Derivative of the Area of a Circle

husky88

1. The problem statement, all variables and given/known data

Show that the rate of change of the area of a circle with respect to its radius is the same as the circumference of the circle. Can you suggest why?

2. Relevant equations

A = [tex]\pi[/tex]r[tex]^{2}[/tex] = f(r)
L = 2[tex]\pi[/tex]r = g(r)

3. The attempt at a solution

I have showed that the derivative of f(r) is equal to g(r).
But I have no idea why the area and the circumference of the circle are related in such a way. Any suggestions greatly appreciated.
Thank you.

Quisquis

Start by thinking about what any derivative of a function is describing in general, and then how it applies here specifically.

husky88

The derivative describes the slope of a tangent to the circle which is perpendicular to the radius... but I don't seem to go anywhere from here... hmmm