. On Cauchy's modulus surfaces. value in 2) we get \^lHAy-y^ityj-I^Vi^^ ^ - 0 or,z^t^^t^^Tiy^^ ~^-2N(r:7^Yy^7^ = 0 cpancling: 2Z^2ii\lP75^VX^^ -2y2Yl\Ay-y-Ay= = _:.>a\1^:^^ri^_2\r^>^^Or, collecting terms: 2(Z-1) ^)iy(Z-l) r: -(2Zt2)YH?^y5(24X7)(Z-l!) = -Y^vAy^y^ squaring: 0 2Z|-2 igvAy^Cz-i)^ = it^ly-y^ or, (44 ^Xy^-XS^Cz^-sz fi) = (i+Xy-y^4(z^2Z4i) Or 4Z%4AyZtZy*A-8Z-cS>Z-2;f:^Z+4:i4;|yf?ly^ CollectirfT terns: we get for tlio reqmirecl equation(r) Z = 0 The projection of a curve of fastest descent upon the yZ plane istherefore a quartlc. That the curve must generally be a quartic is
. On Cauchy's modulus surfaces. value in 2) we get \^lHAy-y^ityj-I^Vi^^ ^ - 0 or,z^t^^t^^Tiy^^ ~^-2N(r:7^Yy^7^ = 0 cpancling: 2Z^2ii\lP75^VX^^ -2y2Yl\Ay-y-Ay= = _:.>a\1^:^^ri^_2\r^>^^Or, collecting terms: 2(Z-1) ^)iy(Z-l) r: -(2Zt2)YH?^y5(24X7)(Z-l!) = -Y^vAy^y^ squaring: 0 2Z|-2 igvAy^Cz-i)^ = it^ly-y^ or, (44 ^Xy^-XS^Cz^-sz fi) = (i+Xy-y^4(z^2Z4i) Or 4Z%4AyZtZy*A-8Z-cS>Z-2;f:^Z+4:i4;|yf?ly^ CollectirfT terns: we get for tlio reqmirecl equation(r) Z = 0 The projection of a curve of fastest descent upon the yZ plane istherefore a quartlc. That the curve must generally be a quartic is also gerametricallyapparent. As both surfaces pass thru the absolute at infinity, theircurve of intersection being of the sixth , degenerates into aquartic and the infinite imaginary circle, absolute. To determine the asymptotes to this curve, put y = mZ f 1b, Then (y(+4)Z^(inZtb)^-2U-4)Z(mZfbf - 16Az(mZfb) V (^^4) (mZ+bf-lGZ = 0. Expanding and collecting terms. The above figure illustrates the circles and line of fastest show that the curve of intersection of the surface and yZ planeis tangent to xy plane at (x=l,y=0). V/e have previously shown that the surface is tangent at this point. Therefore if the curve passesthru this point it must be a point of tangency. The coordinates(1,0) satisfy the equation of the curve, therefore (1,0) is a pointof tangcnc;^. —--—=====—= The tangent rlane at (0,0,1) ir, 304x4Z-l= 0 for^ = 2(x+l)Z-2(x-1> for x = 0,Z= 1 this heroines2Z,+2=4 = 2Zv-2v - 0, for y= 0, Z= 1. f =[(xflf-f-yfl X = G, y= 0, hecomes 1. Z ^,4xfZrl = 0 is the tangent plane at (0,(1,1). To getthe equation of the projection of the lines of fastestdescent in the xZ plane eliminate y hetween 1) z[()^^-y^- ^x-lf+-y3- 0 and 2) x^ (y- X )^= f n From 1) -tUH)^-^ (x-1)^ - Z-1 Z(x-»lf 4 (x-lyZ-1 And suh8titutin<j this value in 2) we get x^f -Ux^if^ (x-1)^- xltz(^i)^M^zir^ = 1- Or, expanding, removing parentheses, collectin
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