Question

question 3
4 pts
each side of a square is increasing at a rate of 3 cm/s. at what rate is the area of the square increasing when the area of the square is 9 cm²?
9 $\frac{cm^{2}}{s}$
the correct answer is not listed.
3 $\frac{cm^{2}}{s}$
27 $\frac{cm^{2}}{s}$
18 $\frac{cm^{2}}{s}$

Explanation:

Step1: Find the side - length of the square

Given the area formula of a square $A = s^{2}$, when $A = 9\ cm^{2}$, we solve for $s$:
$A=s^{2}=9$, so $s = 3\ cm$ (since $s>0$ as it represents a length).

Step2: Differentiate the area formula with respect to time

We have $A = s^{2}$. Differentiating both sides with respect to time $t$ using the chain - rule, we get $\frac{dA}{dt}=2s\frac{ds}{dt}$.

Step3: Substitute the known values

We know that $\frac{ds}{dt}=3\ cm/s$ and $s = 3\ cm$. Substitute these values into the equation $\frac{dA}{dt}=2s\frac{ds}{dt}$:
$\frac{dA}{dt}=2\times3\times3$.
$\frac{dA}{dt}=18\ cm^{2}/s$.

Answer:

E. $18\frac{cm^{2}}{s}$