# High School Calculus/Related Rates

## Related Rates

When you need to find the rate of change of two or more related variables, you can use chain rule to find these related rates.

Remember, you can think of derivatives as rates of change for the function you are deriving.

In the case of related rates, you are performing differentiation with respect to time, t.

When solving related rates, here are some steps you can take to aid solving the problem.

1. Identify all the variables you are given, and all of the variables to be determined.

2. Draw a picture of the scenario.

3. Find an equation related to all variables, known or unknown.

4. Use chain rule to differentiate implicitly with respect to time.

5. Substitute in all given information.

6. Solve for the unknown variable.

### Example

Imagine a balloon that is having air pumped inside. If we knew the rate of change of the volume to be 8 inches per second, what would the rate of change be for the radius of the balloon when the radius is 4 inches?

These are related rates.

The first step is to determine all the known and unknown variables.

We know that the change in volume is 8 inches per second, and that the radius of the balloon will be 4 inches. We are trying to find the rate of change of the radius.

The second step is to draw a diagram that shows as air is being pumped into the balloon, its radius will be expanding.

Third, we need an equation relating these variables.

Since we are dealing with the volume of a sphere, we know this equation to be: ${\displaystyle V={\frac {4}{3}}{\pi }r^{3}.}$

Fourth, we need to differentiate with respect to time, this looks like:

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

Fifth, we need to substitute in all known information, which is the rate of change of the volume, and the radius of the sphere.

${\displaystyle 8=4{\pi }(4)^{2}{\frac {dr}{dt}}}$

${\displaystyle 8=64{\pi }{\frac {dr}{dt}}.}$

Lastly, we need to solve for the rate of change of the radius. Since all the units were inches, no changes need to be made beforehand.

${\displaystyle {\frac {8}{64{\pi }}}={\frac {dr}{dt}}}$

${\displaystyle {\frac {1}{8{\pi }}}={\frac {dr}{dt}}.}$

The rate at which the radius is changing is ${\displaystyle {\frac {1}{8{\pi }}}}$ per second.

Sometimes related rates can be much more difficult and involve multiple unknown variables. In these cases you may need to use multiple equations to find unknown variables before finding the initial unknown you were looking for.

The situation of a related rate can be many things, from volumes, to airplanes and velocity, to basketballs, etc.

Something else to watch is the units of your variables. In some cases you may need to convert your variables to match units.

Here are some equations that are frequently used for related rates.

Circles/Spheres: Area = ${\displaystyle {\pi }r^{2}}$ Volume = ${\displaystyle {\frac {4}{3}}{\pi }r^{3}}$ Surface Area = ${\displaystyle 4{\pi }r^{2}}$

Distance Formula: ${\displaystyle d={\sqrt {x^{2}+y^{2}}}}$

Cones: A = area of base = ${\displaystyle {\pi }r^{2}}$ Volume = ${\displaystyle {\frac {Ah}{3}}}$

Cylinder: Volume = ${\displaystyle {\pi }r^{2}h}$

### Practice Problems

A person is watching a plane fly by. Let z be the distance from the person to the plane. Let x be the horizontal distance between the plane and the person. If the rate at which z is decreasing is 200 miles per hour when z = 5 miles, what is the actual speed of the plane?

*Hint* Use distance formula, and draw a diagram of what is happening.

A small child is pouring sand into a conical pile. If sand is being dropped at a rate of 5 cubic centimeters a second, if the diameter of the base is two times the height of the pile, at what rate is the height of the pile changing when it is 10 centimeters tall?