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Math Help - Confused about partial derivatives of higher order...

  1. #1
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    Confused about partial derivatives of higher order...

    After literally hours of "no, and what and how", I need help with the following task:

    f(u,v)= g(x(u,v),y(u,v))

    x(u,v)= u^2- v^2
    y(u,v)= 2uv

    Expressing f'u and f'v as partial derivatives of g I get:

    f'u= (dg/dx)*(dx/du) + (dg/dy)*(dy/du)
    f'v= (dg/dx)*(dx/dv) + (dg/dy)*(dy/dv)

    (written in crooked d's)

    Now I'm supposed to find all functions f that satisfy u*f'u - v*f'v=0

    The hardest part (which confuse me the most because there are so many steps involving the chain rule is the following): "Express f''uu+ f''vv through partial derivatives of g. Simplify as much as possible"

    Please help me with either of the tasks, I need this by monday and I'm so confused and tired of staring at this task!
    Last edited by tinyone; September 25th 2010 at 11:42 AM.
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  2. #2
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    *bump*, please help
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  3. #3
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    Quote Originally Posted by tinyone View Post
    After literally hours of "no, and what and how", I need help with the following task:

    f(u,v)= g(x(u,v),y(u,v))

    x(u,v)= u^2- v^2
    y(u,v)= 2uv

    Expressing f'u and f'v as partial derivatives of g I get:

    f'u= (dg/dx)*(dx/du) + (dg/dy)*(dy/du)
    f'v= (dg/dx)*(dx/dv) + (dg/dy)*(dy/dv)

    (written in crooked d's)

    Now I'm supposed to find all functions f that satisfy u*f'u - v*f'v=0
    So you know that f'_u = \frac{\partial g}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial g}{\partial y}\frac{\partial y}{\partial u}. You can write \frac{\partial g}{\partial x} as g'_x, and \frac{\partial g}{\partial y} as g'_y (to save writing). You can also find \frac{\partial x}{\partial u} and \frac{\partial y}{\partial u} (from the formulas for x(u,v) and y(u,v)). That has the effect of writing f'_u as

    f'_u = 2ug'_x + 2vg'_y\qquad(*).

    Now do the same for the other partial derivative f'_v.

    Having done that, multiply the formula for f'_u by u, multiply the formula for f'_v by v, and subtract. That will tell you something about the possible functions that satisfy uf'_u - vf'_v = 0.

    Quote Originally Posted by tinyone View Post
    The hardest part (which confuse me the most because there are so many steps involving the chain rule is the following): "Express f''uu+ f''vv through partial derivatives of g. Simplify as much as possible"
    Differentiate the formula (*) partially with respect to u, using the chain rule and the product rule: f''_{uu} = 2g'_x + 2u\frac{\partial}{\partial u}(g'_x) + 2v\frac{\partial}{\partial u}(g'_y). Now you need to figure out what \frac{\partial}{\partial u}(g'_x) and \frac{\partial}{\partial u}(g'_y) are. To do that, use the formula (*), but replace f by g'_x. That tells you that \frac{\partial}{\partial u}(g'_x) = 2ug''_{xx} + 2vg''_{xy} (and you can get a similar expression for \frac{\partial}{\partial u}(g'_y)).

    That way, you get an expression for f''_{uu} in terms of the second partial derivatives of g. Do the same for f''_{vv}, add, and simplify ... .
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