Question

A pebble is dropped into a calm pond, causing ripples in the form of concentric circles. The radius of the outer ripple is increasing at a constant rate of 6 inches per second. When the radius is 4 feet, at what rate (in ft2/sec) is the total area A of the disturbed water changing

Answer 1

Answer:

The rate of area of the disturbed water changing is 12.56 $\frac{ft^{2} }{sec}$

Explanation:

Given:

Radius increasing rate $\frac{dr}{dt} = 6$ $\frac{in}{sec}$

Radius $r = 4$ ft

Now we convert radius increasing rate into feet per second,

   $\frac{dr}{dt} = \frac{6}{12} \frac{ft}{sec}$

Here we have to find total area rate $\frac{dA}{dt}$

    $A = \pi r^{2}$

   $\frac{dA}{dt} = \frac{d(\pi r^{2} )}{dt}$

   $\frac{dA}{dt} =\pi \frac{d( r^{2} )}{dt}$

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

   $\frac{dA}{dt} = 2\pi \times 4 \times \frac{6}{12}$

   $\frac{dA}{dt} = 12.56$ $\frac{ft^{2} }{sec}$

Therefore, the rate of area of the disturbed water changing is 12.56 $\frac{ft^{2} }{sec}$