C = 2πr
C' = 2π
[This equals the rate at which the circumference is changing]
A = πr^2
A' = 2πr
[This equals the rate at which the area is changing]
A' = 2C'
2πr = 2(2π)
2πr = 4π
r = 4π/2π
r = 2
So, the answer would be D.
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?
A. 1/2
B. 1
C. Sqrt 2
D. 2
E. 4
1 answer