Page 199 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 199

190                   APPLICATIONS OF PARTIAL DERIVATIVES                  [CHAP. 8



                      8.2. In what point does the normal line of Problem 8.1(b) meet the plane x þ 3y   2z ¼ 10?
                              Substituting the parametric equations of Problem 8.1(b), we have
                                             1   6t þ 3ð2 þ 11tÞ  2ð14t   1Þ¼ 10  or  t ¼ 1
                          Then x ¼ 1   6t ¼ 7; y ¼ 2 þ 11t ¼ 9; z ¼ 14t   1 ¼ 15 and the required point is ð7;  9;  15Þ.

                                             2
                                                      3
                      8.3. Show that the surface x   2yz þ y ¼ 4is perpendicular to any member of the family of surfaces
                           2
                                         2
                                              2
                          x þ 1 ¼ð2   4aÞy þ az at the point of intersection ð1;  1; 2Þ:
                              Let the equations of the two surfaces be written in the form
                                           2
                                                                                       2
                                                   3
                                                                       2
                                                                                   2
                                       F ¼ x   2yz þ y   4 ¼ 0  and  G ¼ x þ 1  ð2   4aÞy   az ¼ 0
                          Then
                                                    2
                                         rF ¼ 2xi þð3y   2zÞj   2yk;  rG ¼ 2xi   2ð2   4aÞyj   2azk
                          Thus, the normals to the two surfaces at ð1;  1; 2Þ are given by
                                               N 1 ¼ 2i   j þ 2k;  N 2 ¼ 2i þ 2ð2   4aÞj   4ak
                              Since N 1   N 2 ¼ð2Þð2Þ  2ð2   4aÞ ð2Þð4aÞ  0, it follows that N 1 and N 2 are perpendicular for all a,
                          and so the required result follows.
                      8.4. The equation of a surface is given in spherical coordinates by Fðr; ; Þ¼ 0, where we suppose
                          that F is continuously differentiable. (a) Find an equation for the tangent plane to the surface at
                          the point ðr 0 ;  0 ;  0 Þ.(b) Find an equation for the tangent plane to the surface r ¼ 4 cos   at the
                                 p ffiffiffi
                          point ð2 2; =4; 3 =4Þ.(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
                                                         1 @     1 @     1 @
                                                                             e 3
                                                   r  ¼      e 1 þ   e 2 þ
                                                        h 1 @u 1  h 2 @u 2  h 3 @u 3
                                                    1 @r         1 @r        1 @r
                              where              e 1 ¼   ;   e 2 ¼   ;   e 3 ¼
                                                    h 1 @u 1     h 2 @u 2    h 3 @u 3
                              (see Pages 161, 175).

                                 In spherical coordinates u 1 ¼ r; u 2 ¼  ; u 3 ¼  ; h 1 ¼ 1; h 2 ¼ r; h 3 ¼ r sin   and r ¼ xi þ yjþ
                              zk ¼ r sin   cos  i þ r sin   sin  j þ r cos  k.
                              Then
                                                   8
                                                    e 1 ¼ sin   cos  i þ sin   sin  j þ cos  k
                                                   <
                                                    e 2 ¼ cos   cos  i þ cos   sin  j   sin  k       ð1Þ
                                                    e 3 ¼  sin  i þ cos  j
                                                   :
                              and
                                                         @F   1 @F     1  @F
                                                                            e 3
                                                         @r   r @    r sin   @
                                                    rF ¼   e 1 þ  e 2 þ                              ð2Þ
                                 As on Page 183 the required equation is ðr   r 0 Þ rFj P ¼ 0.
                                 Now substituting (1) and (2), we have

                                               @F            1 @F             sin   0 @F
                                                                                        i

                                               @r    P       r 0 @     P     r 0 sin   0 @     P
                                        rFj P ¼    sin   0 cos   0 þ  cos   0 cos   0

                                                 @F            1 @F            cos   0 @F
                                                                                         j
                                                 @r    P      r 0 @     P     r 0 sin   0 @     P
                                              þ      sin   0 sin   0 þ  cos   0 sin   0 þ

                                                 @F        1 @F
                                                                  sin   0 k
                                              þ      cos   0
                                                 @r    P   r 0 @     P
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