There will be a rise and fall twice in every lunar day, which will be the lunar semi-diurnal tide.

When the Moon is not on the equator and also in consequence of the unsymmetrical distribution of land and water the two tides in the lunar day will not be equal, so that to produce the resultant effect a tide of period a lunar day must be superimposed on the lunar semi-diurnal tide, and this is the lunar diurnal tide.

In consequence of the variation of the Moon's declination there will be a variation with a period of a fortnight of the average height of water at any place, and this effect can be produced by the superposition of a lunar fortnightly tide.

To take account of the distribution of land and water, the level surface of the sea must be such that the part of it which bounds the sea encloses the same volume of water.

[Thomson and Tait's Natural Philosophy, § 807, 808, 809.]

上

=

1. FIND the polar equation of a conic section referred to the focus as pole, and the polar equation of a chord and a tangent.

r

A hyperbola is described similar to the given hyperbola

=1+e cos having the same focus and touching it at the

point = a, prove that the length of its latus rectum will be

and the equation of the common chord of the hyperbolas will be

then the equation of a pair of common chords of the hyperperbolas is

2. If a, b, c be the sides, A, B, C the angles, and E the spherical excess of a spherical triangle, prove that

If C=A+B, prove that the chord triangle is rightangled, the angular radius of the circumscribing small circle a b = tan- tan 2

is and sin

2

E

2

2

If the arc CD be drawn making the angle ACD=A, then the angle BCD = B and AD=DC= DB.

Hence D, the middle point of the arc B, is the centre of the circumscribing small circle, and if the radius to D meet the chord AB in d, d is the centre of the plane circle circumscribing the chord triangle ABC, and therefore the chord triangle is right angled.

(This spherical triangle in which one angle is equal to the sum of the other two is more analogous to the rightangled plane triangle than the spherical triangle which has one angle a right angle.)

3. Explain Horner's method of approximating to the real roots of an equation.

Find to two places of decimals the roots of the equation x2-6x2+9x 3 = 0.

The roots are 0:47, 165, 3.88.

4. If x, y be the rectangular, r, the polar coordinates of a point which moves once round the perimeter of a closed curve in a certain direction, prove that the area of the curve is expressed by faxdy or - fydx or 1 fr2 d0.

are

Interpret these expressions when the perimeter cuts itself. Prove that the areas of the two loops of the curve

r2-2ar cos 0-8ar + 9a2 = 0,

{32π+24 √√(3)} a* and {16π- 24 √(3)} a2.

As the point travels round the perimeter in the direction so as to have the area of the curve on the left hand, the sum of all the elements xdy for the same value of y (fig. 46) will be the area of the curve cut off by two straight lines parallel to the axis of x at distances y and y+dy, and therefore the whole area is fxdy.

Similarly the area is - fyda (fig. 47), and 1 fr2dė (fig. 48). (Since xdy+ydx=d.xy is a complete differential, therefore f(xdy+ydx) = 0 round a closed curve; but xdy — ydx = r2do is not a complete differential, therefore f(xdy - ydx) = fr2do depends on all the intermediate value of x and y, and is equal to twice the area of the curve.

An integrating factor of xdy-ydx is

fxdy

1- ydx = √do =

+ y2

2

1

x2 + y2 1

and then

=0 or 2π, according as the origin is

outside or inside the curve).

If the perimeter cuts itself (fig. 49) the value of any one of the integrals taken round a loop will be numerically equal to the area of the loop, and positive or negative, according as the area of the loop is to the left or the right of the point as it travels round the perimeter of the loop.

Therefore, if the point travel once round the perimeter the value of any one of the integrals will be the sum of the areas of the loops which were on the left hand diminished by the sum of the areas of the loops which were on the right hand. The curve is a limaçon or nodal Cartesian oval (fig. 50)

=

4 + cos @ ± √ {( 1 + cos 0) (7 + cos 0)},

= 24 +16 cos + cos 20 ± 2 (4+cos Ø) √ {(1+cos @) (7 + cos 0)}.

The area of a loop is "de, the upper sign being taken with the radical for the outer loop, the lower sign for the inner loop.

motion of a particle under the action of a central force P.