and let x, y, z be the currents in the wires supposed to be going from P to Q. The electromotive force in the wire A is (a'-') — (α — λ), and by Ohm's law this is equal to the product of the resistance into the current. or Therefore by symmetry, with Similarly Therefore therefore x + y + z=0. (ca+ab) zab (x + y) = a (b+c) 1 + b (c+a) Is, 2= a a (b + c) I + b (c + a) I¿ bc + ca+ab 1. FROM the consideration of a conic and its director circle, prove that the condition that it may be possible to circumscribe quadrilaterals to a conic S such that the ends of two diagonals shall lie on another conic S', is Θ* - 400' Δ + 8Δ Δ' = 0, where k3A+k2® + kO' + A' is the discriminant of kS+ S'. Prove that the two diagonals of any such quadrilateral intersect in a fixed point, that the third diagonal will be a fixed straight line, and that the two tangents drawn from any point of S to S will divide this straight line in an involution whose double points lie on S'. Prove also that the points of contact of the tangents drawn from these double points will lie on S'. and since this is homogeneous in all senses (when A, O, O', A' are supposed to be of dimensions 3, 2, 1, 0; 0, 1, 2, 3; or 1, 1, 1, 1), it is the relation expressing the projective relation between the two conics, that quadrilateral can be circumscribed to S, such that the ends of two diagonals lie on S'. In the case of a conic and its director circle, any such quadrilateral is a rectangle, the two diagonals intersect in the centre, and the third diagonal is at infinity; therefore in general the two diagonals intersect in a fixed point, and the third diagonal is a fixed straight line. Any two tangents from a point on the director circle to the conic are at right angles, that is, divide the distance between the circular points harmonically, and the circular points are the points in which the line at infinity meets the director circle; therefore in general the two tangents drawn from any point on S' to S divide the third diagonal in an involution, the double points of which are the points in which the third diagonal meets S. At the points of intersection of a conic and its director circle that is, they are the points where the directrices meet the conic, or the points of contact of the tangents drawn from the foci, which are also the points of contact of the tangents drawn to the conic from the circular points; hence, in general, the tangents drawn to S from the double points lie on S'. 2. Prove that the normals drawn at different points of a small portion of a surface pass through two focal lines at right angles to each other. Deduce Gauss' measure of curvature at any point of a surface. If the part of a screw surface of uniform pitch be taken which is formed by one complete revolution of a generating line and the axis of this part of the surface be bent into a circle, prove that this part of the surface supposed inextensible will assume the form of a surface generated by the revolution of a catenary about its directrix. Normals along a line of curvature ultimately intersect; hence if C, C' be the centres of principal curvature at the point of a surface (fig. 82), and if a small square PQRS formed by lines of curvature be drawn enclosing O, then the normal planes through PQ and RS will ultimately pass through C, and the normal planes through PS and QR will ultimately pass through C', and these planes will pass through the two focal lines AB, A'B', which are parallel to the lines of curvature at 0; and therefore any normal drawn at a point near O will pass through these two focal lines. If P'Q'R'S' be the small square cut out on the unit sphere by the normals parallel to the normals to the surface along PQRS, then it P'S' 1 - PS ρ and lt P'Q' the principal radii of curvature at 0. area P'Q'R'S' 1 PP area PQRS of curvature at the point 0. which is Gauss' measure A screw surface of uniform pitch is generated by the motion of a straight line which intersects at right angles a fixed axis, about which it twists with an angular velocity which bears a constant ratio, 1 с suppose, to the velocity of the point of intersection with the axis; and therefore when the generating line has made a complete revolution, the point of intersection with the axis will have moved through a distance 2c, and a point on the generating line at a distance σ from the axis will have described a helix of length 2π √(o2+c2). When this length 27c of the axis is bent into a circle of radius c, the helix will also be bent into a circle of radius √(o+c2), and the generating lines will be bent into meridian curves on a surface of revolution. If y be the distance of a point from the axis of revolution, then y =√(o2+ c2), and the meridian curve is therefore a catenary, of which the axis of revolution is the directrix. In the screw surface if the axis be taken as the axis of z, then the equation of the surface is and therefore at corresponding points of the two surfaces Gauss' measure of curvature is the same. a. Show how to find y as a function of x so that the value of the integral may be a maximum or mininum, the form of ƒ being given. |