A stone is dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 2 meters per second. How rapidly is the area enclosed by the ripple increasing at the end of 20 seconds?

To find how rapidly the area enclosed by the ripple is increasing at the end of 20 seconds, we need to use the formula for the area of a circle: A = πr^2, where A is the area and r is the radius.

First, let's find the radius of the ripple after 20 seconds. We know that the radius increases at a constant rate of 2 meters per second. So, after 20 seconds, the radius will be:

r = 2 meters/second * 20 seconds = 40 meters

Now we can find the area of the ripple. Plugging the radius into the formula, we get:

A = π * (40 meters)^2

Simplifying this, we have:

A = π * 1600 square meters

To find how rapidly the area enclosed by the ripple is increasing at the end of 20 seconds, we need to differentiate the area with respect to time. Taking the derivative of both sides of the equation, we get:

dA/dt = 2πr * dr/dt

where dA/dt represents the rate of change of the area with respect to time.

We already know that dr/dt (rate of change of the radius with respect to time) is 2 meters per second. So, plugging in the values, we get:

dA/dt = 2π * 40 meters * 2 meters/second

Simplifying this, we have:

dA/dt = 160π square meters/second

Therefore, the area enclosed by the ripple is increasing at a rate of 160π square meters per second at the end of 20 seconds.

To find the rate at which the area enclosed by the circular ripple is increasing at the end of 20 seconds, we need to use the formula for the area of a circle and differentiate it with respect to time.

The formula for the area of a circle is:

A = πr^2

Where A is the area and r is the radius of the circle.

Given that the radius of the ripple is increasing at a constant rate of 2 meters per second, we can write r = 2t, where t is the time in seconds.

Substituting this value into the formula for the area, we get:

A = π(2t)^2
A = 4πt^2

Now, we need to differentiate the area formula with respect to time (t) to find the rate of change of area:

dA/dt = d/dt(4πt^2)
dA/dt = 8πt

To find the rate at the end of 20 seconds, we substitute t = 20 into the derivative:

dA/dt = 8π(20)
dA/dt = 160π

Therefore, the area enclosed by the ripple is increasing at a rate of 160π square meters per second at the end of 20 seconds.

hint

dA = 2 pi r dr so dA/dr = 2 pi r
and
dA/dt = dA/dr * dr/dt