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)
Best Answer
Since $\pi$ is a constant, all you have to do is $$ \frac {dA}{dt} = 2 \pi r\frac{dr}{dt}$$
You do not need to do the product rule in this case.