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Tangent plane of a surface

  1. Aug 12, 2010 #1
    1. The problem statement, all variables and given/known data
    See figure.


    2. Relevant equations



    3. The attempt at a solution

    Rearranging my equation,

    [tex]z = \sqrt{\frac{x^{3}+3y^{2}-3}{3}}[/tex]

    Let [tex]f(x,y) = \sqrt{\frac{x^{3}+3y^{2}-3}{3}} [/tex]

    Then,

    [tex]f_{x}(x,y) = \sqrt{x^{2}}[/tex]

    [tex]f_{y}(x,y) = \sqrt{2y}[/tex]

    So,

    [tex]f_{x}(3,1) = \pm 3[/tex]

    [tex]f_{y}(3,1) = \sqrt{2}[/tex]

    Therefore the tangent plane is defined as,

    [tex]z - 3 = 3(x-3) + \sqrt{2}(y-1)[/tex]

    Does this look correct?
     

    Attached Files:

  2. jcsd
  3. Aug 12, 2010 #2

    vela

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    No, it's not correct since, for one thing, you didn't differentiate correctly.
     
  4. Aug 12, 2010 #3

    Mark44

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    No. You are forgetting that you need to use the chain rule.
     
  5. Aug 12, 2010 #4

    Mark44

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    Darn, vela, you beat me by a minute!
     
  6. Aug 12, 2010 #5
    Is it that,

    [tex]f_{x}(x,y) = \frac{1}{2}x^{\frac{1}{2}}[/tex] or [tex]f_{x}(x,y) = \frac{1}{2}\sqrt{x}[/tex]

    and,

    [tex]f_{y}(x,y) = 1[/tex]

    ?
     
  7. Aug 12, 2010 #6

    Mark44

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    It's
    [tex]f_{x}(x,y) = \frac{1}{2}(x^3/3 + 3y^2/3 - 1)^{\frac{-1}{2}}* x^2[/tex]

    and similar for the other partial.
     
  8. Aug 12, 2010 #7

    vela

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    Usually it's the other way around!
     
  9. Aug 12, 2010 #8
    Okay so,
    [tex]f_{x}(x,y) = \frac{1}{2}(x^3/3 + 3y^2/3 - 1)^{\frac{-1}{2}}* x^2[/tex]

    and

    [tex]f_{y}(x,y) = \frac{1}{2}(x^3/3 + 3y^2/3 - 1)^{\frac{-1}{2}}* 2y[/tex]

    That being said,

    [tex]f_{x}(3,1) = \frac{1}{2}(3^3/3 + 3^2/3 - 1)^{\frac{-1}{2}}* 3^2 = \frac{3}{2}[/tex]

    [tex]f_{y}(x,y) = \frac{1}{2}(x^3/3 + 3y^2/3 - 1)^{\frac{-1}{2}}* 2y = \frac{1}{3}[/tex]


    So the tangent plane is defined as,

    [tex]z-3 = \frac{3}{2}(x-3) + \frac{1}{3}(y-1)[/tex]

    Is this any closer?
     
  10. Aug 12, 2010 #9

    vela

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    Yup, that matches what I got.
     
  11. Aug 13, 2010 #10

    ehild

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    There is an easier way to get the tangent plane by using the gradient vector to the 3-variable function

    F(x,y,z)= x3+3y2-3z2.

    ∇F(x0, y0, z0) is perpendicular to the tangent vectors at (x0, y0, z0) to any curve passing through (x0, y0, z0) and lying on the surface F(x,y,z)=constant, so the gradient vector is normal to any vector lying in the tangent plane:

    ∇F(r-r0)=0

    r0=3i+j+3k

    ∇F=∂F/∂xi+∂F/∂yj+∂F/∂zk=3x02i+6y0j-6z0k.

    ehild
     
  12. Aug 13, 2010 #11
    Okay I'm going to see if I can follow with what you're explaining ehild.

    So the gradient of our function at the specified point (x0, y0, z0) will be perpendicular to all the tangent vectors at that point, as well as any curve passing through that point.

