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Tangent plane of a parametric function

  1. Apr 21, 2014 #1

    BiGyElLoWhAt

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    Gold Member

    Ok, so I'm really hoping someone can help me logic my way through this.

    I have a function to the effect of: ##r(u,v)=f(u,v)\hat{i} + g(u,v)\hat{j} +h(u,v)\hat{k}##

    I need to find an equation of a tangent plane at a point ##(u_{0},v_{0})##
    and quite frankly I'm at a loss on how to do this.

    So what I ultimately want is 2 vectors tangent to the surface at that point, I'm not sure that with what I have I can simply chain some derivatives together and cross them...

    If I had z(x,y) I would simply take ##\frac{\partial z}{\partial x} (cross) \frac{\partial z}{\partial y}## as that would give me 2 (presumably different) vectors and crossing them will give me my normal to use in my tangent function...

    Now how to apply this to a parametric function in 3-space...

    Ok... I think I just answered my question as I was typing this... (It happens that way quite a bit)
    ##\frac{\partial r}{\partial u} (cross) \frac{\partial r}{\partial v}##

    But what if I had ##r(u,v,w)## ?
    In my head , whether I chose to cross
    ##\frac{\partial r}{\partial u} (cross) \frac{\partial r}{\partial v}##
    or
    ##\frac{\partial r}{\partial u} (cross) \frac{\partial r}{\partial w}##
    or
    ##\frac{\partial r}{\partial w} (cross) \frac{\partial r}{\partial v}##
    It really shouldn't matter, I guess at this point I just want to hear somebody's input other than my own. All of these will give me a normal, no?

    (once I plug in my point that is...)
     
    Last edited: Apr 21, 2014
  2. jcsd
  3. Apr 21, 2014 #2

    LCKurtz

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    A surface only has two parameters. So you will only run into the ##r_u\times r_v## case for a surface. And you are correct, that is how you find a normal.
     
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