Well part A gives us the radius of 25M so we have enough to find how fast the area is changing.
We will use the equation > dA/dt = 2pirdr/dt
dA/dt = 2pi(25)(2) = 100pi = 314.6
So the area would be increasing by 314.6 square meters per minute or 100pi square meters per minute.
For part B we have t = 0 which then means t = 4, the value will then have increased to 8m. So we plug it into the equation.
dA/dt = 2pi(8)(2) = 32pi = 100.53
This means the area is increasing at 100.53 square meters per minute at t = 4!
