Using calculus to derive a rate of change given two related rates of change The surface area of a closed cylinder is given by $$A = 2\pi r^2 + 2\pi rh$$
where h  is height
r is the radius of the base.
At the time where the surface are s increase at the rate of $20\pi\space cm^2 s^{-1}$ the radius is 4 cm and the height is 1 cm, the height is decreasing at a rate of $2 \space cm^{-1}$ Find the rate of change of radius at this time.
I thought about differentiating the equation for surface area with respect to r but seeing as its a unit with no indication of time, differentiating it would be pointless; how should I approach this question?
 A: Edited for more explanation
You need to do the total derivative
$$\frac{dA}{dt} = \frac{d}{dt} \left( 2\pi r^2 + 2\pi r h \right)$$
$$\frac{dA}{dt} = \frac{d}{dt} \left( 2\pi r^2 \right) + \frac{d}{dt} \left( 2\pi r h \right)$$
Both $r$ and $h$ change with time, therefore from the chain rule
$$\frac{d}{dt} \left( 2\pi r h \right) = 2 \pi h \frac{dr}{dt} + 2 \pi r \frac{dh}{dt}$$
So we're back at the original first line.
$$\frac{dA}{dt} = 4 \pi r \frac{dr}{dt} + 2 \pi h \frac{dr}{dt} + 2 \pi r \frac{dh}{dt}$$
$$\frac{dA}{dt} = 2\pi\left(2 r + h \right) \frac{dr}{dt} + 2 \pi r \frac{dh}{dt}$$
So
$$20\pi\ \frac{\mathrm{cm}^2}{\mathrm{s}} = 2\pi \times \left(2 \times (4\ \mathrm{cm}) + (1\ \mathrm{cm}) \right) \frac{dr}{dt} + 2 \pi \times (4\ \mathrm{cm}) \left(2\ \frac{\mathrm{cm}}{\mathrm{s}} \right)$$
$$10 \frac{\mathrm{cm}^2}{\mathrm{s}} = \left(2 \times (4\ \mathrm{cm}) + (1\ \mathrm{cm}) \right) \frac{dr}{dt} + (4\ \mathrm{cm}) \left(2\ \frac{\mathrm{cm}}{\mathrm{s}} \right)$$
$$10\ \frac{\mathrm{cm}^2}{\mathrm{s}} = \left(9\ \mathrm{cm} \right) \frac{dr}{dt} + 8\ \frac{\mathrm{cm}^2}{\mathrm{s}}$$
$$2\ \frac{\mathrm{cm}^2}{\mathrm{s}} = (9\ \mathrm{cm}) \frac{dr}{dt}$$
$$\frac{dr}{dt} = \frac{2}{9}\ \frac{\mathrm{cm}}{\mathrm{s}}$$
