Question

A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, how rapidly is the area enclosed by the ripple increasing?

Answer 1

Answer:

After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

Explanation:

Area of a circle = πr²

where;

r is the circle radius

Differentiate the area with respect to time.

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

dr/dt = 4 ft/sec

after 12 seconds, the radius becomes = $\frac{dr}{dt} X 12 = 4 \frac{ft}{sec} X 12 sec = 48 ft$

To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$

$\frac{dA}{dt} = 2\pi (48)(4)$

    dA/dt = 1206.528 ft²/sec

Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec