To find the instantaneous rate of change of a function at a point, we take the first derivative of the function and evaluate the derivative at that point.

`A(r)=pi r^2`

`(dA)/(dr)=2pir` so `A'(r)=2pir`

When the radius is 10:

`A'(10)=2pi(10)=20pi~~62.83`

So when the radius is 10ft, the area is increasing by `20pi"sqft"` or approximately 63 sq ft