The Marble and the Vase

A fairly common max/min problem asks the student to find the point on the parabola f\left( x \right)={{x}^{2}} that is closest to the point A\left( 0,1 \right).  The solution is not too difficult. The distance, L(x), between A and the point \left( x,{{x}^{2}} \right) on the parabola  is given by

\displaystyle L\left( x \right)=\sqrt{{{\left( x-0 \right)}^{2}}+{{\left( {{x}^{2}}-1 \right)}^{2}}}=\sqrt{{{x}^{4}}-{{x}^{2}}+1}

And the minimum distance can be found when

\displaystyle \frac{dL}{dx}=\frac{4{{x}^{3}}-2x}{2\sqrt{{{x}^{4}}-{{x}^{2}}+1}}=0

This occurs when x=0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}. The local maximum is occurs when x = 0. The (global) minimums are the other two values located symmetrically to the y-axis.


Somewhere I saw this problem posed in terms of a marble dropped into a vase shaped like a parabola. So I think of it that way. This accounts for the title of the post. The problem is, however, basically a two-dimensional situation.

In this post I would like to expand and explore this problem. The exploration will, I hope, give students some insight and experience with extreme values, and the relationship between a graph and its derivative. I will pose a series of questions that you could give to your students to explore. I will answer the questions as I go, but you, of course, should not do that until your students have had some time to work on the questions.

Graphing technology and later Computer Algebra Systems (CAS) will come in handy.


1. Consider a general point A\left( 0,a \right) on the y-axis. Find the x-coordinates of the closest point on the parabola in terms of a.

The distance is now given by

\displaystyle L\left( x \right)=\sqrt{{{\left( x-0 \right)}^{2}}+{{\left( {{x}^{2}}-a \right)}^{2}}}=\sqrt{{{x}^{4}}+\left( 1-2a \right){{x}^{2}}+{{a}^{2}}}

\displaystyle \frac{dL}{dx}=\frac{2{{x}^{3}}+\left( 1-2a \right)x}{\sqrt{{{x}^{4}}+2\left( 1-2a \right){{x}^{2}}+{{a}^{2}}}}

And \frac{dL}{dx}=0 when x=0,\frac{\sqrt{2\left( 2a-1 \right)}}{2},-\frac{\sqrt{2\left( 2a-1 \right)}}{2}

The (local) maximum is at x = 0. The other values are the minimums. The CAS computation is shown at the end of the post. This is easy enough to do by hand.

2. Discuss the equation {{L}^{2}}={{x}^{2}}+{{\left( x-a \right)}^{2}} in relation to this situation.

This is the equation of a circle with center at A with radius of L. At the minimum distance this circle will be tangent to the parabola.

3. What happens when a=\tfrac{1}{2} and when a<\tfrac{1}{2}?

When  a=\tfrac{1}{2}, the three zeroes are the same. The circle is tangent to the parabola at the origin and a is the minimum distance.

When a<\tfrac{1}{2}, the circle does not intersect the parabola. Notice that in this case two of the roots of \frac{dL}{dt}=0 are not Real numbers.

4. Consider the distance, L(x), from point A to the parabola. As x moves from left to right describe how this length changes. Be specific. Sketch the graph of this distance y = L(x). Where are its (local) maximum and minimum values, relative to the parabola and the circle tangent to the parabola?

The clip below illustrates the situation. The two segments marked L(x) are congruent. The graph of y = L(x) is a“w” shape similar to but not quartic polynomial. The minimums occur directly under the points of tangency of the circle and the parabola. The local maximum is directly over the origin. Is it coincidence that the graph goes through the center of  the circle? Explain.

