The area of a circle is increasing at a rate of 8pi square meters per hour. At a certain instant, the area is 36 pi square meters. What is the rate of change of the circumference of the circle at that instant (in meters per hour)? Choose 1 answer: (A) 4sqrt2 (B) 12 pi (C) (pi)/(6) (D) (4pi)/(3)

Find Radius of Circle: First, let's find the radius of the circle using the area formula A = pi * r^2. We have A = 36pi, so 36pi = pi * r^2.Calculate Circumference: Divide both sides by pi to get r^2 = 36. Then take the square root of both sides to find the radius r = 6 meters.Use Rate of Change Formula: Now, let's find the circumference using the formula C = 2 * pi * r. Plugging in r = 6, we get C = 2 * pi * 6 = 12pi meters.Solve for dr/dt: To find the rate of change of the circumference, we'll use the relationship between the rate of change of the area and the rate of change of the radius. Since dA/dt = 8pi, and A = pi * r^2, we can write dA/dt = 2 * pi * r * (dr/dt).Calculate dC/dt: We know dA/dt = 8pi and r = 6, so we can solve for dr/dt: 8pi = 2 * pi * 6 * (dr/dt). Simplify to get dr/dt = (8pi) / (12pi) = 2/3 meters per hour.Finally, the rate of change of the circumference (dC/dt) is given by dC/dt = 2 * pi * (dr/dt). Plug in dr/dt = 2/3 to get dC/dt = 2 * pi * (2/3) = (4pi)/3 meters per hour.

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The area of a circle is increasing at a rate of
8pi square meters per hour.
At a certain instant, the area is
36 pi square meters.
What is the rate of change of the circumference of the circle at that instant (in meters per hour)?
Choose 1 answer:
(A)
4sqrt2
(B)
12 pi
(C)
(pi)/(6)
(D)
(4pi)/(3)

The area of a circle is increasing at a rate of $8 \pi$ square meters per hour. $\newline$ At a certain instant, the area is $36 \pi$ square meters. $\newline$ What is the rate of change of the circumference of the circle at that instant (in meters per hour)? $\newline$ Choose $1$ answer: $\newline$ (A) $4 \sqrt{2}$ $\newline$ (B) $12 \pi$ $\newline$ (C) $\frac{\pi}{6}$ $\newline$ (D) $\frac{4 \pi}{3}$

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Grade 6

Area of quadrilaterals and triangles: word problems

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

8 \pi

square meters per hour.

\newline

At a certain instant, the area is

36 \pi

square meters.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

4 \sqrt{2}

\newline

(B)

12 \pi

\newline

(C)

\frac{\pi}{6}

\newline

(D)

\frac{4 \pi}{3}

Find Radius of Circle: First, let's find the radius of the circle using the area formula $A = \pi \cdot r^2$ . We have $A = 36\pi$ , so $36\pi = \pi \cdot r^2$ .
Calculate Circumference: Divide both sides by $\pi$ to get $r^2 = 36$ . Then take the square root of both sides to find the radius $r = 6$ meters.
Use Rate of Change Formula: Now, let's find the circumference using the formula $C = 2 \cdot \pi \cdot r$ . Plugging in $r = 6$ , we get $C = 2 \cdot \pi \cdot 6 = 12\pi$ meters.
Solve for $\frac{dr}{dt}$ : To find the rate of change of the circumference, we'll use the relationship between the rate of change of the area and the rate of change of the radius. Since $\frac{dA}{dt} = 8\pi$ , and $A = \pi r^2$ , we can write $\frac{dA}{dt} = 2 \pi r \left(\frac{dr}{dt}\right)$ .
Calculate $\frac{dC}{dt}$ : We know $\frac{dA}{dt} = 8\pi$ and $r = 6$ , so we can solve for $\frac{dr}{dt}$ : $8\pi = 2 \cdot \pi \cdot 6 \cdot \left(\frac{dr}{dt}\right)$ . Simplify to get $\frac{dr}{dt} = \frac{8\pi}{12\pi} = \frac{2}{3}$ meters per hour.
Calculate $\frac{dC}{dt}$ : We know $\frac{dA}{dt} = 8\pi$ and $r = 6$ , so we can solve for $\frac{dr}{dt}$ : $8\pi = 2 \times \pi \times 6 \times (\frac{dr}{dt})$ . Simplify to get $\frac{dr}{dt} = \frac{8\pi}{12\pi} = \frac{2}{3}$ meters per hour.Finally, the rate of change of the circumference ( $\frac{dC}{dt}$ ) is given by $\frac{dC}{dt} = 2 \times \pi \times (\frac{dr}{dt})$ . Plug in $\frac{dr}{dt} = \frac{2}{3}$ to get $\frac{dC}{dt} = 2 \times \pi \times (\frac{2}{3}) = \frac{4\pi}{3}$ meters per hour.