# Path of particle in field

1. Apr 25, 2008

### gop

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

The temperature is given by

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

Calculate the path of a heat-seeking particle.

2. Relevant equations

3. The attempt at a solution

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

$$g(t)=\left[\begin{array}{c} g_{1}(t)\\ g_{2}(t)\end{array}\right]$$

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

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)

2. Apr 25, 2008

### olgranpappy

what's "f"?

3. Apr 26, 2008

### HallsofIvy

Staff Emeritus
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.

Last edited: Apr 26, 2008
4. Apr 26, 2008

### gop

thanks that did the trick