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[music] Here is a classic related rates problem.

So here's the question. You're walking away from a light source,

at some speed, and you don't know how fast is your shadow's length changing.

There's a standard method to solving a related rates problem, four steps.

So, step 1 is draw a picture, alright? You're told some sort of story, you should

convert it into some diagram. Step 2, find an equation.

Given that picture, you should label everything in the picture so you can write

down some equation. Step 3 is to differentiate that equation

which would probably involve the chain rule.

And step 4 is to solve that equation for whatever you're interested in finding out.

I mean, hopefully you're trying to find some rate of something so, you're going to

try to find one of these derivatives in there.

Let's draw a picture to represent this person moving away from the light source.

Here's my picture, I've got a light source right here and I've got this person right

here. And I imagine that person walking away

from the light and I've drawn in their shadow.

And I've drawn this beam of light just glancing across this person's head.

Now we'll turn this picture into an equation using similar triangles.

You might suppose that the original problem told us that the lamppost was 3

meters tall, and the person was, say, two meters tall.

And I can label the other relevant lengths on this diagram.

I'll call x the distance from the bottom of the lamppost to the person's feet, and

the distance from the feet to the tip of the shadow, I'll call that s for shadow.

So, now I've got a nicely labeled diagram, and I can replace that nicely labeled

diagram with a slightly more extract diagram, where the person is now a

vertical line, the lamppost is now a vertical line, and I've got these

triangles. The up shot here is that I've got similar

triangles. This little triangle is similiar to this

big triangle. Well, because the angles are all the same,

right? Both of these triangles have a right angle

here, they have the same angle here, consequently these angles are the same.

They're two triangles that have the same angles, they're similar triangles.

And as a result, I get this equation out of the diagram.

This distance to this distance, x plus s to 3, must be the same as the

corresponding ratio in to the other triangle, which is this distance to this

distance s over 2. With an equation in hand we can now

differentiate. Thinking of my position and the shadow's

length as a function of t, I could rewrite this equation as say, x of t plus s of t

divided by 3 equals s of t divided by 2, right?

My position is a function of time and my show's length is a function of time, and I

can differentiate both sides of this equation.

The derivative of this is 1 3rd, the derivative of x plus the derivative of s,

and the derivative of the other side is one half the derivative of s.

And I could solve for the derivative of s. Let me first expand this out.

I've got 1/3 derivative of x plus 1/3 derivative of s, is 1/2 derivative of s.

I'll subtract 1/3 s prime from both sides. Got 1 3rd x prime is 1 6th s prime.

'Because a half minus a 3rd is a 6th. And then I multiply both sides by 6.

So I find out that s prime. This is the rate of change in the shadow's

length is twice. X prime this is the speed for which the

person is moving. There's a bit of a surprise to this

answer. Take a look at this equation, It's telling

you that the speed that your shadows length is changing is just twice the speed

that your walking. But it doesnt matter where your standing.

The value of x doesn't appear on this side only x prime appears on this side.

So that's kind of interesting right? The speed with which your shadow is

growing only has to do with how fast you're walking.

In fact, the speed at which your shadow is growing is exactly twice the speed at

which you are walking. Assuming of course, you are two meters

tall and the lamp post is three meters tall.

I think the important thing to take away from this example is that similar

triangles are hugely helpful. If you see similar triangles, use them.

There's also one thing that I think is very funny about this particular solution

to the problem. Here's the challenge for you to think

about. If your velocity is 90% the speed of

light, so you're almost going the speed of light.

Then this formula would be telling you that your shadow's length would be moving

faster than the speed of light, is that really possible?