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Introduction to Related Rates

Distinguish fixed constraints from snapshots, then differentiate all changing quantities with respect to the shared clock.

This is shared AB/BC instruction. Open the AB edition of Topic 4.4. Your BC practice and progress stay in this book.

College Board Topic 4.4 · CED p. 85 (PDF p. 90) · CED effective Fall 2020; project course outline 2026–27.

Related rates problems have several moving quantities but one shared clock. Let x(t), y(t), and V(t) stay as functions until you have differentiated their relationship. The chain rule is what creates dx/dt, dy/dt, and dV/dt; a letter does not suddenly become a constant because the problem gives its value at one instant.

For a circle, A = πr² gives dA/dt = 2πr·dr/dt. For a rectangle, A = xy gives dA/dt = x·dy/dt + y·dx/dt by the product rule. For a ratio R = x/y, dR/dt = (y·dx/dt − x·dy/dt)/y² by the quotient rule. The same chain rule sits inside all three calculations. Differentiate with respect to time even when t does not appear explicitly.

Separate permanent constraints from snapshots. A ladder of fixed length L satisfies x² + y² = L² at every instant. “The foot is 3 meters from the wall” supplies x at one instant; it does not make x a constant. You may substitute a genuinely constant dimension before differentiating. You must keep changing dimensions symbolic until afterward.

Mark the information before differentiating
StatementRole
A ladder is always 5 m longConstant constraint: L = 5
The foot is 3 m from the wall right nowSnapshot: x = 3 after differentiation
The foot moves away at 0.2 m/sRate: dx/dt = 0.2
dAdt=x·dydt+y·dxdt

Let's work through it

  1. A rectangle has changing sides x and y. Write A = xy.

    Why: Geometry connects the quantities for every instant.

  2. Differentiate: A′ = x′y + xy′.

    Why: Both factors change, so the product rule is necessary; each prime is with respect to t.

  3. At x = 4 cm, y = 7 cm, x′ = 2 cm/s and y′ = −1 cm/s, A′ = 2(7) + 4(−1) = 10 cm²/s.

    Why: Substitute the snapshot only after the rate equation exists.

  4. The area increases at 10 cm²/s at that instant.

    Why: A positive signed area rate means growth even though one side shrinks.

Try it from memory

Pause & retrieve

Try it before you peek.

Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.

Question 1 of 30 self-assessed

Question 1

Differentiate x² + y² = 25 with respect to t.

See every recall question with its answer

These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.

Retrieval 1

Differentiate x² + y² = 25 with respect to t.

Worked answer and reason

2x·dx/dt + 2y·dy/dt = 0.

The chain rule differentiates x(t) and y(t); the fixed right side has derivative zero.

Retrieval 2

Differentiate z = x/y when both x and y change.

Worked answer and reason

dz/dt = (y·dx/dt − x·dy/dt)/y², for y ≠ 0.

A ratio of changing quantities requires the quotient rule.

Retrieval 3

Can a fixed cylinder radius be substituted before differentiating volume?

Worked answer and reason

Yes, if it stays fixed for the whole process.

A permanent constant and an instantaneous measurement play different roles.

Catch the tempting mistake

Put every given number into the geometric equation first.
Doing that to changing quantities erases the very dependence you need to differentiate.
d(xy)/dt = x′y′.
The product rule adds x′y and xy′; it does not multiply the derivatives.
Calculator workspace and limits

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Put it to work

For the teacher
  • Use two colors for permanent constraints and instantaneous values.
  • Before allowing substitution, ask each learner to name the differentiation variable and identify every changing quantity.

Unit planning and formative guidance →

Glossary and deeper reading · 4 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Chain rule

    The chain rule differentiates a composition by multiplying the outer derivative at the inner output by the inner derivative. For differentiable functions, (f(g(x)))′=f′(g(x))g′(x). For example, the derivative of sin(x²) is cos(x²)·2x. Nested compositions require a factor for each successive layer.

    Read more: 3.1 · Chain rule →

  • Product rule

    The product rule differentiates a product of two differentiable functions. It gives (fg)′=f′g+fg′, changing one factor at a time and adding the results. For example, the derivative of x sin x is sin x+x cos x. The product of the two derivatives is generally not the derivative of the product.

    Read more: 2.8 · Product rule →

  • Quotient rule

    The quotient rule differentiates a quotient where the denominator is nonzero and both functions are differentiable. It gives (f/g)′=(f′g−fg′)/g². For example, differentiating x/(x+1) produces 1/(x+1)² for x≠−1. The order of the subtraction and the denominator square both matter.

    Read more: 2.9 · Quotient rule →

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