Originally Posted by

**TKHunny** Consider everything a function of time (t). For your circle:

$\displaystyle A = \pi\cdot r^{2}$ suggests $\displaystyle A(t) = \pi\cdot r(t)^{2}$

We find the derivative with respect to t = time.

On the left-had side, we have simply $\displaystyle dA(t)/dt$.

On the right-hand side, we have $\displaystyle \pi\cdot (2r(t)\cdot (dr(t)/dt))$

Putting it all together (and dispensing with the cumbersome '(t)' notation), we get

$\displaystyle dA/dt = \pi\cdot (2r\cdot (dr/dt))$

Knowing this, one is armed with enough to solve many problem types.

Actually, we may also want to dispose of the cumbersome 'dt' notation and just go with the differential usage.

$\displaystyle dA = \pi\cdot (2r\cdot dr)$

In words, this says, "The change in the area is $\displaystyle 2\pi r$ multiplied by the change in the radius.

We are given dr = 15 cm/s.

Let's see what you get.