Related Rates

By the end of this lesson, you will be able to solve related rates problems by relating changing quantities through an equation, differentiating implicitly with respect to time, and interpreting the results.

The Big Idea

Related rates problems ask: given that one quantity is changing at a known rate, how fast is a related quantity changing? The connection is made through a geometric or physical equation; implicit differentiation with respect to time $t$ reveals how the rates relate.

The Procedure

1. Draw a diagram and label all variables.
2. Write an equation relating the variables.
3. Differentiate both sides with respect to $t$ (implicit differentiation — every variable is a function of $t$).
4. Substitute the known values and rates.
5. Solve for the unknown rate.
6. Interpret with units.

Example 1 — Expanding Circle

A circular oil spill expands at $20$ m²/min. How fast is the radius growing when $r = 5$ m?

$A = \pi r^2$

$\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}$

$20 = 2\pi(5)\dfrac{dr}{dt}$

$\dfrac{dr}{dt} = \dfrac{20}{10\pi} = \dfrac{2}{\pi} \approx 0.637$ m/min

Example 2 — Ladder Problem

A 13-foot ladder leans against a wall. The bottom slides away at 2 ft/s. How fast is the top sliding down when the bottom is 5 feet from the wall?

$x^2 + y^2 = 169$ (Pythagorean theorem)

When $x = 5$: $y = 12$.

Differentiate: $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$

$2(5)(2) + 2(12)\dfrac{dy}{dt} = 0$

$\dfrac{dy}{dt} = -\dfrac{20}{24} = -\dfrac{5}{6}$ ft/s (negative: top is moving down)

Example 3 — Water Tank (Conical)

Water drains from a conical tank (radius 4 m, height 10 m, vertex down) at $3$ m³/min. How fast is the water level dropping when $h = 6$ m?

Volume of cone: $V = \dfrac{1}{3}\pi r^2 h$

By similar triangles: $\dfrac{r}{h} = \dfrac{4}{10} \implies r = \dfrac{2h}{5}$

Substitute: $V = \dfrac{1}{3}\pi\left(\dfrac{2h}{5}\right)^2 h = \dfrac{4\pi h^3}{75}$

Differentiate: $\dfrac{dV}{dt} = \dfrac{4\pi \cdot 3h^2}{75}\dfrac{dh}{dt} = \dfrac{4\pi h^2}{25}\dfrac{dh}{dt}$