. On Cauchy's modulus surfaces. = i as the section made l)y the yz plane. Therefore the line l,x=0) lies in the surface. To find the cross seccio . xz plane with the surface, put y = 0, and we have a cubic. (5) Z(x.^ir(x^l)^=: 0, or Z = To determine the asymptotes to this curve, let Z = mx^b,the equation of the asymptote. Then TTixfh = ^x-l\a. \x-vl) or (mx^h)(x-^ir= (x-1) and mx 2r!X%riixi%hx^2hy^h=x^2x+l mx^ ^ (2m^h-l) x \ 42) xl|f)-i- C m=0, 2iT!4h-l=rG,/,h=0. The qquation of the asymptote is therefore Z = 1. To determine the points of inflexion of the cuhib Z =|x--lj ,we have from dz = (2x
. On Cauchy's modulus surfaces. = i as the section made l)y the yz plane. Therefore the line l,x=0) lies in the surface. To find the cross seccio . xz plane with the surface, put y = 0, and we have a cubic. (5) Z(x.^ir(x^l)^=: 0, or Z = To determine the asymptotes to this curve, let Z = mx^b,the equation of the asymptote. Then TTixfh = ^x-l\a. \x-vl) or (mx^h)(x-^ir= (x-1) and mx 2r!X%riixi%hx^2hy^h=x^2x+l mx^ ^ (2m^h-l) x \ 42) xl|f)-i- C m=0, 2iT!4h-l=rG,/,h=0. The qquation of the asymptote is therefore Z = 1. To determine the points of inflexion of the cuhib Z =|x--lj ,we have from dz = (2x-2) (X v3 )^ - (x-Xx^l) 2 ) = 4(x-l) (x-1) 4 Ufl)! the condition d\ = 4(xvl? -4(>:-l)(x^l)^3 = -S(x-^2) . TMs liecorcef; O if Sx-in = 0, or x = 2, therefore x=2, y==0, is a pointof inflexion, for f(Z) cl anges sign as x passes thr?i this value. If Z = 0, then (x*\ y^-D^V 47*^ 0 is the equation oft fee curveof intersection of the surface and the xy plane. x^^ 1 = ^=t-2iy. »^y = 0, x^ 1, X =, Points only. Figure The ahove figure will give a general idea of the appearance ofthe surface. To determine whether or not the xy plane is tangent to thesurface at (1,0,0). AL (z-^i^ \ (x-Xy)-v8f (y-y.) = 0 is the equation of the tangentdz, c)x, ciYi plane to f(x,y,z) = ;fj_,- 2yz - 2y; ?>f = C. f_ =^ (xvif ^ y^jdf = 4. Therefore the tangent plane at (1,C,C) is 4Z = 0, or Z = 0, tbe xy plane. To determine the minimun; points of the surface. Write theequation in the explicit form. Taking we get 2(x-l)f(x41 )\/?- fx-lfyj2(xVl) Andb\ =r2[(xnf t ^-f a(x-l)a(xvl)-X(x-l)2()-[(x^lf V ^ f [2f(xvlf t y-)2(x>l)^ . ^x^lH y^ Computin , similarly,and^\ and substituting (1,0,0), the <5y ST^ coordinates of the point of tangency, for x,y,z, in these express-ions; we see that the first partial cerivatives vanish, while thesecond partial derivatives are positive: therefore (l,0,n) is aminimum point of the
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