Water flows from a conical tank with circular orifice at the rate

dx/dt = -0.6*pi*r^2*sqrt(2g)*sqrt(x)/A(x)

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.

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, y(0) 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)?