The area of a circle is changing at a rate of 2 in.^(2)/sec. At what rate is its radius changing when the radius is 9 in.?

a. (9/2)in./sec.

b. (2/9)in./sec.

c. (2/9pie)in./sec.

d. (1/9pie)in./sec.

e. (1/9)in./sec.

A = πr^2

d(A)/dt = 2πr dr/dt

given: dA/dt = 2
find : dr/dt when r = 9

2 = 2π(9) dr/dt
dr/dt = 1/(9π) inches/sec

To find the rate at which the radius is changing, we need to use the relationship between the area and the radius of a circle. The formula for the area of a circle is A = πr^2, where A is the area and r is the radius.

We are given that the area is changing at a rate of 2 in^2/sec. So, dA/dt = 2 in^2/sec.

We need to find the rate at which the radius is changing, dr/dt, when the radius is 9 in.

To do this, we can use the chain rule of differentiation. Since A = πr^2, we can differentiate both sides of the equation with respect to time:

d/dt(A) = d/dt(πr^2)

dA/dt = 2πr(dr/dt)

Now we can plug in the given values: dA/dt = 2 in^2/sec and r = 9 in.

2 in^2/sec = 2π(9 in)(dr/dt)

Simplifying the equation:

2π(9 in)(dr/dt) = 2 in^2/sec

Dividing both sides by 2π(9 in):

(dr/dt) = 2 in^2/sec / (2π * 9 in)

Simplifying further:

(dr/dt) = 1 / (9π) in/sec

Therefore, the rate at which the radius is changing when the radius is 9 in. is (1/9π) in/sec.

So the correct answer is (d) (1/9π) in/sec.