Each side of a square is increasing at a rate of 3 cm/s3 cm/s . At what rate is the area of the square increasing when the area of the square is 9 cm29 cm2 ?

The problem involves rates of change, so we'll use differential calculus to solve it. Here are the steps:

1. Let's denote the side of the square as s and the area of the square as A. We know that A = s^2.

2. We're given that ds/dt = 3 cm/s, which is the rate at which the side of the square is increasing.

3. We're asked to find dA/dt, the rate at which the area of the square is increasing, when A = 9 cm^2.

4. First, we need to find the value of s when A = 9 cm^2. Since A = s^2, we can solve for s to get s = sqrt(A) = sqrt(9) = 3 cm.

5. Now we can find dA/dt. We know that dA/dt = 2s * ds/dt (this comes from differentiating the equation A = s^2 with respect to t).

6. Substituting the given values, we get dA/dt = 2 * 3 cm * 3 cm/s = 18 cm^2/s.

So, the area of the square is increasing at a rate of 18 cm^2/s when the area of the square is 9 cm^2.