Then if x, y, z, x', y', ' be the co-ordinates of the extremities of the least distance (S),

d2 = (x − x′)2 + (y − y')2 + (≈ − x′)2

is to be a minimum, the variables being subject to the conditions (1) and (2). Differentiating,

0 = (x − x') (dx − d x') + (y − y') (dy − dy') + (x − x') (d≈ − dx').

But from (1) and (2) we have

Therefore, substituting these values, and, as

r and r are independent, equating to zero the coefficients of the differentials, we have the two conditions.

Now multiply each term of (3)

and (4) by the corre

sponding members of (1) and (2), subtract the one result from the other and transpose; then observing that

(x − x')2 + (y - y' )2 + (≈ − x′)2 = 82,

-

-

S2 = (a − a') (x − x') + (b − b′) (y − y') + (c − c ́) (≈ − x′).

--

Dividing the first member of this by (6), and each term of the second member by each member of (5), we find

(a — a′) (mn' — m'n) + (b − b') (n l' — n't) + (c − c') (l m' − I'm)

{(mn' - m'n)2 + (nl′ — n'l)2 + (I'm' — I'm)2 } }

Equations (5) are the equations to the line of least distance, and it appears that it is perpendicular to both the lines (1) and (2), since we have

1 (mn' — n' n) + m (n l′ − n' l) + n (lm' — l′m) = 0,

and l'(mn' - m'n) + m' (nl' — n' l) + n' (lm' − l′ m) = 0,

which are the conditions of perpendicularity.

(15) To find the maximum and minimum radii of a section of the surface, the equation to which is

(x2 + y2 + x2)2 = a2x2 + b2 y2 + c2x2

x

made by the plane la+my+ n≈ = 0.

We have here to find

p2 = x2 + y2 + ", a maximum,

x, y, z, being connected by the equations

r2 = a2x2 + b2y" + c2x2,

0 = lx + my + nz.

(1) +λ (3) - μ (2) = 0 gives, on equating to zero the coefficients of each differential,

x + \ l = μ a2 x, y + λm = μb3y, ≈ + λn = μc2 z.

Multiply by x, y, z, and add, then by the original con

ditions

Substituting this value, and transposing,

(-1)=x, (-1) - xm, (-1) = -x

=

ψ λm,

= λη.

a quadratic equation for determining 2, and consequently r.

This is the equation in the Wave Theory of Light by which the velocities of a wave propagated in a crystalline medium are determined. The surface ra2x2 + b2 y2 + c2x2 is called the surface of elasticity. See Fresnel, Mémoires de l'Institut, Vol. VII. p. 130, and Herschel's Light, Sect. 1012.

(16) To find the area of a section of the ellipsoid,

made by the plane la + my + n≈ = 0.

By the same method as in the last example we obtain as the equation for determining the principal axes,

a212 7.2 a2

b2 m2 2
+
b2

+

c2 n2

The last term of this when arranged according to powers of 2 is

a2 b2 c2

a2 l2 + b2 m2 + c2 n2

and this being equal to the product of the roots, the area of the section is

(17) tion is

πανε

(a2 l2 + b2 m2 + c2 n2)}'

To find the volume of the ellipsoid whose equa

ax2 + a'y2 + a′′z2 + 2 by z + 2b′x z +2b′′xy = c.

As in the preceding examples we have first to find the value of the principal axes, or rather of their product; and if this be aẞy, then the volume of the ellipsoid will be

Now the principal axes are maxima or minima values of the radius; we therefore have

r2 = x2 + y2 + x2 a maximum;

x, y, being subject to the equation of condition

a x + a'y2 + a′′x2 + 2byx + 2 b'x z + 2b′′xy = c. Differentiating,

(ax+b'z+b′′y) dx+(a'y+bz+b′′x) dy+(a′′z+by+b'x)dz=0, (2) λ (1)+(2)=0 gives, on equating to zero the coefficients of each differential,

λ x + a x + b'z + b′′"y = 0

λy + a'y + bx + b′′x = 0

λ x + a′′x + by + b'x = 0

(3)

Multiply these equations by x, y, z respectively and add, then by the equation of condition,

On substituting this value of A in the equations (3) they become

To eliminate x, y, z, multiply the first of these by

-a

(a)(-a)-; the second by {b'+"a")};

a cubic equation in

= 0 ;

C

If it be arranged according to

powers of 2, the last term with its sign changed will be equal to the product of the roots, that is, to the product of the squares of the principal axes; and its square root is the quantity which we seek.

Multiplying it therefore by

we find that the volume of the ellipsoid is equal to

(18)

4.π

3 (aa'a" - ab a′b" - a"b"" + 2bb'b'')

4.π

3

To find the least ellipse which will circumscribe a given triangle.

Let ABC (fig. 10) be the triangle. Take C as the origin, CA, CB as the axes of x and y.

AC = a, BC

=

b, ACB = 0.

The general equation to an ellipse is

Ax2 + Bxy + Cy2 + Dx + Ey + 1 = 0,

which involves five arbitrary constants; three of these may be determined by the conditions that the ellipse shall pass through the three points A, B, C. Instead however of directly expressing the undetermined coefficients in terms of those which are determined, it will conduce to the symmetry of our analysis to assume two indeterminate quantities of which the coefficients of the equation are functions which can be determined by the conditions of the ellipse passing through the three given points; and then the actual values of the indeterminate quantities may be found by the condition of the minimum. The two quantities which we shall assume are the co-ordinates of the centre