Parabolic mirrors. So this is a particular shape - parabola shape - and these are used to focus
signals. So they can be used in satellite dishes, that's why they have this shape, it's a parabola.
You can use them in listening devices, to focus a weak sound signal - it's used in espionage; and
then amplify the signal that you're getting. It's used in torches, spotlights, to focus the beam.
Really cool stuff and we're going to go through the maths of it, the geometry, to see how that
focusing works. Like why does it have to be a parabola shape to get this focusing behaviour?
All right, so big parabola, something like that. So let's say this is x, this is y, and my parabola
here is going to be - oh I've drawn it this way, so it's y squared equals 4ax is the equation. So
you might be used to it being- y equals x squared would be that way up; because I've rotated it. So
let's now take point in our parabola, let's call this point P and this is going to be a distance x
on the x-axis. This is going to be at the position a times t squared. Now a is important because a
here is controlling how steep the parabola is; so as we make a bigger these sort of become steeper
closer, and as we make a smaller these become narrower. So a just controls the shape so we're
covering all possible parabolas by including it. Of course we could just set it to 1 but let's be
general. And then t, as we increase and decrease t we slide up and down the x-axis. Now if this is at
x equals at squared then we can work out the y value of this. So if y squared is 4ax then I think
that's going to mean that y can be plus or minus 2at and there's going to be a -2at down
here. We're going to focus on the plus point. So they're the two points on the parabola when x is
this satisfying that equation. Now, how a parabolic mirror works is you have light, sound, coming in
and the parabolic shape acts to focus it onto a single point. So that's how you get your TV signal;
you have all these bits of the signal, they get focused, give you a nice strong signal; same for
listening devices, torches, whatever it might be. You can even use them to collect sunlight and create
a little bit of heat a bit of fire, a bit of fire if you're in the desert for example. I am going to claim that this
is going to go to a point on the x-axis (a,0). So a distance of a inside parabola - and again remember
a just controls the shape. So obviously this will move as the shape changes, as you would expect but
it's definitely at y value 0 so it's on the x-axis. So if we're going to argue that this is
the case then what this is going to mean is we're going to have a wave - light wave/sound wave - coming
in horizontally and we need it to bounce like this. Now we could have another one - say it was here like
this - coming in horizontally, hitting here and I'm claiming it's going to bounce to that same point.
That's what we want, that's the whole purpose of the parabolic mirror. All these horizontal beams
coming in, bouncing off that, and they all converge at the same point; amplifies your signal, gives you
your heat/your light/your sound whatever you want. - (Brady: So that's like where you'd put maybe)
(your sensor or your receiver or-) - Exactly that that's where your sensors or- or your your bulb in a
torch. So- because for a torch it bounces out this way and then you get a nice perfect horizontal
beam. So if you if you have a torch you get like a focused beam- - (Oh imagine what that might look like!)
It's quite a focused beam, okay, it sort of expands a little bit but it's relatively focused compared
to like the light on the ceiling or you know a studio light that we have over here. The light
just kind of goes everywhere it's not in a focused beam. Because the studio light is just a load of
bulbs putting in all directions; here what we're doing is the bulb is pinging everything so it
comes out in this focused beam - it's how spotlights work. - (Where's the parabola there?) - So the parabola
is going to be inside a bit, just inside there. Quite hard to tell it's exactly a parabola,
but if you measured it I can guarantee you it would be. It's having the same the same effect.
So if we're going to show how this works mathematically we need to do some geometry.
