Related Rates Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Related Rates.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

Problems where two or more quantities change with time and are related by an equation. Differentiate the equation with respect to time t and use known rates to find an unknown rate.

If two quantities are linked by an equation, their rates of change are also linked. A balloon inflating: as the radius increases, the volume increases too. How fast does the volume grow if the radius grows at 2 cm/s? The chain rule connects the rates.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Related rates problems always follow the same steps: (1) draw a picture, (2) write an equation relating the quantities, (3) differentiate with respect to t using the chain rule, (4) plug in known values and solve for the unknown rate.

Common stuck point: Don't plug in specific values until AFTER differentiating. The values of the variables change with time, so substituting before differentiating treats changing quantities as constants and gives the wrong equation.

Sense of Study hint: Draw the picture, label every changing quantity with a variable, write the geometric equation, THEN differentiate with respect to t.

Worked Examples

Example 1

easy
A spherical balloon is being inflated so its radius increases at 2 cm/s. How fast is the volume increasing when the radius is 5 cm?

Solution

  1. 1
    Volume of sphere: V = \frac{4}{3}\pi r^3.
  2. 2
    Differentiate with respect to time: \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.
  3. 3
    Given: \frac{dr}{dt} = 2 cm/s, r = 5 cm.
  4. 4
    \frac{dV}{dt} = 4\pi(25)(2) = 200\pi cm³/s.

Answer

\frac{dV}{dt} = 200\pi \approx 628 \text{ cm}^3/\text{s}
The chain rule links rates through their geometric relationship. Write the geometric formula, differentiate both sides with respect to t, then substitute the known values.

Example 2

hard
A 13-ft ladder rests against a wall. The base slides away at 5 ft/s. How fast is the top sliding down when the base is 5 ft from the wall?

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
A circle's radius grows at 3 cm/s. How fast is the area increasing when r = 10 cm?

Example 2

medium
Water fills a cone (apex down) of radius 3 m and height 6 m at 2 m³/min. How fast is the water level rising when h = 4 m?
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Background Knowledge

These ideas may be useful before you work through the harder examples.

chain ruleimplicit differentiationrate of change