710.

Let u be the height of the vessel above the orifice,

y the radius of the base; then the velocity with which the water

1

is projected from the vessel is due to parameter of the para

4

bola described, and also to the altitude u; .. this parameter is

4u.

Hence, by the question, and the equation to the parabola,

Again, cpa, where p is the parameter of the parabola de

scribed at the end of the time t.

Hence the velocity at the vena

the area of the base of the cylinder is known; and its altitude is

711.

Let a denote the axis of the paraboloid, 6 the radius of its base, and p its parameter; then the section parallel to the axis and horizon is a parabola, whose axis is

x being the distance of the section from the lowest point. Hence

712.

Let AN (draw the figure) be the altitude of the

orifice, NP the horizontal distance; produce NA to T making

AT NA and join TP.

Make the TPS PTS; then

since SPST, and TP is a tangent to the parabola described by the fluid, the point S is the focus of the parabola. But SA =

4

of the parameter space due to the velocity at the vena-contracta = altitude of the fluid above the orifice A. Make, therefore, AS, and we have NB NA + AB = the required alti

AB

tude of the fluid.

713.

Let y be the radius of any horizontal surface of water in the required vessel, distant from the bottom by the interval x; then if A be the area of the orifice, the velocity of the orifice is

Now the surface being supposed to begin its descent with the given velocity V, we have (a being the given value of x)

When a = 0, the content is 0, and C will be 0,

714.

Let A be the area of either orifice, A' that of the descending surface; then the surface being distant from the upper orifice by the interval x, and from the lower by a + x, the velocities at these orifices are

√(9x) and √ (g.a+x)

respectively. Hence the rate of afflux is

√g × W√x + √a+x).

and the velocity of the descending surface is..

2A'

3aAg

a

× {(a+x)3 − x3+ C}

Let x = 0; then to and Ca3.

.. when x = a, we get the time of emptying the first half of the prism, viz.

715.

t =

2A'

3aAg

4A' a 3A 9

× (√2 − 1).

The semi-cube thus cut off may evidently be a rightangled triangular prism, whose ends are each equal to the semiside of the cube. In this case there will be no difficulty in finding the

time of discharge. If, however, the bisection of the cube be effected by a plane passing through the diagonal of the cube, and equally inclined to the pairs of faces which meet at the extremities of that diagonal, then the semicube will be a kind of non-descript solid, having two faces rectangular trapeziums, two others equal rightangled A, another side a square, (the face of the cube,) and the last a rhombus made by the cutting plane.

Moreover, all sections parallel to this rhombus will be segments of this same rhombus, being deficient by an isosceles ▲. If the student cut a cubical piece of an apple, or any other substance, as above directed, he will more clearly understand the form of this solid, than by inspecting any diagram.

Now the semi-cube being filled with water, and its face (the rhombus) being placed parallel to the horizon, let an orifice, whose area is A, be made in the lowest point, viz., in the solid subtended by the rhombus. Again, let a be the side of the cube, a the distance of the lowest point from the rhombus, and a the distance of that point from the surface of the water at any time during its descent; then since the diagonals of the rhombus are those of the cube and its base, they are

Also the part cut off from these legs by the water when de