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Vector Calculus

                        1 −                                              (b) 2i +  4 j +  k
                   (a)                   (b) 0
                       6
                                                                                1    

                       1                                                 (c)            (2i +  4 j −  ) k
                   (c)                   (d) 6                                 21  
                       6
                                                                         (d) none
                                               [IISC 2006]
                                                                  73.    If the temperature at any point in space
                             3                                           is given by T =  xy +   +
                                 R
            70.    Let  : f R →  be defined                                                   yz zx , then the
                     ( , , y z =
                                  2
                                                    4
                                               2
                    f x      ) x +    2xy + 5y −   z −  1.               directional derivative of T in the
                                                                         direction of the vector 3i −  4k  at
                   The unit vector u which gives the
                   maximum value for the directional                     (1, 1, 1) is
                   derivative  D f  at the point (1, 0, 1) is
                                u                                              2                  1
                   _______                                               (a) −                 (b)
                                                                               5                  5
                   (a) u = (1, 0, 0)
                                                                             5                       5
                   (b) u = (0, 0, 1)                                     (c)                   (d) −
                                                                             2                       2
                            − 1
                                      )
                   (c) u =     (1,0,1                                    In what direction from (3, 1, −
                             2                                    74.                                       ) 2 ,
                                                                         the directional derivative of
                            1                                                           2  2 4
                                                                            x
                   (d) u =     (1,1, 2−  )            [IISC 2006]         ( , , y z =
                                                                                   ) x y z  is maximum?
                             6
                                                                         (a)  (
            71.    Find the gradient of the function                        96 i +  3j −  3k )
                    =  x −  2xy +  z at the point
                         2
                                     2
                                                                             96 i −
                   (2, 1,1−  ) .                                         (b)  (     3j +  3k )


                                                                            96 i −
                   (a) 6i −  4 j +  2k    (b) 4i −  6 j +  2k            (c)  (     3j −  3k )


                   (c) 6i +  4 j −  2k    (d) 6i −  4 j −  2k            (d)  (
                                                                             48 i −  3 j +  3k )
            72.    A unit normal to the surface  z =  2xy  at

                   the point (2, 1, 4) is                                         Divergence

                                                                  75.    The divergence of
                                                                                         +
                                                                         x zi −  2  2y z j xy k  at the point
                                                                                              2
                                                                                    3
                                                                                      2

                                                                         (1,1, 1−  )  is
                               −
                   (a) 2i +  4 j k

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