Hi, I am doing a Related Rates problem and I got stuck on finding the derivative with respect to time of the equation for the volume of a cylinder:

V = (pi)(r^2)h

Question

Hi, I am doing a Related Rates problem and I got stuck on finding the derivative with respect to time of the equation for the volume of a cylinder:

V = (pi)(r^2)h

anonymous · Answer

V = (pi)(r^2)h this does not depend on time

anonymous · Answer

so derivative is =0

anonymous · Answer

it doesn't? hmm.. here lemme write out the problem because i thought this would be the right equation to use

anonymous · Answer

piston is seated at the top of a cylindrical chamber with radius 5 cm when ti starts moving into the chamber at a constant speed of 8 cm/s. What is the rate of change of the volume of the cylinder when the piston is 6cm from the base of the chamber?

anonymous · Answer

so i copied this in my notes:

$$V = \pi r^2h$$
$$dV/dt = d/dt (\pi r^2h)$$
$$= \pi r^2dh/dt$$

i don't get how she went from the 2nd step to the 3rd

anonymous · Answer

h is function of time. It is h(t)=$h_0$-8t. Where $h_0$ is cylinder height.
So dh/dt=-8

anonymous · Answer

and rate of change of volumen will be $\pi$25(-8)=?