Final answer:
The area enclosed by the ripple is increasing at a rate of 68.544 ft^2/s when the radius is 4 feet, and after 8.4 seconds, the rate of increase is 385.543 ft^2/s.
Step-by-step explanation:
To solve part (a) of the question involving the circular ripple in a pond with a radius increasing at a constant rate, we will use the formula for the area of a circle, A = πr^2, where r is the radius and π is the constant 3.14159. Since the radius increases at a rate of 2.7 ft/s, we use the chain rule to find how quickly the area is increasing. We differentiate both sides with respect to time (t) to get dA/dt = 2πr(dr/dt). When the radius is 4 feet, we have dA/dt = 2π(4 ft)(2.7 ft/s), which gives us a rate of area increase of 68.544 ft^2/s.
For part (b), to find the rate at which the area is increasing at the end of 8.4 seconds, we first calculate the radius at that time which is r = 2.7 ft/s × 8.4 s = 22.68 ft. Plugging this radius into the derivative equation for area, we get dA/dt = 2π(22.68 ft)(2.7 ft/s) which gives us a rate of area increase of 385.543 ft^2/s.