Let A(t) be the (changing) area of a circle with radius r(t), in feet, at any time t in min.

If the radius is changing at the rate of dr/dt =3ft/min, find the rate of change
of the area(dA/dt) at the moment in time when r = 16 ft

dA/dt=

A = πr^2

dA/dt = 2πr dr/dt

so when r = 16 ft and dr/dt = 3 ft/min

dA/dt = 2π(16)(3) ft^2/min
= 96π ft^2/min

To find the rate of change of the area with respect to time (dA/dt), we can use the formula for the derivative of the area of a circle with respect to its radius.

The formula for the area of a circle is A = πr^2, where A is the area and r is the radius.

To find dA/dt, we need to take the derivative of both sides of the equation with respect to time (t). However, we don't have an explicit equation for A in terms of t. Instead, we have A in terms of r and r in terms of t.

By differentiating the area formula, we get:

dA/dr = 2πr

This equation gives us the rate of change of the area with respect to the radius (dA/dr).

Now, we need to find the rate of change of the radius with respect to time (dr/dt). The problem states that dr/dt = 3 ft/min.

Finally, to find dA/dt, we can use the chain rule, which states that dA/dt = (dA/dr) * (dr/dt).

Substituting the known values, we get:

dA/dt = (2πr) * (dr/dt)

Given that r = 16 ft and dr/dt = 3 ft/min:

dA/dt = (2π(16)) * 3
= 96π ft^2/min

So, the rate of change of the area at the moment in time when r = 16 ft is 96π square feet per minute.