Question

Each side of a square is increasing at a rate of $6\mathrm{cm}\mathit{/}s$. At what rate is the area of the square increasing when the area of the square is $16{\mathrm{cm}}^2$?

Answer 1

Find the rate at which the area of the square increases:

Consider $x$ as the side length of the square.
Then the area of the square is $A\left(x\right)=x^2$.
The area of the square is $16{\mathrm{cm}}^2$ when the side length$\left(x\right)$ is $4\mathrm{cm}$.
The rate at which the area is increasing must be determined, when $x=4$.

It's known that each side is increasing at the rate of $6\mathrm{cm}/\mathrm{s}$, so that $\frac{dx}{dt}=6$.
Differentiate the two sides based on $t$:
$$\begin{gathered} \frac{dA}{dt}=\frac{d\left(x^2\right)}{dt} \\ \Rightarrow \frac{dA}{dt}=\frac{d\left(x^2\right)}{dx}\times \frac{dx}{dt} \\ \Rightarrow \frac{dA}{dt}=2x^{2-1}\times \frac{dx}{dt} \\ \Rightarrow \frac{dA}{dt}=2x\times \frac{dx}{dt}\left(\therefore x=4and\frac{dx}{dt}=6\right) \\ \Rightarrow \frac{dA}{dt}=2\left(4\right)\times 6 \\ \Rightarrow \frac{dA}{dt}=48{\mathrm{cm}}^2/s \end{gathered}$$
.
Therefore, the area is increasing at the rate of $48{\mathrm{cm}}^2/s$.