How to solve a related rates problem with an accumulation function?

Question

So I have been trying to solve this related rates problem. I am not having much trouble with accumulation functions, but more with the related rates part of the problem. I am having a hard time figuring out where to begin. Any suggestions?

Answer 1

Given the volume of the cup as a function of $y$:
$$ V(y)=\pi y^2/6$$ and the rate of change of water volume: $$\frac{dV}{dt} = 1 \text{ cm}^3/\text{min}$$ Note that $y=f(x)=3x^2$, so the units of volume, $\text{cm}^3$ is encapsulated by $y^2$. Related rates can be found by $$ \frac{dV}{dt} = \frac{dV}{dy}\frac{dy}{dt}=\frac{\pi y}{3}\frac{dy}{dt}$$ such that the rate of change of water height at any given height is $$ \frac{dy}{dt} = \frac{3}{\pi y} \frac{dV}{dt}$$ When $y=12$ cm, the rate of change of height is $$\frac{dy}{dt} = \frac{3\cdot 1\text{ cm}^3/\text{min}}{\pi\cdot 12\text{ cm}}$$ The units come out as $\text{cm}^2/\text{min}$, but the reason for that is related to the fact that units of volume is encapsulated by $y^2=3x^2y$.