# Shortest path between two points on a paraboloid

1. Mar 3, 2016

### RubinLicht

1. The problem statement, all variables and given/known data
I am only currently in multivariate calculus, so i haven't even touched differential geometry yet, but a question that i had while learning about gradients came up and it led me to the topic of geodesics and differential geometry, so here goes:

Class problem: Find the equation representing the path of the particle as it moves in the direction of maximum temperature increase on a plate whose temperature at (x,y) is f(x,y) = 20 - 4x^2 - y^2. The particle starts at (2,-3).

Solution: take contour map and the curves just end up being the family of orthogonal trajectories, then plug in the point for the particular solution.

My thought: if you were you graph the temperature function in 3D you would get a paraboloid, and if you graphed the path in 3D it would be a parabolic cylinder (just the path stretched in the z direction). Is the curve of intersection of these two surfaces the shortest path between any point on the that curve and the "source of heat"?

clarification: in my case, the particle is confined to travel on the paraboloid as opposed to just a flat plane for the class problem.
2. Relevant equations
well i mean gradient i guess....

3. The attempt at a solution
I googled geodesics and such, but the rigor of calculus is a little beyond my weak high school abilities heh.

thanks, I did my best.

2. Mar 3, 2016

### Brian T

Interesting question. the two concepts are related to a variational problem, i.e. to find a function which produces a stationary value a given integral. Consider the following:
Consider a surface parametrized by $\vec{\phi}(x,y)$
A curve on the surface can be parametrized by $$\vec{\alpha}(t)=\vec{\phi}(x(t),y(t))$$
Geodesics on the surface are curves which minimize the arclength, which (assuming you are familiar with this) is the time integral of the speed:
$$L = \int\Big{|}\frac{d\vec{\alpha}(t)}{dt} \Big{|}dt$$
So, the geodesics have the property that they minimize this arclength integral (to be precise, there are some more technical details about completeness which I won't go into here due to your limited background). So, loosely speaking,
$$\vec{\alpha}(t) \ geodesic \rightarrow \vec{\alpha} \ minimizes\ L = \int\Big{|}\frac{d\vec{\alpha}(t)}{dt} \Big{|}dt$$
Now, consider your problem. Let the curve $\vec{\beta} = (x(t), y(t))$ give the x and y coordinates with time. The rate of change of temperature along the path is given by taking the time derivative of the temperature composed with the path parametrization:
$$\frac{d(f \circ \vec{\beta})}{dt}$$
So, the total change in temperature along the path is found by integrating
$$\Delta f = \int \frac{d(f \circ \vec{\beta} )}{dt} dt$$
So, in both cases, you're trying to maximize or minimize a given integral along a curve (these kinds of problems would fall into the domain of the calculus of variation). The problems are similar however, if i've read your question correctly, you're trying to maximize the change in temperature integral as opposed to minimize it. If you want, you could completely convert your temperature function into a surface, like you said, by using a Monge patch $\phi(x,y) = (x, y, f(x,y))$.

3. Mar 3, 2016

### RubinLicht

I will chew on this when I get more time or take a break from homework. Thanks.

I lied, I got curious so I read it. I think my question is exactly a geodesic problem, since, disregarding the fact that I said that it was a temperature model, I simply wanted to find out if the curve was the shortest path between a point on the surface and the "vertex" (assuming there is a better word).

Maximizing temperature change would be more along the lines of "find the shortest distance between this point and the equitemperature ellipse on the surface"

But yea, I understand geodesics more now thanks to you, but would still like to be provided with any solution curve that isn't one a line of symmetry. I am thankful that you have spent time answering this, and would really appreciate it if you could find a particular solution

Last edited: Mar 3, 2016
4. Mar 3, 2016

### ehild

Try to do what the problem says: find the path of the particle on the plate, that is, a 2D path. Find the temperature at the starting point (2, -3). Draw contour map, that is curves of equal temperature. How do you get the direction of the maximum change of the temperature? (Yes, it is the same as the direction of the gradient vector, which is perpendicular to the contour line. )

5. Mar 3, 2016

### RubinLicht

Yes, that is the solution to the Class question, but my question was whether or not that would also be the geodesic if you were to "project" the class question solution onto the 3D surface that represents temperature.

6. Mar 3, 2016

### ehild

Yes, the particle would move along a geodesic line on the 3D curve 4x2+y2+z=20.

7. Mar 3, 2016

### RubinLicht

So the geodesic line that passed through both the vertex and any given point is also the projection of the class problem solution for the given point onto the 3D graph?

8. Mar 4, 2016

### Ray Vickson

This is NOT a 3-dimensional problem; it asks for a path in the x-y plane. No fancy geodesics or variational methods are needed.

9. Mar 4, 2016

### Brian T

Of course it's not necessary, but op's asking out of interest how it's related. Here's another way you can think of it. Consider two contour lines (say f=a and f=b, b>a). You're currently on a point on the f=a contour. Assume you travel at arclength speed (one unit distance per unit time), then getting to contour b as fast as possible (i.e greatest rate of change of temperature) is the geodesic from contour a to a point on contour b

10. Mar 4, 2016

### Ray Vickson

There is a difference between time and distance. If the speed of travel is related to the function, the path that minimizes the time to get from contour 1 to contour 2 may not be a shortest-distance path.

For example, a light-beam shining from point A in one medium to point B in another, takes the shortest-time path, leading to refraction of the beam; but the shortest path would be straight, with no refraction.

11. Mar 4, 2016

### Brian T

I know, that's why I said assume arclength parametrization (as is usual done when discussing geodesics, since they travel at constant speed anyway)

12. Mar 4, 2016

### Ray Vickson

OK. I missed that.