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Normal to parameterised surface

  1. Jun 24, 2009 #1
    1. The problem statement, all variables and given/known data
    surface : $(u,v) is (e^u,(v^2)*(e^(2u)),2e^(-u) +v)
    0<=u<=3 -4<=v<=4
    find a normal in terms of u and v
    write down a double integral for the surface(dont solve)
    find equation of tangent plane to surface at (1,4,0)

    2. Relevant equations



    3. The attempt at a solution
    I have no idea where to start

    any help please?
     
  2. jcsd
  3. Jun 24, 2009 #2
    Given two vectors, what is one easy way to get a vector normal to both?
     
  4. Jun 24, 2009 #3
    cross product
     
  5. Jun 24, 2009 #4
    Can you find two non-collinear vectors tangent to the surface at a single point? Since both vectors are tangent, their cross-product should point normal to the surface.
     
  6. Jun 24, 2009 #5
    ok got the first part thanks, now for the area, i know in the double integral has |J| in it somehow, how wuld i find it. I know from a mate u can do |Tu X Tv|, but i dont think we learnt this, and i was wondering how wuld u do this by using jacobians, if possible.

    im confused coz we abs(del(x,y,z)/del(u,v)| and i dunno how to solve this
     
  7. Jun 24, 2009 #6

    HallsofIvy

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    One very good way to do all three of those questions is to find the "fundamental vector product". To do that, you find the vector derivatives of "$(u,v)" with respect to u and v and take the cross product of those two vectors.

    The derivative vectors are, of course, tangent vectors to the "coordinate curves" v= constant and v= constant at each point and so lie in the tangent plane at each point. Therefore, their cross product is normal to the tangent plane and so to the surface at that point. Further, the length of that vector is the Jacobian (it is the |Tu x Tv| you mention). And, of course, if you know a normal at a point, it is easy to write down the equation of the tangent plane at that point.

    By the way, have you determined u and v for the point (1, 4, 0)?
     
  8. Jun 24, 2009 #7
    the points are u=0 and v=-2, so is the tangent plane is Tu(0,2)*(u-0)+Tv(0,2)*(v-2)=0
     
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