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Thread: Partial Derivative Proof

  1. #1
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    Partial Derivative Proof

    Hi all,

    I have been running through this problem but can't get anywhere near it:

    $\displaystyle x = \frac {sin(u)}{cos(v)}$ prove:

    $\displaystyle \frac {\partial u}{\partial x} = \frac{cos^3(v)cos(u)}{cos(u+v)cos(u-v)} $

    I start off:

    $\displaystyle \frac {\partial x}{\partial u} = \frac {cos(u)}{cos(v)}=> \frac {\partial u}{\partial x} = \frac {cos(v)}{cos(u)}$ multiply top and bottom by $\displaystyle cos^2(v)cos(u)$ to give the numerator. Whats left is to prove the denominator:

    $\displaystyle cos^2(u) cos^2(v) = cos(u+v)cos(u-v)$

    I have tried whatever I can think of, double angle identities,product to sum and even a messy Euler exponential expansion, but fail... is there something fundamental I have missed here? Maybe this should be in the trig sub forum.....

    Thanks for reading!
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  2. #2
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    I'm glad you couldn't prove $\displaystyle \cos^{2}(u)\cos^{2}(v)=\cos(u+v)\cos(u-v),$ because if you plug in $\displaystyle u=\pi/4=v,$ you don't get equality!

    I think your method of differentiation is incorrect. As in, perhaps you're supposed to view $\displaystyle v$ as a function of $\displaystyle x$ as well. What's independent here, and what's dependent?
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  3. #3
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    Thanks for looking Ackbeet.... At least I am not going out of my mind!

    The wording is suppose $\displaystyle x $ & $\displaystyle y$ are related to $\displaystyle u $ & $\displaystyle v $ via

    $\displaystyle x = \frac {sin u}{cos v} ; y = \frac{sin v}{cos u}$

    show that (expression in original post) and find a similar expression for $\displaystyle \frac{\partial u}{\partial y}$

    I didnt consider that u&v themselves to be functions- that maybe what I needed. I'll give that ago and see what I get.
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  4. #4
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    In that case, it's pretty clear that u and v are both functions of x and y. You'll need to use the appropriate formulas for their derivatives.
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  5. #5
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    Thanks Ackbeet.

    I am rather confused by this problem. Can you by any chance point me in the dirction of the method that will be used so I can look this up..... It looks like a change of variable problem, but it lacks the relation between $\displaystyle x$ and $\displaystyle y$..... do we assume an unknown relation, say $\displaystyle z(x,y)$ define the partials and eliminate the $\displaystyle \partial z$ or is it somrthing different to this?

    Any input will be much appreciated...
    Last edited by padawan; Aug 17th 2010 at 01:26 PM.
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  6. #6
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    Find u as a function of x and y. Do that by eliminating parameters and using the Pythagorean theorem. I think that path might work for you.
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  7. #7
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    This path does work. You can find $\displaystyle \cos^{2}(u)$ in terms of $\displaystyle x$ and $\displaystyle y$, and then use implicit differentiation. You then plug in what $\displaystyle x$ and $\displaystyle y$ are, and simplify to get your final answer. You'll need a trig identity or two, but it does work out.
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  8. #8
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    I have had a go at this and get to an identity of:

    $\displaystyle cos^2(u) = $$\displaystyle 1- x^2\over{1 - x^2y^2}$

    Is this along the lines you took Ackbeet? I guess it would be a case of implicitly differentiating and then plugging original $\displaystyle x$ & $\displaystyle y$ in... is that the route you took? Big thanks to you for helping so far..
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  9. #9
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    You should clean up your code a bit. I realize that with the Preview feature taking so long, this isn't very desirable. However, I got the following:

    $\displaystyle \displaystyle{\cos^{2}(u)=\frac{1-x^{2}}{1-x^{2}y^{2}}},$

    which is, I think, what you're getting at. The route you're suggesting is indeed where I went with it.
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  10. #10
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    And we're back. What did you get?
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