To find the rate at which the area of a square is increasing when the area of the square is 81 cm², we need to use the relationship between the side length of the square and its area.

Let $s$ be the side length of the square, and $A$ be the area of the square. The area of the square is given by: $A = s^2$

We are given that the side length $s$ is increasing at a rate of 2 cm/s, which means: $\frac{ds}{dt} = 2 \, \text{cm/s}$

We need to find the rate at which the area $A$ is increasing, which is $\frac{dA}{dt}$. We use the chain rule to relate $\frac{dA}{dt}$ to $\frac{ds}{dt}$: $\frac{dA}{dt} = \frac{dA}{ds} \cdot \frac{ds}{dt}$

Since $A = s^2$, we have: $\frac{dA}{ds} = 2s$

Substituting this into our chain rule equation gives: $\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}$

We are given that the area of the square is 81 cm². To find the corresponding side length $s$, we solve for $s$ in the equation: $s^2 = 81$ $s = \sqrt{81}$ $s = 9 \, \text{cm}$

Now we substitute $s = 9$ cm and $\frac{ds}{dt} = 2 \, \text{cm/s}$ into the equation for $\frac{dA}{dt}$: $\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}$ $\frac{dA}{dt} = 2 \cdot 9 \cdot 2$ $\frac{dA}{dt} = 36 \, \text{cm}^2/\text{s}$

Therefore, the rate at which the area of the square is increasing when the area is 81 cm² is $36 \, \text{cm}^2/\text{s}$.

Would you like more details or have any questions about this problem? Here are some related questions to expand your understanding:

1. How would the rate of change of the area be different