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The side length of a square is decreasing at a rate of 2 kilometers per hour.
At a certain instant, the side length is 9 kilometers.
What is the rate of change of the area of the square at that instant (in square kilometers per hour)?
Choose 1 answer:
(A) -324
(B) -4
(C) -36
(D) -81

The side length of a square is decreasing at a rate of $2$ kilometers per hour. $\newline$ At a certain instant, the side length is $9$ kilometers. $\newline$ What is the rate of change of the area of the square at that instant (in square kilometers per hour)? $\newline$ Choose $1$ answer: $\newline$ (A) $-324$ $\newline$ (B) $-4$ $\newline$ (C) $-36$ $\newline$ (D) $-81$

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Calculate unit rates with fractions

Full solution

Q. The side length of a square is decreasing at a rate of

2

kilometers per hour.

\newline

At a certain instant, the side length is

9

kilometers.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

-324

\newline

(B)

-4

\newline

(C)

-36

\newline

(D)

-81

Square Area Formula: The area of a square is given by the formula $A = s^2$ , where $s$ is the side length.
Side Length Rate: If the side length $s$ is decreasing at a rate of $2$ km/h, we can represent this as $\frac{ds}{dt} = -2$ km/h, where $\frac{ds}{dt}$ is the rate of change of the side length with respect to time.
Rate of Area Change: To find the rate of change of the area $\frac{dA}{dt}$ , we use the chain rule: $\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}$ .
Calculation: Substitute $s = 9$ km and $\frac{ds}{dt} = -2$ km/h into the equation: $\frac{dA}{dt} = 2 \times 9 \times (-2) = -36$ km $^2$ /h.

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