I'm here in London, in the middle of the city, and behind me you can see the amazing building at 20 Fenchurch Street.

you can see that phenomenal curved front on it.

and when they were originally putting it together,

during the construction

something mysterious started to happen.

Nearby cars started to melt;

paint was blistering,

a bike seat caught on fire,

someone unexpectedly scorched their lemon.

and that's not even a euphemism.

You're in London,

you're out taking your lemon for a walk.

sunlight is hitting it, but you're not worried.

even though sunlight carries a lot of energy,

not enough is hitting the lemon at once for it to become scorched.

But then

you walk past a reflective building

sunlight is now bouncing off the windows on that building onto your lemon, and it is

still fine. Not enough is hitting it at once.

Only if that building is accidentally curved

in a certain way

will it take sunlight over a large area

and focus in onto this one little lemony point

and then suddenly

there it goes

you've scorched your lemon.

Now I will admit the whole lemon thing is a bit ridiculous

The only citation I could find for it was a Daily Mail article.

But the BBC did confirm that the curved surface on the 20 Fenchurch Street Building

did melt someone's car.

And the Independent Newspaper reported that that fantastic curved surface

focused enough light

to burn this doormat

in a nearby barber shop.

Thankfully, the sun was only in the right position

to hit the curved surface of the buildling

and cause an actual problem

for a couple of weeks of the year

and even then

it was only an issue for a few hours each day

and even then

it was only an issue if it was actually sunny

which, in London,

is not a major concern.

And just to make sure it's completely safe

for that occasional sunny day

the building has since been fitted with a very large sunglass-style monocle, I guess

I mean, it is in London, after all.

Could it have been worse?

Was the 20 Fenchurch Street the worst possible shape to focus sunlight in?

Well, I couldn't find out what shape it actually is,

but I can confirm, having looked at it

for a while

that it's probably not the worst possible shape.

That would be a shape known as the paraboloid.

To make a parabaloid, you start with a parabola

and then you rotate it around the vertical axis.

and there you are,

instant paraboloid.

The great thing about a shape like this

is all the parallel light

that goes straight into it will be focused in

on one focal point.

It's incredibly useful.

If you want to focus radiation,

this is the shape for you.

So, something like a satellite dish

where you want to get lots of incoming radiation

onto one receiver.

Satellite dishes are actually paraboloids

as you can see on my satellite dish here

and if you're concerned that that is a ridiculously big satellite dish

of course, I'm kidding.

That's not my satellite dish.

THAT is my satellite dish.

Hypothetically, the 20 Fenchurch Street could have been worse

but surely, nobody is going to make a massive shiny structure

that is shaped like a paraboloid?

Here in Nottingham, at the Nottingham Playhouse Theatre

There is actually a massive paraboloid sculpture.

Over five metres tall

it's known as the sky mirror

you can see, it's actually not that wildly different to mine

there we go

in fact, on a sunny day

it's even been rumoured

that this thing

will set pigeons on fire.

Yeah, I may have bought some paraboloids

but let me explain.

I'm part of a nerd comedy group called Festival of the Spoken Nerd.

I'm approximately one third

and on our tour with the show "Just for graphs"

I actually have a paraboloid or two in the show itself.

We're actually about to do the final performance on the tour of "Just for Graphs"

We're here in the Queen Mary University of London's People's Palace

which is an amazing theatre

and given it is the last show on the tour

I thought I would pop down to the stage

and show you what we use these paraboloids actually for.

On the right hand side of the stage

I'm putting one paraboloid on a tripod

pointing over towards the left hand side of the stage

where I have a second paraboloid

pointing perfectly back at it.

On the right hand side of the stage

I have an incredibly bright heat lamp

and on the left hand side, I have some flash cotton

which is incredibly flammable

Fine, I'll turn the heat lamp on

oh goodness, right.

In theory, if I get this perfectly in the focal point on this side

it will focus all the light in on the flash cotton

and once it heats up

YES!

