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[music].
A classic related wraith problem involves

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a ladder leaning up against the side of a
building.

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

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

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

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

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

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

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of the wall?
That's the question.

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

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How fast is this side moving down this
wall?

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

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

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

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

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

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

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

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

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asking how quickly this distance is
changing.

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

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picture.
So the equation here is just x squared

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

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

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

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change.
It's always 5 meters long.

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

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

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

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So let's take this equation and
differentiate it.

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

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

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

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The derivative of this constant is just
zero.

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How do I differentiate x squared plus y
squared?

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

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

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

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a chain rule.
The derivative of something squared is

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

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

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dt.
That's zero.

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

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

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right?
At this particular moment x is 3 meters,

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

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

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

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

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

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make sense?
So, yeah.

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

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

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

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

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moves down the side of the building.
Right?

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

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

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dy dt is a negative.