The circumference of a circle is increasing at 11.6 feet/second. When the radius is 8 feet, at what rate(feet/sec) is the radius changing, and at what rate (feet^2/ sec) is the area changing?
One way to do this problem is to write a formula for the circumference in terms of time. So, you could do
C = 2π(8) + 11.6t
Differentiating this equation shows that dC/dt = 11.6, meaning it increases by 11.6 feet/second. We can rewrite the above equation as
2πr = 2π(8) + 11.6t
(divide by 2π)
r = 8 + 1.8462t
You differentiate that, and you find dr/dt = 1.846. So, the radius changes at 1.846 feet/second.
Now, in order to find the area, we can do this:
A = πr^2
A = π(8 + 1.8462t)^2
A = π(3.40845 t^2 + 29.5392 t + 64)
A = 10.708t^2 + 92.8001t + 64π
Now, when you differentiate that, you get 21.416t + 92.8001. Recall that we want the rate when r = 8. So, we plug in 8 into our r(t) equation. 8 = 8 + 1.8462t, 0 = 1.8462t, t = 0. So, the rate at which the radius changes is 92.8 feet^2/sec