Calculus problem- application of derivatives?

A = πr²

Finding the circumference is equivalent to finding the radius, right? If we know the radius, then we can find the circumference:

C = 2πr

So find the radius that satisfies this:

Implicitly differentiate the area equation (since it talks about area and radius):

dA/dt = 2πr * dr/dt

radius is increasing HALF as quickly as area:

dr/dt = ½ * dA/dt

--> plug that in

dA/dt = 2πr * ½ * dA/dt

--> now the dA/dt's cancel leaving

1 = 2πr * ½

--> 2 * ½ = 1

1 = πr --> r = 1/π

Now plug into circumference equation:

C = 2πr = 2π * (1/π) = 2

V = c r^2 h the place c (for convenience is shorthand for the consistent a million/3 pi) Now the two r and h are purposes of time t. So as quickly as we differentiate the two aspects with admire to t we'd desire to apply the chain rule. dV/dt = c r^2 dh/dt + c 2rh dr/dt = c(40 9 x 2 + 2x7x12x5) = c 938 = 938 x pi/3 = 982.3

dr/dt = 1/2*(dA/dt)

dr/dt = 1/2*(2*pi*r* (dr/dt))

pi*r = 1 --> r = 1/pi

Circumference = pi*2/pi = 2