Question

when the area in square units of an expanding circle is increasing twice as fast as its radius in linear units, the radius is

Explanation:

Step1: Recall the formula for the area of a circle

The area \( A \) of a circle with radius \( r \) is given by \( A = \pi r^2 \).

Step2: Differentiate the area with respect to time \( t \)

Using the chain rule, we differentiate both sides of \( A = \pi r^2 \) with respect to \( t \). The derivative of \( A \) with respect to \( t \) is \( \frac{dA}{dt} \), and the derivative of \( \pi r^2 \) with respect to \( t \) is \( 2\pi r \frac{dr}{dt} \) (by the chain rule, since \( r \) is a function of \( t \)). So we have \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \).

Step3: Use the given condition

We are given that the area is increasing twice as fast as the radius, which means \( \frac{dA}{dt} = 2 \frac{dr}{dt} \).

Step4: Substitute the given condition into the differentiated equation

Substitute \( \frac{dA}{dt} = 2 \frac{dr}{dt} \) into \( \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \). We get \( 2 \frac{dr}{dt} = 2\pi r \frac{dr}{dt} \).

Step5: Solve for \( r \)

Assuming \( \frac{dr}{dt}
eq 0 \) (if \( \frac{dr}{dt} = 0 \), the radius is not changing, so the area wouldn't be changing either, which contradicts the given condition), we can divide both sides of the equation \( 2 \frac{dr}{dt} = 2\pi r \frac{dr}{dt} \) by \( 2 \frac{dr}{dt} \). This gives \( 1 = \pi r \), and then solving for \( r \) we get \( r = \frac{1}{\pi} \).

Answer:

\(\frac{1}{\pi}\)