Path of particle in field

  • Thread starter gop
  • Start date
  • #1
gop
58
0

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.


Homework Equations





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

  • #2
olgranpappy
Homework Helper
1,271
3
what's "f"?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
956
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:
  • #4
gop
58
0
thanks that did the trick
 

Related Threads on Path of particle in field

Replies
2
Views
372
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
7K
Replies
3
Views
718
Replies
1
Views
5K
Replies
2
Views
398
Replies
1
Views
829
Replies
2
Views
1K
Replies
4
Views
438
Top