Problem Solving > Related Rates

**Related rate **problems involve functions where a relationship exists between **two or more derivatives**. For example, you might want to find out the rate that the distance is increasing between two airplanes.

Solving related rate problems has many real life applications. For example, a gas tank company might want to know the rate at which a tank is filling up, or an environmentalist might be concerned with the rate at which a certain marshland is flooding.

Solving the problems usually involves knowledge of geometry and algebra in addition to calculus. Often, the “hard” part is the geometry or algebra—not the calculus, so you’ll want to make sure you brush up on those skills. Solving related relate problems also involves applications of the chain rule and implicit differentiation—where you differentiate both sides of the equation.

## Related Rates Example problem #1

Q. A rock is dropped into the center of a circular pond. The ripple moved outward at 4 m/s. How fast does the area change, with respect to time, when the ripple is 3m from the center?

Step 1: **Draw a picture of the problem **(this *always* helps, especially when geometry is involved).

Step 2: **Write out what you know about the problem**, using equations. You know that the rate at which the ripple (r, the radius of the circle) is moving, with respect to time *t*, is 4 m/s, so:

- dr/dt = 4 m/s

Step 3: **Write out what you want to know **(what you are trying to solve for). You want to know how fast the area

*A*is changing with respect to time

*t*:

- dA/dt = x, when r = 3m

Step 4: **Use the chain rule to find a solution for your Step 3 equation.** The chain rule tells us that:

- dA/dt = dA/dr * dr/dt

Step 5: **Figure out what dA/dr is. **From geometry, we know that A = πr^{2}. So dA/dr is just the derivative of A = πr^{2}.

- d/dr[A] = d/dr[πr
^{2}] - = dA/dr = 2πr
- So: dA/dt = 2πr * dr/dt

Step 6: **Solve the Step 5 equation.** You know that dr/dt (from Step 2) is 4 m/s, and r is 3 m (from the question), so:

- dA/dt = 2π(3) * 4 m/s
- = 24π m
^{2}/s

*That’s it!*

## Related Rates Example problem #2:

The length of a rectangular drainage pond is changing at a rate of 8 ft/hr and the perimeter of the pond is changing at a rate of 24 ft/hr. At what rate is the width changing?

Step 1: **Figure out which geometric formulas are related **to the problem. From basic geometry, the formula for perimeter is P = (2*l) + (2*w) and that A = l*w.

Step 2: **Differentiate the perimeter equation**:

dp/dt = (2*dl/dt) + (2*dw/dt)

Step 3: Substitute in the information you know from the question. You know that the rate at which the perimeter is changing is 24 ft/hr and the length is changing at 8 ft/hr, so (using algebra):

- 24 = (2*8) + (2*dw/dt)
- Dividing by 2 gives us: 12 = 8 + dw/dt
- Subtracting 8 from both sides: 4 = dw/dt

The width of the drainage pond is changing at 4 ft/hr.

*That’s it!*

**CITE THIS AS:**

**Stephanie Glen**. "Related Rates: Example Problems Step by Step" From

**CalculusHowTo.com**: Calculus for the rest of us! https://www.calculushowto.com/related-rates/

**Need help with a homework or test question? **With Chegg Study, you can get step-by-step solutions to your questions from an expert in the field. Your first 30 minutes with a Chegg tutor is free!