Related Rates: Interactive Word Problem Guide & Solver

To solve a related rates problem, write an equation relating the quantities, differentiate both sides implicitly with respect to time, then substitute the known values and rates to solve for the unknown rate. Plug in fixed (constant) quantities only after differentiating.

Related Rates is widely considered one of the most difficult topics in calculus. This is because you're given word problems where you have to create the equations yourself.

Here is the best way to solve related rate problems:

1. Figure out the shape & visualize

First, figure out what shape you're dealing with so you can visualize what's going on. Once you know the shape, you can grab the equations related to it that will help you solve the problem.

2. Find what's not moving (constants)

Next, figure out what is not moving in the scenario. Whatever isn't moving is a constant. You can plug in that constant value right away to remove that variable, because when you take the derivative, it disappears.

3. Take the derivative implicitly

Finally, when you take the derivative, be very careful to differentiate implicitly with respect to time.

How do you solve a related rates problem?

\underbrace{\frac{dV}{{\color{#60a5fa}dt}} = 4\pi r^2 \frac{dr}{{\color{#60a5fa}dt}}}_{\text{Example Formula (Sphere Volume)}}

Related rates worked example: inflating spherical balloon

A spherical balloon is inflated with gas at a rate of 20 cm³/s. Find the rate at which the radius is increasing when the radius is 5 cm.

1. Identify the given rate of change of volume: dV/dt = 20. We want to find dr/dt when r = 5.

2. The volume of a sphere is V = 4/3 ℼ r³.

3. Differentiate implicitly with respect to time:

\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

4. Substitute r = 5 and dV/dt = 20:

20 = 4\pi (5)^2 \frac{dr}{dt} \implies 20 = 100\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{5\pi} \approx 0.064 \text{ cm/s}

Another worked example

A 13-foot ladder slides down a wall; its base moves away at 2 feet per second. Find how fast the top slides down when the base is 5 feet out. Step 1: x squared plus y squared equals 169, so y equals 12 when x equals 5. Step 2: differentiate to get 2x times dx-dt plus 2y times dy-dt equals 0. Step 3: substitute x = 5, y = 12, dx-dt = 2 to get dy-dt equals negative 5 over 6 feet per second.

Aligned with the College Board AP Calculus CED — Unit 4 (Contextual Applications of Differentiation), Topics 4.4 and 4.5. College Board ↗

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