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 is given by \( A = \pi r^2 \), where \( r \) is the radius.

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

Using the chain rule, we differentiate both sides 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, so \( \frac{dA}{dt} = 2 \frac{dr}{dt} \). Substitute this into the equation from Step 2: \( 2 \frac{dr}{dt} = 2\pi r \frac{dr}{dt} \).

Step4: Solve for \( r \)

Assuming \( \frac{dr}{dt}
eq 0 \) (since the circle is expanding, the radius is changing, so its rate of change is not zero), 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 \( 1 = \pi r \), so \( r = \frac{1}{\pi} \).

Answer:

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