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The radius of a circle is decreasing at a rate of 6.5 meters per minute.
At a certain instant, the radius is 12 meters.
What is the rate of change of the area of the circle at that instant (in square meters per minute)?
Choose 1 answer:
(A) 
-156 pi
(B) 
-144 pi
(C) 
-288 pi
(D) 
-42.25 pi

The radius of a circle is decreasing at a rate of 66.55 meters per minute.\newlineAt a certain instant, the radius is 1212 meters.\newlineWhat is the rate of change of the area of the circle at that instant (in square meters per minute)?\newlineChoose 11 answer:\newline(A) 156π -156 \pi \newline(B) 144π -144 \pi \newline(C) 288π -288 \pi \newline(D) 42.25π -42.25 \pi

Full solution

Q. The radius of a circle is decreasing at a rate of 66.55 meters per minute.\newlineAt a certain instant, the radius is 1212 meters.\newlineWhat is the rate of change of the area of the circle at that instant (in square meters per minute)?\newlineChoose 11 answer:\newline(A) 156π -156 \pi \newline(B) 144π -144 \pi \newline(C) 288π -288 \pi \newline(D) 42.25π -42.25 \pi
  1. Circle Area Formula: The formula for the area of a circle is A=πr2A = \pi r^2, where AA is the area and rr is the radius.
  2. Rate of Change Derivation: To find the rate of change of the area, we need to differentiate the area with respect to time tt. So we get dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}.
  3. Radius Rate of Change: We know the radius is decreasing at a rate of 6.56.5 meters per minute, so drdt=6.5\frac{dr}{dt} = -6.5 meters/minute.
  4. Instant Calculation: At the instant when the radius is 1212 meters, we plug r=12r = 12 meters and drdt=6.5\frac{dr}{dt} = -6.5 meters/minute into the formula dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}.
  5. Final Result: So, dAdt=2π×12 meters×(6.5 meters/minute)=156π square meters per minute.\frac{dA}{dt} = 2\pi \times 12 \text{ meters} \times (-6.5 \text{ meters/minute}) = -156\pi \text{ square meters per minute}.

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