Related rates

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.

1. Find given and missing values
2. Related them in an equation
3. Implicitly derive both sides with respect to time
4. 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 mi²/h. To find the rate the radius is changing when the area is 9 mi², 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.
$\frac{\frac{dA}{dt}}{2\pi r}=\frac{dR}{dt}$

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

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