1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)

    x0=(2,-1,ln7)



    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

    z=ln(u)

    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

    uart

    User Avatar
    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?

    z2+x2+y2=1

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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

    ??

    Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help me find this normal vector
  1. Finding Normal Vector (Replies: 1)

Loading...