    That last part is where I get confused.

    Is it that any curve that lies in the surface F(x,y,z) is simply a constant?

    and I'm not sure what this last part is,

    I understand the mechanics of the gradient, but I don't understand what you're trying to do with the two vectors r and r0. Are they position vectors of some sort? Where are they positioned?

    Could you go into more detail as required?

    Thanks again!
     
  13. Aug 13, 2010 #12

    jambaugh

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    FWIW:
    Another (equivalent) way to tackle tangent line/plane problems is by using differentials (this is one way to define them).

    Given a constraint surface, e.g. [tex]x^2 + y^2 + z^2 = R^2[/tex] (R a constant.)
    Take the differential of the constraint equation:
    [tex]2x\cdot dx + 2y \cdot dy + 2z \cdot dz = 0[/tex]
    (which you can clean up by dividing by 2)
    This is just like the rules for implicit differentiation except you are indeed taking differentials instead of derivatives.

    The (modern) definition of the differentials is that they are local coordinates for the tangent plane at the point of tangency. So given one point on the surface:[tex]x=x_0, y=y_0, z=z_0[/tex], you get:
    [tex]x_0\cdot dx + y_0 \cdot dy + z_0 \cdot dz = 0[/tex]

    Now to convert the differentials to the original coordinates, rewrite:
    [tex]dx =(x-x_0) , dy=(y-y_0), dz = (z-z_0)[/tex]
    and you have for this example:
    [tex]x_0(x-x_0)+ y_0(y-y_0) + z_0(z-z_0) = 0[/tex]

    Thus for a sphere of radius 5, at the tangent point (3,4,0) you get the tangent plane:
    [tex]3(x-3) + 4(y-4) = 0[/tex]

    [edit] or [tex] 3x + 4y = 25[/tex] [end edit]

    One may or may not prefer this method but it, I think, gives one a clearer picture of what differentials are and how they are distinct from derivatives. A differential of a variable is a new variable with constraints tied to the constraints on the old variables, namely that they define local coordinates for points on the tangent plane/line/hyperplane.
     
  14. Aug 13, 2010 #13

    ehild

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    F(x,y,z)=constant is valid for any point of the surface, so also for any lines on the surface.

    Well, I forgot to make clear that r is the position vector of an arbitrary point on the tangent plane, and r0 is the given point where the plane is tangent to the surface F(x,y,z)=constant.

    ehild
     
  15. Aug 13, 2010 #14

    vela

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    Whenever you have a surface that can be expressed in the form F(x,y,z)=c where c is a constant, you can find the normal to the surface merely by calculating ∇F. In this case, it's much simpler than solving for z=f(x,y) and using the method you chose.

    Once you have the normal, regardless of which way you found it, you just write down the equation of the plane the same way.
     
  16. Aug 13, 2010 #15
    So,

    [tex]\nablaf(x,y,z) = 3x^2 + 6y -6z = 0[/tex]

    So what are the components of my normal vector, [tex]\vec{n}[/tex]

    Is it,

    [tex]\vec{n} = <3,6,-6>[/tex]?

    Or,

    [tex]\vec{n} = <3x^2,6y,-6z>[/tex]?

    EDIT: Actually, if we take the point P(3,1,3) then,

    [tex]\vec{n} = <27,6,-18>[/tex]

    Then,

    [tex]27(x-3) +6(y-1) -18(z-3) = 0[/tex]

    Would this be correct?
     
  17. Aug 13, 2010 #16

    vela

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    You mean [itex]\nabla f(x,y,z)=(3x^2, 6y, -6z)[/itex], right?
    Yes. If you divide this equation through by 18, you'll see you'll get the equation you derived previously.
     
  18. Aug 13, 2010 #17
    Whoops, yes you're right!

    Thanks for showing that to me, that's alot more effective!
     
  19. Aug 15, 2010 #18

    vela

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    Just wanted to point out, this method is what ehild was talking about. I just posted since you seemed to be getting lost in complications due to the notation. If you go back and reread what he posted, it may make more sense to you now.
     
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