I'm having trouble with a question for a Numerical Analysis class. It's a textbook question, but I can't seem to wrap my head around it. I've viewed over the Runge-Kutta method in my book, but I have no idea how to take what the question gives and translate it into what I need. Here's the question itself:
"Water flows from ana inverted conical tank with circular orifice at the rate
$dx/dt = -0.6 \pi r^2\sqrt{2g}\frac{\sqrt{x}}{A(x))}$
where r is the radius of the orifice, x is the height of the liquid level from the vertex of the cone and A(x) is the area of the cross section of the tank x units above the orifice. Suppose $r=.1ft$, $g=32.1ft/s^2$ and the tank has an initial water level of 8 feet and initial volume of $512(\pi/3)ft^3$. Use the Runge-Kutta method of order four to find the following:
a) The water level after 10 minutes (600s) with h=20s (-> 30 steps)
b) When the tank will be empty, to within a minute."
Among other things, I don't see how the area of a cross section should impact the rate of flow; if the size of the orifice is constant, then so too should be the rate of flow, as we have no exterior forces (other than gravity) that impact how fast the water flows out. If A(x) is just the area of some circle at some height x, then as x ->0, A(x) gets larger, so the flow rate gets smaller, but does that make sense? Perhaps I'm missing something.
As a side note, I'm using Maple, so any code blocks/suggestions for that could be helpful.
Thank you in advance.
Best Answer
Hint: We know that the surface area of the cone, $A= \Pi r^2$ and the volume, $V = \frac{1}{3}\Pi r^2x$. Since $V$ is given in the question, we can find that $r^2 =\frac{512}{x}$ and substitute the $r^2$ in the area equation and write it in terms of $x$, which gives us $A(x)=\frac{512\Pi}{x}$.
Lastly, you use $\frac{dx}{dt}$ on Maple to find the solution. Look into p.290 of Burden and Faires's textbook Numerical Analysis (ninth edition) for an idea on how to write the Maple code. I assume you're using this text b/c the question you asked is right from that very textbook.