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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) -81
(B) -36
(C) -4
(D) -324

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) $-81$ $\newline$ (B) $-36$ $\newline$ (C) $-4$ $\newline$ (D) $-324$

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Write equations of parabolas in vertex form using properties

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)

-81

\newline

(B)

-36

\newline

(C)

-4

\newline

(D)

-324

Area Formula: The formula for the area of a square is $A = s^2$ , where $s$ is the side length.
Rate of Change: The rate of change of the area with respect to time is given by the derivative $\frac{dA}{dt}$ .
Chain Rule Application: Using the chain rule, $\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}$ , since the derivative of $s^2$ with respect to $s$ is $2s$ .
Given Side Length and Rate: We know $\frac{ds}{dt} = -2 \, \text{km/h}$ (the side length is decreasing).
Substitution and Calculation: Substitute $s = 9 \, \text{km}$ and $\frac{ds}{dt} = -2 \, \text{km/h}$ into the derivative to find $\frac{dA}{dt}$ . $\frac{dA}{dt} = 2 \times 9 \times (-2) = -36 \, \text{km}^2/\text{h}$ .
Final Rate of Change: The rate of change of the area of the square at that instant is $-36$ square kilometers per hour.

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