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Math Help - How to know if implicit definition of a variable possible?

  1. #1
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    How to know if implicit definition of a variable possible?

    The point P = (-2,1,1) satisfies z^3 + x y^2 z + 1 = 0
    Can we implicitly define z in terms of x and y using this equation in a region about P?
    Find dz/dx, dz/dy, d2z/dx2 if it makes sense to do so. (note: these are partial derivatives)

    I am not sure how to know if we can implicitly define, but is it correct to find partial differentials as follows?
    z^3 + x y^2 z + 1 = 0
    3z^2 dz/dx + y^2 (x dz/dx + z) = 0
    3z^2 dz/dx + xy^2 dz/dx + zy^2 = 0-------------(1)
    dz/dx = -zy^2/(3z^2 + xy^2)
    ------------------------
    z^3 + x y^2 z + 1 = 0
    3z^2 dz/dy + x(2yz + y^2 dz/dy) = 0
    dz/dy = -2xyz/(3z^2 + xy^2)
    -----------------------

    I can get d2z/dx2 by differentiating (1) and replacing dz/dx by its expression.

    Is this method correct? Please let me know how to decide if we can implicitly define z in terms of x and y in a region around P.
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  2. #2
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    Re: How to know if implicit definition of a variable possible?

    Is this correct?
    Let F(x,y,z) = z^3 + xy^2 z + 1
    F(x,y,z) is continuous at P.
    Partial derivative dF/dz = 3z^2 + xy^2
    At P: dF/dz = 3(1)^2 + (-2)1^2 = 1 (not= 0)
    Therefore the implicit definition is possible.

    dz/dx = -(dF/dx)/(dF/dz) = -(y^2 z)/(3z^2 + xy^2)
    dz/dy = -(dF/dy)/(dF/dz) = -2xyz/(3z^2 + xy^2)

    Is this correct? But how to find the d2z/dx2 ?

    Note: I have used 'd' here but it is actually 'del' (partial derivative)
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  3. #3
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    Re: How to know if implicit definition of a variable possible?

    I think that's correct, you need \frac{\partial{F}}{\partial{z}}\ne{0} for the function to be defined. The situation is more complicated if you're trying to define a function of more than one variable.

    To get \frac{\partial^2z}{\partial{x}^2}, I think you can do an implicit differentiation on your equation (1), or you could also differentiate \frac{\partial{z}}{\partial{x}} directly using the quotient rule. It's pretty much the same calculation.

    - Hollywood
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