Answers for Tutorial 2

							Answers for Tutorial 2
[M283 - Field theory and PDEs]

Thursday 18th October

  1. The gradients of the following functions:
     (a) T (x, y, z) = x2 + xy + 3z 3
     (b) T (x, y, z) = x2 y 3 + z 4
     (c) T (x, y, z) = ez sin x + ln(y)
     are respectively
     (a) ∇T = (2x + y, x, 9z 2 )
     (b) ∇T = (2xy 3 , 3x2 y 2 , 4z 3 )
     (c) ∇T = (ez cos x, 1/y, ez sin x).

  2. The unit inward pointing normal to the sphere defined by the equation T (x, y, z) =
     x2 + y 2 + z 2 = 1 at a general point (x, y, z) is (note the minus sign)

                             ∇T        (2x, 2y, 2z)          (x, y, z)
                    ˆ
                    n=−           = −√ 2       2 + 4z 2
                                                        = −√ 2          .
                           | ∇T |     4x + 4y               x + y2 + z2

     At point a = (1, 0, 0) it can be recast as:

                                          ˆ
                                          n = −(1, 0, 0) .

     Hence, the cartesian equation for the tangent plane which touches the sphere at that
     point is given by

                                                ˆ
                                      (r − a) · n = 0
                                      (r − (1, 0, 0)) · (1, 0, 0) = 0
                                      (x − 1, y, z) · (1, 0, 0) = 0 .

     Or
                                               x=1.

  3. The unit outward pointing normal to the ellipsoid T (x, y, z) = x2 + 2y 2 + 3z 2 = 6 at a
     general point (x, y, z) is (note that the sign is now positive)

                           ∇T       (2x, 4y, 6z)         (x, 2y, 3z)
                    ˆ
                    n=          =√ 2        2 + 36z 2
                                                      =√ 2              .
                         | ∇T |   4x + 16y              x + 4y 2 + 9z 2

     At point a = (1, 1, 1) it can be recast as:

                                                   (1, 2, 3)
                                           ˆ
                                           n=        √       .
                                                       14

                                               1
   Hence, the cartesian equation for the tangent plane which touches the ellipsoid at that
   point is given by

                                          ˆ
                                (r − a) · n = 0
                                                 (1, 2, 3)
                                (r − (1, 1, 1)) · √        =0
                                                     14
                                (x − 1) + 2(y − 1) + 3(z − 1) = 0

   Or
                                       x + 2y + 3z = 6 .

4. The directional derivative of the function
                                                                1
                                    φ(x, y, z) = x2 + y 2 +
                                                                z
   at a general point (x, y, z) in the direction given by the vector b = (−1, −1, 1) writes
   as
                                      b                         (−1, −1, 1)
                     Db Φ = ∇Φ ·          = (2x, 2y, −1/z 2 ) ·    √        .
                                    |b|                               3
   At the point a = (1, 0, 1), it reduces to

                                 b                 (−1, −1, 1)  −2 − 1   √
            Db Φ(a) = ∇Φ(a) ·       = (2, 0, −1) ·    √        = √     =− 3.
                                |b|                      3         3

5. Given that r = (x, y, z), r = |r|, and that r = r/|r|, we see that
   (a) ∇(r 2 ) = (∂/∂x, ∂/∂y, ∂/∂z)(x2 + y 2 + z 2 ) = (2x, 2y, 2z).
   This can be further simplified as

                                                    (x, y, z)               r
                  ∇(r 2 ) = 2 x2 + y 2 + z 2 √                    = 2| r |     = 2r r
                                                   x2 + y 2 + z 2          |r|

   (b) ∇ r12 = (∂/∂x, ∂/∂y, ∂/∂z)(x2 + y 2 + z 2 )−1 = − (x(2x,2y,2z))2 .
                                                            2 +y 2 +z 2

   This can be further simplified as
                          √ 2
                 1          x + y2 + z2     (x, y, z)                     r       2
            ∇ 2 = −2 √ 2          2 + z2 √ 2
                                                          = −2| r |         5 = − 3r
                r           x +y          x +y    2 + z24               |r|      r

6. The divergence and curl of the function

                                    v = (2xy, −yz, 3zx) .

   are respectively

                ∇ · v = (∂/∂x, ∂/∂y, ∂/∂z) · (2xy, −yz, 3zx) = 2y − z − 3x ,

                                               2
  (which is a scalar field) and

             ∇ × v = (∂/∂x, ∂/∂y, ∂/∂z) × (2xy, −yz, 3zx)
                         ∂           ∂            ∂         ∂
                   = ˆ
                     i      (3zx) − (−yz) − ˆ   j    (3zx) − (2xy)
                         ∂y         ∂z            ∂x        ∂z
                     ˆ ∂ (−yz) − ∂ (2xy)
                   + k
                         ∂x           ∂y
                   = yˆ − 3zˆ − 2xk
                       i      j    ˆ                                                (1)

  which is a vector field.

7. The vector field
                                                 j ˆ
                            F1 = (14x + 3y)ˆ + 3xˆ + k ,
                                           i
  has a curl given by

             ∇ × F1 = (∂/∂x, ∂/∂y, ∂/∂z) × (14x + 3y, 3x, 1)
                          ∂        ∂            ∂        ∂
                    = ˆ
                      i      (1) − (3x) − ˆ  j    (1) − (14x + 3y)
                          ∂y      ∂z           ∂x       ∂z
                      ˆ ∂ (3x) − ∂ (14x + 3y)
                    + k
                          ∂x         ∂y
                                        ˆ
                    = 0ˆ − 0ˆ − (3 − 3)k = 0 .
                        i    j                                                      (2)

  It can thus be expressed as the gradient of a scalar field since for any scalar field f
  ∇ × ∇f = 0 This scalar field is f (x, y, z) = 7x2 + 3yx + z. Indeed, we check that

           ∇f = (∂/∂x, ∂/∂y, ∂/∂z)(7x2 + 3xy + z) = (14x + 3y, 3x, 1) = F1 .




                                         3