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:

Step 1: Recall the formula for the area of a circle

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

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

Step 3: Use the given condition

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

Step 4: Substitute and solve for \( r \)

Substitute \( \frac{dA}{dt}=2\frac{dr}{dt}\) into the equation \( \frac{dA}{dt}=2\pi r\frac{dr}{dt}\). We get \( 2\frac{dr}{dt}=2\pi r\frac{dr}{dt}\). Assuming \( \frac{dr}{dt}
eq0 \) (since the circle is expanding, the radius is changing, so its rate of change is non - zero), we can divide both sides of the equation by \( 2\frac{dr}{dt}\). This gives \( 1=\pi r\), and then solving for \( r \) we have \( r = \frac{1}{\pi}\).

Answer:

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