Page 138 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 138

CHAP. 6]                       PARTIAL DERIVATIVES                              129

                                  @    @  3   xy 2  2   xy 2  2  2   2 xy 2
                                        ðx y þ e  Þ¼ 3x y þ e    y ¼ 3x y þ y e
                                  @x  @x
                           ðaÞ    x ¼  ¼
                                  @   @  3    xy 2  3  xy 2    3      xy 2
                                        ðx y þ e  Þ¼ x þ e    2xy ¼ x þ 2xy e
                                  @y  @y
                           ðbÞ    y ¼  ¼
                                   2
                                   @    @    @     @  2  2 xy 2     2  xy 2  2     4 xy 2
                           ðcÞ    xx ¼  2  ¼  ¼  ð3x y þ y e  Þ¼ 6xy þ y ðe    y Þ¼ 6xy þ y e
                                   @x  @x @x   @x
                                    2
                                   @    @  3     xy 2        @  xy 2  xy @
                                                                      2
                                         ðx þ 2xy e           ðe  Þþ e
                           ðdÞ   yy ¼  2  ¼        Þ¼ 0 þ 2xy            ð2xyÞ
                                   @y  @y                   @y         @y
                                                             2 2 xy
                                      ¼ 2xy   e xy 2    2xy þ e xy 2    2x ¼ 4x y e  2  þ 2xe xy 2
                                    2
                                   @     @    @     @  2  2 xy 2  2  2  xy 2    xy 2
                                                  ð3x y þ y e  Þ¼ 3x þ y   e    2xy þ e    2y
                           ðeÞ    xy ¼  ¼      ¼
                                   @y @x  @y @x  @y
                                                  2     xy 2
                                               3 xy
                                          2
                                       ¼ 3x þ 2xy e  þ 2ye
                                    2
                                    @    @    @     @  3   xy 2  2      xy 2  2  xy 2
                                                   ðx þ 2xy e  Þ¼ 3x þ 2xy   e    y þ e    2y
                                   @x @y  @x @y  @x
                           ð f Þ    yx ¼  ¼    ¼
                                                   2    xy 2
                                                3 xy
                                           2
                                       ¼ 3x þ 2xy e  þ 2ye
                                  Note that   xy ¼   yx in this case.  This is because the second partial derivatives exist and are
                              continuous for all ðx; yÞ in a region r. When this is not true we may have   xy 6¼   yx (see Problem 6.41,
                              for example).
                                                 2   2   2  1=2
                     6.11. Show that Uðx; y; zÞ¼ ðx þ y þ z Þ  satisfies Laplace’s partial differential equation
                           2     2    2
                           @ U  @ U   @ U
                                         ¼ 0.
                                   þ
                              þ
                           @x 2  @y 2  @z 2
                              We assume here that ðx; y; zÞ 6¼ð0; 0; 0Þ.  Then
                                 @U
                                       1  2  2  2  3=2       2   2  2  3=2
                                       2
                                 @x  ¼  ðx þ y þ z Þ    2x ¼ xðx þ y þ z Þ
                                 2
                                @ U   @    2   2  2  3=2      3  2  2   2  5=2    2   2  2  3=2
                                                              2
                                 @x 2  ¼  @x  ½ xðx þ y þ z Þ  мð xÞ½  ðx þ y þ z Þ    2xŠþðx þ y þ z Þ   ð 1Þ
                                                     2
                                                                   2
                                                                       2
                                                         2
                                                            2
                                          3x 2      ðx þ y þ z Þ  2x   y   z 2
                                       2  2   2 5=2  2  2  2 5=2  2   2  2 5=2
                                    ¼                          ¼
                                     ðx þ y þ z Þ  ðx þ y þ z Þ  ðx þ y þ z Þ
                                                         2
                                                      2
                                                                    2
                                                                           2
                                                                              2
                                               2
                                              @ U   2y   x   z 2   @ U   2z   x   y 2
                              Similarly          ¼             ;      ¼             :
                                              @y 2  2   2  2 5=2   @x 2  2   2  2 5=2
                                                   ðx þ y þ z Þ         ðx þ y þ z Þ
                                                                   2
                                                               2
                                                          2
                                                         @ U  @ U  @ U
                              Adding,                       þ    þ    ¼ 0:
                                                         @x 2  @y 2  @z 2
                                              2
                                       y      @ z
                                 2    1
                     6.12. If z ¼ x tan  , find    at ð1; 1Þ.
                                       x     @x @y
                                              @z  2    1     @      2   x 2  1   x 3
                                                               y
                                                                       2
                                                           2
                                                                                2
                                              @y  ¼ x   1 þð y=xÞ @y x  ¼ x    x þ y 2     x  ¼  x þ y 2
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