The area of a square is increasing at a rate of 20 square meters per hour. At a certain instant, the area is 49 square meters. What is the rate of change of the perimeter of the square at that instant (in meters per hour)? Choose 1 answer: (A) (40)/(7) (B) 2sqrt5 (C) 28 (D) 7

Calculate Side Length: First, let's find the length of one side of the square using the area. The area of a square is given by side^2. So, side = sqrt(area).Find Perimeter: Now, calculate the side length when the area is 49 square meters. side = sqrt(49) = 7 meters.Differentiate Perimeter: The perimeter of a square is given by 4 times the side length. So, the perimeter at that instant is 4 * 7 = 28 meters.Substitute Rate of Change: To find the rate of change of the perimeter, we need to differentiate the perimeter with respect to time. Since the perimeter P = 4 * side, and the side s = sqrt(area), we have P = 4 * sqrt(area).Simplify Expression: Differentiate P with respect to time t. dP/dt = 4 * (1/2) * area^(-1/2) * d(area)/dt.Substitute the rate of change of the area, which is 20 square meters per hour, and the area at the instant, which is 49 square meters. dP/dt = 4 * (1/2) * 49^(-1/2) * 20.Simplify the expression. dP/dt = 2 * (1/7) * 20 = (40/7) meters per hour.

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The area of a square is increasing at a rate of 20 square meters per hour.
At a certain instant, the area is 49 square meters.
What is the rate of change of the perimeter of the square at that instant (in meters per hour)?
Choose 1 answer:
(A)
(40)/(7)
(B)
2sqrt5
(C) 28
(D) 7

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) $\frac{40}{7}$ $\newline$ (B) $2 \sqrt{5}$ $\newline$ (C) $28$ $\newline$ (D) $7$

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Math Problems

Grade 6

Area of quadrilaterals and triangles: word problems

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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)

\frac{40}{7}

\newline

(B)

2 \sqrt{5}

\newline

(C)

28

\newline

(D)

7

Calculate Side Length: First, let's find the length of one side of the square using the area. The area of a square is given by $\text{side}^2$ . So, $\text{side} = \sqrt{\text{area}}$ .
Find Perimeter: Now, calculate the side length when the area is $49$ square meters. $\text{side} = \sqrt{49} = 7$ meters.
Differentiate Perimeter: The perimeter of a square is given by $4$ times the side length. So, the perimeter at that instant is $4 \times 7 = 28$ meters.
Substitute Rate of Change: To find the rate of change of the perimeter, we need to differentiate the perimeter with respect to time. Since the perimeter $P = 4 \times \text{side}$ , and the side $s = \sqrt{\text{area}}$ , we have $P = 4 \times \sqrt{\text{area}}$ .
Simplify Expression: Differentiate $P$ with respect to time $t$ . $\frac{dP}{dt} = 4 \times \left(\frac{1}{2}\right) \times \text{area}^{-\frac{1}{2}} \times \frac{d(\text{area})}{dt}$ .
Simplify Expression: Differentiate $P$ with respect to time $t$ . $\frac{dP}{dt} = 4 \times \left(\frac{1}{2}\right) \times \text{area}^{-\frac{1}{2}} \times \frac{d(\text{area})}{dt}$ .Substitute the rate of change of the area, which is $20$ square meters per hour, and the area at the instant, which is $49$ square meters. $\frac{dP}{dt} = 4 \times \left(\frac{1}{2}\right) \times 49^{-\frac{1}{2}} \times 20$ .
Simplify Expression: Differentiate $P$ with respect to time $t$ . $\frac{dP}{dt} = 4 \times \left(\frac{1}{2}\right) \times \text{area}^{-\frac{1}{2}} \times \frac{d(\text{area})}{dt}$ .Substitute the rate of change of the area, which is $20$ square meters per hour, and the area at the instant, which is $49$ square meters. $\frac{dP}{dt} = 4 \times \left(\frac{1}{2}\right) \times 49^{-\frac{1}{2}} \times 20$ .Simplify the expression. $\frac{dP}{dt} = 2 \times \left(\frac{1}{7}\right) \times 20 = \left(\frac{40}{7}\right)$ meters per hour.