So this is just from some college exam and I was trying to do it and got stuck on this question because I've never really encountered related rates before. I read up a bit about related rates but still having some trouble. How would you guys approach this exactly? This is what I was thinking: Water is poured into the cup at a rate of 4 cubic cm per second so this is telling us $dh/dt = 4$ or $d'(t) = 4$. We're given that the height of the cup is 18cm and the radius of the top is 6 cm. Then the questions ask how fast is the water level rising when the water is 9cm deep? So to me this sounds like $dh/dt$ at $h=9$? Now I'm just not sure how exactly to approach this because I saw some other examples of people solving questions where they just say $r$ is a function of time or something and so they just replace $r$ with $r(t)$ (which makes sense to me) can I just do that? Say $V = (\pi r^2*h)/3$ can I just say that $r = r(t)$ and $h = h(t)$? Or what should I do? I've seen some example videos but didn't really understand any of them too well.
Note that at any moment of time $r(t)=h(t) \frac{6}{18}$. Thus, $V=\frac{1}{27}\pi h^3 $, differentiating by time we come to $$\frac{dV}{dt}=\frac{1}{9} \pi h^2 \frac{dh}{dt}$$ You are also given that $\frac{dV}{dt}=4$ and you need to find $\frac{dh}{dt}$ when $h=9$ so...