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

Answer:

$\frac{1}{\pi}$

Explanation:

Step1: Recall area formula of circle

The area formula of a circle is $A = \pi r^{2}$, where $A$ is the area and $r$ is the radius.

Step2: Differentiate with respect to time

Differentiate both sides with respect to time $t$ using the chain - rule. $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$.

Step3: Use given condition

We are given that $\frac{dA}{dt} = 2\frac{dr}{dt}$.

Step4: Substitute and solve for $r$

Substitute $\frac{dA}{dt} = 2\frac{dr}{dt}$ into $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$. We get $2\frac{dr}{dt}=2\pi r\frac{dr}{dt}$. Since $\frac{dr}{dt}\neq0$ (the circle is expanding), we can cancel out $2\frac{dr}{dt}$ from both sides of the equation. Then $1=\pi r$, and solving for $r$ gives $r = \frac{1}{\pi}$.