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Normal vector to a curve

  1. Feb 12, 2006 #1
    Hi I'm having trouble finishing this problem and I'd appreciate it if someone could help me out.

    I need to find the unit vector normal to x=y^2=z^3 at the point (1,1,1) and it can't have components along the line x=y=z

    Heres what I did so far:
    1. dx=2ydy=3z^2dz
    2. At (1,1,1) dx=2dy=3dz
    3. differential length vector dl=dxax+(1/2)dxay+(1/3)dxaz
    (ax is the unit vector in the x-direction etc)
    4. Rewrote dl as dl=[ax+(1/2)ay+(1/3)az)]dx

    If I use the dot product and I let an=normal vector then the dot product of dl and an should equal 0. OK that's as far as I can get :(
     
  2. jcsd
  3. Feb 12, 2006 #2
    Let's see. If you if parametrize given curve you would get [tex]\gamma(t)=(t, t^\frac{1}{2}, t^\frac{1}{3})[/tex]. Tangent vector to a curve is given by [tex]\vec{\tau}=\frac{d\gamma(t)}{dt}=[1, \frac{1}{2}t^\frac{1}{2},\frac{1}{3}t^{-\frac{2}{3}}][/tex]. At point M=(1,1,1) it becomes [tex]\vec{\tau_M}=[1,\frac{1}{2},\frac{1}{3}][/tex]. Vector that is along x=y=z direction is given by [tex]\vec{a}=[1,1,1][/tex]. So, your vector should be normal to the curve at point M (which means it should be normal to [tex]\vec{\tau_M}[/tex] and should not be in direction of [tex]\vec{a}[/tex]. To be sure it is so, set your vector to be [tex]\vec{b}=\vec{\tau_m}\times \vec{a}[/tex]. Then just divide resulting vector by its magnitude and this is it.
     
    Last edited: Feb 12, 2006
  4. Feb 12, 2006 #3
    Oh yes, excuse me. My fingers were faster then brain. Ill fix it immidiatelly
     
    Last edited: Feb 12, 2006
  5. Feb 12, 2006 #4

    arildno

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    To remain annoying, please add a minus sign to the differentiated y-component's exponent..:smile:
     
  6. Feb 13, 2006 #5
    Hmm, I don't see edit button under my post.
     
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