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The area of a circle is decreasing at a constant rate of 205 square inches per minute. At the instant when the area of the circle is $25 \pi$ square inches, what is the rate of change of the radius? Round your answer to three decimal places.

Step 1 ：Given that the area of the circle is decreasing at a constant rate of 205 square inches per minute, we can write this as \(\frac{dA}{dt} = -205\).

Step 2 ：We are asked to find the rate of change of the radius, \(\frac{dr}{dt}\), at the instant when the area of the circle is 25π square inches.

Step 3 ：The area of a circle is given by the formula \(A = \pi r^2\), where r is the radius of the circle.

Step 4 ：We can use the chain rule to differentiate the area with respect to time, t. This gives us: \(\frac{dA}{dt} = 2\pi r \frac{dr}{dt}\).

Step 5 ：Substitute the given values into this equation: \(-205 = 2\pi r \frac{dr}{dt}\) when \(A = 25\pi\).

Step 6 ：From \(A = \pi r^2\), we can solve for r when \(A = 25\pi\): \(25\pi = \pi r^2\) so \(r = \sqrt{25} = 5\).

Step 7 ：Substitute \(r = 5\) into the equation: \(-205 = 2\pi*5\frac{dr}{dt}\) so \(-205 = 10\pi \frac{dr}{dt}\).

Step 8 ：Solving for \(\frac{dr}{dt}\) gives: \(\frac{dr}{dt} = -205 / 10\pi = -20.5/\pi\).

Step 9 ：Rounding to three decimal places, \(\frac{dr}{dt} = -6.527\) inches per minute.

Step 10 ：So, the radius of the circle is decreasing at a rate of \(\boxed{-6.527}\) inches per minute when the area of the circle is 25π square inches.