Question

A rock is thrown into a lake. The radius of the ripple formed by the rock grows at a rate of 60 cm/sec. Find the rate at which the area is growing 2 seconds after the rock hits the lake

Answer 1

Nice question. So formula of the area with respect to time is\[A = \pi*(t*60) ^{2}\] where radius of the ripple is t*60. To find the growth rate you need to take its derivative.
\[A \prime(2) = \pi*2*(t*60)*60\] at t=2. So the answer is\[\pi*2*(2*60)*60 = 14400\pi\]

Answer 2

Thank you so much!