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Finding the Gradient of an Integral?

  1. Jun 12, 2012 #1
    1. The problem statement, all variables and given/known data
    https://dl.dropbox.com/u/64325990/Capture.PNG [Broken]

    I'm not even sure where to start :O
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 12, 2012 #2
    How about

    [tex]f(0+\Delta x,0+\Delta y)\approx f(0,0)+\frac{\partial f}{\partial x}(0,0)\Delta x+\frac{\partial f}{\partial y}(0,0)\Delta y[/tex]

    This uses the gradient in the sense of

    [tex]df\approx \nabla f\cdot(dx,dy)[/tex]

    Though I'm sorting of putting together whatever notation comes to mind, so let us know if you have notes close to this but aren't sure how they relate, or any other questions.
     
  4. Jun 12, 2012 #3

    Curious3141

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    Start with the Fundamental Theorem of Calculus to work out [itex]\frac{\partial f}{\partial y}[/itex] and [itex]\frac{\partial f}{\partial x}[/itex].

    For the latter, you might find it more helpful to switch the bounds and put a negative sign on the integral.
     
    Last edited by a moderator: May 6, 2017
  5. Jun 12, 2012 #4

    SammyS

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    You have received two very good tips.

    I'll make my response more concrete.

    cos(t2) is integrable, but not in closed form with elementary functions.

    Let G(t) be an anti-derivative of cos(t2).

    Then of course, cos(t2) is the derivative of G(t).

    Using this to evaluate your integral gives:

    [itex]\displaystyle \int_{x}^{y}{\cos(t^2)}\,dt=G(y)-G(x)\ .[/itex]

    Now take the gradient of that.

    Can you take it from here?
     
    Last edited by a moderator: May 6, 2017
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