Question

How does the diameter of each ripple change as time passes

Answer 1

The diameter of the ripple changes by the following equation = $\frac{dA}{dt} = \frac{\pi }{2} D . \frac{d(D)}{dt}$

Explanation:

When a pebble or a rock is dropped in the water then the ripples or waves are formed which starts moving radially outward. As the time passes, the radius of the circle increases.

Let D be the diameter of the circle.

$\frac{dD}{dt}$ will be the rate of change of diameter of the circle with respect to time.

Let A be the area of the circle.

$\frac{dA}{dt}$ will be the rate of change of area of the circle with respect to time.

Area of the circle = $\frac{\pi }{4} D^{2}$

So,

$\frac{dA}{dt} = \frac{d [\frac{\pi }{4} D^{2}] }{dt}$

$\frac{dA}{dt} = \frac{\pi }{4} \frac{d[ D(t)^{2}] }{dt}$

Using chain rule,

$\frac{dA}{dt} = \frac{\pi }{4} X \frac{d[D(t)^{2} ]}{d[D(t)]} X \frac{d[D(t)]}{dt}$

$\frac{dA}{dt} = \frac{\pi }{4} . 2D(t) . \frac{dD}{dt}$

$\frac{dA}{dt} = \frac{\pi }{2} X D(t) X \frac{d(D)}{dt}$

$\frac{dA}{dt} = \frac{\pi }{2} D . \frac{d(D)}{dt}$

Thus, the following equation determines the change in area in terms of diameter with respect to time :  $\frac{dA}{dt} = \frac{\pi }{2} D . \frac{d(D)}{dt}$