Question

At what rate, in cm/s, is the radius of the circle increasing when the radius is 5 cm if oil is poured on a flat surface, and it spreads out forming a circle and the area of this circle is increasing at a constant rate of 5 cm2/s?

Answer 1

You can think that both your area `A` and radius `r` are function of time so basically:
`A(t)=pi*[r(t)]^2` (area of the circle)

deriving you get that your area changes and your radius changes as:

`(dA(t))/dt=2pir(t)(dr(t))/dt`

But:

`(dA(t))/dt=5(cm^2)/s`
Considering the instant when `r(t)=5cm` you get:
`5=2pi*5(dr(t))/dt` you can simplify `5` and take `2pi` to the side dividing;
and finally the rate of change of your radius will be:
`(dr(t))/dt=1/(2pi)=0.16(cm)/s`