- #1
coolxal
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Homework Statement
Water flows from a conical tank with circular orifice at the rate
[tex]\frac{dx}{dt} = -0.6*\pi*r^2\sqrt{2g}\frac{\sqrt{x}}{A(x)}[/tex]
r is the radius of the orifice, x is the height of the liquid from the vertex of cone, A(x) is area of the cross section of the tank x units above the orifice. Suppose r = 0.1 ft, g = 32.1 ft/s, tank has initial water level of 8ft and initial volume of 512*(pi/3). Computer water level after 10 min with h = 20s.
Homework Equations
Modified Euler's Method:
[tex]w_{0} = \alpha[/tex]
[tex]w_{i+1} = w_{i} + \frac{h}{2}[f(t_{i}, w_{i}) + f(t_{i+1}, w_{i} + hf(t_{i}, w_{i}))][/tex]
The Attempt at a Solution
Is A(8) = (512*(pi/3))/(8/3) = 201.0619?
What do I set for y and t? If the question was f(y, t) = y' = -y + t + 1, 0 <= t <= 1, y(0) = 1, h is the step size, [tex]w_{0} = \alpha = y(0)[/tex] is the initial condition, t is variable between 0 to 1 with step size h, ...
But if I input all those numbers into the equation I get dx/dt = -0.6*pi*0.1^2*sqrt(2*32.1)*sqrt(x)/201.0619 which leaves me with just x. I assume x = t in this case but what is y? How do I get it into the form f(y, t)?