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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been
doing related rates problems.

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I've got another
example for you, here.

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So this one's a
really tricky one.

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So I'm going to give you
some time to work on it.

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But I really think you
should try and work

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through this one yourself.

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It'll be well worth your effort.

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So we've got here--

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OK, so this is a mouthful.

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We've got a 20-foot-long
ladder and it's

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leaning against a 12-foot wall.

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But it's leaning over the wall.

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So 5 feet of the ladder project
over the top of the wall.

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And then the bottom
of the ladder

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is being pulled away from the
wall at 5 feet per second.

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The question is, while
this is going on,

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how quickly is the top of
the ladder going downwards?

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How quickly is it
approaching the ground?

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So why don't you
take a few minutes.

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Well, maybe more than
a few for this one.

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It took me a while to work
it out the first time.

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Take a few minutes, work
this one out, come back,

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and we'll see how it went.

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All right.

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So hopefully you've had
some luck working out

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this problem on your own.

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Now let's work it out together.

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So OK, so this is a sort
of classic, really tricky

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related rates problem in that
there's a lot of geometric work

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that you have to
do at the beginning

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in order to get this right.

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And then once you get
the geometry down,

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the calculus is basically
totally straightforward.

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You know, you compute a
couple of derivatives,

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you use the chain
rule once, whatever.

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You know.

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So OK, so let's start
off by trying and drawing

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a careful picture.

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And then we'll
have to figure out

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what the quantities
that we're interested in

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are and the relationships
between them.

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So we have here,
here's the ground.

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And we have a 12-foot wall.

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And we have a 20-foot ladder
leaning against the wall

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and extending over it.

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So, my ladder.

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So here's my ladder.

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Here's my wall.

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All right.

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And this, the bottom
of the ladder here,

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is what's getting pulled
away from the wall at 5

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units per second.

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5 feet per second.

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So 5 feet per second is
how fast that's going.

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And we're interested--

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so while you do that,
you know, the ladder

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is changing position.

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So the question is, how fast
is the top of the ladder

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descending?

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So you know, it
might also be moving

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in some other
directions, but we're

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just interested in how quickly
it's going straight down.

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So, OK, so first, let's
talk about the things

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that don't change.

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So the height of this
wall is not changing.

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The wall is staying
put the whole time.

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And the length of the
ladder is not changing.

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That's staying put, as well.

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But what--

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OK, so those are the fixed
quantities in this problem.

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Basically, everything
else is changing.

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So for example, this distance
between the base of the wall

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and the base of
the ladder, that's

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changing because the
ladder is being pulled away

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from the wall.

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So we can give that a name.

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Let's call that-- that's
a horizontal distance,

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so we can call that x, say.

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And the height of the top
of the ladder is changing,

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so let's draw that in.

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That's what we're interested in.

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It's the height, so
we can call it y.

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And let's see.

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What else?

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Well OK, the whole ladder
isn't changing in length,

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but the amount, but
as this gets pulled,

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you know, this top point
is sort of sliding down

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towards the wall, the amount
of the ladder that extends

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over the wall is changing.

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So I guess we could choose.

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We could give this
a variable name

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or we could give this
part a variable name.

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I think I'm going to give the
little one a variable name.

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I'll call it d for
distance, I guess.

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And, OK.

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So if this is d, because the
whole ladder has length 20,

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we have that this part, this
segment has length 20 minus d.

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OK so, I think that
those quantities

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describe all the
possible lengths

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of interest in this picture.

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So now, OK.

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So that's just the
first, the set up.

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That's the first thing
that has to happen.

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Now after we set up,
we need to figure out

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what the relationships between
these different variables are.

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Well, let's see.

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What have we got?

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Well one thing we've
got is that we've

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got a right triangle here.

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So the wall and the
ground and the ladder.

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Those three segments
form a right triangle.

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So OK.

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So we can apply the
Pythagorean Theorem here.

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So that's one
relationship we have,

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and that's going to give
us a relationship between x

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and well, OK, 20 minus d.

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So that'll give us a
relationship between x and d.

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So this, so OK, so what does
the Pythagorean Theorem say?

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So by the Pythagorean
Theorem, we

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have x squared plus-- that's
a 1-- 12 squared is equal

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to the quantity 20
minus d squared.

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So that's one identity that
we have, identity we have that

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relates these three quantities.

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And so that's good.

