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: Define area and radius relation

The area of a circle is given by $A = \pi r^2$, where $A$ = area, $r$ = radius.

Step2: Set up rate condition

The area increases twice as fast as the radius, so $\frac{dA}{dt} = 2\frac{dr}{dt}$.

Step3: Differentiate area w.r.t time

Differentiate $A = \pi r^2$:
$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

Step4: Substitute rate condition

Substitute $\frac{dA}{dt} = 2\frac{dr}{dt}$ into the derivative:
$2\frac{dr}{dt} = 2\pi r \frac{dr}{dt}$

Step5: Solve for radius $r$

Divide both sides by $2\frac{dr}{dt}$ (assuming $\frac{dr}{dt}
eq 0$):
$1 = \pi r$
Rearrange to solve for $r$:
$r = \frac{1}{\pi}$
$r = \sqrt{\frac{1}{\pi}}$ or $r = \frac{1}{\sqrt{\pi}}$

Answer:

$\sqrt{\frac{1}{\pi}}$ or $\frac{1}{\sqrt{\pi}}$