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THe problem says that the level of liquid [tex]h = y_2 - y_1[/tex] drops at a rate

[tex]\frac{dh}{dt} = -\sqrt\frac{2gh{A_1}^2}{{A_2}^2-{A_1}^2}[/tex]

where [tex]A_1[/tex] and [tex]A_2[/tex] are the areas of the opening (water spilling out) and the top of the surface respectively. Viscosity is ignored...

So the problem asks to solve this differential equation for h(t), letting [tex]h = h_0[/tex] at [tex]t = 0[/tex]

So I notice from my diff eq class that this equation is seperable.

Thus I wrote

[tex]{(\frac{dh}{dt})}^2 = \frac{2gh{A_1}^2}{{A_2}^2-{A_1}^2}[/tex]

Then I isolate h:

[tex]\frac{(dh)^2}{h} = \frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}(dt)^2[/tex]

[tex]\frac{dh}{\sqrt h} = \sqrt\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}dt[/tex]

[tex]2\sqrt h = \sqrt\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}t + C[/tex]

[tex]\sqrt h = \frac{1}{2}\sqrt\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}t + C_1[/tex]

So finally:

[tex]h = (\frac{1}{4})\frac{2g{A_1}^2}{{A_2}^2-{A_1}^2}t^2 + C_2[/tex]

Would anyone be willing to confirm if I did indeed do this correctly? And how do I solve for the arbitrary constant?

In addition, the 2nd part asks how long it would take to fill a 9.4cm tall cylinder filled with 1.0 L of water if the opening is at the bottom and has a 0.50 cm diameter.

Would i just use my newly acquired h(t) equation and simply isolate t, plug in constants and solve?

Thanks a lot!!!