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Math Help - involved derivative

  1. #1
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    involved derivative

    Let x, y and z be given parametrically by the arbitrary variables n & m:

    x = ncos(m) + f'(m)sin(m);
    y = nsin(m) - f'(m)cos(m);
    z = n + f(m);

    where f is a function of m, assumed to be derivable twice.

    Then what are the partial derivates of z, with respect to the two variables x and y?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by tombrownington View Post
    Let x, y and z be given parametrically by the arbitrary variables n & m:

    x = ncos(m) + f'(m)sin(m);
    y = nsin(m) - f'(m)cos(m);
    z = n + f(m);

    where f is a function of m, assumed to be derivable twice.

    Then what are the partial derivates of z, with respect to the two variables x and y?
    Maybe someone more knowedgable will come and answer this, but I have a question of clarification. You do not have z explicitly as a function of x,y. Is that intentional? Are we supposed to solve for z?
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  3. #3
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    yup. that's the way it is. z is given in function of u and v, not x and y.
    solving for u and v is impracticable since the function f(m) is introduced in z while its derivative is given in x and y.
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  4. #4
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    Can one at least write:

    \frac {\partial z} {\partial x} = \frac {\partial z} {\partial u} . \frac {\partial u} {\partial x} + \frac {\partial z} {\partial v} . \frac {\partial v} {\partial x}
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