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, where we first differentiate \( \pi r^{2}\) with respect to \( r \) to get \( 2\pi r \) and then multiply by \( \frac{dr}{dt} \) which is the derivative of \( r \) with respect to \( t \)). 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 its radius, which means \( \frac{dA}{dt} = 2\frac{dr}{dt} \).

Step4: 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 \( \frac{dr}{dt}
eq0 \) (because if \( \frac{dr}{dt} = 0 \), the radius is not changing and the area also won't be changing, which doesn't fit the "expanding circle" context), we can divide both sides of the equation by \( 2\frac{dr}{dt} \). This gives \( 1=\pi r \), and then solving for \( r \) we get \( r = \frac{1}{\pi} \).

Answer:

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