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[music]. A classic related wraith problem involves

a ladder leaning up against the side of a building.

Well, here's the setup. I've got a wall and the ground, and I've

got a ladder resting on one side on the ground, the other side, against the wall.

Ladder is 5 meters long and the bottom of the ladder is 3 meters from the base of a

wall. I'm going to start pulling the bottom of

the ladder away at a speed of one meter per second.

Just as I do this, how fast does the top of the ladder start moving down the side

of the wall? That's the question.

I'm moving the bottom of the ladder. The ladder's sliding down the wall.

How fast is this side moving down this wall?

So, remember we've got this four step process.

The first step is to draw picture. Of course, I think this is already a

pretty good picture. The second step is to write down an

equation. So, I'm label everything in my diagram and

then figure out what the equation is. So, now I want to label this, instead of

3, I'll call this distance from the wall to the base of the ladder, x.

And I'll call this distance from the top of the ladder to the bottom of the wall y.

And this speed, the speed with which I'm pulling the ladder away, that's really

asking how quickly this distance is changing.

So that's dx dt. So now I've labeled everything on my

picture. So the equation here is just x squared

plus y squared equals 5 squared because this ladder, even as it slides down the

side of a wall, it still makes a right triangle with this leg, this leg and this

is the hypotenuse. And the length of the ladder doesn't

change. It's always 5 meters long.

So, by Pythagorean theorem, tells me that x squared plus y squared is 5 squared.

The third step is to differentiate. I'm going to differentiate the equation,

regarding the variables as functions that depend upon t, time.

So let's take this equation and differentiate it.

So I'm going to be differentiating with respect to time.

So d dt of x squared plus y squared is d dt of the length of the ladder.

Now the length of the ladder is not changing, right?

The derivative of this constant is just zero.

How do I differentiate x squared plus y squared?

Well, that's the derivative of a sum. So, it's the sum of the derivatives.

It's d dt x squared and d dt y squared. Now, I'm thinking of x and y as functions

of t. So, I'm going to differentiate these with

a chain rule. The derivative of something squared is

twice the inside times the derivative of the inside plus, and now I gotta

differentiate y thinking of it as a function of t, and that's twice y times dy

dt. That's zero.

Now I'll evaluate. So, now I just want to evaluate, all

right? I know the values of x, dx, dt, and y,

right? At this particular moment x is 3 meters,

and dx dt is 1 meter per second, and y is 4 meters.

And what I'm trying to figure out is dy dt.

So, once I plug in what I know, I get this equation.

And then I can easily solve for dy dt. And when I solve for dy dt, I get that dy

dt is a negative 3/4 meters per second. Does the sign, the sign of that answer

make sense? So, yeah.

Does it make sense that this is negative 3/4 of a meter per second?

Well, take a look at our picture, right? I'm pulling the bottom of the ladder this

way, right? So I'm moving the bottom of the ladder

this way. And as I do that, the top of the ladder

moves down the side of the building. Right?

So, as this distance, which is x, is increasing, this distance, which is y, is

decreasing. And so, yeah, it totally makes sense that

dy dt is a negative.