Thank you

Thank you for watching my video all about paraboloids

Now as I mentioned, I did do that demonstration

fire in our Spoken Nerd show

"Just for Graphs"

which, by the time you watch this, will be over.

but don't worry, I'm sure we're doing shows somewhere.

in our previous show

"Full Frontal Nerdity".

is available for download

and as a DVD

if you want to see more Spoken Nerd stuff,

just click on, well, actually,

the logo on the screen.

in the background

like there.

Good video this week everybody

we have the building that sets things on fire,

that is a fantastic hook to get people interested

oh, we had the hilarious lemon

the one that magically appears

and disappears

at random

there was satellite dishes

fantastic. Practical application. Good job.

we had the proof for why parabolas all focus

oh wait...

people. We forgot the proof!

Oh, my goodness

there's some very upset viewers, look

ok, look

ok

we will get a proof together

just give me a moment

I have a proof

now, interesting, when most people meet parabolas,

they come across them as quadratics

a plot an an πΒ² equation

however, parabolas are also intimately linked to the concept of a focal point.

one definition of a parabola

which ignores quadratics completely,

is that it's all the points which are the same distance

from another point

and a line

If you want to, I'll have a link to the example here

so you can have a play with it

and see how that works

but I wasn't happy with that

I wanted to show

starting with an πΒ² equation

can we prove a parabola

has a focal point?

but for that

it looks like we need

a chalkboard

and some working out

I'm going to start with a standard vertical π¦-axis

horizontal π-axis

and onto this

I'm going to put everyone's standard issue parabola

π¦=πΒ²

now

we can have a look at what happens

if we have a single beam of light

coming directly down

parallel to the vertical axis

impacts the reflective surface of our parabola

and goes flying off in a new direction

to see what happens

at that point of impact

you're going to have to zoom right in

and I'm talking uncomfortably close

into this point here

well, this really is uncomfortably close

but here we are - the point where it hits our parabola

I'm actually going to take off the arrowhead for now

just to make it a little bit clearer what's going on

and I'm going to draw the line

if it continues to go on straight down and hits the π-axis

and because we're going to do this in general, we're going to hit the π-axis at any point

π, which I do apologise, is slightly out of shot

there. We can now approximate the parabola because we don't care about how it curves over there

we only care about the gradient where it impacts

so we can replace that with a tangent

this pink line here

that is the tangent

right at that point where it hits

the line and it is now going to bounce off in an all-new direction.

and so, here comes our light beam, it hits the surface there

and it flies off such that the angle onto, I guess we could say, mirror

it goes on the same angle out, so there it goes off into it's new direction

to calculate the angle it's going on

we need to know initially what angle our mirror is

and we know that for a π¦=πΒ² parabola

that the gradient

is the derivative at that point

and the derivative of π¦=πΒ²

is 2π

and so the gradient is always twice what the value of π happens to be

so, let's label that in vaguely

so for every 1 we go across, we go up that little distance there

will be 2π

for whatever π may be

which means - let's go for blue -

in there

we have some angle π

now actually, we have got two straight lines here

we've got one straight line of the beam as if it carried straight on

the other straight line here which is the tangent

so we know that opposite angle over there

that must also be π

and that that angle in equals the angle out, relative to the mirror

so when it bounces off

the parabola there

this angle it makes with the tangent is also π

so in fact, it exits on an overall angle so that entire angle all the way around there

is simply twice whatever the angle of the tangent is at the point where it impacts

so where has that left us?

