Related Rates: A Repeatable 6-Step Problem Strategy

Related rates problems sound abstract — "a ladder slides down a wall, how fast is the top falling?" — but they all follow the same six-step pattern. Master the recipe and these problems shift from terrifying to mechanical.

The 6-step recipe

1. Read the problem twice and identify every quantity. Sketch it.
2. Label quantities that change with letters; constants with numbers.
3. Find an equation relating the changing quantities (geometry, Pythagorean, similar triangles, area, volume…).
4. Differentiate both sides with respect to time $t$ implicitly. Every changing quantity contributes a $\frac{d \cdot}{dt}$ term.
5. Plug in the snapshot values only after differentiating. Substituting too early kills the rate information.
6. Solve for the unknown rate and double-check units.

Example 1: the sliding ladder

A 13 ft ladder leans against a wall. Its base slides outward at 2 ft/sec. How fast is the top sliding down when the base is 5 ft from the wall?

1. Variables: $x$ = base distance, $y$ = top height. Both change with $t$.
2. Constraint: $x^2 + y^2 = 169$ (Pythagorean — ladder length is constant).
3. Differentiate: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$.
4. Snapshot: $x = 5$, so $y = \sqrt{169 - 25} = 12$. Given $\frac{dx}{dt} = 2$.
5. Solve: $2(5)(2) + 2(12)\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6}$ ft/sec.

The top falls at $5/6$ ft/sec. Negative sign means the height is decreasing — sanity check passes.

Example 2: the cone filling with water

Water pours into a cone (vertex down) at $3 \text{ ft}^3/\text{min}$. The cone has height 10 ft and top radius 4 ft. How fast is the water level rising when the depth is 6 ft?

1. Variables: $V$ = water volume, $h$ = water depth, $r$ = water surface radius.
2. Volume of cone: $V = \frac{1}{3}\pi r^2 h$. Use similar triangles: $r/h = 4/10 \Rightarrow r = 0.4h$.
3. Substitute to one variable: $V = \frac{1}{3}\pi (0.4h)^2 h = \frac{0.16\pi}{3} h^3$.
4. Differentiate: $\frac{dV}{dt} = 0.16\pi h^2 \frac{dh}{dt}$.
5. Plug $h = 6$, $\frac{dV}{dt} = 3$: $3 = 0.16\pi (36) \frac{dh}{dt}$.
6. Solve: $\frac{dh}{dt} = \frac{3}{5.76\pi} \approx 0.166$ ft/min.

Common mistakes

• Plugging numbers in too early — derivatives "freeze" the relationship; you lose information about how things change.
• Forgetting the chain rule when differentiating something like $r^2$ — it becomes $2r \frac{dr}{dt}$, not $2r$.
• Not eliminating extra variables with similar triangles before differentiating.

Try with the AI Derivative Solver

Use the Derivative Calculator to verify any related-rate differentiation step — particularly the implicit ones.

Related references:

• Limit Calculator — derivatives are limits underneath
• Integral Calculator — antiderivative companion
• Triangle Solver — for the geometry setup of many problems