# Implicit Differentiation for calculating area of a circle

I am working on a implicit differentiation equation of a circle where the radius grows by 1/per sec.

Can someone advise if I have differentiate the equation correctly?

The radius of the circle grows by 1/s given: $$\frac{dr}{dt} = 1$$

The area of the circle given: $$A = \pi r^2$$

Differentiate both sides of the equation with respect to the third variable: time $$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$$

The differentiation of the left hand side of the equation: $$\frac{d}{dt}[A] = \frac{dA}{dt}$$

The differentiation of the right hand side of the equation: $$\frac{d}{dt} [\pi r^2] = [f(x)g'(x)+g(x)f'(x)]\frac{dr}{dt}$$

$$= [\pi\frac{d}{dr}[r^2] + r^2 \frac{d}{dr}[\pi]]\frac{dr}{dt}$$

$$= 2\pi r \frac{dr}{dt}$$

$$= 2\pi r$$ (since dr/dt = 1)

• its correct. Continuing with the given values will give the answer you require Commented Jun 3, 2018 at 11:52
• @TheIntegrator "A" should not be treated as a constant since it is dependent on $$A = \pi r^2$$ Am I right? Commented Jun 3, 2018 at 11:54
• Yes , if you want you can write it as $A(r)$ to be more clear Commented Jun 3, 2018 at 11:55
• What's $\frac 1 s$ ?
– BCLC
Commented Jun 3, 2018 at 12:02
• @BCLC OP means one per second ie $1 s^{-1}$ Commented Jun 3, 2018 at 12:03

Since $\pi$ is a constant, all you have to do is $$\frac {dA}{dt} = 2 \pi r\frac{dr}{dt}$$