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

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anonymous · Answer

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$$\pi*2*(2*60)*60 = 14400\pi$$

anonymous · Answer

Thank you so much!