Toricelli's Law, Water draining out off a tank

[Idealistic]

A water tank has the shape obtained by revolving the parabola x[sup:2p91lfhh]2[/sup:2p91lfhh] = b(y) around the y-axis. The water depth is 4 ft at noon then the plug is removed. At 1 pm, the depth of the water is 1 ft.

find y(t), then find when the tank will be empty.

I know:

A(y)dy/dt = -k(y)[sup:2p91lfhh](1/2)[/sup:2p91lfhh].

I'm just having trouble finding A(y)

A(y) = pi(r)^2, but I need to find "r" as a function of "y" and all I know is that y = x[sup:2p91lfhh]2[/sup:2p91lfhh]/b.

I tried x = r = (yb)[sup:2p91lfhh](1/2)[/sup:2p91lfhh] in:

pi(yb)[sup:2p91lfhh](1/2)[/sup:2p91lfhh] dy/dt = -k(y)[sup:2p91lfhh](1/2)[/sup:2p91lfhh]

but that just gives me a linear function of y(t) which doesn't makes sense seeing as how the parabola will be narrower towards the botom of the tank thus making the rate at which the depth decreases faster.

 

[Deleted member 4993]

A water tank has the shape obtained by revolving the parabola x[sup:4jcp8n7y]2[/sup:4jcp8n7y] = b(y) around the y-axis. The water depth is 4 ft at noon then the plug is removed. At 1 pm, the depth of the water is 1 ft.

find y(t), then find when the tank will be empty.

I know:

A(y)dy/dt = -k(y)[sup:4jcp8n7y](1/2)[/sup:4jcp8n7y].

I'm just having trouble finding A(y)

A(y) = pi(r)^2, but I need to find "r" as a function of "y" and all I know is that y = x[sup:4jcp8n7y]2[/sup:4jcp8n7y]/b.

I tried x = r = (yb)[sup:4jcp8n7y](1/2)[/sup:4jcp8n7y] in:

pi(yb)[sup:4jcp8n7y](1/2)[/sup:4jcp8n7y] dy/dt = -k(y)[sup:4jcp8n7y](1/2)[/sup:4jcp8n7y]

but that just gives me a linear function of y(t) which doesn't makes sense seeing as how the parabola will be narrower towards the botom of the tank thus making the rate at which the depth decreases faster.

That is the correct answer.

As height decreases - the surface area also decreases - thus opposing the tendency to reduce the flow due to reduced height.

V = b/2 * h[sup:4jcp8n7y]2[/sup:4jcp8n7y]

dV/dt = kh = b/2 * 2 * h * (dh/dt) ? dh/dt = k/b

dh/dt = -3 ft/hr = k/b

h = -3*t + 4

h = 0 at t = 4/3 = 1 hr 20 min