Vector Calculus Solution by DynamiteKegs

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									                         Math 223 VECTOR CALCULUS

               SOLUTION FOR MIDTERM EXAM – I (09/29)

                                                                    September 29 (Mon), 2008

Instructor:              Yasuyuki Kachi

  Line #: 32702.

[I] (20pts)    In R3 , with the coordinate system             x, y, z ,

(1)   the defining equation of the plane Π passing through P =                  1, 0, −1 , and

  having the normal vector n =           3, 6, 10 , is

                                  3 x + 6 y + 10 z = −7.


(2)   Let Q =          0, −1, 0 . Then

                      −
                     −→
                 b = PQ =          0 − 1,      −1           − 0,   0−     −1


                            =      − 1, −1, 1 .

(3)   The distance between the plane Π and the point Q is

                           n ·b                    3, 6, 10 · − 1, −1, 1
       projn b    =                  =               √
                            n                          32 + 62 + 102

                                              1
                                     =   √              .
                                             145

[II] (20pts)     (1)     We may rewrite the equation which is originally written in

cylindrical coordinate system        r, θ, z       as

                                                  1
(∗)                                  z =                       ,
                                               sin 2θ

                                               1
using Cartesian coordinate system         x, y, z , as follows. First, recall

                           sin 2θ       = 2       cos θ          sin θ


  double angle formula    . Accordingly, we may “artificially” rewrite the right-hand

side of the given equation (∗) as

                        1                                  1
                                    =
                     sin 2θ                2 ·      cos θ            sin θ

                                                               r2
                                    =                                              .
                                           2 ·     r cos θ               r sin θ


Recall that the conversion formulas between the two coordinate systems are

                                        x = r cos θ,

                                        y = r sin θ,

                                        z =        z.

Also recall that a part of the reverse conversion formulas is
                                    x2 + y 2 =            r2 .

Hence we may rewrite the right-hand side of (∗), using x, y, z, as

                                         x2 + y 2
                                                           .
                                           2xy

In sum, the equation (∗) becomes
                                              x2 + y 2
                                z =                              ,
                                                2xy

or the same
                                2xyz = x2 + y 2                      .


                                              2
(2)     We may prove that the surface in R3 as in (1) is ruled , as follows:

   Let z0 be an arbitrary real number such that z0 > 1.                        Substitute z = z0
                                1
into the equation   z =               ,    thus obtain
                             sin 2θ

                                                    1
                                    z0 =                         ,
                                              sin 2θ
or, the same to say,
                                                        1
                                    sin 2θ     =                     .
                                                        z0

             1
Since              < 1,        there are four θs in          0 ≤ θ < 2 π.         We may confirm
             z0
this by drawing the graph of u = sin 2θ in the                   θ, u    coordinate plane, and use

the horizontal line test.        For an arbitrary choice of θ = θ1 satisfying the

highlighted equation, there is another θ = θ2 satisfying the highlighted equation,

such that

                                   θ1 − θ2         =         π       .

In other words, we may list the complete set of solutions to the highlighted equation as

  θ1 , θ2 , θ3 , θ4 ,    in such a way

                        θ1 − θ2 = π,          and            θ3 − θ4 = π

hold. Accordingly, the half-lines θ = θ1 and θ = θ2                         make a pair, and the

half-lines   θ = θ3      and     θ = θ4      make a pair, and each of the two pairs of

half-lines makes a flat angle, thus makes a whole line.

  To conclude, the intersection of the surface with the plane z = z0 is the union

of two lines. This proves that the surface is ruled.

                                               3
[III] (20pts)    We make the triplet of vectors                   a, b, c        in R3   below an

orthonormal frame by filling in appropriate real numbers in the blanks:

                                  25              144
                 a =                      , −         ,                     ,
                                 169              169
                                                   25
                 b =                      ,           ,                    ,
                                                  169
                               √
                             60 2                                 119
                 c =     −        ,                         , −            .
                              169                                 169

We rely on the classification of orthonormal frames in R3 :

         a =     a2 + b2 − c2 − d2 ,          2 bc + ad ,               2 bd − ac    ,

         b =     2 bc − ad ,           a2 − b2 + c2 − d2 ,              2 cd + ab    ,

         c =     2 bd + ac ,           2 cd − ab ,            a2 − b2 − c2 + d2 ,


where a, b, c, d ∈ R such that a2 + b2 + c2 + d2 = 1. First, observe

       the first entry of a + the second entry of b + the third entry of c

   =    3 a2 − b2 − c2 − d2

   =    4 a2 −     a2 + b2 + c2 + d2

   =    4 a2 − 1                    since         a2 + b2 + c2 + d2 = 1 .

