Hence the point, where fracture will take place, is given

23. A vessel, of given capacity, in the form of a surface of revolution with two circular ends, is just filled with inelastic fluid which revolves about the axis of the vessel, and is supposed to be free from the action of gravity: investigate the form of the vessel that the whole pressure which the fluid exerts upon it may be the least possible, the magnitudes of the circular ends being given.

Shew that, for a certain relation between the radii of the circular ends, the generating curve of the surface of revolution is the common catenary.

Let the axis of the vessel be taken as the axis of x (fig. 57). Then

Pressure on curve surface = [2πwyds.‡pw°y" = πpw" fy" (1+p3)* dx.

But V = Pp + C, C being a constant: hence

The values of y at the circular ends being given, there is Sy, = 0, Sy,, = 0: thus the equation for the limits becomes

бу

(V., - PP.) dx,, - (V, — P,p,) dx, = 0, or Cdx,, - Cdx, = 0:

but Sx, Sx, are independent of each other: hence C=0, and the equation for the generating curve becomes

dx

x + c = log (y+(y-a), c being a constant:

y=a(1+p2), y2 - a2 = a2

=

α (y2 — a2)1

...(1),

a

a and m being unknown constants.

Let the origin be in one of the ends, of which the radius=b:

Also let 3 denote the capacity of the vessel: then, from (1), b' denoting the radius of the other end,

Thus (3) and (4) determine the constants a and m. The equation (2) defines the generating curve.

COR. If m = 1, the curve is the common catenary, the conditions being, from (3) and (4),

a = b, k3

=

4′′πb {b′ (b” — b")* + ba log

b' + (b' — b*)*)
b

24. If a, ß, y, be the direction-cosines of one of the two lines of vibration of the plane front of a wave in a biaxal crystal, and a', B', y', those of either of the two lines of vibration of a plane front intersecting the former plane front at right angles and passing through the line (a, B, y), prove that

Let (a, B, y) be a line of vibration in the front

lx + my + nz = 0 ........

..(1).

The equation to a plane front perpendicular to (1) and passing through (a, ß, y), is

Then, (a, ẞ, y) being a line of vibration in (1),

Also, (a', B', y') being a line of vibration in (2),

The equations (5) and (6) become therefore

(6).

TUESDAY, Jan. 3, 1854. 9...12.

1. THE complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.

If K be the common angular point of these parallelograms, and BD the other diameter, the difference of the parallelograms is equal to twice the triangle BKD.

Since BK (fig. 58) bisects EG, and KD bisects HF, the two GBK, FKD are together equal to EBK, HKD; to these equals add the unequals GF, EH; then the difference of the parallelograms GF, EH is equal to the difference of the figures CBKD, ABKD: but the latter difference is evidently equal to twice the triangle KBD; for CBKD exceeds CBD or ABD by the triangle KBD, and CBD or ABD exceeds ABKD by the triangle KBD; therefore the difference of the parallelograms GF, EH is equal to twice the triangle KBD.

2. Divide a given straight line into two parts so that the rectangle contained by the whole line and one of the parts shall be equal to the square of the other part.

Produce a given straight line to a point such that the rectangle contained by the whole line thus produced and the part produced shall be equal to the square of the given straight line.

In Euclid's figure, the rectangle contained by CF and FA is proved to be equal to the square on CA.

If therefore CA be the given line, describe a square on CA, and proceed as in Euclid: F will be the point required.