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G x1 ; x2 ; . . . ; xn  F 1 1 2 2 k k subject to the (necessary) conditions @G @G @G 0; 0; . . . ; 0 @x1 @x2 @xn where 1 ; 2 ; . . . ; k , which are independent of x1 ; x2 ; . . . ; xn , are the Lagrange multipliers. APPLICATIONS TO ERRORS
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The theory of di erentials can be applied to obtain errors in a function of x; y; z, etc., when the errors in x; y; z, etc., are known. See Problem 8.28.
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TANGENT PLANE AND NORMAL LINE TO A SURFACE 8.1. Find equations for the (a) tangent plane and (b) normal line to the surface x2 yz 3y2 2xz2 8z at the point 1; 2; 1 .
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(a) The equation of the surface is F x2 yz 3y2 2xz2 8z 0. A normal to the surface at 1; 2; 1 is N0 rFj 1;2; 1 2xyz 2z2 i x2 z 6y j x2 y 4xz 8 kj 1;2; 1 6i 11j 14k Referring to Fig. 8-1, Page 183: The vector from O to any point x; y; z on the tangent plane is r xi yj zk. The vector from O to the point 1; 2; 1 on the tangent plane is r0 i 2j k. The vector r r0 x 1 i y 2 j z 1 k lies in the tangent plane and is thus perpendicular to N0 . Then the required equation is r r0 N0 0 i:e:; f x 1 i y 2 j z 1 kg f 6i 11j 14kg 0 or 6x 11y 14z 2 0 6 x 1 11 y 2 14 z 1 0
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(b) Let r xi yj zk be the vector from O to any point x; y; z of the normal N0 . The vector from O to the point 1; 2; 1 on the normal is r0 i 2j k. The vector r r0 x 1 i y 2 j z 1 k is collinear with N0 . Then    i j k    x 1 y 2 z 1 0 i:e:; r r0 N0 0    6 11 14  which is equivalent to the equations 11 x 1 6 y 2 ; These can be written as x 1 y 2 z 1 6 11 14 often called the standard form for the equations of a line. By setting each of these ratios equal to the parameter t, we have x 1 6t; y 2 11t; z 14t 1 14 y 2 11 z 1 ; 14 x 1 6 z 1
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called the parametric equations for the line.
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[CHAP. 8
8.2. In what point does the normal line of Problem 8.1(b) meet the plane x 3y 2z 10
Substituting the parametric equations of Problem 8.1(b), we have 1 6t 3 2 11t 2 14t 1 10 or t 1
Then x 1 6t 7; y 2 11t 9; z 14t 1 15 and the required point is 7; 9; 15 .
8.3. Show that the surface x2 2yz y3 4 is perpendicular to any member of the family of surfaces x2 1 2 4a y2 az2 at the point of intersection 1; 1; 2 :
Let the equations of the two surfaces be written in the form F x2 2yz y3 4 0 Then rF 2xi 3y2 2z j 2yk; rG 2xi 2 2 4a yj 2azk and G x2 1 2 4a y2 az2 0
Thus, the normals to the two surfaces at 1; 1; 2 are given by N1 2i j 2k; N2 2i 2 2 4a j 4ak
Since N1 N2 2 2 2 2 4a 2 4a  0, it follows that N1 and N2 are perpendicular for all a, and so the required result follows.
8.4. The equation of a surface is given in spherical coordinates by F r; ;  0, where we suppose that F is continuously di erentiable. (a) Find an equation for the tangent plane to the surface at the point r0 ; 0 ; 0 . (b) Find an equation for the tangent plane to the surface r 4 cos  at the p point 2 2; =4; 3=4 . (c) Find a set of equations for the normal line to the surface in (b) at the indicated point.
(a) The gradient of in orthogonal curvilinear coordinates is r 1 @ 1 @ 1 @ e e e h1 @u1 1 h2 @u2 2 h3 @u3 3 e2 1 @r ; h2 @u2 e3 1 @r h3 @u3
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