What is the path of a heat-seeking particle in a given temperature field?

[gop]

Homework Statement

The temperature is given by

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

Calculate the path of a heat-seeking particle.

Homework Equations

The Attempt at a Solution

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

g(t)=\left[\begin{array}{c}<br /> g_{1}(t)\\<br /> g_{2}(t)\end{array}\right]

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

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)

 

[olgranpappy]

what's "f"?

 

[HallsofIvy]

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
\frac{dy}{dx}= \frac{-\sqrt{2}e^{-y}sin(x)}{-\sqrt{2}e^{-y}cos(x)}
or just dy/dx= tan(x). Integrate.

 

[gop]

thanks that did the trick