## Trying to find math inside everything else

### Parallel to a Parabola?

A while back, I was working on a lesson about average rate of change and wondered the following question: “Could you use the word ‘parallel ‘to describe two non-linear functions that have the same rate of change/don’t intersect?” Jonathan’s response, though, made me think about what it actually means to be parallel. Often when you ask students, they will respond “two lines that never intersect,” which I usually push back against because 1) how do you know they never ever intersect? and b) skew lines never intersect, either. So when I explain parallel lines, I use the fact that they have the same slope/go in the same direction as the actual definition, which has the consequence of never intersecting. So I looked it up on Wikipedia.

Given straight lines l and m, the following descriptions of line m equivalently define it as parallel to line l in Euclidean space:

1. Every point on line m is located at exactly the same minimum distance from line l (equidistant lines).
2. Line m is on the same plane as line l but does not intersect l (even assuming that lines extend to infinity in either direction).
3. Lines m and l are both intersected by a third straight line (a transversal) in the same plane, and the corresponding angles of intersection with the transversal are congruent. (This is equivalent to Euclid‘s parallel postulate.)

I don’t think statement 3 was particularly useful to me, but the idea of being equidistant was interesting. A vertically shifted parabola is not equidistant from the original – though they never touch, the distance between them gets smaller and smaller.

So that raised the next question – how do I actually measure the distance between two parabolas at a given point? I asked my boyfriend and he responded, “Well, you definitely need calculus….” And who better to swoop in and help with that than Sam Shah. So now that I know how to find the minimum distance between two functions, all I need to do is find a function that whose minimum distance to my original function is constant, and then I should have something that you could call parallel.

I made a little Desmos graphs with sliders, to help me visualize the process (click to access): So I have the equation of the perpendicular line $y = \frac{-1}{f'(a)}(x-a)+f(a)$

But that wasn’t really helping me see what the parallel function would actually look like. So then I turned to Geogebra. I needed to make a point on the perpendicular line that was a certain distance away from the function, say a distance of 1. So to figure out the coordinates of that point (x,y), I just used the distance formula, plugging in y from above.: $\sqrt{(x-a)^2+([\frac{-1}{f'(a)}(x-a)+f(a)] - f(a))^2} = 1$

That gave me the coordinates of the point that is a distance of 1 away from f(x) at a: $(a + \frac{f'(a)}{\sqrt{1+(f'(a))^2}},\frac{-1}{\sqrt{1+(f'(a))^2}}+f(a))$

So I made that point in Geogebra and activated the trace, which gave me this: Lastly, I thought, well, what exactly is this function that I’ve traced? It looks kinda quartic, but that can’t be, because any quartic like this would intersect the parabola, right? So I tried to write the function for it, using parametric equations. Using $f(t) = t^2$, I made the parametric equation $(t + \frac{2t}{\sqrt{1+4t^2}},\frac{-1}{\sqrt{1+4t^2}}+t^2)$.

I tried to plug that into Wolfram-Alpha to get the closed form, but it was a mess, so I still don’t really know what the closed form would look like. But who says a parametric form isn’t a solution?

#### Comments on: "Parallel to a Parabola?" (4)

1. I don’t think you’re going to get a closed-form solution there: you’ve also got the “inside” distance-1 curve which ends up looping back on itself, so it won’t be a function of x.

I came across a more general version of your question a while back: instead of finding the shortest distance from g(x) to the single point (a, f(x)), what is the shortest distance between g(x) and *any* point on f(x)?
I’ve lost my work for that exploration, but I remember stumbling across an explanation for why the point with perpendicular slope is closest: imagine enlarging a circle from your point until it is tangent to g. The radius of this circle will be your smallest distance, since it is the first distance from your point to encounter the function, and the tangent of a circle is always perpendicular to the radius!

It was fun generalizing your Desmos graph. I added a slider for the distance, and parameterized both the ‘inside’ and ‘outside’ curves for arbitrary functions. I like looking at the sinusoidals — when you hit ‘play’ it looks like they layout for a roller coaster!

2. This investigation is really similar to one that came up at the Math Forum recently! We had an old Calculus Problem of the Week about a rug in the shape of an ellipse, that had a border with a width of foot. All that we wanted to know about was the area of the rug, but one of the students wondered, can both the inner and outer ovals be ellipses if the width of the border is a constant 1 foot. I made a sketch to illustrate it here: http://mathforum.org/~max/ovalrug.html or you can get a trial account to the Math Forum and then read all about the whole problem here: http://mathforum.org/pows/documents/packets/packet990.pdf

It was really neat to find such similar thinking about perpendiculars to tangent lines for other conic sections!

3. Excellent write-up. I definitely appreciate
this site. Stick with it!

4. Steve Wait said:

From the better late than never files, this may be of some help for an offset or parallel curve to the Parabola.

In the case of:
y = a x^2

Xd = x ± ( ( 2 d x ) / ( 2 √ ( .25 a^2 + x^2 ) ) )
Yd = y ± ( ( .5 d a ) / ( √ ( .25 a^2 + x^2 ) ) )

Where:
x = coordinate for point on the Parabola
y = coordinate for point on the Parabola
a = dilation coefficient of the Parabola
Xd = coordinate for point displaced normal to a tangent of the Parabola
Yd = coordinate for point displaced normal to a tangent of the Parabola
d = displaced amount