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392    CHAPTER 12  Vector Integral Calculus

                                 Then
                                                                        √
                                                              N(P 0 ) = i +  2j + k.
                                 This result is consistent with the fact that a line from the origin through a point on this hemisphere
                                 is normal to the hemisphere at that point.

                                 12.5.2  Tangent Plane to a Surface
                                 If a surface   has a normal vector N(P 0 ) at a point then it has a tangent plane at P 0 .This
                                 is the plane through P 0 : (x 0 , y 0 , z 0 ) having normal vector N(P 0 ). The equation of this tangent
                                 plane is

                                                    N(P 0 ) ·[(x − x 0 )i + (y − y 0 )j + (z − z 0 )k]= 0,
                                 or

                                        ∂(y, z)              ∂(z, x)              ∂(x, y)
                                                   (x − x 0 ) +         (y − y 0 ) +         (z − z 0 ) = 0.
                                        ∂(u,v)               ∂(u,v)               ∂(u,v)
                                               (u 0 ,v 0 )          (u 0 ,v 0 )          (u 0 ,v 0 )
                                 If   is given by z = S(x, y), this tangent plane has equation

                                                   ∂S                ∂S
                                               −           (x − x 0 ) −      (y − y 0 ) + z − z 0 = 0.
                                                   ∂x  (x 0 ,y 0 )   ∂y  (x 0 ,y 0 )
                         EXAMPLE 12.18
                                                                                           √
                                 For the elliptical cone of Example 12.16, the tangent plane at ( 3a/4,b/4,1/2) has
                                 equation
                                                 √         √
                                                   3b       3a    a      b    ab     1
                                                −      x −      −    x −    +     z −   = 0.
                                                   4        4     4      4     2     2


                         EXAMPLE 12.19
                                                                                  √
                                 For the hemisphere of Example 12.17, the tangent plane at (1, 2,1) has equation
                                                                √      √
                                                        (x − 1) +  2(y −  2) + (z − 1) = 0,
                                 or
                                                                  √
                                                               x +  2y + z = 4.

                                 12.5.3 Piecewise Smooth Surfaces


                                   A curve is smooth if it has a continuous tangent. A smooth surface is one that has a contin-
                                   uous normal. A piecewise smooth surface is one that consists of a finite number of smooth
                                   surfaces. For example, a sphere is smooth and the surface of a cube is piecewise smooth,
                                   consisting of six smooth faces. The cube does not have a normal vector (or tangent plane)
                                   along an edge.



                                     In calculus it is shown that the area of a smooth surface   given by z = S(x, y) is

                                                                              2       2

                                                                         ∂S      ∂S
                                                     area of   =    1 +       +       dA
                                                                 D       ∂x       ∂y

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                                   October 14, 2010  14:53  THM/NEIL   Page-392        27410_12_ch12_p367-424
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