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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. I've

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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, OK so
this is a mouthful.

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

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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

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the top of the wall.

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And then the bottom of the
ladder is being pulled away

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from the wall at 5
feet per second.

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The question is, while this is
going on, how quickly is the

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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, and we'll

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see how it went.

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All right.

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So hopefully you've had some
luck working out this problem

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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 related

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

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you have to do at
the beginning in

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order to get this right.

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And then once you get the
geometry down, the calculus is

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basically totally straight
forward.

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You know, you compute a couple
derivatives, you use the chain

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rule once, whatever.

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You know.

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

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drawing a careful picture.

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And then we'll have to figure
out what the quantities that

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we're interested in are and the

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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 and

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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, is what's getting

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pulled away from the wall
at 5 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, so while
you do that, you know, the

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ladder 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 in some other

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directions, but we're just
interested in how quickly it's

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going straight down.

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

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things 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, OK, so those
are the fixed

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qualities 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 changing because the

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ladder is being pulled
away from the wall.

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So we can give that a name.

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Lat's call that, that's a
horizontal distance, so we can

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call that x, say.

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

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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 tit 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, but

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the amount, but as this gets
pulled, you know, this top

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point is sort of sliding down
towards the wall, the amount

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of the ladder that extends 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 or we could give this

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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 so this is d
because the whole

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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 describe all the

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possible lengths 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 what the

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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 got a right

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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, and that's going to give

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us a relationship between x
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 have x squared plus--

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that's a 1--

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12 squared is equal the to
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 we can use related

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rates to figure out how fast
d 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, but what we

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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 asking for-- how

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quickly is the top of the ladder
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 to figure out

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what the relationship with y is
and these other variables.

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And there are a couple
of different

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ways to go about this.

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

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right triangle here.

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

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then you could do another
right triangle.

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But then you'd need a third
relationship relating this

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segment to d or something
like that.

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So I'm not going to go that
route, but I'm going to go do

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something related to that,
which is that this bigger

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right triangle, right, which has
the height of the top of

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the ladder from the ground and
the whole length of the ladder

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as two of its sides, and a piece
of the ground, so that

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bigger right triangle
is similar to this

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smaller right triangle.

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Right?

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I mean, they have, they're right
triangles and they have

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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 I have to

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remember all my geometry,
right?

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So the ratio of corresponding
sides are equal, so in this

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case, the ratio of the
hypotenuses is equal to the

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ratio of these legs.

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And that'll relate.

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So the hypotenuses involve d and
the vertical legs, well,

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one of them is just constant
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 and I'll

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have d linked to y.

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And so then we can use related
rates to figure out what the

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relationship between y and x
is and figure out the thing

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we're after, 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,et's look at
the hypotenuses.

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So the big hypotenuse has
length 20, and the small

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hypotenuse has length
20 minus d.

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Good.

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So that's the ratio of the
hypotenuses, and so that has

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to be equal to the ratio 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, oh, I
guess, actually, we

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need a few more things.

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Take it back.

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So now we need, we're after, as
I said, a particular moment

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in time, and at our particular
moment in time, so I haven't

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used this one given, yet, that
5 feet of the wall--

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oh, sorry--

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5 feet of the ladder projects
over the wall.

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So OK, so at our moment, so at
the key moment, we have that d

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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, x in this right

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triangle is the third leg in a
right triangle with hypotenuse

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15 and one leg 12.

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So that's one of your three,
four, five triangles, but

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blown up a little bit.

209
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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, we can use this

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other relationship that we have.
So 20 over 15 is equal

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to y over 12, 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 are the values that

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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, x is

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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

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going to plug in.

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All right, good.

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00:10:55,962 --> 00:10:57,620

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So now we are after dy dt at
this moment. and we don't

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have, and we know dx dt.

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

229
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but 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

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dt, and because of this
relationship,

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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

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with respect to t.

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x and d are functions of t.

240
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So OK.