Because what we know from physics; and we do this all the time as mathematicians, we borrow
from laws of physics. We know that the light coming in, the angle here - and this is an angle
with the tangent to the parabola right. So we draw the gradient of the parabola there, put
its tangent in, we know the angle there is the angle there. Because it comes in and reflects at
the same angle. It might not look like it on my diagram - this is a sketch. So what we need to do,
what we need to show mathematically, is that this angle here and this angle are the same. Because
if we can show that, if we can show the angle coming in and coming out based on going from
a general point always to here that would prove, from a mathematical standpoint, that everything is
going to be focused at this point. - (That same theta applies) - Yes exactly, exactly. So so the angle will
change; this theta will be different to the theta up here, but the key thing is in both situations
- so that's- call that one theta 1 and let's call this one theta 2. So in both situations we know it
has to come in and go out at the same angle, that's just- that's the law of physics we have
to follow. But the key thing is it's coming in at a certain angle, bouncing out at the same one, going
to the focus - our amplifier. This one comes in at a different angle, bounces out at the same different
angle, but again it ends up at the same point. And if I had another one here it would come in, bounce
out. Every time, no matter what angle it comes in at, it goes out at the same one but always into the
same point. - (Is this only for beams that are coming in at a straight line?) - This is only beams that come
in horizontally, yes. So it is just for horizontal beams coming in. Now of course there's going to be
some error, there's some you know- I don't know what to call it, noise, in some of these signals but
this is the sort of theory behind the parabolic shape. It's only the parabola that will focus to
this point and this is what we're going to show. - (And if your target's very far away like a star or)
(something in space, for all intents and purposes they all are coming in at the same straight line.)
Exactly yeah, yes. So this is- it also is used in telescopes exactly for that. Because stars,
they're so far away you need to amplify the light that's coming in so you have to focus it all into
the same point; so you use again parabolic mirrors to do that. So the way we do this is going to be
using vectors. We want to describe our positions; so this is the focus, our Point P, we want to be able
to describe these with vectors and construct an equation to argue why these two angles are
the same. So to start with we're going to assume they're not the same. We're going to work out this
one, so let's call this theta 2 hat and this one theta 2. Just to be clear but we're going to end up showing
that theta 2 hat is theta 2 where specifically one of them is the horizontal coming in and one
of them is going to the focus. And the key aspect to this is remember I introduced this parameter
t which meant there was nothing special about the point I picked. Because as I vary T the whole
problem just moves around; so that's where we get the generality and that's why we are ultimately
showing how these things work. First of all we need to know the tangent to the curve up here. We
fortunately have an equation; we know that y squared is equal to 4ax, we want to know the gradient of
P, so the gradient at that point is found by differentiating our function - dy/dx. So we
know the coordinates of P as well don't we, we know that's at squared and y is 2at. So we have to do
implicit differentiation which tells us 2y dy/dx equals 4a. So that's just differentiating
both sides with respect to x. And then we rearrange that so that tells us that dy/dx is 4a
over 2y. We want to know it at this point where this is x and this is y so we sum in that y value
into here. So we can say the gradient dy/dx at the point P is 4a over 2 times the y value
there which is 2 at. 4a over 4at is 1 over t. So the first thing to check is: is that
sensible? As I change t my gradient here changes. So this is our gradient at this point: now as I
decrease t I'm moving closer and becomes steeper. Right here it's vertical, so when t is naught I've got
1 over naught, that's tending to infinity which is a vertical gradient. So we see that the gradient
actually gets steeper as we move in this way, and it's going to get flatter as I move out here. So it
does make sense that it would come out like this, but it's good that t is in there because we want
this to be general for any point in the diagram. So now we've got the gradient at the point P,
we now want to work out well what is this direction vector here between this line and this?
So we're trying to get theta 2. So we want the vector from P to our focus which is (a,0). So a vector
between those two points is just the vector here minus the value here. Doesn't matter whether we're
going out or in, we just need something in that direction. So the vector from P to that point is
simply just P minus that; so that's P subtract that, so that vector will be- let's call it l. So we can
say l is going to be the x value, at squared minus a, that one minus that one and then that y value
minus that one. So this is a vector, x component y component, which describes this direction here. Now,
we need a vector to describe the direction of our tangent, so let's call that t. So the tangent vector
t: we know dy/dx is 1 over t, this is the change in y over the change in x. So this tells
you that every time x changes by t, y changes by 1. That's what a gradient means basically. So if
we put every time x changes by t, y changes by 1. So our tangent vector is given by this: (t, 1)
corresponds to this particular gradient up here. So what we've now got is we know the direction
here, and we know the direction here, as vectors so if we take the dot product of those two vectors
gives us the angle. I'm just looking, you're nodding like, okay if you say so Tom. Okay so let's
do that. So if we do l dot t, then by definition the dot product is the length of your first vector
times the length of your second vector times cos of the angle between them - so that will be theta 2.
