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Math Help - Jacobian determinant question

  1. #1
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    Question Jacobian determinant question

    The equations,

    x^{3} + xy - uy = v + 4, y^{2} - xy^{3} + uv = 1,

    define x and y as functions of u and v. Find, \frac{\partial{x}}{\partial{v}}<br />

    See figure attached for my attempt. It get's a little messy with the bottom determinant, so I'm not entirely convinced I've done this correctly.

    Did I do anything wrong?
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  2. #2
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    Quote Originally Posted by jegues View Post
    The equations,

    x^{3} + xy - uy = v + 4,\;\;y^{2} - xy^{3} + uv = 1,

    define x and y as functions of u and v. Find \frac{\partial{x}}{\partial{v}}<br />

    See figure attached for my attempt. It get's a little messy with the bottom determinant, so I'm not entirely convinced I've done this correctly.

    Did I do anything wrong?
    Doing it by a slightly different method, I got almost the same answer. In fact, if you use implicit differentiation to differentiate each of the given equations partially with respect to v then you get

    3x^2\frac{\partial{x}}{\partial{v}} + y\frac{\partial{x}}{\partial{v}} + x\frac{\partial{y}}{\partial{v}} -u\frac{\partial{y}}{\partial{v}} = 1,
    2y\frac{\partial{y}}{\partial{v}} -y^3\frac{\partial{x}}{\partial{v}} - 3xy^2\frac{\partial{y}}{\partial{v}} + u = 0.

    Write these equations as

    (3x^2+y)\frac{\partial{x}}{\partial{v}} + (x-u)\frac{\partial{y}}{\partial{v}} = 1,
    y^3\frac{\partial{x}}{\partial{v}} + (3xy^2-2y)\frac{\partial{y}}{\partial{v}} = u.

    Multiply the first equation by 3xy^2-2y and the second by x-u and subtract, to get

    \bigl((3x^2+y)(3xy^2-2y) - y^3(x-u)\bigr)\frac{\partial{x}}{\partial{v}} = 3xy^2-2y - (x-u)u.

    That is the same as your answer except that one of us has missed a minus sign somewhere, so that my denominator is the negative of yours.
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