241
00:11:38,090 --> 00:11:42,500
So differentiating this
identity I get

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2x times dx dt plus--

243
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well, OK, derivative of 144
with respect to t is 0--

244
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is equal to--

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I guess I could expand this out,
but it's easier just to

246
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use the chain rule right away--
so this is equal to 2

247
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times 20 minus d times the
derivative of 20 minus d with

248
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respect to t, which
is minus dd dt.

249
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All right.

250
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I beg the gods of
math notation to

251
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forgive me for dd dt.

252
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But, OK.

253
00:12:27,020 --> 00:12:30,280
So, allright, so now,
this is always true.

254
00:12:30,280 --> 00:12:33,910
And what we want is the value
of dd dt at our particular

255
00:12:33,910 --> 00:12:35,310
moment in time.

256
00:12:35,310 --> 00:12:40,423
So at our particular moment in
time we have, well let's

257
00:12:40,423 --> 00:12:42,760
see,so OK, we have x, we have
the dx dt, and we have d.

258
00:12:42,760 --> 00:12:47,030
So we can just plug all
those things in.

259
00:12:47,030 --> 00:12:50,030
So, at our moment, let me see
how I'm going to do this.

260
00:12:50,030 --> 00:12:52,200
I'll do it over here.

261
00:12:52,200 --> 00:13:06,790
At our moment, so 2x is 18
times dx dt is 5 equals 2

262
00:13:06,790 --> 00:13:17,620
times 20 minus d is 15
times minus dd dt.

263
00:13:17,620 --> 00:13:21,500
And dd dt is what we're after.

264
00:13:21,500 --> 00:13:21,760
So OK.

265
00:13:21,760 --> 00:13:25,980
So divide 2 into this is 9.

266
00:13:25,980 --> 00:13:34,490
45 divided by 15 is 3, so dd dt
is equal to minus 3 at the

267
00:13:34,490 --> 00:13:36,140
moment that we're
interested in.

268
00:13:36,140 --> 00:13:36,480
All right.

269
00:13:36,480 --> 00:13:37,800
Great.

270
00:13:37,800 --> 00:13:41,250
So now we have dd dt, and so
now we can go back to the

271
00:13:41,250 --> 00:13:44,900
second relationship 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 --> 00:13:52,240
We can take a derivative, use
the chain rule, use, and get a

274
00:13:52,240 --> 00:13:57,160
relationship between dy
dt and d and dd dt.

275
00:13:57,160 --> 00:14:00,660
And then we'll be able to plug
in all these values we have

276
00:14:00,660 --> 00:14:04,310
for our particular moment.

277
00:14:04,310 --> 00:14:06,530
So OK, so I think I'm
just going to

278
00:14:06,530 --> 00:14:07,870
differentiate this straight.

279
00:14:07,870 --> 00:14:10,390
There are some simplifications
I could do first, but it'll

280
00:14:10,390 --> 00:14:12,770
work fine if we just
do it straight.

281
00:14:12,770 --> 00:14:15,880
So OK, so I need to compute
the derivative.

282
00:14:15,880 --> 00:14:16,630
This is an identity.

283
00:14:16,630 --> 00:14:20,070
I can take its derivative, so I
differentiate the left hand

284
00:14:20,070 --> 00:14:28,270
side with respect to t, so I
get 20 over-- so this is a

285
00:14:28,270 --> 00:14:33,640
minus first power-- so I get
minus 20 over 20 minus d

286
00:14:33,640 --> 00:14:35,590
quantity squared times--

287
00:14:35,590 --> 00:14:44,140
I need the derivative of the
bottom-- 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 --> 00:14:59,280
So I have this relationship
between d and y, I take its

290
00:14:59,280 --> 00:15:02,230
derivative, now I have a
relationship that involves dy

291
00:15:02,230 --> 00:15:05,150
dt, which is the thing
that I'm after.

292
00:15:05,150 --> 00:15:07,290
So, whew.

293
00:15:07,290 --> 00:15:09,760
So now, I'm back at this
plugging in stage.

294
00:15:09,760 --> 00:15:13,470
So at our moment in time, I
know what d is, and I just

295
00:15:13,470 --> 00:15:17,420
figured out what dd dt is, so
I can just plug those values

296
00:15:17,420 --> 00:15:21,050
in to figure out
what dy dt is.

