Question

Each side of a square is increasing at a rate of 2 cm/s. At what rate is the area of the square increasing when the area of the square is 9 cm^2?

Answer 1

Set up the geometry here. Write the Area as a function of side length, and then observe what occurs when you differentiate the function with respect to side length.

Answer 2

You're trying to find the rate at which the Area of the square is increasing with respect to time.

Answer 3

That's your square.

Answer 4

So each side is x, and x is increasing with time, so

x = x(t)

rate at which each side is increasing \[\rm \frac{\partial x}{\partial t}=2~\frac{\text{cm}}{\text{sec}}\]

Answer 5

Once you obtain this, you can use the chain rule with the abovementioned relationship between dA and dx to get dA/dt.

Answer 6

The Area of a square is: \(A_s = x^2\) , \(\therefore\) if the side is changing with respect to time, the area is changing with respect to the side, as well as with respect to time.\[\frac{\partial A}{\partial t} =\frac{\partial A}{\partial x} \cdot \frac{\partial x}{\partial t} \]

Answer 7

\[A =x^2\]\[9 ~ \text{cm}^2 = x^2\]\[x = \sqrt{9} ~\text{cm}^2 = 3~ \text{cm}^2\]

Answer 8

\[\frac{\partial A}{\partial t} =\frac{\partial A}{\partial x} \cdot \frac{\partial x}{\partial t}\]\[\frac{\partial A}{\partial t} = 2x \cdot \frac{\partial x}{\partial t}\]plug in your values.