Welcome to another Mathologer video. Have a look at this. What do you see? Yes, some cool Mathologer toys in the background but ignore those. Yes, I know, it's hard but ignore the toys :) Focus on the black UFO
at the bottom. On top of the UFO is a circular mirror and on top of the mirror
is a coin featuring pretty Princess Leia and her trusty robot r2d2. Now let me
pick up the coin. What!? My fingers are passing right through the
coin. Is this some kind of Jedi mind trick? What I want to do today is to explain
our ghostly Princess Leia as well as a closely related way to conjure up
ghostly voices. It's really cool to be able to create these ghosts but today's
maths is also super applicable. In fact, if this maths ceased to exist from one day to
the next you wouldn't recognize the world you live in anymore. Well let's get
to it. It's all got to do with our high school friend
good old x-squared. Sticking with movie references, x squared is a little bit
like Clark Kent. Most of Clark's friend think they know
all as to know about him but only a select few are aware that he's actually
Superman. Same with x squared which also has some hidden superpowers that hardly
anybody knows about. Okay, here we go. Did you know that the point (0,1/4)
and the horizontal at y= -1/4 are super special for the
parabola y = x^2. The point is called the focus and the line
is called directrix of the parabola. What's special is that every point on
the parabola is exactly the same distance from the focus and the
directrix. So these two distances there are always the same.
That looks tricky but to show it we just need help from our other school friend
Mr. Pythagoras. Here we go. If our parabola point has coordinates x and x
squared, then Pythagoras tells us the square of the green distance. And going
straight down to the directrix the square of the yellow distance is this.
Now it's just a matter of going on algebra autopilot to check that these
two expressions are equal. And that shows the two distances are the same.
Easy-peasy. The directrix is the secret ingredient for lots of parabolic magic.
For the first magic trick, let's position the parabola on a piece of paper so that
the red directrix coincides with the bottom edge of the paper. Now look at any point on the bottom edge, that one there. Fold the paper so that the black point ends up on the focus. So there fold, fold, fold. Right
on top. And unfold again. Okay it looks as if the paper crease is a tangent of the
parabola and that the touching point is right above the black point. And looks
are not deceiving. Starting with any point at the bottom, folding results in a
crease that is tangent to the parabola. If you do this for all the points of the
directrix, you get all the tangents of the parabola. So why does this work?
All those tangents suggests calculus but you really don't need it. All you need is
a little middle school OWL-gebra :) Anyway I'll leave it as a challenge for
you to give a proof in the comments and if you're desperate
I'll give one possible proof at the very end of this video. Anyway for the record
let's note that proving the second super property of the parabola is also easy
peasy. The second property gives a really pretty way to create a parabola from
scratch without having to calculate anything. Start with a piece of paper,
mark a point close to the middle of one of the sides and perform the folding
action for a bunch of points on the side. Then the parabola materializes as if by
magic. Super super nice :) Okay after this little piece of paper magic we're almost
ready to conjure our ghosts. We just need one more super property of the parabola
and actually you probably all know this one, although I'm guessing that only a
few of you will have seen a proper explanation. Looking again at our paper
folding notice that this green triangle there
gets folded smack on top of this identical pink triangle and that means
that the green angle and the pink angle are the same. Right?
Then the angle opposite the green is also of the same size. Now for what we
are after we just need that these two angles here are the same. We also don't
need the directrix anymore, so let's get rid of that too. There's not much left of
our picture but it tells us something super interesting.
Imagine the parabola is a mirror and the vertical line is a ray of light hitting
the mirror. Then this ray of light will be reflected like this and the reflected
ray will pass through the focus. But of course the same is true for any vertical
ray of light and so all the vertical rays get focused on the, well, focus :) Lucky
that that's what we chose to call it. Of course, this also works in the opposite
direction: any ray emanating from the focus will be reflected into a vertical
ray. I'm sure that many of you are aware of this focusing property of the
parabola and it's myriad applications in the guise of parabolic reflectors and
mirrors. Now finally we are ready to conjure some ghosts. To begin let's add
another parabola to this picture like so. Then it's clear that a whole sector of
rays emanating from the red focus will end up passing through the green focus.
Chances are you've seen the setup before in the guise of the mysterious
whispering dishes at science museums. A whispering dish is exactly a circular
paraboloid, the shape you get when you spin a parabola around its axis of
symmetry. As such the paraboloid inherits all the nice reflective properties of
the parabola. The whispering dish in the picture is located at Scienceworks the
Science Museum in my hometown of Melbourne. The focus of the dish is
located inside the ring I'm pointing at. Okay, now take two of these dishes and
place them 50 meters apart. Then if the junior mathologeress Lara
whispers at the green focus Mathologer junior Karl will hear her disembodied voice at the read focus. Really quite a stunning
effect. It's a great experiment but what's not
so great is the explanation on the whispering dish. What it says there is:
"The other person hears you clearly because the curved shape of the dish
focuses the sound into the ring at their end." Pretty damn nothingy, isn't it?
