Trajectory using gradient and differential equations

1. Feb 27, 2008

issisoccer10

[SOLVED] Trajectory using gradient and differential equations

1. The problem statement, all variables and given/known data
A heat-seeking particle is located at the point P on a flat metal plate whose temperature
at a point (x, y) is T(x, y). Find parametric equations for the trajectory of the particle if
it moves continuously in the direction of maximum temperature increase.

T(x, y) = 5 − 4x$$^{2}$$ − y$$^{2}$$; P(1, 4).

2. Relevant equations
velocity vector = $$\left \langle$$x'(t), $$\right y'(t)\rangle$$

$$\nabla T(x,y)$$ = $$\left\langle$$$$T_{x}$$, $$\right T_{y}\rangle$$
3. The attempt at a solution
Let x = x(t), y = y(t) be parametric equations for the trajectory of the particle.
We want the particle to move continuously in the direction of maximum temperature
increase. Equivalently, we want the velocity vector $$\left \langle$$x'(t), $$\right y'(t)\rangle$$ to point in the direction
of $$\nabla T(x(t),y(t))$$. Since $$\nabla T(x,y)$$= $$\left \langle$$ -8x, $$\right -2y\rangle$$, the previous condition will be satisfied if
$$\left \langle$$x'(t), $$\right y'(t)\rangle$$ = $$\left \langle$$ -8x(t), $$\right -2y(t)\rangle$$ for all t. To find such functions x(t) and y(t), we must
solve the differential equations
x'(t) = -8x(t), y'(t) = -2y(t)

but this is where I get stuck...I know that to determine solution, we need the initial condition of x(0) = 1, y(0) = 4. But as to solve the differential equations, I have no idea. This is from my multivariable differential calculus class, and I think they assume that I know how to do this... sadly I do not. Any help would be greatly appreciated... thanks

2. Feb 27, 2008

Pere Callahan

You might want to introduce a constant factor here:

$$\left \langle x'(t),\right y'(t)\rangle =\alpha \nabla T(x(t),y(t))$$.

There are two reason for this. First, you only know that the particle moves in the direction of the tempertaure gradient, you do not know how strong it is coupled to that gradient. Second, in modelling physical situations you should be aware of the units you are using. Your velocity vector has units length/time while the temperature gradient is given in temperature/length; so in order for your equation to make sense $\alpha$ had better have units of (length/time) / (temperature/length) = length^2 / (time*temperature). You need not worry about this explicity but it is a point to consider.

Looks like x(t) and y(t) hardly change when you differentiate them, they just seem to pick up a factor -8 and -2 ... do you know any functions showing this strange behavior with respect to differentiation?

Last edited: Feb 27, 2008
3. Feb 27, 2008

issisoccer10

if x(t) = e$$^{-8t}$$ and y(t) = e$$^{-2t}$$ that would work right? so differential equations are just logic? at least at this level anyways..

but as for the constant $$\alpha$$ you were talking about.. that would change my parametrizations to x(t) = e$$^{-8t}$$ and y(t) = 4e$$^{-2t}$$ if it is to go through the point (1,4). But i'm not sure if that covers the $$\alpha$$.

4. Feb 28, 2008

Pere Callahan

Yes, the exponential's are solutions to your equations. Well, of course there are actual methods to solve this kind of equation but I don't know how deeo you want to go into this. In easy cases it's often possible to ust guess an solution and that's perfectly acceptable - the only problem being that this method doesn't work in more complicated situations.

If you include the constant your differential equations become

$$x'(t)=-8\alpha x(t)\qquad y'(t)=-2\alpha y(t)$$

I wrote above that alpha must have units of length^2 / (time*temperature) so the units in these equations are

$$\frac{\tex{length}}{\text{time}}=[x'(t)]=-8[\alpha] [x(t)]=\frac{\text{length}^2}{\tex{time*temperature}}\text{length}=\frac{\text{length}^3}{\tex{time*temperature}}$$

So this seems not to work.. the reason for this is that the 8 in this equation is not dimensionless either (although in the previous calcuation I took it to be so).

If you consider your defintion of the temperature field

T(x, y) = 5 − 4x^2 − y^2,

then on the right hand side it should be temperatures, but the left hand side is not. Also note that being pedantic with units would make it impossible to actually have something like 5+x^2 .. so you could have yet another constant here to convert the left hand side to temperature

$$T(x, y) = 5\beta-\gamma(4x^2-y^2)$$

with $$[\beta]=\text{temperature},\qquad [\gamma]=\frac{\text{temperature}}{\text{length}^2}$$

then the equation would allow you to consistently measure distance in units of length and temperature in -well - units of temperature.

Then including this constant as well we get:
$$x'(t)=-8\gamma\alpha x(t)\qquad y'(t)=-2\gamma\alpha y(t)$$

note that $[\gamma\alpha]=\frac{1}{\text{time}}$ so thats fine now.

So solving as you did gives

$$x(t)=e^{-8\gamma\alpha t}\qquad y(t)=e^{-2\gamma\alpha t}$$

Here the argument of the exponential is now dimensionless which is very good

Including your initial condidion you multiply this solution for x(t) [y(t)] by 1 [4]. Considering that this 1 [4] has units of length, your final x(t) [y(t)] will also have units of length because the exponential is just a dimensionless factor.

That's it.

But actually nobody cares about that.. the most practical way is to just assume that these constants accounting for conversion of units are all set to one and hence don't actually appear, or you ignore units alltogether and hope that it works out fine...however when it comes to interpreting your result in a physical way you must consider the issue of units in one way or another.

Last edited: Feb 28, 2008
5. Feb 28, 2008

issisoccer10

wow...units aren't cool...actually they make everything make sense and since they make sense, all is well... anyways thanks again