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Path of particle in field

  1. Apr 25, 2008 #1


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    1. The problem statement, all variables and given/known data

    The temperature is given by

    [tex]T(x,y)=\sqrt{2}e^{-y}\cos x[/tex]

    Calculate the path of a heat-seeking particle.

    2. Relevant equations

    3. The attempt at a solution

    [tex]\nabla f(x,y)=\left[\begin{array}{c}
    -\sqrt{2}e^{-y}\sin x\\
    -\sqrt{2}e^{-y}\cos x\end{array}\right][/tex]


    [tex]\dot{g}_{1}(t)=-\sqrt{2}e^{-g_{2}(t)}\sin g_{1}(t)[/tex]
    [tex]\dot{g}_{2}(t)=-\sqrt{2}e^{-g_{2}(t)}\cos g_{1}(t)[/tex]

    That's where I'm stuck. I have to solve the differential equations but they depend very heavily on each other so I can't get them decoupled.
    Also tried to solve them in maple but maple just complains that the numverator of the ODE depens on the highest derivative.

    Do I missing something obvious? (we did differential equations only briefly)
  2. jcsd
  3. Apr 25, 2008 #2


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    Homework Helper

    what's "f"?
  4. Apr 26, 2008 #3


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    Science Advisor

    Since you are only interested in the path- i.e. the curve in xy space, independent of t", you are better off not introducing "g1(t)" and "g2(t)". You can rewrite the equations as
    [tex]\frac{dy}{dx}= \frac{-\sqrt{2}e^{-y}sin(x)}{-\sqrt{2}e^{-y}cos(x)}[/tex]
    or just dy/dx= tan(x). Integrate.
    Last edited by a moderator: Apr 26, 2008
  5. Apr 26, 2008 #4


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    thanks that did the trick
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