Because these are our two vectors, angle between them there it's theta 2. It could of course be
the same one up here but we're- doesn't matter for our purposes, it's the same angle. Right so we
sub these in. Right so, l dot t is size of l; so when you work out the size of a vector you square each
term, add them together and then square root. So the size of l is this first one squared, at squared
minus a squared, plus second one squared 4at squared t squared. So that is the size of l. Size of
t a bit easier thankfully, it's the square root of the first one squared plus the second one squared
and then we've got cos theta 2. Now of course we can do the dot product, we've got vectors so you
do the first components multiplied together added to the second two multiplied together.
So that tells us that one times that one, so at cubed minus at; plus that one times that one,
plus 2at. Equals the two square roots times cos theta 2. Because what we want is cos theta 2; so
if I rearrange cos theta 2 is equal to this thing - we've got an -at plus 2at so that gives me
an at cubed plus at on the top. And then we've got these things on the bottom. Now I'm going to
simplify it a little bit up here; so if we expand out the square, a squared t to the 4, minus two
lots of the cross term so -2 a squared t squared plus a squared and then plus 4 a
squared t squared. So we've got -2 lots of it plus 4 lots of it, so that becomes a plus
2. So we've got a squared brackets t to the four plus 2 t squared plus 1 in that first bracket. And
then we've got second bracket, t squared + 1. Now this in here, this is t squared plus 1 squared.
Expand that out, t to the fourth plus two cross terms plus 1. So that's that. So you've got an
a squared times something else squared, we lose the square root. So what we finally end up with as
cos theta 2 is - we've got an a factor here as well, an at factor at the top.
at times t squared + 1 divided by square root of this, so it gives us a times t squared
plus one times the square root t squared plus one. And now everything cancels which is lovely, cancels
cancels cancels cancels, and we're left with t over the square root of t squared plus one.
So we have a formula for cos of theta 2 - this angle. What we now need to show is that we get
the same formula for theta 2 hat and then we've figured it out. Yeah another piece of paper.
So to work out cos of theta 2 hat here, same approach, let's figure out the vectors coming
in. So we've got t is one vector and the other one is a horizontal vector here, h. And we want
the angle between the two of them, so we want the dot product between t and h. So we know
what t is, we figured that one out to be (t,1). h is a horizontal vector so that means x changes
and y doesn't so it's just (1,0), it's your basis vector. So when we take the dot product
t dot h is equal to that one times that one: t plus (1 times 0) is uh 0, add them
together. That is then equal to the size of t times the size of h times cos of the angle
between them which is that one - theta 2 hat. So so we've got t is equal to the size of this,
square root of 1 + t squared; size of h, well that's size 1; times cos of theta 2 hat. So we
rearrange and say so cos of theta 2 hat is t over square root 1 + t squared. But what did we get
cos of theta 2 was? t over the square root of one plus t squared. So that allows us to conclude this
is the same, it's cos of theta 2, so we can take our cos of both sides. So in other words theta 2
is theta 2 hat which is exactly what we wanted. So rather than just taking my word for the terrible
sketch, we've been able to show mathematically using vectors using geometry the angle coming
in is the same as the angle going out, where we're coming in horizontal and we always bounce to the
focus. Because there was nothing special about the point P, I could just change the value of t
and go here. The whole working out works; so we've actually showing that any horizontal
ray coming in and hitting our parabolic mirror will always go to that central point because
these two angles are always the same, and light reflection or sound wave reflection means the
angle in needs to be the angle out. So that's how parabolic mirrors work. - (Lovely)
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