This equals
                        25          25                119                 69
                             +                −               =      −       .
                       169         169                169                169
In short,
                                                             69
                                 4 a2 − 1         =     −        .
                                                             169

                                                  4
From this it follows
                                                 1                69
                                  a2 =                      1−
                                                 4                169

                                                 25
                                         =           .
                                                 169
Next, observe

       2 b2 − 2 c2     =      the first entry of a                 −     the second entry of b
                              25                      25
                       =                     −
                              169                    169
                       =     0.

Hence b2 = c2 . Accordingly,
                                                                                  25
                  a 2 − d2        =      the first entry of a                  =       .
                                                                                  169
                         25
Since we had a2 =            , it follows d = 0. Now we substitute d = 0,
                        169
and also b2 = c2 , into the above expression of the triplet a , b , c :

                      a =         a2 ,           2 bc,       − 2 ac      ,

                      b =     2 bc,                  a2 ,        2 ab     ,

                      c =     2 ac,          − 2 ab,        a2 − 2 b2 .

In particular,
                                                                                        √
                                                                                      60 2
           the third entry of a          =       − the first entry of c =                   ,
                                                                                       169

and also
                                                                                          144
            the first entry of b          =           the second entry of a =         −        .
                                                                                          169

                                                        5
The latter also reads
                                                         144
                                          2 bc = −                     < 0.
                                                         169

This and b2 = c2 prove c = −b.

Hence
                                                                                                √
                                                                                              60 2
            the third entry of b              =       the third entry of a =                       ,
                                                                                               169

and
                                                                                                √
                                                                                              60 2
        the second entry of c                 =   − the third entry of b =                  −      .
                                                                                               169

To conclude,
                                                                                       √
                              25                         144                         60 2
          a =                                      −                                                  ,
                              169             ,          169             ,            169

                                                                                       √
                              144                        25                          60 2
          b =             −                                                                       ,
                              169             ,         169              ,            169

                                √                         √
                              60 2                      60 2                          119
          c =             −                       −                              −                .
                               169            ,          169             ,            169



 Note       :     More generally, if

                                  2                            2
      a =                 cos α       ,           − sin α          ,                              ,

                                                               2
      b =                                 ,            cos α       ,                              ,

                      1
      c =       − √           sin 2α ,                                       ,       cos 2α
                      2
                                                        6
form an orthonormal frame, then they are

                               2                                     2               1
    a =                cos α       ,                   − sin α           ,          √          sin 2α   ,
                                                                                      2
                               2                                     2                  1
    b =        −       sin α       ,                         cos α       ,          √          sin 2α   ,
                                                                                        2

                   1                                     1
    c =     − √              sin 2α ,              − √            sin 2α ,                  cos 2α          .
                   2                                     2


[IV] (20pts)   (1)       For three vectors

                                          a =            a1 , a2 , a3 ,

                                              b =        b1 , b2 , b3 ,

                                              c =        c1 , c2 , c3


in R3 , their scalar triple product                       a × b              · c    coincides with the following
quantity:

                        a1     a2        a3
                        b1     b2        b3        =     a1 b2 c3 + a2 b3 c1 + a3 b1 c2
                        c1     c2        c3
                                                         − a3 b2 c1 − a1 b3 c2 − a2 b1 c3 .

This last quantity is called the determinant .                                 The determinant possesses the
following property:
                                    a1        a2   a3             c1         c2    c3
                                    b1        b2   b3        =    a1         a2    a3 ,
                                    c1        c2   c3             b1         b2    b3

which can be rewritten in terms of the scalar triple product of vectors in R3 , as

                                       a ×b         ·c =             c ×a          · b.


                                                              7
(2)   For three vectors a , b and c in R3 , their vector triple product                                               a ×b       ×c
coincides with
                                     a   ·           c        b −                 b     ·    c           a.

