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Plane Tangent

  1. Feb 21, 2007 #1
    1. The problem statement, all variables and given/known data
    Find an eqn for the plane tangent to the given surface at the specified point.

    x = u^2 - v^2
    y= u +v
    z = u^2 + 4v

    At (-1/4, 1/2, 2)

    2. Relevant equations
    A(x-x_0) + B(y-y_0) + C(z-z_0)

    3. The attempt at a solution
    I thought that I could simply use x = u^2 - v^2 as A, and y = u + v as B, etc. But yeah, it's wrong obviously. I looked through my notes already and my professor didn't go over this so I'm entirely lost. How do I find the plane tangent without knowing what u and v are? Pls help, having a headache over this. The examples in the book didn't help at all either....it didn't have any examples on this.:confused:
     
  2. jcsd
  3. Feb 21, 2007 #2

    AKG

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    You have a function f : RxR -> RxRxR defined by:

    f(u,v) = (u^2-v^2, u^v, u^2+4v)

    The surface is thus

    {f(u,v) : (u,v) is in RxR}

    You want to find the tangent plane to this surface at the point (-1/4, 1/2, 2). First, find (u', v') such that f(u', v') = (-1/4, 1/2, 2). Then the plane is given by:

    [tex](-0.25,\, 0.5,\, 2)\ +\ \mbox{Span}\left \{\frac{\partial f}{\partial u} (u',\, v'),\ \frac{\partial f}{\partial v}(u',\, v')\right \}[/tex]
     
  4. Feb 21, 2007 #3
    How do you find f(u',v')? Is that just taking the derivative of all the components?

    Edit: Eh nvm. >< I'm not thinking clearly. Thanks for the help, I got it now.
     
    Last edited: Feb 21, 2007
  5. Feb 21, 2007 #4

    AKG

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    u' and v' aren't derivatives. It might have been better if I wrote u0 and v0, it was just easier to do it the way I did. What final answer did you get? You should have got (u',v') = (0,1/2). If you got that you probably got the rest right too.
     
    Last edited by a moderator: Feb 22, 2007
  6. Feb 21, 2007 #5
    Yeah, that's what I thought. So I took u + v = 1/2 and solve for v in terms of u, then plugging in v for the u^2 - v^2 = -1/4.

    My final answer is z = 3x + 4y + 3/4?
     
  7. Feb 22, 2007 #6

    AKG

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    [itex]f_u(u',v') = (2u', 1, 2u') = (0, 1, 0)[/itex]
    [itex]f_v(u',v') = (-2v', 1, 4) = (-1, 1, 4)[/itex]

    So P = (-0.25, 0.5, 2) + {(-t, t+s, 4t) : t, s in R}.

    Can you get for any x and y, a z such that (x, y, z) is in P?

    -0.25 - t = x
    0.5 + t + s = y

    t = -(0.25 + x)
    s = y - t - 0.5 = y + x - 0.25

    Let z = 2 + 4t = 1 - 4x.

    So P = {(x,y,z) : z = 1 - 4x, x,y,z in R} = {(x,y,1-4x) : x,y in R}.
     
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