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Find Differential Length

  1. Jan 31, 2010 #1
    I got an answer for this, but it just doesn't seem correct. Maybe someone could tell me what I am doing wrong, or that I am just over thinking the problem and I am right.

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

    Write an expression for the differential length vector dl at the point (1,2,8) on the straight line y=2x, z=4y, and having the projection dx on the x-axis


    2. Relevant equations

    dl=dxax+dyay+dzaz


    3. The attempt at a solution

    y=2x, z=4y, therefore z=8x
    therefore

    dxax+2dxay+8dxaz.

    since there are no variables remaining after you take the derivative, dl is then
    1ax+2ay+8az

    Any help would be amazing, and please excuse me if I am just being a complete idiot.

    Thanks
     
  2. jcsd
  3. Jan 31, 2010 #2

    Dick

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    Science Advisor
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    That seems ok to me. So dl=dx(ax+2*ay+8*az). ax, ay and az are the unit vectors in each direction, right?
     
  4. Jan 31, 2010 #3
    Yes, ax, ay, az are the direction unit vectors. Thanks
     
  5. Dec 1, 2010 #4
    ..... I haven't really understood the problem, but i do understand the solution.. can anyone explain more about this problem especially the "projection dx on the x-axis" part..

    please help me.... i badly needed the explanation..

    thanks in advance....
     
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