Radius of a cylinder grows with speed = $2\ \mathrm{cm/s}$ and its height grows with speed = $3\ \mathrm{cm/s}$.
What is the instantaneous volume growth rate of the cylinder when radius = $5\ \mathrm{cm}$ and height = $15\ \mathrm{cm}$?
With only one variable changing , I can take the volume derivate to respect to it and then substitute it, but I can not figure out how to do it with radius and height increasing at the same time with different rates.
I suggest you try to write the volume increase: $$V=\pi R^2h\\V+dV=\pi (R+dR)^2(h+dh)$$ Then, keeping only first order terms $$dV=\pi(R+dR)^2(h+dh)-\pi R^2 h\\\approx\pi R(2dR)h+\pi R^2dh$$ Then $$\frac{dV}{dt}=2\pi Rh\frac{dR}{dt}+\pi R^2\frac{dh}{dt}$$
This is similar to the other answer but emphasises the use of product and chain rules.
$V = \pi r^2 h$
$$\frac {dV} {dt} = \pi \left(\frac{d r^2} {dt } h + r^2\frac {dh} {dt}\right ) $$
$$\ \ \ \ \ \ \ \ \ \ \ = \pi \left(\frac{d r^2} {dr } \frac {dr} {dt} h + r^2\frac {dh} {dt} \right)$$
$$\ \ \ \ \ \ \ \ = \pi \left(2r h \frac {dr} {dt} + r^2\frac {dh} {dt}\right)$$
And now you can substitute the relevant parameters.
