CTK Insights

13 Sep

The shortest path between two points in a plane

Number Crunching

One of the first challenge problems Paul H. Nahin offers in his new book comes from his experience as a freshman at Stanford. This is a nice yarn.

When I was a freshman at Stanford I did well enough during the first two terms of calculus to be allowed to transfer into the honors section of the course. (That's when I found lots of my fellow freshmen were at least as good at math as I was!) One of the homework problems in that course (Math 53, Spring Quarter 1959) was the following: prove that the shortest path length between any two given points in a plane is that of the straight line segment connecting the two points. This is, of course, "obvious" to anybody with a body temperature above that of an ice cube, but the point of the problem was to construct a proof.

Nahin the freshman hit on the idea of using the arc length integral

L=\int^{a}_{b}\sqrt{1+(f')^2}dx

He reasoned that if a curve is not a graph of a function it could be made such by choosing a proper system of Cartesian coordinates. The length of a curve does not depend on the coordinate system and thus could be computed in any suitable one.

Now it is clear that to achieve the required minimum the curve needs to have f' = 0 throughout, meaning that it is a straight line.

The homework has been returned with a big red "NO" and a grade 1 (for trying) out of 10.

The problem is to answer the question, "Why? What was wrong with Nahin's solution?"

Curiously, 50 years later Nihin located his old professor - John Lamerti - and turned to him to explain where Nahin the freshman went wrong.

In his gracious reply the professor admitted that a score of 1 out of 10 seem unreasonably negative and found in the submitted homework some justification to raise the score to 7 out of 10 for a good but not quite right try.

He also pointed out that it is not always possible for a curve to be represented in the form y = f(x) and suggested that the problem was given at the time they discussed parametric equations so that a reference to the parametric form

L=\int^{a}_{b}\sqrt{\dot{x}^{2}+\dot{y}^2}dt

would have been corrected and expected.

As far as I am concerned, my difficulty was not whether to use the parametric form but in the definition of the path. If the problem has implicitly assumed that the paths in question had to be smooth - which may be so due to the course context - then the parametric form was certainly the way to go. But the problem formulation does not say that much and the author does not mention that until a few steps into the yarn. Still, it was a pleasure to read that story.

References

  1. P. H. Nahin, Number-Crunching, Princeton University Press, 2011

2 Responses to “The shortest path between two points in a plane”

  1. 1
    Scott Brodie Says:

    This story reminds me of a similar problem which I encountered as an undergraduate:

    Can one "prove" using the techniques of modern Calculus the second of following Axioms of Archimedes:

    1. Of all lines which have the same extremities, the straight line is the least.
    2. Of other lines in a plane and having the same extremities (any two) such are unequal whenever both are concave in the same direction and one of them is either wholly included between the other and the straight line which has the same extremities with it, or is partly included by, and is partly common with, the other; and that [line] which is included is the lesser of the two.

    I never found a proof in terms of the simple function representation and arc-length integral mentioned above. For a complete discussion, see:

    Archimedes' Axioms for Arc-Length and Area
    Author(s): Scott E. Brodie Source: Mathematics Magazine, Vol. 53, No. 1 (Jan., 1980), pp. 36-39Published

    Scott.

  2. 2
    admin Says:

    Hi, Scott:

    I too do not remember seeing a proof that only depends on the arc-length integral. On the other hand, as I mentioned in the post, my quandary was elsewhere: why the restriction to smooth paths? Polygonal paths should have been a legitimate object of investigation. A proof of an analogue of 2. for convex polygons - one inside the other - does not use anything beyond a plain integral and is based on this statement: if the sum of projections of one group of vectors on any direction is greater than such a sum for another group of vectors then the sum of lengths of the first is greater than the sum of the lengths of the second. (V. V. Prasolov, Essays on Numbers and Figures, AMS, 1997, 27-32)

    To me it appeared strange that polygonal paths have been excluded from consideration.

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