Related Rates: Interactive Word Problem Guide & Solver

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.

Formula

dVdt=4πr2drdtExample Formula (Sphere Volume)\underbrace{\frac{dV}{{\color{#60a5fa}dt}} = 4\pi r^2 \frac{dr}{{\color{#60a5fa}dt}}}_{\text{Example Formula (Sphere Volume)}}

Easy Example Problem

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:

dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

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

20=4π(5)2drdt    20=100πdrdt    drdt=15π0.064 cm/s20 = 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}
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