Related rates in AP Calculus

By Ashraf Bhura, Co-Founder, STEM-Tutor · Updated October 2026

Related rates problems ask how fast one quantity changes when another changes. They follow a reliable method.

The method

1. Draw a picture and label quantities that change with variables.
2. Write what you know and what you want as derivatives with respect to time, such as dx/dt = 3.
3. Find an equation linking the variables, such as Pythagoras or a volume formula.
4. Differentiate both sides with respect to time, using the chain rule.
5. Substitute values for the instant asked about, then solve.

Example: the sliding ladder

A 10 m ladder slides down a wall. The base moves away at 1 m/s. How fast is the top sliding down when the base is 6 m from the wall?

• x² + y² = 100, so 2x(dx/dt) + 2y(dy/dt) = 0.
• When x = 6, y = 8. Substitute: 2(6)(1) + 2(8)(dy/dt) = 0.
• dy/dt = −0.75, so the top slides down at 0.75 m/s.

Example: filling a cone

For a cone, V = ⅓πr²h. Use similar triangles to write r in terms of h first, so V depends on one variable. Then differentiate.

Common mistakes

• Substituting values before differentiating. Constants become zero only if they truly stay constant.
• Forgetting the chain rule on every variable.
• Sign errors: decreasing quantities have negative rates.
• Missing units in the final answer.

More worked examples

Example 3: an expanding circle

A circle's radius grows at 2 cm/s. How fast is its area growing when r = 5 cm?

• A = πr², so dA/dt = 2πr(dr/dt) = 2π × 5 × 2 = 20π cm²/s.

Example 4: inflating a balloon

Air enters a spherical balloon at 100 cm³/s. How fast is the radius growing when r = 5 cm?

• V = (4/3)πr³, so dV/dt = 4πr²(dr/dt).
• 100 = 4π × 25 × dr/dt, so dr/dt = 1/π ≈ 0.32 cm/s.

Example 5: filling a cone

Water pours at 3 m³/min into an inverted cone 10 m tall with top radius 5 m. How fast is the depth rising when h = 2 m?

• Similar triangles give r = h/2, so V = (1/3)π(h/2)²h = πh³/12.
• dV/dt = (πh²/4)(dh/dt): 3 = π × 4/4 × dh/dt, so dh/dt = 3/π ≈ 0.95 m/min.

Example 6: a moving shadow

A 1.8 m person walks away from a 6 m lamppost at 1.5 m/s. How fast does their shadow lengthen?

• Similar triangles: 6/(x + s) = 1.8/s, so 4.2s = 1.8x and s = 3x/7.
• ds/dt = (3/7)(1.5) ≈ 0.64 m/s, the same at every distance.

More mistakes to avoid

• Substituting values before differentiating.
• Forgetting the chain rule on each variable, for example writing d/dt(r²) = 2r instead of 2r(dr/dt).
• Not eliminating an extra variable, such as r in the cone problem, before differentiating.
• Sign errors: decreasing quantities have negative rates.
• Leaving out units in the final answer.

Practice questions

Try each one before opening the answer.

Key takeaways

• Draw a diagram and name every changing quantity.
• Find an equation linking the variables before differentiating.
• Differentiate with respect to time, using the chain rule.
• Substitute values only after differentiating.
• Give the final answer with units and the correct sign.

Why related rates work

In a related rates problem, several quantities change over time, and an equation links them at every instant. Because the equation is true at every moment, you can differentiate both sides with respect to time. That turns a relationship between quantities into a relationship between their rates. The chain rule is the engine: every variable that depends on time picks up its own rate, such as dx/dt.

Recognising the equation to use

• Right triangles (ladders, planes, cars at junctions): Pythagoras, x² + y² = z².
• Shapes changing size (circles, spheres, cubes): area or volume formulas.
• Cones and similar shapes: volume formulas plus similar triangles to remove a variable.
• Shadows and lights: similar triangles.
• Angles: trigonometric ratios, such as tanθ = opposite/adjacent.

Example: an angle changing

A plane flies horizontally at 200 m/s at a height of 1000 m, directly toward an observer. How fast is the angle of elevation θ changing when the horizontal distance is 1000 m?

• tanθ = 1000/x, so sec²θ(dθ/dt) = −1000/x² × dx/dt.
• At x = 1000: θ = 45°, sec²θ = 2, and dx/dt = −200 (distance shrinking).
• 2(dθ/dt) = −1000/10⁶ × (−200) = 0.2, so dθ/dt = 0.1 rad/s, increasing.

Exam tips

• Write "given" and "find" rates with units before starting.
• State which quantities are constant (like the ladder's length) and which change.
• In free-response answers, show the differentiated equation; it carries a mark of its own.
• Interpret your answer in a sentence, including whether the quantity is increasing or decreasing.

Frequently asked questions

Why do some answers come out negative?

A negative rate means the quantity is decreasing, such as the top of a sliding ladder moving down.

When can I substitute numbers?

Only after differentiating. Before that, only quantities that are truly constant may be replaced with numbers.

Are related rates on the calculator section?

They can appear in either section. The method is the same; the calculator only helps with arithmetic.

One more worked example: two cars

Car A drives north from a junction at 60 km/h. Car B drives east from the same junction at 80 km/h, starting at the same time. How fast is the distance between them growing after 1 hour?

• Positions after t hours: A is 60t north, B is 80t east. Distance z satisfies z² = x² + y².
• Differentiate: 2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt).
• At t = 1: x = 80, y = 60, z = 100. So 100(dz/dt) = 80 × 80 + 60 × 60 = 10,000.
• dz/dt = 100 km/h. In this special case the distance grows at a constant rate.

Study tip

Keep a page of related rates diagrams you have solved. Before each new problem, check whether it matches one of your saved shapes.

Where to go next

Related rates depend on confident differentiation, so if any step felt shaky, revisit the rules first. Then put the skill into a full review plan for the exam.

• AP Calculus derivative rules
• AP Calculus AB review and study plan
• What to memorise for AP Calculus

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