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Angle between two surfaces

  1. Aug 26, 2011 #1
    1. The problem statement, all variables and given/known data
    find the angle between the normals to the surface xy=z2 at the points (1,4,2) and (-3,-3,3)


    2. Relevant equations

    none

    3. The attempt at a solution
    del S = 2zy i + 2zx j
    and at the two points, del S = 16i+4j and del S = -18i-18j

    using the dot product, i got cos theta = -5/(4*sqrt 34)

    which is NOT the answer. Is my partial derivative equation right ??? Please help.
     
  2. jcsd
  3. Aug 26, 2011 #2

    LCKurtz

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    No, it isn't right. Write f = xy - z2 = 0 and calculate

    ∇f = <fx,fy,fz>
     
  4. Aug 26, 2011 #3

    Mark44

    Staff: Mentor

    It doesn't look right to me. I don't understand how you got what you did.

    The equation for your surface is xy = z2, or equivalently, xy - z2 = 0. The implied function here is F(x, y, z) = xy - z2. Your surface is the set of points (x, y, z) for which F(x, y, z) = 0.

    For the gradient, calculate [itex]\nabla F[/itex] and evaluate it at the two points. Each will give you a vector. From these vectors you can determine the angle, using the approach that you showed above.
     
  5. Aug 26, 2011 #4
    sorry .. should have checked the partial derivative part. I thought i got the idea wrong. Its heartening to know I'm on the right track atleast :) thankyou
     
  6. Aug 27, 2011 #5

    SammyS

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    stallionx,

    Please don't post complete solutions.

    See the rules for posting Homework Help:
    "On helping with questions: Any and all assistance given to homework assignments or textbook style exercises should be given only after the questioner has shown some effort in solving the problem. If no attempt is made then the questioner should be asked to provide one before any assistance is given. Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made."

    It may still be possible to Edit your post.
     
  7. Aug 27, 2011 #6
    I am sorry, Sir.
     
  8. Aug 27, 2011 #7
    For a tangent plane equation to the z=f(x,y))

    z-z0=(delz/delx)(x-x0)+(delz/dely)(y-y0)

    which says the normal vector is < delz/delx , delz/dely , -1>

    Find those two vectors for 2 different points

    Dot product these vectors

    V1 * V2 = length(V1)*length(V2) * Cos(theta)

    Theta = ACOS ( V1*V2)/ ( product of lengths of perpendicular/ orthogonal vectors )
     
  9. Aug 27, 2011 #8
    DOT PRODUCT can be found from coordinates multiplication and addition

    CASE : (x,y,z) dot (a,b,c) is xa+by+cz
     
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