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Vector Function

  1. Nov 21, 2009 #1
    Okay. My reason for posting this is that I need help actually formulating the 'math part' of it. I can get the right answer by 'inspection.' And from the way the book is written, I believe that is how the authors expect you to find it. But for self gratifying reasons, I wish to generalize the answer a little bit:

    1. The problem statement, all variables and given/known data

    Write the formula for a vector function in 2 dimensions whose direction makes an
    angle of 45 degrees with the x-axis and whose magnitude at any point (x,y) is (x+y)2.


    3. The attempt at a solution

    Now just looking at it, we can say that to insure that the vector at any point (x,y) is at 45 degrees to the x-axis, we can let the funciton equal something to the effect of:

    [tex]\mathbf{F}(x,y) = k(\bold{i} + \bold{j})[/tex]

    Now k(x,y) is the 'scalar portion' of F and must have the effect that its product with the magnitude of the
    vector portion of F, namely v=(i + j), must equal (x+y)2.

    Now since the magnitude of v=(i + j) is [itex]\sqrt{2}[/itex], k(x,y) must have [itex]\sqrt{2}[/itex] in its denominator.

    This will make 'unit vector' in the v direction or

    [tex]\bold{u}_v = \frac{\bold{v}}{\sqrt{2}}=\frac{\bold{i} + \bold{j}}{\sqrt{2}}[/tex]

    Therefore, multiplying v by k(x,y) = (x+y)2/[itex]\sqrt{2}[/itex] gives the desired result.

    [tex]\Rightarrow \bold{F}(x,y) = \frac{(x+y)^2}{\sqrt{2}}(\bold{i} + \bold{j})[/tex]


    I am just curious how other would approach this problem, or if this is the most efficient method from a mathematical standpoint.

    ~Casey
     
    Last edited: Nov 21, 2009
  2. jcsd
  3. Nov 21, 2009 #2
    Dunno if it helps but....

    I assumed a vector initially completely in the i direction: [tex]\mathbf{F} = (x+y)^2 \mathbf{i}[/tex].

    Then just did a coordinate rotation:

    [tex]\left(\begin{array}{cc}cos(45) & -sin(45)\\sin(45) & cos(45)\end{array}\right) \left(\begin{array}{cc}(x+y)^2\\0\end{array}\right) = \frac{(x+y)^2}{\sqrt{2}}\left(\begin{array}{cc}\mathbf{i}\\\mathbf{j}\end{array}\right)[/tex]
     
  4. Nov 21, 2009 #3
    Thanks Feldoh! That's a very interesting approach. I like it!

    :smile:
     
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