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Finding the equation of a plane

  1. Sep 6, 2007 #1
    1. The problem statement, all variables and given/known data

    This isn't actually a homework question, but I'm revising for my resits and can't do this past paper question. And of course if i don't figure out how to do it it will obviously come up on the exam

    A plane is tangential to the surface xexp(-x^2 - y^2) - z = 0 at the point where z takes on its maximum value. Find the equation of the plane

    2. Relevant equations



    3. The attempt at a solution

    I think I have to find where z is at its maximum and then find the normal to the tangent at that point to enable me to write the plane equation in the form r.n=0
    z will be at it's maximum when xexp(-x^2 - y^2) is at its maximum but i don't know when that will be?
     
  2. jcsd
  3. Sep 6, 2007 #2

    NateTG

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    Do you know how to find the maximum of a function?
     
  4. Sep 6, 2007 #3

    HallsofIvy

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    As NateTG suggested, first find the maximum point by setting the gradient of z equal to 0. Of course, the tangent plane at a maximum (or minimum) is parallel to the xy-plane, just as the tangent line at a maximum (or minimum) of y= f(x), in two dimensions, is parallel to the x-axis.
     
  5. Sep 6, 2007 #4
    so since z = x exp(-x^2 - y^2) if i differentiate this and set it to zero i should get my point. But what do i differentiate z with respect to since there's x and y in the function.
    Since I'm trying to find the equation of the plane does that just mean that it's a plane parallel to the x-y axis?
     
  6. Sep 6, 2007 #5

    NateTG

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    Actually, it's parallel to the x-y plane.

    Since you're apparently not familiar with partial derivatives, I'd suggest looking at
    [tex]x e^{-x^2-y^2}[/tex]
    and trying to find what [itex]y[/itex] is at the maximum by inspection.
     
  7. Sep 6, 2007 #6
    oops yeh i meant plane

    i am familiar with partial derivatives.

    so the partial derivative wrt x is -2x^2[exp(-x^2-y^2)] + exp(-x^2-y^2)
    and wrt y is -2y^2[exp(-x^2-y^2)]

    but I'm still confused as to where i'm going with this, i don't really get the bigger picture of what i'm trying to do
     
  8. Sep 6, 2007 #7

    Dick

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    If you are at a maximum point, the partial derivatives wrt x and y will both vanish.
     
  9. Sep 6, 2007 #8
    ok now i'm really confused. how do you recomend i start this question as i don't seem to be getting anywhere with my current route =(
     
  10. Sep 6, 2007 #9

    Dick

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    Just set your two partial derivatives equal to zero. That's two equations in two unknowns, x and y. Once you've found your critical points test them to see if you can find which one could give you a maximum for z.
     
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