. The water is a smaller cone of depth h and surface radius r. — LevlPrep original diagram.

4. Which relation between the water's radius $r$ and depth $h$ holds at every instant?

$h = \frac{r}{2}$
$r = h - 3$
$r = \frac{h}{2}$
$r = 3$

Show the answer

The water is a cone similar to the tank: $\frac{r}{h} = \frac{3}{6}$ , so $r = \frac{h}{2}$ at every depth.

$h = \frac{r}{2}$ : The ratio is upside down: the depth is twice the radius, not half of it.
$r = h - 3$ : Similar triangles give a ratio, not a difference.
Correct: $r = \frac{h}{2}$ : Similar triangles: $\frac{r}{h} = \frac{3}{6}$ .
$r = 3$ : 3 ft is the radius of the full tank; the water's radius changes with $h$ .

5. Water is poured in at 4 cubic feet per minute. How fast is the water level rising when the water is 2 feet deep? Include units.

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

Show the answer

$V = \frac{\pi h^3}{12}$ , so $\frac{dV}{dt} = \frac{\pi h^2}{4}\frac{dh}{dt}$ . At $h = 2$ : $4 = \pi\frac{dh}{dt}$ , so $\frac{dh}{dt} = \frac{4}{\pi} \approx 1.27$ feet per minute.

Answer: $\frac{4}{\pi}$ ≈ 1.273 feet per minute

6. In a related rates problem, which step comes right after writing an equation that links the changing quantities?

Substitute the values given for that instant first
Differentiate both sides with respect to time
Solve the equation for one variable and stop
Set each rate equal to 0

Show the answer

Draw and label, write the relation, differentiate with respect to $t$ , substitute the instant, solve, and answer with units.

Substitute the values given for that instant first: Substituting first turns changing quantities into constants.
Correct: Differentiate both sides with respect to time: Then substitute the values of the instant.
Solve the equation for one variable and stop: The question asks for a rate, which needs a derivative.
Set each rate equal to 0: Rates are what the problem is about; they are rarely 0.

Part 9 · Summary

Summary

To solve a related rates problem: draw and label, write an equation linking the quantities at every instant, differentiate it with respect to time, then substitute the instant's values and solve for the unknown rate. Use geometry to reduce to one changing quantity, and give the answer with its sign and units.

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 ladder sliding down a wall

Water filling an inverted cone

Summary

What comes next

Connections