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Help me find this normal vector

  1. Jul 28, 2012 #1
    help me find this normal vector!!!

    1. The problem statement, all variables and given/known data

    Find a normal vector n and the equation of the tangent plane to the surface S at the point x0

    S: z=ln(x2+3y2)


    2. Relevant equations

    3. The attempt at a solution

    ok so I have this general formula from the textbook.

    z= z0+Fx(x0,y0)(x-x0)+Fy(x0,y0)(y-y0)

    and n=(Fx(x0,y0) , Fy(X0,Y0) , -1) respectively.

    so using these

    S: z= ln(x2+3y2)

    let u=x2+3y2


    dz/du=(1/u) (u')

    Fx(x,y) = d/dx(u)/u = 2x/(x2+3y2) = 4

    Fy = d/dy (u)/u = 6y/ (x2+3y2) = -6

    z= ln7 + 4(x-2)-6(y+1)

    z= ln7 +4x -6y -14

    ln7 +4x -6y -14 -z = 0

    so n =(4,-6,-1)T

    but they have 4x-6y-7z-14+7ln7=0

    and n= (4, -6, -7)T

    ?? Im not sure what I have done wrong but it looks very close to what they have.

    any help is appreciated.
  2. jcsd
  3. Jul 28, 2012 #2


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    Science Advisor

    Re: help me find this normal vector!!!

    Check your numerical results of the substitution of x0,y0 here. :)
  4. Jul 28, 2012 #3
    Re: help me find this normal vector!!!

    ohh no (-1)2≠-1.

    Thanks a bunch uart!!!
  5. Jul 28, 2012 #4
    Re: help me find this normal vector!!!

    Do you think you could give me a hint in the right direction for this one?


    x0=(1/3 , 1/2, √23/ 6)

    If I do it the usual way and I solve for z, I get a negative sqrt term so I don't think that is the right approach.

    there is some inequalities I can see but I don't think that helps
    I think its a sphere but that doesn't really help me either.

    please help. Thanks.
  6. Jul 28, 2012 #5


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    Science Advisor

    Re: help me find this normal vector!!!

    If [itex]z^2+ x^2+ y^2= 1[/itex] then [itex]z= \pm\sqrt{1- x^2- y^2}[/itex]. That is not a "negative squareroot" it simply means that [itex]x^2+ y^2[/itex] cannot be larger than 1. And that must be true geometrically because this is the surface of a sphere of radius 1.

    But it is not necessary to solve for z. If a surface is given by F(x, y, z)= constant, then the gradient, [itex]\nabla F[/itex] is a normal vector at each point on the surface. Here [itex]F(x, y, z)= x^2+ y^2+ z^2[/itex] so that [itex]\nabla F= 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}[/itex]. At the given point, that is [itex](2/3)\vec{i}+ \vec{j}+ \sqrt{34}/3\vec{j}[/itex] so the tangent plane at that point is [itex](2/3)(x- 1/3)+ (y- 1/2)+ \sqrt{34}{3}(z- \sqrt{34}/6)= 0[/itex].
  7. Jul 28, 2012 #6
    Re: help me find this normal vector!!!

    ok I think that should be a √23 /6 not √34 /6

    but that is a really convenient way of doing that. thanks a lot.

    but for some reason in their answer they have multiplied it by three i.e.

    2x +3y+√23z -6 = 0

    does this not change the normal vector in a geometric way?

    since they have n=(2,3,√23)T

    and not n=(2/3,1,√23 /3)T


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