(3)   We may prove the identity
                                                                                      a ·c       b ·c
                                 a ×b            ·       c ×d             =
                                                                                      a ·d       b ·d

for vectors a , b , c and d in R3 , as follows:

      a ×b       ·       c ×d


      =          c ×d            ×a          ·b                                         by the formula in (1)


      =              c       ·       a       d −                  d       ·       a      c       ·b

                                                                                        by the formula in (2)


      =          c       ·       a               d        ·       b           −          d       ·       a        c     ·    b


      =          a       ·       c               b       ·        d           −          a       ·       d        b     ·    c


             a ·c            b ·c
      =                           .
             a ·d            b ·d


 Note 1      :           We may use (3) to prove

                                             2                    2           2                      2
                                 a ×b                =        a       b           −     a ·b                  ,


for vectors a and b in R3 , as follows:                                   In the formula obtained in (3), substitute
c with a , and substitute d with b , and obtain

                                                                      8
                                                                 a ·a             b ·a
                        a ×b     ·    a ×b           =                                 .
                                                                 a ·b             b ·b

This is the same as
                                                             2
                                       2                 a          a ·b
                              a ×b           =                                     ,
                                                                             2
                                                     a ·b                b

or
                                 2               2           2                           2
                         a ×b        =       a           b        −          a ·b            .


 Note 2       :       An instant application of the identity in “Note 1” above is the

following Cauchy–Schwarz inequality for vectors in R3 :

                                 a ·b ≤              a        b               .



[V] (20pts)       Consider the equation

                                           2/5                           2/5           −5/2
(#)                   r =        cos θ           +           sin θ                                   ,


written in the polar coordinate system .

(1a)   Substitute (#) into       x = r cos θ,                and         y = r sin θ                 each:


                                                     cos θ
                     x =                                                                5/2
                                                                                                 ,
                                             2/5                             2/5
                                     cos θ           +           sin θ


                                                     sin θ
                     y =                                                                5/2
                                                                                                 .
                                             2/5                             2/5
                                     cos θ           +           sin θ

                                                 9
(1b)     x2/5 + y 2/5

                                              2/5                                                                        2/5
                                cos θ                                                                     sin θ
        =                      2/5                      2/5
                                                                      +                                   2/5                            2/5
                      cos θ           +     sin θ                                        cos θ                     +    sin θ

                                      2/5                                   2/5
                         cos θ                +          sin θ
        =                             2/5                         2/5
                                                                                                 =         1.
                              cos θ         +       sin θ

(2)     Know that the equation (#) in the previous page is modified as

                 1 −6/5                   −3/5                     −3/5
(##)               r             cos θ                  sin θ                        1 − r2
                 5
                                            −1/5                   −1/5                                                 1/5                              1/5
             =        r −2/5     cos θ                  sin θ                    −           r 2/5             cos θ                 sin θ                     .

Substitute             x = r cos θ,               and      y = r sin θ                       in (##):

       1                                                                                 5                     5             1       5           1       5
                                 1 − x2 − y 2
                                                                                −1                   −1
         x−3/5 y −3/5                                             =         x                    y                     − x                   y                 .
       5

(3)     Take the 5-th power of the both sides of the identity in (2):

            1                                                 5
                  x−3 y −3        1 − x2 − y 2
            55
                                                                            −3       5           −3       5              3       5           3       5
                  =        x−1 y −1 − x y                − 5            x                    y                     − x                   y
                                                                            −1       5           −1        5             1       5           1       5
                                                         + 10           x                    y                     − x                   y

                  =       x−1 y −1 − x y

                                  −       5        x−2/5 y −2/5                 x−1/5 y −1/5 − x1/5 y 1/5

                                                                                                                                     2
                                  +       5        x1/5 y 1/5               x−1/5 y −1/5 − x1/5 y 1/5                                    .

