Bridges 2014 — Math and Art

bridges 2014The 2014 Bridges Math and Art conference will be held August 14th through 19th in Seoul, South Korea.

The Bridges organization has been hosting this international conference highlighting the connections between art, mathematics, and computer science since 1994.  I have participated in several Bridges conferences, and my experiences there have greatly influenced me as a mathematician and a teacher.

I am proud to have two photographs in this year’s art exhibit.  You can see my work here, and you can peruse the entire 2014 Bridges Art Gallery here.

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By MrHonner, ago

The Closest Point on a Parabola to a Point on its Axis

closest point to x^2While playing around with distances from points to graphs, I discovered the following interesting property of parabolas.

Suppose you have a standard parabola, y = x^2, and a point on its axis of symmetry, (0,a).  Then the point on the graph of the parabola closest to the point (0,a) either has y-coordinate equal to a - \frac{1}{2}, or is the vertex of the parabola.

A calculus-based proof is fairly straightforward.

The distance between a point on the parabola, say (x, x^2) and the point (0,a) is given by

d(x) = \sqrt{x^2 + (x^2-a)^2}

To find the minimum distance, differentiate d(x) to get

d'(x) = \frac{2x + 2(x^2-a)2x}{2\sqrt{x^2 + (x^2-a)^2}}

This derivative is undefined when the denominator is zero, which only happens when the point (0,a) is the vertex.  The derivative is zero when

2x + 2(x^2-a)2x = 2x ( 1 + 2(x^2 - a)) = 2x(1 + 2x^2-2a)=0

By the zero-product property, either x = 0 or 1 + 2x^2-2a = 0.  

The critical value x = 0 corresponds to the vertex of the parabola.  The distance from (0,a) to the vertex is a minimum if a \leq \frac{1}{2}, or a relative maximum when a > \frac{1}{2}. 

Now, if 1 + 2x^2 - 2a = 0, we have x^2 = a - \frac{1}{2}.  Since y = x^2 on the parabola, the point associated with this critical value has y-coordiante a - \frac{1}{2}.  Some simple analysis shows that the distance to this point is minimal.

As an alternative to optimization, one could approach the problem geometrically.  Since the shortest path from a point to a line is the perpendicular line segment connecting them, the shortest path between a point and a curve should be a line segment perpendicular to a tangent to the curve.  At a point on the parabola (x,x^2), the slope of the tangent line is 2x, and so the slope of the line perpendicular to the tangent at (x,x^2) would be -\frac{1}{2x}.  This is equivalent to -\frac{\frac{1}{2}}{x}, and since x is the distance to the axis of symmetry, it follows that this line intersects the axis of symmetry at a point a half-unit up in the y-direction.

This fun and surprising result would make a nice exploration in a Calculus class.  It easily extends to general parabolas, and the situation naturally suggests some compelling extension questions.  

You can explore the problem, and create variations, in this interactive Desmos demonstration I put together.

By MrHonner, ago
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