In differential calculus, **Related Rates** problems are an application of derivatives, where one uses given lengths and rates to find missing ones. The rate of change is usually respective to time.

## Steps

Generally, the steps are as follows.

- Find given and missing values
- Related them in an equation
- Implicitly derive both sides with respect to time
- Substitute known quantities and solve

### Common equations used

- Rectangles: $ A=bh $

- Triangles: $ a^2+b^2=c^2, A=1/2bh $

- Spheres: $ V=4/3\pi r^3, SA=4\pi r^2 $

- Cylinders: $ V= \pi r^2h $

- Cones: $ V= 1/3\pi r^2h $

**Note**: With cones, usually one must do some reasoning with $ r $ and $ h $ by setting them in a proportion; i.e. since $ r=x $ when $ h=y $, $ r/h = x/y $, etc. Solve for whatever variable is not given in terms of the other one.

## Example

Oil is spilling from a ruptured tanker. The area of the spill is changing at a rate of 6 mi^{2}/h. To find the rate the radius is changing when the area is 9 mi^{2}, begin by listing which values are known and which are not.

$ A=9 $

$ \frac{dA}{dt}=6 $

$ R=\frac{3}{\sqrt{\pi}} $ (obtained by geometric formulas)

$ \frac{dR}{dt}=? $

Now take the derivative of the formula for the area of circle.

$ A=\pi r^2 $

$ \frac{dA}{dt}=2\pi r \frac{dR}{dt} $

From here, it is only a simple matter of solving for $ \frac{dR}{dt} $ and substituting the known values.

$ \frac{\frac{dA}{dt}}{2\pi r}=\frac{dR}{dt} $

$ \frac{6}{\frac{6\pi}{\sqrt{\pi}}}=\frac{dR}{dt} $

$ \frac{dR}{dt}=\frac{1}{\sqrt\pi}\approx0.5642 mi/h $