. A rock drops into a still lake, creating a circular ripple that expands so that the radius grows at a rate of

1.5 feet per second. At the moment when the area is 16π square feet, find the rate of change of the
following: a) the area and b) the circumference

when A = 16π, r=4, C = 2πr = 8π

dA = C dr, so
(a) dA/dt = 8π * 1.5 = 12π ft^2/s
(b) dC/dt = 2π dr/dt = 3π ft/s

Thanks oobleck

To find the rate of change of the area and circumference of the circular ripple, we can start by using the formulas for the area and circumference of a circle.

a) The formula for the area of a circle is A = πr^2, where A is the area and r is the radius. Since we know that the radius is growing at a rate of 1.5 feet per second, we can differentiate the formula with respect to time to find the rate of change of the area (dA/dt):

dA/dt = d/dt (πr^2)

To find dA/dt, we need to know the rate of change of the radius (dr/dt). Since the radius grows at a constant rate of 1.5 feet per second, dr/dt is simply 1.5 ft/s.

Now, let's differentiate the formula for the area:

dA/dt = d/dt (πr^2)
= 2πr (dr/dt)

Substituting the given value of dr/dt:

dA/dt = 2πr (1.5 ft/s)

At the moment when the area is 16π square feet, we can find the radius by rearranging the formula for the area:

A = πr^2
16π = πr^2
r^2 = 16
r = 4 feet

Substituting this value of r into the formula for dA/dt:

dA/dt = 2π(4 ft) (1.5 ft/s)
= 12π ft^2/s

Therefore, at the moment when the area is 16π square feet, the rate of change of the area is 12π square feet per second.

b) The formula for the circumference of a circle is C = 2πr, where C is the circumference and r is the radius. To find the rate of change of the circumference (dC/dt), we can differentiate the formula with respect to time:

dC/dt = d/dt (2πr)

Again, we need to know the rate of change of the radius (dr/dt). Since the radius grows at a constant rate of 1.5 feet per second, dr/dt is 1.5 ft/s.

Differentiating the formula for the circumference:

dC/dt = d/dt (2πr)
= 2π (dr/dt)

Substituting the given value of dr/dt:

dC/dt = 2π (1.5 ft/s)
= 3π ft/s

Therefore, at the moment when the area is 16π square feet, the rate of change of the circumference is 3π feet per second.