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What is the step-by-step method for solving a full related-rates word problem from start to finish?

Topic 4.5 Solving Related Rates Problems: solve complete related-rates problems using a structured method.

A focused answer to AP Calculus AB Topic 4.5, presenting a structured method for full related-rates problems - draw, relate, differentiate, substitute - with worked ladder and cone examples and the order-of-operations that avoids common errors.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The method
A worked ladder problem
Eliminating extra variables
Signs, units and the final sentence

What this topic is asking

The College Board (Topic 4.5) is about solving complete related-rates word problems: read the situation, set up the linking equation, differentiate with respect to time, and solve for the unknown rate at a specific instant. The skill is procedural discipline - following a fixed sequence so that the algebra stays organized and the answer carries the right sign and units.

The method

A worked ladder problem

The sliding ladder

A $10$ m ladder leans against a wall. The base is pulled away at $3$ m/s. How fast is the top sliding down when the base is $6$ m from the wall?

step 1 Draw and label

Let $x$ be the base's distance from the wall and $y$ the top's height. Both change with time. Given $\frac{dx}{dt} = 3$ ; find $\frac{dy}{dt}$ when $x = 6$ .

step 2 Relate with the Pythagorean theorem

The ladder length is constant: $x^2 + y^2 = 10^2 = 100$ .

step 3 Differentiate with respect to time

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \quad\Rightarrow\quad x\frac{dx}{dt} + y\frac{dy}{dt} = 0.

step 4 Find the missing value, then substitute

When $x = 6$ : $y = \sqrt{100 - 36} = 8$ . Substitute $x = 6$ , $y = 8$ , $\frac{dx}{dt} = 3$ :

6(3) + 8\frac{dy}{dt} = 0 \quad\Rightarrow\quad 18 + 8\frac{dy}{dt} = 0.

step 5 Solve and interpret

$\frac{dy}{dt} = -\frac{18}{8} = -\frac{9}{4} = -2.25$ m/s. The negative sign means the top is sliding down at $2.25$ m/s.

Eliminating extra variables

Many problems start with an equation in more variables than you have rates for, and the fix is to reduce before differentiating. The classic case is the draining cone: $V = \frac{1}{3}\pi r^2 h$ involves both $r$ and $h$ , but you are usually given only $\frac{dV}{dt}$ and asked for $\frac{dh}{dt}$ . Similar triangles relate $r$ and $h$ (for a cone with top radius $R$ and height $H$ , $\frac{r}{h} = \frac{R}{H}$ ), letting you write $r$ in terms of $h$ and substitute, so $V$ becomes a function of $h$ alone. After that substitution the differentiation produces only $\frac{dV}{dt}$ and $\frac{dh}{dt}$ , exactly the two rates you care about. Always look for a way to express the formula in just the variables whose rates appear; carrying an extra variable with no known rate leaves you unable to solve.

Signs, units and the final sentence

A complete answer does three things the careless student skips. First, it gets the sign right and explains it: a negative rate for a falling height or a shrinking distance is physically meaningful and should be stated as "decreasing" or "downward". Second, it carries units throughout, so the final rate reads as "meters per minute" or "cubic meters per second", not a bare number. Third, on free-response questions it ends with a sentence in context answering the question asked. Examiners award separate points for the correct setup, the correct differentiation, and the correct interpreted answer, so even a small arithmetic slip late in the problem can preserve most of the marks if the structure is clearly laid out. The single most common avoidable error remains substituting a changing value before differentiating; guard against it by differentiating the general equation first, every time, and only then plugging in the instant.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (style)1 marksSection I (multiple choice). A 13 m ladder leans on a wall; its base slides away at 2 m/s. When the base is 5 m from the wall (top at 12 m), the top slides down at a rate of (A)

\frac{5}{6}

m/s (B)

\frac{6}{5}

m/s (C)

2

m/s (D)

\frac{12}{5}

m/s

Show worked answer →

The correct answer is (A), $\frac{5}{6}$ m/s (downward).

From $x^2 + y^2 = 169$ , differentiating gives $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ . At $x = 5$ , $y = 12$ , $\frac{dx}{dt} = 2$ : $2(5)(2) + 2(12)\frac{dy}{dt} = 0 \Rightarrow 20 + 24\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6}$ . The top moves down at $\frac{5}{6}$ m/s.

AP 2023 (style)4 marksSection II (free response). Water drains from an inverted cone (height 12 m, top radius 6 m) so the depth falls. The volume drops at 8 cubic meters per minute. (a) Express the radius in terms of depth

h

using similar triangles. (b) Write the volume as a function of

h

alone. (c) Find how fast the depth falls when

h = 6

Show worked answer →

A 4-point related-rates word problem.

(a) (1 point) Similar triangles: $\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$ , so $r = \frac{h}{2}$ .
(b) (1 point) $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi}{12}h^3$ .
(c) (2 points) $\frac{dV}{dt} = \frac{\pi}{4}h^2\frac{dh}{dt}$ . With $\frac{dV}{dt} = -8$ and $h = 6$ : $-8 = \frac{\pi}{4}(36)\frac{dh}{dt} = 9\pi\frac{dh}{dt}$ , so $\frac{dh}{dt} = -\frac{8}{9\pi}$ meters per minute.

Related dot points

Sources & how we know this

AP Calculus AB and BC Course and Exam Description — College Board (2020)