Assuming at t = 0, C = 8 ft
then r(t = 0) = 1.2732 ft

C = 2(pi)r

dC/dt = 2(pi)(dr/dt) = 7 ft/min

solve for dr/dt = 1.1141 ft/min

this seems o.k. given C is a linear function of r

Integrate to get r(t) = (1.1141 ft.min)t + C

Evaluating r(t = 0) C = 1.2732 ft

r(t) = (1.1141 ft.min)t + 1.2732 ft

A(t) = (pi)[r(t)]^2 = (pi)[(1.1141 ft.min)t + 1.2732 ft]^2

Expand

A(t) = (pi)[(1.2412 ft^2/min^2)t^2 + (2.8369 ft/min)t + 1.6210]

dA/dt = (pi)[(2)(1.2412 ft^2/min^2)t + (2.8369 ft/min)]

reducing

dA/dt = (7.7987 ft^2/min^2)t + (8.9125 ft^2/min)

this shows the area rate of change is changing with time.

The functions r(t) and A(t) can be double checked with the original information. Calculate C for various minutes and calculate the corresponding radius (C=2*pi*r) and area (A = pi*r^2) and then calculate the radius and area using the functions r(t) and A(t).

References :
From very dusty memories…