                                                                  10
(4)   Substitute the result of (2) into the identity you obtained in (3):
      1                                            5
         x−3 y −3         1 − x2 − y 2
      55
           =      x−1 y −1 − x y

                                                       1
                      − 5 x−2/5 y −2/5                   x−3/5 y −3/5        1 − x2 − y 2
                                                       5
                                                                                                          2
                                  1/5       1/5        1                            2         2
                      + 5 x             y                x−3/5 y −3/5       1 − x       − y                   .
                                                       5
Clean up the right-hand side:
      1                                            5
         x−3 y −3         1 − x2 − y 2
      55

           =      x−1 y −1 − x y

                                                                  1                                               2
               − x−1 y −1        1 − x2 − y 2               +       x−1 y −1     1 − x2 − y 2                         .
                                                                  5
Multiply x3 y 3 to the both sides of the identity:
      1                            5
               1 − x2 − y 2
      55
           =      x2 y 2 − x4 y 4
                                                                1                                 2
               − x2 y 2     1 − x2 − y 2               +          x2 y 2    1 − x2 − y 2              .
                                                                5
This last identity can be rewritten as
                            5
       1 − x2 − y 2             − 3125            x2 y 2 − x 4 y 4
                                                                                                  2
       + 3125 x2 y 2            1 − x2 − y 2             − 625 x2 y 2       1 − x2 − y 2                  = 0.

The left-hand side is indeed a polynomial in x, and y.

(5)    The graph of the curve prescribed by the equation in (4) is the asteroid.

It has cusps at each of           1, 0 ,           0, 1 ,        − 1, 0 ,      0, −1 .

                                                        11
[VI] (Extra 20pts)           (1)   The Jacobi Identity for vectors in R3 is as follows:


             a       ×       b     ×        c


                             +         b    ×       c         ×       a


                                                +         c       ×       a        ×   b        =   0.


(2)   We may calculate the Lie bracket                            A, B        = AB − BA for matrices
                                                                                                 
                0                −a3        a2                           0                −b3    b2
          A =  a3                0        −a1          and       B =  b3                0    −b1  ,
               −a2                a1        0                           −b2               b1     0

as follows. First,

                                                                                   
                  0                −a3           a2      0            −b3          b2
           AB =  a3                0           −a1   b3             0          −b1 
                 −a2               a1            0      −b2            b1          0

                                                                                                  
                     − a2 b2 − a3 b3                         a2 b1                      a3 b1
                 =        a1 b2                        − a1 b1 − a3 b3                 a3 b2      .
                           a1 b3                             a2 b3                 − a1 b1 − a2 b2

Second,

                                                                                   
                  0                −b3           b2      0            −a3          a2
           BA =  b3                0           −b1   a3             0          −a1 
                 −b2               b1            0      −a2            a1          0

                                                                                                  
                     − a2 b2 − a3 b3                         a1 b2                      a1 b3
                 =        a2 b1                        − a1 b1 − a3 b3                 a2 b3      .
                           a3 b1                             a3 b2                 − a1 b1 − a2 b2

                                                          12
Accordingly,


        A, B     = AB − BA
                                                                                                                 
                                            0                       a2 b1 − a1 b2                a3 b1 − a1 b3
                                                                                                                 
                                                                                                                 
                 =             a1 b2 − a2 b1                               0                    a3 b2 − a2 b3    .
                                                                                                                 
                                                                                                                 
                                                                                                                 
                                a1 b3 − a3 b1                       a2 b3 − a3 b2                        0

(3)     We may paraphrase the result of (2) as
                                                                                   
                                                         0              −c3     c2
                                    A, B             =  c3              0      −c1  ,
                                                        −c2             c1       0

where
               a2          a3                              a3           a1                         a1        a2
      c1 =                          ,           c2 =                                ,   c3 =                      .
               b2          b3                              b3           b1                         b1        b2

Agree
                     c1 , c2 , c3               =        a1 , a2 , a3           ×       b1 , b2 , b3 .

(4)     By the result of (3), the cross product for vectors in R3 and the Lie bracket for
skew-symmetric 3 × 3 matrices are parallel in structure. We may use this fact to
prove the Jacobi Identity as in (1), as follows:

  Clearly it suffices to prove


                 A     ,        B       ,        C

                           +                    B    ,     C        ,       A

                                                     +              C   ,       A       ,    B           =   O.

                                                               13
The left-hand side equals


       A     B    −    B     A     C    −     C      A     B    −      B   A

+      B     C    −    C     B     A    −     A      B     C    −      C   B

+      C     A    −    A     C     B    −    B       C     A    −      A   C   .

This is simplified as


                        ABC − BAC − CAB + CBA

                       + BCA − CBA − ABC + ACB

                       + CAB − ACB − BCA + BAC


                   =        O,


where we have used the associativity law for matrix multiplications:

                                 AB C = A BC .




                                       14

								
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