Thread: Relating rates (please help)

Water in a paper conical filter drips into a cup as shown in the attachment. Let x denote the height of the water in the filter. Let y denote the height of the water in the cup. If 10 in^3 of water is poured into the filter, find the relationship between dy/dt and dx/dt

First, we need to find an equation which connects all the variables in this problem. From your description and the figure, it is easy to see that at any given time the volume of the water in the conical filter plus the volume of water in the cup is 10 in^3 (assume that it takes no time for the water to drip from the filter to the cup). Mathematically, we have the equation:

$\displaystyle \frac{1}{3}\pi r^2 x + \pi (4)^2 y = 10$

Also we are interested in the variables $\displaystyle x$ and $\displaystyle y$, so it is better to rewrite the third variable $\displaystyle r$ in terms of $\displaystyle x$ or $\displaystyle y$. By the properties of similar figures, it is easy to see that $\displaystyle \frac{r}{2}=\frac{x}{4} \rightarrow r=\frac{x}{2}$. Plug it into the above equation, we have:

$\displaystyle \frac{1}{3}\pi (\frac{x}{2})^2 x + \pi (4)^2 y = 10$ or $\displaystyle \frac{\pi}{12} x^3 + 16\pi y = 10$

Now differentiate both sides with respect to time $\displaystyle t$, we have:

$\displaystyle \frac{\pi}{4}x^2 \frac{dx}{dt}+16\pi\frac{dy}{dt}=0 \Rightarrow x^2\frac{dx}{dt}+64\frac{dy}{dt}=0$

We have the relations between $\displaystyle \frac{dx}{dt}$ and $\displaystyle \frac{dy}{dt}$