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Because we know how
fast x is changing.

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And so with this
identity, that means

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we can use related rates
to figure out how fast d

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is changing, as well.

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So that's good.

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That'll be a step in
the right direction,

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but what we actually need to
know is how fast y is changing.

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So-- right?

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Because that's what
the question is

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asking for-- how quickly
is the top of the ladder

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approaching the ground?

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So how, what's the
rate of change in y

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with respect to time?

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So I need another
identity here in order

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to figure out what the
relationship with y is

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and these other variables.

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And there are a couple
of different ways

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to go about this.

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I think the one that I'm
going to do-- so there

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is another right triangle here.

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So you could, if you wanted
to name this segment,

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as well, then you could
do another right triangle.

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But then you'd need
a third relationship

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relating this segment to
d or something like that.

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So I'm not going
to go that route,

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but I'm going to go do
something related to that, which

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is that this bigger
right triangle, right,

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which has the height of the top
of the ladder from the ground

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and the whole length of the
ladder as two of its sides,

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and a piece of the ground, so
that bigger right triangle is

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similar to this
smaller right triangle.

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Right?

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I mean, they have,
they're right triangles

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and they have the
same base angle there.

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So they're similar triangles.

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So OK, so by similar
triangles-- now

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I have to remember all
my geometry, right?

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So the ratio of corresponding
sides are equal,

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so in this case, the
ratio of the hypotenuses

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is equal to the
ratio of these legs.

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And that'll relate-- so
the hypotenuses involve d.

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And the vertical legs, well,
one of them is just constant

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and one of them involves y.

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So that'll set up a
relationship between d and y,

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and so then I'll
have x linked to d

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and I'll have d linked to y.

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And so then we can
use related rates

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to figure out what the
relationship between y and x

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is and figure out the
thing we're after,

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which is the rate
of change in y.

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So OK, so I haven't actually
written anything down, yet.

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So what is the similar triangle?

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So here, let's look
at the hypotenuses.

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So the big hypotenuse
has length 20,

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and the small hypotenuse
has length 20 minus d.

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Good.

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So that's the ratio
of the hypotenuses,

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and so that has to
be equal to the ratio

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of the corresponding legs.

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And so the big leg is y,
and the small leg is 12.

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OK.

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All right, so there's
Pythagorean Theorem.

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There's similar triangles.

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So we now have the
relationships that we're after,

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relating x and y and
d all to each other.

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Great.

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So now we can do the calculus
part of this problem, right?

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So, now we know--

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oh, I guess, actually, we
need a few more things.

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Take it back.

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So now we need the condition.

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We're after this at a
particular moment in time,

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and at our particular
moment in time--

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so I haven't used
this one given yet,

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that 5 feet of the wall-- oh,
sorry-- 5 feet of the ladder

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projects over the wall.

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So OK, so, at our moment,
so at the key moment,

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we have that d is equal to 5.

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OK, so that means 20
minus d is equal to 15.

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What else does that mean?

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So that means that
at that moment,

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x in this right triangle
is the third leg

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in a right triangle with
hypotenuse 15 and one leg 12.

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So that's one of
your 3-4-5 triangles,

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but blown up a little bit.

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So we have at that moment,
that x is equal to 9.

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And OK, so at that
moment, to figure out y,

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we can use this other
relationship that we have.

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So 20 over 15 is
equal to y over 12,

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so y is equal to 20
over 15 times 12.

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So that's 4 over
3, so that's 16.

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So at our key moment
in time, these

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are the values that we're
going to end up plugging in.

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And also, dx/dt is equal to 5.

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x is being increased.

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Increased?

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Yeah x is, right, so as
the foot gets pulled away,

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x is getting bigger.

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So x is being increased
at 5 units per second.

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All right, so these are all the
things we're going to plug in.

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All right, good.

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So now we are after
dy/dt at this moment.

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And we don't have a
relation-- so, we know dx/dt.

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We don't have a direct
relationship between y and x,

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but we do have a direct
relationship between y and d.

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So if we got dd/dt-- sorry--
dd/dt, then we could get dy/dt,

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and because of this
relationship, we can get dd/dt.

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All right.

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So let's start
off by doing that.

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So OK, so this is an identity.

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It holds always, so we
can differentiate it.

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And we're going to differentiate
it with respect to t.

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x and d are functions of t.

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So OK.

