Question 596252
We have


*[tex \LARGE \frac{dr}{dt} = 8t + 3 = 11 \Rightarrow t = 1] so when dr/dt = 11, t = 1.


The area of the circle as a function of time is


*[tex \LARGE A = \pi (r(t))^2]


*[tex \LARGE A = \pi (4t^2 + 3t + 1)^2]


We differentiate both sides with respect to t:


*[tex \LARGE \frac{dA}{dt} = 2\pi (4t^2 + 3t + 1)(8t + 3)]


Replace t = 1 to obtain


*[tex \LARGE \frac{dA}{dt} = 2\pi (8)(11) = 176 \pi] (units squared per unit of time)