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) 12 pi (B) (pi)/(6) (C) 4sqrt2 (D) (4pi)/(3)

Remember Circle Area Formula: First, let's remember the formula for the area of a circle, A = πr^2, where r is the radius.Find Radius from Area: Given that the area A is 36π square meters, we can find the radius by rearranging the formula: r^2 = A/π. So, r^2 = (36π)/π = 36.Calculate Circumference: Now, we take the square root of both sides to find r. r = √36 = 6 meters.Differentiate Circumference: The formula for the circumference of a circle is C = 2πr. So, the circumference at that instant is C = 2π(6) = 12π meters.Rate of Area Change: To find the rate of change of the circumference, we need to differentiate the circumference with respect to time. dC/dt = 2π(dr/dt).Solve for dr/dt: We know the area is increasing at a rate of 8π square meters per hour, so dA/dt = 8π. Since A = πr^2, then dA/dt = 2πr(dr/dt).Substitute into Circumference Rate: We can now solve for dr/dt: 8π = 2π(6)(dr/dt). So, dr/dt = 8π / (2π * 6) = 2/3 meters per hour.Finally, we substitute dr/dt back into the rate of change of the circumference: dC/dt = 2π(dr/dt) = 2π(2/3) = (4π)/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)
12 pi
(B)
(pi)/(6)
(C)
4sqrt2
(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) $12 \pi$ $\newline$ (B) $\frac{\pi}{6}$ $\newline$ (C) $4 \sqrt{2}$ $\newline$ (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)

12 \pi

\newline

(B)

\frac{\pi}{6}

\newline

(C)

4 \sqrt{2}

\newline

(D)

\frac{4 \pi}{3}

Remember Circle Area Formula: First, let's remember the formula for the area of a circle, $A = \pi r^2$ , where $r$ is the radius.
Find Radius from Area: Given that the area $A$ is $36\pi$ square meters, we can find the radius by rearranging the formula: $r^2 = \frac{A}{\pi}$ . So, $r^2 = \frac{36\pi}{\pi} = 36$ .
Calculate Circumference: Now, we take the square root of both sides to find $r$ . $r = \sqrt{36} = 6$ meters.
Differentiate Circumference: The formula for the circumference of a circle is $C = 2\pi r$ . So, the circumference at that instant is $C = 2\pi(6) = 12\pi$ meters.
Rate of Area Change: To find the rate of change of the circumference, we need to differentiate the circumference with respect to time. $\frac{dC}{dt} = 2\pi\left(\frac{dr}{dt}\right)$ .
Solve for $\frac{dr}{dt}$ : We know the area is increasing at a rate of $8\pi$ square meters per hour, so $\frac{dA}{dt} = 8\pi$ . Since $A = \pi r^2$ , then $\frac{dA}{dt} = 2\pi r(\frac{dr}{dt})$ .
Substitute into Circumference Rate: We can now solve for $\frac{dr}{dt}$ : $8\pi = 2\pi(6)\left(\frac{dr}{dt}\right)$ . So, $\frac{dr}{dt} = \frac{8\pi}{(2\pi \cdot 6)} = \frac{2}{3}$ meters per hour.
Substitute into Circumference Rate: We can now solve for $\frac{dr}{dt}$ : $8\pi = 2\pi(6)\frac{dr}{dt}$ . So, $\frac{dr}{dt} = \frac{8\pi}{(2\pi \cdot 6)} = \frac{2}{3}$ meters per hour.Finally, we substitute $\frac{dr}{dt}$ back into the rate of change of the circumference: $\frac{dC}{dt} = 2\pi(\frac{dr}{dt}) = 2\pi(\frac{2}{3}) = \frac{4\pi}{3}$ meters per hour.