Vase 15. Graph y=\frac{dL}{dx}  and compare its graph with the graph of y=L(x)

vase 4

L(x) is the blue graph and and L'(x) is the orange graph.
Notice the concavity of L'(x)

6.  The graph of y=\frac{dL}{dx} appears be concave up, then down, then (after passing the origin) up, and then down again. There are three points of inflection. Find their x-coordinates in terms of a. How do these points relate to y = L(x) ? (Use a CAS to do the computation)

The points of inflection of the derivative can be found from the second derivative of the derivative (the third derivative of the L(x)). The abscissas are x=-\sqrt{a},x=0,\text{ and }\sqrt{a}. The CAS computation is shown below

Vase 2a

CAS Computation for questions 1 and 6.


Stamp Out Slope-intercept Form!

Accumulation 5: Lines

Ban Slope Intercept

If you have a function y(x), that has a constant derivative, m, and contains the point \left( {{x}_{0}},{{y}_{0}} \right) then, using the accumulation idea I’ve been discussing in my last few posts, its equation is

\displaystyle y={{y}_{0}}+\int_{{{x}_{0}}}^{x}{m\,dt}

\displaystyle y={{y}_{0}}+\left. mt \right|_{{{x}_{0}}}^{x}

\displaystyle y={{y}_{0}}+m\left( x-{{x}_{0}} \right)

This is why I need your help!

I want to ban all use of the slope-intercept form, y = mx + b, as a method for writing the equation of a line!

The reason is that using the point-slope form to write the equation of a line is much more efficient and quicker. Given a point \left( {{x}_{0}},{{y}_{0}} \right) and the slope, m, it is much easier to substitute into  y={{y}_{0}}+m\left( x-{{x}_{0}} \right) at which point you are done; you have an equation of the line.

Algebra 1 books, for some reason that is beyond my understanding, insist using the slope-intercept method. You begin by substituting the slope into y=mx+b and then substituting the coordinates of the point into the resulting equation, and then solving for b, and then writing the equation all over again, this time with only m and b substituted. It’s an algorithm. Okay, it’s short and easy enough to do, but why bother when you can have the equation in one step?

Where else do you learn the special case (slope-intercept) before, long before, you learn the general case (point-slope)?

Even if you are given the slope and y-intercept, you can write y=b+m\left( x-0 \right).

If for some reason you need the equation in slope-intercept form, you can always “simplify” the point-slope form.

But don’t you need slope-intercept to graph? No, you don’t. Given the point-slope form you can easily identify a point on the line,\left( {{x}_{0}},{{y}_{0}} \right), start there and use the slope to move to another point. That is the same thing you do using the slope-intercept form except you don’t have to keep reminding your kids that the y-intercept, b, is really the point (0, b) and that’s where you start. Then there is the little problem of what do you do if zero is not in the domain of your problem.

Help me. Please talk to your colleagues who teach pre-algebra, Algebra 1, Geometry, Algebra 2 and pre-calculus. Help them get the kids off on the right foot.

Whenever I mention this to AP Calculus teachers they all agree with me. Whenever you grade the AP Calculus exams you see kids starting with y = mx + b and making algebra mistakes finding b.

Show me the Math!

Is God a Mathematician? by Mario Livio begins

When you work in cosmology … one of the facts of life becomes the weekly letter, e-mail, or fax from someone who wants to describe to you his own theory of the universe (yes, they are invariably men). The biggest mistake you can make is to politely answer that you would like to learn more. This immediately results in an endless barrage of messages. So how can you prevent the assault? The particular tactic I found to be quite useful (short of the impolite act of not answering at all) is to point out the true fact that as long as his theory is not precisely formulated in the language of mathematics, it is impossible to assess its relevance. This response stops most amateur cosmologists in their tracks. … Mathematics is the solid scaffolding that holds together any theory of the universe.

Is God a Mathematician? discusses the question of whether mathematics was invented or discovered. Dr. Livio’s other popular books include The Accelerating Universe (cosmology), The Golden Ratio: The Story of Phi, the World’s most Astounding Number, and The Equation that Couldn’t be Solved: How Mathematical Genius Discovered the Language of Symmetry. All are excellent reads for teachers and students.