Well most science museums try a little harder and at least feature this
suggestive drawing here but there's one very obvious question about this effect
and it's a question that is seldom asked: Why isn't a sound muffled? Specifically,
why doesn't a sound wave leaving the green focus in different directions, then
arrive at the red focus at different times. Well that amounts to asking
whether all the yellow parts in this diagram are the same length.
They don't look it but surprisingly they are. And there's an easy explanation just
using our focal-directrix super property. Don't believe me? Just watch! Let's bring
back the two directrices of the two paraboli and let's take a careful look
at one of the yellow paths. What can we say about the length of this path? Well,
let's see. Because of the focus-directrix super property the red distance from the
focus to the reflecting point on the parabola is the same as the distance
from the reflecting point to the directrix. And, of course, the same is true
up on top. But this means that the length of the path from the red focus to the
green focus is exactly the distance between the two directrix lines. And
since this is the case for all paths all paths have the same lengths. How easy and how pretty a proof is that. This equal length property is also important for
many other really significant applications of parabolic reflectors, but
strangely unlike the focusing property the equal lengths property is rarely
mentioned by anybody. Okay, now what about ghostly Princess Leia? How do we conjure
her? For this we use proper parabolic mirrors
and instead of moving them apart, we move them close together. We
then place Leia and her robot friend in the middle of the bottom mirror, at the
green focus. Then we cut a hole in the middle of the upper mirror just above
the red focus. Then a hologram of Leia materializes at the second focus. Now, this is real mathematical magic :) !! And on that happy note I will declare I am NOT happy. It's time for a Mathologer
sermon. These days here in Melbourne my junior Mathologers Karl and Lara seem to
spend half their time in maths class torturing quadratics but they never get
to see any of the beautiful maths I've shown you today. Much less figure out why it works. If they didn't happen to have an annoying Mathologer for a father
they'd never find out about any of this. Well except for the science museum's
explanation which turns out to be a masterpiece of explaining nothing. This
is especially puzzling and especially annoying because the simple maths that
you need to explain all this super important and super applicable stuff
properly is exactly the school maths that it's done ad nauseam. At the same
time Victoria's maths textbooks are
chock-a-block with pseudo applications like parabolic bike paths, quadratic
types of cheese, and so on (I'm not making this up :) And this is just the tip of a
parabolic iceberg. As my colleague Marty likes to say: our educational authorities
never miss an opportunity to miss an opportunity. I'd be very interested in
finding out from you guys what's the state of educational affairs where you
are. Do kids learn about the things I talked about today in maths class.
Properly? At all? Let us know in the comments. And that's it from me for today.
Okay, let's end on a satisfying mathematical note and tie up a loose end.
Here's a simple explanation for why our paper crease from before just touches
the parabola. If you want to figure this out for yourself now is the time to
avert your eyes and ears. Well, it's easy to read the equation of the yellow
creased line off this diagram. For this we just need its slope and its y-intercept.
The slope of the creased line is 2a, the negative reciprocal of the slope of the
thin segment and it's green y-intercept is minus a squared (may take you a second or two to convince yourself of this.) And so we get this for the equation of
the line. And then where does this crease line hit the parabola? Well we just
equate the line with x squared and solve for x. Okay, so there's exactly one
solution at x=a and that means that the crease touches the
parabola at exactly the point we predicted. All is good, no loose ends,
I'm happy again and we'll all be able to sleep tonight. And that's really it for
today :)
And I thought I knew everything there is to know about x2.
I like the name of the segment from one "side' of the parabola to the other which passes through the focus and is parallel to the directrix.
3Blue1Brown and Mathologer posting on the same day. That's like a blue moon for math.
Let F(0,1/4) be the focus of x2 and P(t,-1/4) be the point we wish to fold onto F. Consider the point A(t, t2) which is the intersection of the parabola and the vertical line going through P.
In the triangle AFP, we send down a height from A to FP, intersecting at a point B. Since AF = AP the height AB is a perpendicular bisector for FP and thus the line of symmetry for the fold.
Consider B, the midpoint between P and F, and take the average of P and F to find B(t/2, 0). Finding the slope of AB gives us m=2t, which means it is indeed the tangent line to x2 at x=t.