Related Rates

You’ll be able to

Set up a related-rates problem by relating variables with an equation
Differentiate an equation implicitly with respect to time
Solve for an unknown rate at a specific instant

Rates linked through a shared equation

A related-rates problem features two or more quantities that change with time and are tied together by an equation. Because they share the equation, their rates of change (derivatives with respect to t) are linked too. Differentiating the equation implicitly with respect to time produces a relationship among the rates dV/dt, dr/dt, and so on. Knowing all but one rate lets you solve for the missing one at a chosen instant.

The reliable five-step setup

Every related-rates problem yields to the same routine: (1) draw and label, naming the changing variables; (2) write an equation relating those variables (geometry formulas, the Pythagorean theorem, similar triangles); (3) differentiate both sides with respect to t, attaching each variable’s rate; (4) substitute the known values and rates; (5) solve for the unknown rate. Substitute numbers only after differentiating — plugging in first freezes variables that are actually changing.

Differentiating a volume relation in time

V = (4/3)πr³ ⟹ dV/dt = 4πr² · (dr/dt)

Every geometric formula becomes a rate equation once differentiated with respect to t via the chain rule.

Worked example

A spherical balloon is inflated so its volume increases at 100 cm³/s. How fast is the radius increasing when r = 5 cm?

1.Relate the variables with the sphere volume: V = (4/3)πr³.
2.Differentiate with respect to time: dV/dt = 4πr² · (dr/dt).
3.Substitute the known values dV/dt = 100 and r = 5: 100 = 4π(5)²·(dr/dt) = 100π·(dr/dt).
4.Solve: dr/dt = 100 / (100π) = 1/π cm/s.

Answer: dr/dt = 1/π ≈ 0.318 cm/s. The radius grows slowly even though volume climbs fast, because surface area is large at r = 5.

Watch out

Never substitute the changing quantity’s value before differentiating. If you set r = 5 first, its derivative becomes 0 and the whole relation collapses. Differentiate symbolically, then plug in the instant’s numbers.

Checkpoint

A 10 ft ladder leans against a wall; its base slides away at 2 ft/s. With x horizontal and y vertical, which equation correctly relates the rates?

Checkpoint

The area of a circle grows as its radius increases. If A = πr² and dr/dt = 3 cm/s, what is dA/dt when r = 4 cm?

On the exam

Related-rates free-response answers must carry units and be evaluated at the stated instant. Write the differentiated equation, then substitute — showing the un-substituted derivative first is what earns the setup points even if arithmetic slips.

Answer the 2 checkpoints as you read.