A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 2.6 ft/s.

(a) How rapidly is the area enclosed by the ripple increasing when the radius is 4 feet?

The area is increasing at

Thank you

To find how rapidly the area enclosed by the ripple is increasing, we can use the formula for the area of a circle:

A = π*r^2,

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

To find how the area is changing with respect to time, we can differentiate both sides of the equation with respect to time (t). Since both the area and the radius are functions of time, we can use the chain rule to differentiate the equation.

dA/dt = d/dt(π*r^2),

Now, let's substitute the given information into the equation. We know that the radius is increasing at a constant rate of 2.6 ft/s. Therefore, dr/dt = 2.6 ft/s.

Let's differentiate the equation using the chain rule:

dA/dt = 2π*r*(dr/dt).

Now, substitute the values:

dA/dt = 2π*(4)*(2.6) ft^2/s.

Calculating this value gives us:

dA/dt ≈ 32.832 ft^2/s.

Therefore, when the radius is 4 feet, the area enclosed by the ripple is increasing at a rate of approximately 32.832 ft^2/s.

To find how rapidly the area enclosed by the ripple is increasing, we need to find the derivative of the area 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.

To find the rate at which the area is changing with respect to time, we will take the derivative of the area formula.

dA/dt = 2πr(dr/dt)

Here, dA/dt represents the rate of change of the area with respect to time, dr/dt represents the rate at which the radius is changing, and r represents the radius of the circle.

Now we can substitute the given values into the formula:

dr/dt = 2.6 ft/s (given)

r = 4 ft (given)

dA/dt = 2π(4)(2.6) = 20.8π ft^2/s

Therefore, the area is increasing at a rate of 20.8π square feet per second when the radius is 4 feet.

area = π r^2

d(area)/dt = 2π r dr/dt
given: dr/dt = 2.6 , r = 4

d(area)/dt = .....

just plug in your values.
Mind the units