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So differentiating this
identity I get 2x times dx/dt

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plus-- well, OK, derivative
of 144 with respect to t

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is 0-- is equal to-- I guess
I could expand this out,

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but it's easier just to use
the chain rule right away.

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00:12:00,610 --> 00:12:07,250
So this is equal to
2 times 20 minus d

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00:12:07,250 --> 00:12:10,070
times the derivative of
20 minus d with respect

247
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to t, which is minus dd/dt.

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All right.

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I beg the gods of math notation
to forgive me for dd/dt.

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But, OK.

251
00:12:27,020 --> 00:12:30,280
So, all right, so now--
this is always true.

252
00:12:30,280 --> 00:12:34,280
And what we want is the value of
dd/dt at our particular moment

253
00:12:34,280 --> 00:12:35,310
in time.

254
00:12:35,310 --> 00:12:39,460
So at our particular moment in
time we have-- well let's see,

255
00:12:39,460 --> 00:12:42,760
so OK, we have x, we have
the dx/dt, and we have d.

256
00:12:42,760 --> 00:12:44,820
So we can just plug
all those things in.

257
00:12:44,820 --> 00:12:47,030
So, at our moment--

258
00:12:47,030 --> 00:12:50,030
let me see how I'm
going to do this.

259
00:12:50,030 --> 00:12:52,200
I'll do it over here.

260
00:12:52,200 --> 00:13:07,960
At our moment-- so 2x is 18
times dx/dt is 5 equals 2 times

261
00:13:07,960 --> 00:13:17,620
20 minus d is 15
times minus dd/dt.

262
00:13:17,620 --> 00:13:21,260
And dd/dt is what we're after.

263
00:13:21,260 --> 00:13:21,760
So OK.

264
00:13:21,760 --> 00:13:25,980
So divide 2 into this is 9.

265
00:13:25,980 --> 00:13:34,220
45 divided by 15 is 3, so
dd/dt is equal to minus 3

266
00:13:34,220 --> 00:13:35,981
at the moment that
we're interested in.

267
00:13:35,981 --> 00:13:36,480
All right.

268
00:13:36,480 --> 00:13:37,800
Great.

269
00:13:37,800 --> 00:13:40,637
So now we have
dd/dt, and so now we

270
00:13:40,637 --> 00:13:42,220
can go back to the
second relationship

271
00:13:42,220 --> 00:13:44,900
we have, the one
that relates d and y,

272
00:13:44,900 --> 00:13:47,790
and we can do the
same thing here.

273
00:13:47,790 --> 00:13:51,790
We can take a derivative,
use the chain rule,

274
00:13:51,790 --> 00:13:57,160
use, and get a relationship
between dy/dt and d and dd/dt.

275
00:13:57,160 --> 00:13:59,440
And then we'll be able to
plug in all these values we

276
00:13:59,440 --> 00:14:04,310
have for our particular moment.

277
00:14:04,310 --> 00:14:07,661
So OK, so I think I'm just going
to differentiate this straight.

278
00:14:07,661 --> 00:14:09,660
There are some simplifications
I could do first,

279
00:14:09,660 --> 00:14:12,770
but it'll work fine if
we just do it straight.

280
00:14:12,770 --> 00:14:15,797
So OK, so I need to
compute the derivative.

281
00:14:15,797 --> 00:14:16,630
This is an identity.

282
00:14:16,630 --> 00:14:19,190
I can take its derivative,
so I differentiate

283
00:14:19,190 --> 00:14:21,980
the left-hand side
with respect to t,

284
00:14:21,980 --> 00:14:29,431
so I get 20 over-- so this
is a minus first power-- so I

285
00:14:29,431 --> 00:14:34,060
get minus 20 over
20 minus d quantity

286
00:14:34,060 --> 00:14:37,830
squared times-- I need the
derivative of the bottom--

287
00:14:37,830 --> 00:14:44,140
is minus dd/dt.

288
00:14:44,140 --> 00:14:55,150
And on the right-hand
side I get dy/dt over 12.

289
00:14:55,150 --> 00:14:58,806
So I have this relationship
between d and y,

290
00:14:58,806 --> 00:15:00,180
I take its
derivative, now I have

291
00:15:00,180 --> 00:15:03,620
a relationship that involves
dy/dt, which is the thing

292
00:15:03,620 --> 00:15:05,150
that I'm after.

