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

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

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Area of quadrilaterals and triangles: word problems

Full solution

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

\frac{\pi}{2}

meters per hour.

\newline

At a certain instant, the circumference is

12 \pi

meters.

\newline

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

\newline

Choose

1

answer:

\newline

(A)

\frac{\pi}{4}

\newline

(B)

36 \pi

\newline

(C)

3 \pi

\newline

(D)

6

Find Circle Radius: First, let's find the radius of the circle using the circumference. The formula for circumference is $C = 2 \times \pi \times r$ , where $C$ is the circumference and $r$ is the radius.
Calculate Rate of Change: Given $C = 12 \times \pi$ , we can solve for $r$ : $12 \times \pi = 2 \times \pi \times r$ . Dividing both sides by $2 \times \pi$ , we get $r = 6$ meters.
Derivative of Area: Now, we need to find the rate of change of the area. The area of a circle is given by $A = \pi \cdot r^2$ . To find the rate of change of the area, we'll use the derivative $\frac{dA}{dt} = 2 \cdot \pi \cdot r \cdot \frac{dr}{dt}$ , where $\frac{dr}{dt}$ is the rate of change of the radius.
Rate of Change Calculation: We know the circumference is increasing at a rate of $\frac{\pi}{2}$ meters per hour, which means the radius is increasing at a rate of $\frac{\pi}{4}$ meters per hour because $\frac{dr}{dt} = \frac{dC}{dt} / (2 \cdot \pi)$ .
Correcting Mistake: Substitute $r = 6$ meters and $\frac{dr}{dt} = \frac{\pi}{4}$ meters per hour into the derivative of the area: $\frac{dA}{dt} = 2 \times \pi \times 6 \times \frac{\pi}{4}$ .
Correcting Mistake: Substitute $r = 6$ meters and $\frac{dr}{dt} = \frac{\pi}{4}$ meters per hour into the derivative of the area: $\frac{dA}{dt} = 2 \cdot \pi \cdot 6 \cdot \frac{\pi}{4}$ .Simplify the expression: $\frac{dA}{dt} = 2 \cdot \pi \cdot 6 \cdot \frac{\pi}{4} = 3 \cdot \pi^2$ meters squared per hour.
Correcting Mistake: Substitute $r = 6$ meters and $\frac{dr}{dt} = \frac{\pi}{4}$ meters per hour into the derivative of the area: $\frac{dA}{dt} = 2 \cdot \pi \cdot 6 \cdot \frac{\pi}{4}$ .Simplify the expression: $\frac{dA}{dt} = 2 \cdot \pi \cdot 6 \cdot \frac{\pi}{4} = 3 \cdot \pi^2$ meters squared per hour.So, the rate of change of the area of the circle at that instant is $3 \cdot \pi^2$ square meters per hour. But wait, there's a mistake here. We should not have $\pi$ squared in the final answer. Let's correct it.

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