AP Calc AB question radius of a cirlce increasing...

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) sq. root of 2
d) 2
e) 4


answer : d) 2

I keep getting 4 and i know i'm doing it wrong because i forgot how to actually do these problems, so if anyone can explain to me how to do it i would greatly appreciate it..

all i can think of is relating area and circumference

A = (pi)(r)^2

C = 2(pi)(r)

then the problem gave

dA/dt = 2(dC/dt)

but i dont know where to go from there


oo and this is from the AP Calc AB 2008 test question 24 (non-calculator)

I guess the simplest way is to set them equal to each other and solve for r. it's not very 'math-y' in the calculus sense of it but it works

(Pi)r^2 = 2(Pi)r
(Pi)r = 2(Pi)
r = 2

sorry this doesn't really address the concept the question is testing

Edit:

well i guess you can also use implicit differentiation

dA/dt = 2(dC/dt)
d/dt(2(pi)r^2) = 2*d/dt(2(Pi)r)
2(Pi)(r)dr/dt = 4(Pi)dr/dt

then divide both sides by 2(Pi)dr/dt

r = 2

but I'd still use the other way. simpler

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?

1st derivative is rate of change

Area = π * r^2
1st derivative of area with respect to r = 2 * π * r

Circumference = 2 * π * r
1st derivative of circumference with respect to r = 2 * π

1st derivative of area = 2 * 1st derivative of circumference
2 * π * r = 2 * π
divide both sides by 2 * π
r = 1