Question

Water leaking onto a floor forms a circular pool. The radius of the pool increases at a rate of 4 cm/min. How fast is the area of the pool increasing when the radius is 5 cm?

Answer 1

`40pi` `"cm"^2"/ min"`

Explanation:

First, we should begin with an equation we know relating the area of a circle, the pool, and its radius:

`A=pir^2`

However, we want to see how fast the area of the pool is increasing, which sounds a lot like rate... which sounds a lot like a derivative.

If we take the derivative of `A=pir^2` with respect to time, `t`, we see that:

`(dA)/dt=pi*2r*(dr)/dt`

(Don't forget that the chain rule applies on the right hand side, with `r^2`--this is similar to implicit differentiation.)

So, we want to determine `(dA)/dt`. The question told us that `(dr)/dt=4` when it said "the radius of the pool increases at a rate of `4` cm/min," and we also know that we want to find `(dA)/dt` when `r=5`. Plugging these values in, we see that:

`(dA)/dt=pi*2(5)*4=40pi`

To put this into words, we say that:

The area of the pool is increasing at a rate of `bb40pi` cm`""^bb2`/min when the circle's radius is `bb5` cm.