The side length of a square is increasing at a rate of 15 millimeters per second. At a certain instant, the side length is 22 millimeters. What is the rate of change of the area of the square at that instant (in square millimeters per second)? Choose 1 answer: (A) 660 (B) 484 (C) 30 (D) 225

Find Area Formula: First, let's find the formula for the area of a square, which is side length squared (s^2).Derivative of Area: Now, we need to find the derivative of the area with respect to time (dA/dt) since we're looking for the rate of change of the area.Derivative of s^2: The derivative of s^2 with respect to time (ds/dt) is 2s * (ds/dt), where ds/dt is the rate of change of the side length.Plug in Values: Plug in the values: s = 22 mm and ds/dt = 15 mm/s. So, dA/dt = 2 * 22 mm * 15 mm/s.Calculate Rate of Change: Calculate the rate of change of the area: dA/dt = 2 * 22 * 15 = 660 mm^2/s.

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The side length of a square is increasing at a rate of 15 millimeters per second.
At a certain instant, the side length is 22 millimeters.
What is the rate of change of the area of the square at that instant (in square millimeters per second)?
Choose 1 answer:
(A) 660
(B) 484
(C) 30
(D) 225

The side length of a square is increasing at a rate of $15$ millimeters per second. $\newline$ At a certain instant, the side length is $22$ millimeters. $\newline$ What is the rate of change of the area of the square at that instant (in square millimeters per second)? $\newline$ Choose $1$ answer: $\newline$ (A) $660$ $\newline$ (B) $484$ $\newline$ (C) $30$ $\newline$ (D) $225$

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Math Problems

Grade 6

Area of quadrilaterals and triangles: word problems

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

15

millimeters per second.

\newline

At a certain instant, the side length is

22

millimeters.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

660

\newline

(B)

484

\newline

(C)

30

\newline

(D)

225

Find Area Formula: First, let's find the formula for the area of a square, which is side length squared $s^2$ .
Derivative of Area: Now, we need to find the derivative of the area with respect to time $\frac{dA}{dt}$ since we're looking for the rate of change of the area.
Derivative of $s^2$ : The derivative of $s^2$ with respect to time ( $\frac{ds}{dt}$ ) is $2s \cdot \frac{ds}{dt}$ , where $\frac{ds}{dt}$ is the rate of change of the side length.
Plug in Values: Plug in the values: $s = 22\,\text{mm}$ and $\frac{ds}{dt} = 15\,\text{mm/s}$ . So, $\frac{dA}{dt} = 2 \times 22\,\text{mm} \times 15\,\text{mm/s}$ .
Calculate Rate of Change: Calculate the rate of change of the area: $\frac{dA}{dt} = 2 \times 22 \times 15 = 660 \, \text{mm}^2/\text{s}.$