(27) OM, OP (fig. 5) are two arcs of great circles on a sphere, and the arc PM is drawn perpendicular to OM; find when the difference between OP and OM is the greatest. Let POM = a, OP=4, OM=0; then we have while, by Napier's rules for the solution of right-angled spherical triangles, therefore or Whence 1 + tan30 = cos a (1 + tan3Ø), 1 + cos3 a tan3 = cos a (1 + tan2(). (28) To determine the dimensions of a cylinder open at the top, which, under the least surface, shall contain a given volume. Let a be the given volume, ≈ the radius of the cylinder, y its base. Then a = y = a determines the cylinder of least surface. (29) The content of a cone being given, find its form when the surface is a maximum. be the given content, a the radius of the base, y the height of the cone. the cone of greatest surface. (30) To inscribe the greatest cone in a given sphere. Let the radius of the sphere be r, and the distance of the base of the cone from the centre be x. the maximum cone. Then ar gives = (31) To find the point in the line joining the centres of two spheres, from which the greatest portion of spherical surface is visible. If r, r' be the radii of the spheres, a the distance of the centres, and the distance of the required point from the centre of the sphere whose radius is r. (32) A regular hexagonal prism is regularly terminated by a trihedral solid angle formed by planes each passing through two angles of the prism; find the inclination of these planes to the axis of the prism in order that for a given content the total surface may be the least possible. Let ABC abc (fig. 6) be the base of the prism, PQRS one of the faces of the terminating solid angle passing through the angles P, R. Let S be the vertex of the pyramid. Draw SO perpendicular to the upper surface of the prism. Join OM, RP, SQ intersecting each other in N. Then it is easy to see that MN = NO and consequently _SO = QN, and as the triangles POR, PMR are equal, the pyramids PSRO and PMRQ are equal, so that, whatever be the inclination of SQ to OM, the part cut off from the prism is equal to the part included in the pyramid SPR, and the content of the whole therefore remains constant. We have then to determine the angle ONS or OSN so that the total surface shall be a minimum. Let AB, the side of the hexagon, = a, AP, the height of the prism, b, OSN = 0. Then This is the celebrated problem of the form of the cells of bees. Maraldi was the first who measured the angles of the faces of the terminating solid angle, and he found them to be 109°. 28′ and 70°. 32" respectively. It occurred to Réaumur that this might be the form, which, for the same solid content, gives the minimum of surface, and he requested Koenig to examine the question mathematically. That geometer confirmed the conjecture; -the result of his calculations agreeing with Maraldi's measurements within 2'. Maclaurint and L'Huillier, by different methods, verified the preceding result, excepting that they shewed that the difference of 2' was due to an error in the calculations of Koenig-not to a mistake on the part of the bees. (33) To determine the greatest parabola which can be cut from a given cone. Let ABC (fig. 7) be the cone BC-a, AC-b, BN = x. EN. DN, or 3 x (ax-x2), which is to be a maximum. (34) To determine the greatest ellipse which can be cut from a given cone. Let AC (fig. 8) = a, major axis of the ellipse. CD = b, CN= x, BP being the of the ellipse shall be a maximum gives x= 2b (a2 — b2) ± b (a1 — 14a2b2 + b1)* * Mémoires de l'Académie des Sciences, 1712, p. 299. In order that this may be possible we must have that is, the angle of the cone must be less than 15o. When this is not the case, the ellipse increases continually till it coincides with the base. It may happen that the maximum value of the section is less than the base of the cone; and this will be the case unless the vertical angle of the cone be less than 11°. 57′. du da SECT. 2. Implicit Functions of Two Variables. If u be an implicit function of two variables a and y, = 0 will determine the values of x for which y is a maximum or minimum. y will be a maximum for a given positive result, and a minimum when it gives a negative one root of this equation is x=-1, which gives y = 1; x = 1 gives y = 1, neither a maximum nor a minimum. x = 0 gives y = 0, a minimum. The nature of this last value cannot be determined by determine it, differentiate u a second time considering y and a both to vary, y being a function of x; we then get, on . Differentiating a third time on the same supposition, and making x = 0, y = 0, ample we find that y=+1 and y = 1 are neither maxima nor minima. and as these values of a and y when substituted in a negative result, the value of y is a maximum. The preceding examples are taken from Euler's Calc. Diff. Part. II. Cap. XI. |