--- title: "Related Rates" description: "Link two changing quantities by an equation, differentiate the equation with respect to time, and solve for the rate you cannot measure directly." canonical: https://lightmysky.com/learn/mathematics/related-rates-mt_3XTIPt8Vmf source: https://lightmysky.com/learn/mathematics/related-rates-mt_3XTIPt8Vmf.md retrieved: 2026-09-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Related Rates Link two changing quantities by an equation, differentiate the equation with respect to time, and solve for the rate you cannot measure directly. Subject: Mathematics · Area: Calculus & Analysis · Ages 18 to 19 Page: https://lightmysky.com/learn/mathematics/related-rates-mt_3XTIPt8Vmf ## Ready when they can - Set up a relation between two quantities before differentiating anything - Differentiate a geometric formula with respect to time - Substitute the instantaneous values only after differentiating, and say why ## Lesson: How linked quantities change together Some problems give you two quantities that change with time. A balloon has a growing radius and a growing volume. The two rates are linked because one equation links the two quantities. Your job has four moves. Name each changing quantity with its units. Write the equation that links them. Differentiate both sides with respect to time. Substitute the instant values only after that. **Example.** A circle has radius 8 cm, growing at 3 cm per second. Area and radius obey A equals pi r squared. Differentiate with respect to time and the chain rule gives dA dt equals 2 pi r dr dt. Now substitute r equals 8 and dr dt equals 3. The area grows at 48 pi square cm per second. A sphere obeys V equals 4 thirds pi r cubed. Differentiate first and you get dV dt equals 4 pi r squared dr dt. Only now may you substitute numbers such as r equals 3. If you substitute too early, the radius freezes into a constant and its rate silently vanishes. Finish with units, since cubic cm per second tells a different story from square cm per second. **Example.** A 10 m ladder slides down a wall. Its foot sits 6 m from the wall and slides out at 2 m per second. Foot and height obey x squared plus y squared equals 100, so y equals 8. Differentiate to get 2 x dx dt plus 2 y dy dt equals 0. Substitute and solve: dy dt equals negative 1.5 m per second. The minus sign says the top slides down. **Tip.** List each quantity with its units before you start. That bookkeeping turns a confusing paragraph into a routine calculation. And when your answer comes out negative, trust it: the quantity is shrinking. **Recap.** Write the link, differentiate with respect to time, then substitute the instant values. ## Practice 14 questions on this page, each with its working shown. ## Needs first - [Implicit Differentiation and Derivatives of Inverse Functions](https://lightmysky.com/learn/mathematics/implicit-differentiation-and-derivatives-of-inverse-functions-mt_P4ABBGsKFT) - [The Derivative as a Rate of Change](https://lightmysky.com/learn/mathematics/the-derivative-as-a-rate-of-change-mt_RqyLE3jrAW)