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A circle has an initial radius of 50 ft when the radius begins decreasing at the rate of \( 5 \mathrm{ft} / \mathrm{min} \). What is the rate in the change of area at the instant that the radius is 20 ft ? The rate of change of the area is (Type an exact answer in terms of \( \pi \).)

Ask by Murray Ramirez. in the United States
Mar 12,2025

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The rate of change of the area is \(-200\pi\) square feet per minute.

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Bonus Knowledge

To find the rate of change of the area of the circle as the radius decreases, we can use the formula for the area of a circle, \( A = \pi r^2 \). First, we'll differentiate the area with respect to time: \[ \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2\pi r \cdot \frac{dr}{dt} \] Given that \( \frac{dr}{dt} = -5 \, \text{ft/min} \) (the radius is decreasing) and we need to evaluate this when \( r = 20 \, \text{ft} \), we plug in the values: \[ \frac{dA}{dt} = 2\pi (20) \cdot (-5) = -200\pi \, \text{ft}^2/\text{min} \] So, the rate of change of the area when the radius is 20 ft is \( -200\pi \, \text{ft}^2/\text{min} \).

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