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Calculus of Variation Questions

  1. Apr 9, 2012 #1
    Hey guys. In my mechanics course, we have began discussing calculus of variations, and I don't really understand what's going on, entirely. Any help understanding would be great. Our professor gave us an easy problem, but I feel like I am just missing something.

    1. The problem statement, all variables and given/known data
    Consider the function f(y,y',x)=3yy' + x^2 where y(x)=2x^2-3x+1. Compute df/dx and ∂f/∂x


    2. Relevant equations
    So our text says df/dx=df{y,y',x}/dx= (∂f/∂y)(dy/dx) + (∂f/∂y')(dy'/dx) + (∂f/∂x) = y'(∂f/∂y') + y''(∂f/∂y') + (∂f/∂x)

    So, maybe I am oversimplifying it, but here's what confuses me from his lectures. Is y'=d(y)/dx and y''=d(y')/dx=d^2(y)/dx^2 ? In class, sometimes he leaves y' and sometimes he doesn't, because he knows how to get to the answer or objective, but I am just not sure what to do about them.


    3. The attempt at a solution

    Assuming the answer to my above question, what I did was...
    y'=4x-3 y''=4 (∂f/∂y)=3y' and (∂f/∂y')=3y

    It makes sense to go on to to ∂f/∂x, since it is needed for df/dx, so I got ∂f/∂x=∂(3yy' + x^2)/∂x, plugging in and doing some adding/multiplying, I finally go ∂f/∂x=72x^2-106x+39

    Then, df/dx= y'(∂f/∂y') + y''(∂f/∂y') + (∂f/∂x) = (after multiplying and adding) 144x^2 - 214x +78.


    I know I skipped a lot in the end there, but my questions arise before the simple algebra, I can easily do that, but this stuff here is causing me a lot of grief. Any help at all would be great. I have searched around and found nothing really that helped.
     
  2. jcsd
  3. Apr 9, 2012 #2

    Dick

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    ∂f/∂x is just 2x. f is a function of the three variables y, y' and x. When you are taking the partial derivative you don't count the implicit dependence of y and y' on x. You just count the explicit dependence and differentiate the x^2.
     
  4. Apr 9, 2012 #3
    Herp. You're right, I was just thinking it was different here. I have never taken a partial that had any implicit dependence on the independent variable (well, not explicitly stated, anyways). So then,

    df/dx= y'(∂f/∂y) + y''(∂f/∂y') + (∂f/∂x) = (4x -3)(3y') + 4(3y) + 2x, that's all? That's much more simple than I expected.
     
  5. Apr 9, 2012 #4

    Dick

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    That's it. You can substitute your expressions for y' and y into that and express it totally in terms of x. And there's another way to find df/dx. Just substitute your expressions for y' and y into f to express it totally in terms of x, then just differentiate with respect to x. You should get the same thing. It's not that mysterious.
     
  6. Apr 9, 2012 #5
    Awesome, thanks a ton! That clears up so much confusion.
     
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