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Homework Help: Points where normal at point on surface equals line from origin to that point

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the points (x,y,z) on the paraboloid z=x2+y2-1 at which the normal line to the surface is the line from the origin to the point (x,y,z).

    2. Relevant equations

    normal line means take the gradient, <∂F/∂x, ∂F,∂y, ∂F,∂z>, and evaluate at the point

    3. The attempt at a solution

    n= -2xi + -2yj + 1k, where i=(1 0 0)T, j=(0 1 0)T, k=(0 0 1)T.

    So we want an (x,y,z) that such that the normal vector n is a scalar multiple of xi + yj + (z=x2+y2+1)k.

    <-2x,-2y,1> = ß<x,y,x2+y2+1>

    <-2,-2,0> = <ß, ß, ß(x2+y2)>

    Am I doing this right? Something ain't working.
  2. jcsd
  3. Apr 20, 2010 #2
    This is a reasonable approach, yes. Note that I corrected your sign above: [tex]z = x^2 + y^2 - 1[/tex], not [tex]z = x^2 + y^2 + 1[/tex].

    How did you get from the first line to the second? It looks like you divided the first component of your vector equation by [tex]x[/tex], the second by [tex]y[/tex], and then tried to add something or other to the third. While in theory you could organize your manipulations that way (except that whatever you did to the third component is wrong), in practice it's a bad idea.

    The equation [tex]\langle -2x, -2y, 1\rangle = \beta \langle x, y, x^2 + y^2 - 1 \rangle[/tex] translates into the system of equations [tex]\begin{cases}-2x = \beta x & \\ -2y = \beta y & \\ 1 = \beta(x^2 + y^2 - 1) & \end{cases}[/tex]. Proceed from there.
  4. Apr 20, 2010 #3

    I just tried to get the components to match up; for example, -2xi=ßxi. But now I see what's happenin'.
  5. Apr 21, 2010 #4
    Actually, I don't see what's happening. ß would be -2, looking at the first two equations, but then plugging that into the third gives the equation x2 + y2 = -1. Explain how to solve that system.
  6. Apr 21, 2010 #5
    No, it doesn't; check your algebra. Also, [tex]\beta = -2[/tex] is not the only possible solution to the first two equations.

    That's not very polite.
  7. Apr 21, 2010 #6

    D H

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    Staff Emeritus
    Science Advisor

    Incorrect. Check your math.
  8. Apr 21, 2010 #7
    For some reason, I'm not getting this problem. The wires in my brain must be crossed.

    So what I did is just say z = r2 - 1, where r is the radius of the cross section of a plane z=c that cuts through the paraboloid. Then dz/dr=2r, meaning that at r=1/2, a scalar multiple of the normal line will pass through the origin.
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