Each side of a square is increasing at a rate of 2 cm/s. At what rate (in cm2/s) is the area of the square increasing when the area of the square is 81 cm2?

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

Discover how to calculate the rate at which the area of a square increases when its side length grows. Use calculus principles to solve for the area and side length changes, suitable for high school students.

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