Question

A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate of 3 foot per second. When the radius is 8 feet, at what rate is the total area A of the disturbed water changing?

Answer 1

Answer:

150.79 ft²/s

Explanation:

area of the circle = π r²

$\frac{\mathrm{d} r}{\mathrm{d} t} = 3 ft/s$

A = π r²

$\frac{\mathrm{d} A}{\mathrm{d} t} = \pi r^2$

$\frac{\mathrm{d} A}{\mathrm{d} t} = \pi \frac{\mathrm{d}r^2 }{\mathrm{d} t}$

$\frac{\mathrm{d} A}{\mathrm{d} t} = \pi\times 2\times 8\times 3$

$\frac{\mathrm{d} A}{\mathrm{d} t} = 150.79 ft^2/s$

hence the rate of total area of the disturbed water changing is equal to

150.79 ft²/s