Well, what we really care about is the focal point

somewhere over here

when that beam carries on

it is going to hit right there

so let's call that point β

and what do we know? Well, we know where it started.

we know that when the beam hits, at point π

if you were to carry that across, by definition, it's a parabola

that point on the vertical axis is going to be the point πΒ²

so we need to know

how much lower

than πΒ² is it going to be

how far is it going to drop

or what is it's distance π

now bear in mind the drop is not a great way to put that because there are some points where the light would come in

here and then bounce back up

to the focal point, so it's not necessarily dropping

but I'm going to use π as a handy way to do distance

and I find that dropping is an easy way to imagine it

to get that π, we have to start with the gradient of our tangent, and we already know because it's the derivative

that the gradient, I'm going to use a triangle for that, and a subscript T

equals 2π

What we care about now is what's the new gradient of the beam flying off there

so we want to find what is the gradient again B for the beam.

To do that, we need to see what doubling the angle does to the gradient.

because we've gone from one angle from vertical

to the initial tangent

and then we've doubled that angle out to where the beam is

and for this, we're going to need a bit of trigonometry

ok, so originally we had that the tan (opposite / adjacent) 2π

the tangent of the angle is 1/gradient

in a situation like this

we're now going from tan of a single π

to tan of double that angle, 2π

and thankfully, there are a whole lot of double angle formulae

you just look up for any trigonometric function

if you double it's angle what is the result

and of course, you could derive these yourself

I highly recommend it

it's good fun

but in this case,

we're just going to look up

that if you double the angle

of tan

you end up with twice whatever that tan of π was

divided by 1-tan(π), all squared

and when we square a trig function we don't put the squared out here

because it looks like we might be squaring the angle

we hide it in there

it's a bit ridiculous

but it helps avoiding mistakes

okay, now we just plug this in

so that is going to equal

two lots of 1/2π, so 2/2π

under that we have 1 minus the square of that, so

1 squared is still 1, and 2 squared is 4πΒ²

to tidy this up, I'm going to multiply the top and the bottom by 4πΒ²

so I'm effectively multiplying it by 1

I'm not going to change it

but that means that that cancels out, that goes there

4π, that becomes 1

that's going to become 4πΒ²

4πΒ²-1

but that is our tan, that is our opposite over our adjacent

to get it back to just be the gradient, we simply flip it the other way up

so the gradient

of the beam

is definitely going to be

4πΒ²-1/4π

for those of you not used to using chalkboards,

I apologise for the sounds that the chalk makes,

it must sound horrific to your delicate whiteboard ears

Let's keep this derivation party going on the other side

so now we know

when the beam comes down

to any point π

and it hits the parabola,

it gets bounced off

it's new gradient is that

so that for every one step back

towards the vertical axis, it takes, it drops down by that much

and how many steps does it go? Well it goes a distance of π

and so we can work out

that drop, or π distance there

is the gradient

times the distance it's travelled

which is π, there's going to be some fantastic cancelling out going on there

it's simply going to equal πΒ²-ΒΌ

and finally, we're going to bring the whole thing home, what is that height there?

how far above the origin?

is our focal point? for any landing point π

well, it's πΒ² minus the drop

so that height is going to equal πΒ² minus the drop of πΒ²-ΒΌ

Hang on a second!

the drop is exactly one quarter of a unit less than πΒ²

it's less than that whole distance there

this all cancels out

the height above the origin equals exactly one quarter.

the truly marvellous thing about our final result

is even though we started at some generic point π

that could have been anywhere on the horizonal axis

it always hits β on the vertical axis

and there are no πs left in that result

that is independent of π

it doesn't matter where the beam starts

it could be over here, it could be over there

somewhere

all beams of light

parallel to the vertical axis

will all converge

and hit the same focal point right in the centre

and yes, we have just done this for the very simple π¦=πΒ² parabola

we could do it for a far more general one

and yes, the working out would be a lot more complicated

therefore, fun,

but we actually don't have to do it

it turns out

if you do it for one parabola,

you've actually done it for all of them

but that is a different story, for a different video.

wink wink, coming soon, watch this space.

this truly is the end of the video

this time. Normally, I'd say something like subscribe to my channel

but franky, if you've watched this far though this video,

and the entire proof that I came up with

then you must already be subscribed

you require no further convincing

just thank you so much for watching me do a mass proof about parabolas