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Tangent planes and surfaces

  1. Oct 10, 2008 #1
    Just when I thought I got the hang of tangent planes and surfaces there comes a question I haven't quite seen before

    z = ln (x[tex]^{2}[/tex]+3y[tex]^{2}[/tex])

    Find a normal vector n and the equation of the tangent plane to the surface at the point
    (2, -1, ln 7)

    So keeping the cartesian equation in mind:

    z = z[tex]_{0}[/tex] + F[tex]_{x}[/tex](x,y)(x - x[tex]_{0}[/tex]) + F[tex]_{y}[/tex](x,y)(y - y[tex]_{0}[/tex])

    Partial derivative with respect to x: [tex]\frac{2x}{x^{2} + 3y^{2}}[/tex]
    Evaluate with the values and I get 4/11

    Partial derivate with respect to y: [tex]\frac{6y}{x^{2} + 3y^{2}}[/tex]
    Evaluate with the values and I get -6/11

    Have I done this correctly?

    So putting it all in the equation I get z = ln 7 + 4x/11 -6y/11 -14/11

    I get the feeling I did something wrong somewhere, and of course the normal vector would be (4/11, -6/11, -1) if everything was right.
    Last edited: Oct 10, 2008
  2. jcsd
  3. Oct 10, 2008 #2
    just take the gradient of [itex]ln (x ^{2} +3y ^{2} ) - z [/itex] and evaluate it at the point
  4. Oct 10, 2008 #3
    Do you mean dz/dx or dz/dy or something else?
  5. Oct 10, 2008 #4
    have you not learned about gradient? [itex]\nabla=<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}>[/itex] = the defining vector of the tangent plane.
  6. Oct 10, 2008 #5
    Well that's what I did earlier lol.

    (4/11, -6/11, -1)

  7. Oct 10, 2008 #6


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    Then do the arithmetic again. x2+ 3y2= 42+ 3(-1)2 is NOT 11!
  8. Oct 10, 2008 #7
    It should be 2^2 + 3(-1)^2 which is 7...hehe, now where did that rogue 11 come from, lol.
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