The area of a square is increasing at a rate of $20$ square meters per hour. $\newline$ At a certain instant, the area is $49$ square meters. $\newline$ What is the rate of change of the perimeter of the square at that instant (in meters per hour)? $\newline$ Choose $1$ answer: $\newline$ (A) $7$ $\newline$ (B) $\frac{40}{7}$ $\newline$ (C) $\mathbf{2 8}$ $\newline$ (D) $2 \sqrt{5}$

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Q. The area of a square is increasing at a rate of

20

square meters per hour.

\newline

At a certain instant, the area is

49

square meters.

\newline

What is the rate of change of the perimeter of the square at that instant (in meters per hour)?

\newline

Choose

1

answer:

\newline

(A)

7

\newline

(B)

\frac{40}{7}

\newline

(C)

\mathbf{2 8}

\newline

(D)

2 \sqrt{5}

Find Side Length: First, let's find the length of one side of the square when the area is $49$ square meters. Since the area of a square is side squared ( $s^2$ ), we can find the side length by taking the square root of the area. $\newline$ So, $s = \sqrt{49} = 7$ meters.
Calculate Perimeter: Now, the perimeter of a square is $4$ times the side length, so the perimeter $P$ when the area is $49$ square meters is $P = 4 \times s = 4 \times 7 = 28$ meters.
Differentiate Perimeter: To find the rate of change of the perimeter, we need to differentiate the perimeter with respect to time. The perimeter $P = 4s$ , and we know that the area $A = s^2$ is changing at a rate of $20$ square meters per hour. So, we need to find $\frac{ds}{dt}$ when $A = 49$ square meters.
Use Chain Rule: Differentiating $A = s^2$ with respect to time $t$ gives us $2s \cdot \frac{ds}{dt}$ . We know that $\frac{dA}{dt} = 20$ square meters per hour, so we can set up the equation $2s \cdot \frac{ds}{dt} = 20$ .
Calculate Rate of Change: Plugging in the side length $s = 7$ meters into the equation gives us $2 \times 7 \times (\frac{ds}{dt}) = 20$ . Solving for $\frac{ds}{dt}$ gives us $\frac{ds}{dt} = \frac{20}{(2 \times 7)} = \frac{20}{14} = \frac{10}{7}$ meters per hour.
Calculate $\frac{dP}{dt}$ : Finally, since the perimeter $P = 4s$ , the rate of change of the perimeter $\frac{dP}{dt}$ is $4$ times the rate of change of the side length $\frac{ds}{dt}$ . So, $\frac{dP}{dt} = 4 \times \left(\frac{10}{7}\right) = \frac{40}{7}$ meters per hour.