297
00:15:21,050 --> 00:15:22,140
So OK, so let's do that.

298
00:15:22,140 --> 00:15:25,570
So at this moment in
time we have--

299
00:15:25,570 --> 00:15:28,220
so let me just multiply through
by the 12 and that

300
00:15:28,220 --> 00:15:34,510
will let us solve-- so we have
dy dt is equal to 12 times,

301
00:15:34,510 --> 00:15:45,050
well, so it's minus 20 over 15
squared, times minus, minus 3.

302
00:15:45,050 --> 00:15:47,150
Times 3.

303
00:15:47,150 --> 00:15:48,400
So this is--

304
00:15:48,400 --> 00:15:51,110

305
00:15:51,110 --> 00:15:51,610
all right.

306
00:15:51,610 --> 00:15:53,470
Arithmetic.

307
00:15:53,470 --> 00:15:54,850
Not my favorite thing
in the world.

308
00:15:54,850 --> 00:15:57,810
So we've got two 3's here.

309
00:15:57,810 --> 00:16:01,500
We're going to end up with a 5
in the denominator, and then

310
00:16:01,500 --> 00:16:09,850
we've got 4 times minus 4, so
that's minus 16 over 5.

311
00:16:09,850 --> 00:16:12,560
So.

312
00:16:12,560 --> 00:16:13,020
All right.

313
00:16:13,020 --> 00:16:16,160
So that's the answer.

314
00:16:16,160 --> 00:16:18,980
Let's remember what
we've done.

315
00:16:18,980 --> 00:16:23,210
So minus 16 over 5, that's
a negative number.

316
00:16:23,210 --> 00:16:24,880
Why is it a negative number?

317
00:16:24,880 --> 00:16:30,160
Well, as we pull this, the
bottom of this ladder here

318
00:16:30,160 --> 00:16:34,680
away from the wall, the top of
the ladder is going to fall

319
00:16:34,680 --> 00:16:36,970
downwards, it's going to get
closer to the ground.

320
00:16:36,970 --> 00:16:42,820
And y is the vertical distance
between the ground and the top

321
00:16:42,820 --> 00:16:43,250
of the ladder.

322
00:16:43,250 --> 00:16:46,220
So that's shrinking; y is
shrinking and should have a

323
00:16:46,220 --> 00:16:47,290
negative derivative.

324
00:16:47,290 --> 00:16:48,150
OK.

325
00:16:48,150 --> 00:16:51,194
Negative derivative, minus 16
over 5 feet per second.

326
00:16:51,194 --> 00:16:55,330
So it would take, if this
instantaneous rate held up,

327
00:16:55,330 --> 00:16:56,750
which it won't in
this situation--

328
00:16:56,750 --> 00:16:59,760
if it held up it would take 5
seconds to get down, so it's

329
00:16:59,760 --> 00:17:01,060
not falling very quickly.

330
00:17:01,060 --> 00:17:01,440
OK.

331
00:17:01,440 --> 00:17:04,320
That seems about right.

332
00:17:04,320 --> 00:17:05,290
Anything else?

333
00:17:05,290 --> 00:17:09,290
I guess we computed that dd
dt was also negative.

334
00:17:09,290 --> 00:17:12,370
Same thing here-- when you pull
this way, this length is

335
00:17:12,370 --> 00:17:14,860
going to shrink, so d is getting
smaller, so it has

336
00:17:14,860 --> 00:17:16,110
negative derivative.

337
00:17:16,110 --> 00:17:18,600

338
00:17:18,600 --> 00:17:19,660
OK.

339
00:17:19,660 --> 00:17:24,704
So at this moment in time, the
top of the ladder is falling--

340
00:17:24,704 --> 00:17:29,280
the thing we're interested in--
at exactly 16 over 5 feet

341
00:17:29,280 --> 00:17:30,880
per second.

342
00:17:30,880 --> 00:17:32,450
And we're all set.

343
00:17:32,450 --> 00:17:32,920