## 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?