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Calc III - finding tangent plane

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    I thought the problem was easy, but my answer is wrong (aparently; I still disagree.)

    First I defined x = (y^3)(z^3) to be a surface of function F

    F(x,y,z) = (y^3)(z^3) - x = 0

    Then, the gradient of F:

    Partial wrt x = -1
    Partial wrt y = 3y^2(z^3)
    Partial wrt z = 3z^2(y^3)

    Gradient F = -1i + 3y^2(z^3)j + 3z^2(y^3)k

    Gradient F(1,-1,-1) = -1i + 3(-1)^2((-1)^3)j + 3(-1)^2((-1)^3)k
    = -i - 3j - 3k

    Then the tangent plane formula is

    - x - 3y - 3z = 5


    x + 3y + 3z = -5

    Where am I going wrong with this?

    Attached Files:

  2. jcsd
  3. Feb 18, 2012 #2
    I'm 'inclined' to agree with you
    Pic related - graphs of the surface and the tangent plane to surface at 1,-1,-1

    I am pretty tired atm so we could both be falling into the same trap though..

    Attached Files:

  4. Feb 18, 2012 #3


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    That's the same result I get.
  5. Feb 18, 2012 #4
    Darn you wiley plus!

    I'm not about to start guessing through their wrong answers to see which one is "correct." Guess I'll email my professor.
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