Question

The radius of a circle is increasing. At a certain instant, the rate of increase in the area of the circle is numerically equal to twice the rate of increase in its circumference. What is the radius of the circle at that instant?

Answer 1

Let A=pi*r^2 and C=2*pi*r
Then the rate of increases are equal dA/dt = dC/dt
And then the rate of increase of the radius is dr/dt,
So then what you would do is this:
Take the derivative of the Area according to r, dA/dr = 2*pi*r
Take the derivative of the Circumference according to r, dC/dr = 2*pi
Multiply both sides of the equation with the rate of increase of the radius dr/dt to get dA/dr = dC/dr
Since you know both dA/dr and dC/dr equate the two to get 2*pi*r = 2*pi
From there you solve the equation to find that r = 1.
I might be wrong so someone should double check this...

Answer 2

Thanks for the help.