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**1. Homework Statement**

Tank is filled with water to a height h_0 = 1.00 meters

Cross section area of tank (A_1 = .785 m^2)

Hole at bottom of tank with an area, A_2 = .002 m^2

**2. Homework Equations**

How much time does it tank to half empty the tank? find t_(1/2).

Question also provides "a useful antiderivative."

/ = integral sign

/ x^(-1/2) dx = 2x^(1/2)

**3. The Attempt at a Solution**

1.

V = A_2 * v_hole

dV/dt = A_2 * v_hole

2. getting an equation for velocity coming out of hole

v_hole = [2gh + (v_surf)^2] ; got that from using Bernoulli's principle.

3. equation of continuity to get v_surf

v_surf = v_hole * A_hole / A_surf

thus,

v_hole = (2gh)^(1/2) , since [1 - (A_2/A_1)^2] = ~1

4. rate of change of the level of water in the tank

dh/dt = (-v_hole * A_2)/(A_1)

dh/dt = -.0113

5.

h_2 - h_0 = -.0113 * t_(1/2)

a half empty tank means, h goes down 1/2, 1/2 of h = 1, h=-.5

t_(1/2) = -.5m / -.0013 m/s = 44.3 s

44.3 s is wrong though. I don't get how come. I went through the hints for this problem. It mentions

dy/dx = f(x) * g(y) ... f(x) and g(y) are known functions

/ dy/g(y) = / f(x)dx

I don't get which known functions they are referring to. And it's also been probably 2 years since I ever used calculus, so those equations don't make much sense to me.

Any help would be appreciated.

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