293
00:15:05,150 --> 00:15:07,290
So, whew.

294
00:15:07,290 --> 00:15:09,760
So now, I'm back at
this plugging-in stage.

295
00:15:09,760 --> 00:15:13,030
So at our moment in
time, I know what d is,

296
00:15:13,030 --> 00:15:15,390
and I just figured
out what dd/dt

297
00:15:15,390 --> 00:15:17,760
is, so I can just
plug those values in

298
00:15:17,760 --> 00:15:21,050
to figure out what dy/dt is.

299
00:15:21,050 --> 00:15:22,140
So OK, so let's do that.

300
00:15:22,140 --> 00:15:25,730
So at this moment
in time we have-- so

301
00:15:25,730 --> 00:15:27,670
let me just multiply
through by the 12

302
00:15:27,670 --> 00:15:32,260
and that will let us
solve-- so we have dy/dt

303
00:15:32,260 --> 00:15:41,670
is equal to 12 times-- well, so
it's minus 20 over 15 squared,

304
00:15:41,670 --> 00:15:45,050
times minus minus 3.

305
00:15:45,050 --> 00:15:47,150
Times 3.

306
00:15:47,150 --> 00:15:51,610
So this is-- all right.

307
00:15:51,610 --> 00:15:53,392
Arithmetic.

308
00:15:53,392 --> 00:15:54,850
Not my favorite
thing in the world.

309
00:15:54,850 --> 00:15:57,810
So we've got two 3's here.

310
00:15:57,810 --> 00:16:01,240
We're going to end up with
a 5 in the denominator,

311
00:16:01,240 --> 00:16:09,850
and then we've got 4 times minus
4, so that's minus 16 over 5.

312
00:16:09,850 --> 00:16:12,521
So.

313
00:16:12,521 --> 00:16:13,020
All right.

314
00:16:13,020 --> 00:16:16,160
So that's the answer.

315
00:16:16,160 --> 00:16:18,980
Let's remember what we've done.

316
00:16:18,980 --> 00:16:23,210
So minus 16 over 5,
that's a negative number.

317
00:16:23,210 --> 00:16:24,880
Why is it a negative number?

318
00:16:24,880 --> 00:16:30,160
Well, as we pull this, the
bottom of this ladder here,

319
00:16:30,160 --> 00:16:33,825
away from the wall,
the top of the ladder

320
00:16:33,825 --> 00:16:35,804
is going to fall
downwards, it's going

321
00:16:35,804 --> 00:16:36,970
to get closer to the ground.

322
00:16:36,970 --> 00:16:42,167
And y is the vertical
distance between the ground

323
00:16:42,167 --> 00:16:43,250
and the top of the ladder.

324
00:16:43,250 --> 00:16:45,840
So that's shrinking; y
is shrinking and should

325
00:16:45,840 --> 00:16:47,290
have a negative derivative.

326
00:16:47,290 --> 00:16:48,150
OK.

327
00:16:48,150 --> 00:16:51,194
Negative derivative, minus
16 over 5 feet per second.

328
00:16:51,194 --> 00:16:55,240
So it would take, if this
instantaneous rate held up,

329
00:16:55,240 --> 00:16:57,240
which it won't in this
situation-- if it held up

330
00:16:57,240 --> 00:16:59,960
it would take 5 seconds
to get down, so it's not

331
00:16:59,960 --> 00:17:00,940
falling very quickly.

332
00:17:00,940 --> 00:17:01,440
OK.

333
00:17:01,440 --> 00:17:04,320
That seems about right.

334
00:17:04,320 --> 00:17:05,290
Anything else?

335
00:17:05,290 --> 00:17:09,290
I guess we computed that
dd/dt was also negative.

336
00:17:09,290 --> 00:17:10,150
Same thing here.

337
00:17:10,150 --> 00:17:13,189
When you pull this way, this
length is going to shrink,

338
00:17:13,189 --> 00:17:15,480
so d is getting smaller, so
it has negative derivative.

339
00:17:18,600 --> 00:17:19,660
OK.

340
00:17:19,660 --> 00:17:24,130
So at this moment in time,
the top of the ladder

341
00:17:24,130 --> 00:17:27,360
is falling-- the thing we're
interested in-- at exactly

342
00:17:27,360 --> 00:17:30,880
16 over 5 feet per second.

343
00:17:30,880 --> 00:17:32,420
And we're all set.