1. The radius of a circle is changing at the rate of \( \frac{1}{\pi} \) inches per second. At what rate, in square inches per second, is the circle’s area changing when \( r = 5 \) in?






Answer is: option2

10

Solution:

Given the area of a circle:

\( A = \pi r^2 \)

Differentiate both sides with respect to \( t \):

\(\frac{dA}{dt} = \frac{d}{dt} (\pi r^2) \)

\(\frac{dA}{dt} = 2\pi r \frac{dr}{dt} \)

Given:

\( r = 5 \) inches

\(\frac{dr}{dt} = \frac{1}{\pi} \) inches per second

\(\frac{dA}{dt} = ? \)

Substituting the given values:

\(\frac{dA}{dt} = 2 \times \pi \times 5 \times \left(\frac{1}{\pi}\right) \)

\(\frac{dA}{dt} = 10 \) square inches per second

Final Answer: 10 inches per second

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