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Path of particle in field

  • Thread starter gop
  • Start date
gop
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1. Homework Statement

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. Homework 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]g(t)=\left[\begin{array}{c}
g_{1}(t)\\
g_{2}(t)\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)
 

Answers and Replies

olgranpappy
Homework Helper
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3
what's "f"?
 
HallsofIvy
Science Advisor
Homework Helper
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893
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:
gop
58
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thanks that did the trick
 

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