Related rates

When two or more quantities are linked by an equation and both vary over time, their rates of change are linked too. Related rates problems exploit that link: given one rate, you can find the other. The tool is implicit differentiation with respect to time, which turns every variable in the equation into a derivative like dx/dt.

The procedure is consistent. First, write an equation relating the quantities using geometry or a formula. Second, differentiate both sides with respect to t, applying the chain rule so that each variable contributes its own rate. Third, substitute the values at the instant in question and solve for the unknown rate. A common mistake is substituting numbers before differentiating — quantities that change must stay as variables until after the differentiation step.

The classic example uses the Pythagorean theorem. A 10 ft ladder leans against a wall, and its base slides away at 1 ft/s. With x as the distance from the wall and y as the height on the wall, x² + y² = 100. Differentiating gives 2x(dx/dt) + 2y(dy/dt) = 0, so dy/dt = −(x/y)(dx/dt). When the base is 6 ft out, the top is 8 ft up, and dy/dt = −(6/8)(1) = −0.75 ft/s. The negative sign says the top is sliding down.

Other standard setups include a cone-shaped tank whose volume V = (1/3)πr²h relates draining rate to falling water level, an expanding circle or sphere relating radius growth to area or volume growth, and two vehicles moving apart along perpendicular roads. In each case the geometry supplies the equation and a substitution such as a similar-triangles ratio often eliminates an extra variable before differentiating.

Related rates is a named topic on the AP Calculus AB exam, appearing in both multiple-choice and free-response sections. Graders expect the relating equation, the differentiated form, correct units, and a sign interpreted in context — and the same chain-rule reasoning underpins optimization problems, which the exam pairs with it in the contextual and analytical applications units.