Question

A heat seeking particle has the property that at any point (x,y) in the plane it moves in the direction of maximum temperature increase if the temperature at (x,y) is T(x,y)=-e^(-2y)cos(x) find an equation y=f(x) for a path of a heat seeking particle at point(pi/4,0)

Answer 1

I dont know how to get started

Answer 2

You are concerned with how the funciton T increases wrt both x and y, the gradient.

Looking into this i found a similar question.
http://mypages.iit.edu/~maslanka/251notes8.pdf

i'm still working through it to get a proper answer but its a very good starting point

Answer 3

also: "orthogonal trajectory" is the method of finding the maximum gradient

Answer 4

Your function T(x,y) will look like this in 3 space:

Answer 5

But if you look at it projected onto the x,y plane you will see that you obtain a family of curves. The lines of contours indicate places of the function that are at the same height. In both plots the red represents increasing temperature.

Answer 6

In order to follow the path of maximum temperature increase you must take a path that is normal/orthogonal/perpendicular to this family of curves. To do this you must:
1. find the slope of the original family of curves
2. Take the negative reciprocal (orthogonal) slope
3. find the equation that fits this slope but also goes through your original point. This is done by solving a diff eq'n which is separable.

Answer 7

The original family of curves is:

\[\huge e^{-2y}cosx=C\]
and the derivative wrt x becomes:

\[\huge-e^{-2y}sinx-2e^{-2y}\frac{dy}{dx}cosx=0\]
simplifying we obtain:
\[\huge \frac{dy}{dx}=-\frac{1}{2}tanx\]

but what we really need is the orthogonal slope which turns into:
\[\huge \frac{dy}{dx}=2cotx\]

solve the family of curves that will give you that derivative. you'll have a constant, but you can set the constant to match your point (pi/4,0). good luck

Answer 8

@psk981, check to see if this helps