JOEL LEWIS: Hi.
Welcome back to recitation.
In lecture, you've been doing related rates problems. I've
got another example for you, here.
So this one's a really tricky one.
So I'm going to give you some time to work on it.
But I really think you should try and work
through this one yourself.
It'll be well worth your effort.
So we've got here, OK so this is a mouthful.
We've got a 20 foot long ladder and it's leaning
against a 12 foot wall.
But it's leaning over the wall.
So 5 feet of the ladder project over
the top of the wall.
And then the bottom of the ladder is being pulled away
from the wall at 5 feet per second.
The question is, while this is going on, how quickly is the
top of the ladder going downwards?
How quickly is it approaching the ground?
So why don't you take a few minutes.
Well, maybe more than a few for this one.
It took me a while to work it out the first time.
Take a few minutes, work this one out, come back, and we'll
see how it went.
All right.
So hopefully you've had some luck working out this problem
on your own.
Now let's work it out together.
So OK, so this is a sort of classic, really tricky related
rates problem in that there's a lot of geometric work that
you have to do at the beginning in
order to get this right.
And then once you get the geometry down, the calculus is
basically totally straight forward.
You know, you compute a couple derivatives, you use the chain
rule once, whatever.
You know.
So OK, so let's start off by trying and
drawing a careful picture.
And then we'll have to figure out what the quantities that
we're interested in are and the
relationships between them.
So we have here, here's the ground.
And we have a 12 foot wall.
And we have a 20 foot ladder leaning against the wall and
extending over it.
So my ladder.
So here's my ladder.
Here's my wall.
All right.
And this, the bottom of the ladder here, is what's getting
pulled away from the wall at 5 units per second.
5 feet per second.
So 5 feet per second is how fast that's going.
And we're interested, so while you do that, you know, the
ladder is changing position.
So the question is, how fast is the top of the ladder
descending?
So you know, it might also be moving in some other
directions, but we're just interested in how quickly it's
going straight down.
So OK, so first, let's talk about the
things that don't change.
So the height of this wall is not changing.
The wall is staying put the whole time.
And the length of the ladder is not changing.
That's staying put, as well.
But what, OK, so those are the fixed
qualities in this problem.
Basically, everything else is changing.
So for example, this distance between the base of the wall
and the base of the ladder, that's changing because the
ladder is being pulled away from the wall.
So we can give that a name.
Lat's call that, that's a horizontal distance, so we can
call that x, say.
And the height of the top of the ladder is changing, so
let's draw that in.
That's what we're interested in.
It's the height, so we can call tit y.
And let's see.
What else?
Well OK, the whole ladder isn't changing in length, but
the amount, but as this gets pulled, you know, this top
point is sort of sliding down towards the wall, the amount
of the ladder that extends over the wall is changing.
So I guess we could choose.
We could give this a variable name or we could give this
part a variable name.
I think I'm going to give the little one a variable name.
I'll call it d for distance, I guess.
And OK so this is d because the whole
ladder has length, 20.
We have that this part, this segment has length 20 minus d.
Ok so, I think that those quantities describe all the
possible lengths of interest in this picture.
So now, OK.
So that's just the first, the set up.
That's the first thing that has to happen.
Now after we set up, we need to figure out what the
relationships between these different variables are.
Well, let's see.
What have we got?
Well one thing we've got is that we've got a right
triangle here.
So the wall and the ground and the ladder.
Those three segments form a right triangle.
So OK.
So we can apply the Pythagorean Theorem here.
So that's one relationship we have, and that's going to give
us a relationship between x and well, OK, 20 minus d.
So that'll give us a relationship between x and d.
So this, so OK, so what does the Pythagorean Theorem say?
So by the Pythagorean Theorem, we have x squared plus--
that's a 1--
12 squared is equal the to quantity 20 minus d squared.
So that's one identity that we have, identity we have that
relates these three quantities.
And so that's good.
Because we know how fast x is changing.
And so with this identity, that means we can use related
rates to figure out how fast d is changing, as well.
So that's good.
That'll be a step in the right direction, but what we
actually need to know is how fast y is changing.
So right?
Because that's what the question is asking for-- how
quickly is the top of the ladder approaching the ground?
So how, what's the rate of change in y
with respect to time?
So I need another identity here in order to figure out
what the relationship with y is and these other variables.
And there are a couple of different
ways to go about this.
I think the one that I'm going to do-- so there is another
right triangle here.
So you could, if you wanted to name this segment, as well,
then you could do another right triangle.
But then you'd need a third relationship relating this
segment to d or something like that.
So I'm not going to go that route, but I'm going to go do
something related to that, which is that this bigger
right triangle, right, which has the height of the top of
the ladder from the ground and the whole length of the ladder
as two of its sides, and a piece of the ground, so that
bigger right triangle is similar to this
smaller right triangle.
Right?
I mean, they have, they're right triangles and they have
the same base angle there.
So they're similar triangles.
So OK, so by similar triangles-- now I have to
remember all my geometry, right?
So the ratio of corresponding sides are equal, so in this
case, the ratio of the hypotenuses is equal to the
ratio of these legs.
And that'll relate.
So the hypotenuses involve d and the vertical legs, well,
one of them is just constant and one of them involves y.
So that'll set up a relationship between d and y,
and so then I'll have x linked to d and I'll
have d linked to y.
And so then we can use related rates to figure out what the
relationship between y and x is and figure out the thing
we're after, which is the rate of change in y.
So OK, so I haven't actually written anything down, yet.
So what is the similar triangle?
So here,et's look at the hypotenuses.
So the big hypotenuse has length 20, and the small
hypotenuse has length 20 minus d.
Good.
So that's the ratio of the hypotenuses, and so that has
to be equal to the ratio of the corresponding legs.
And so the big leg is y, and the small leg is 12.
OK.
All right, so there's Pythagorean Theorem.
There's similar triangles.
So we now have the relationships that we're after
relating x and y and d all to each other.
Great.
So now we can do the calculus part of this problem, right?
So now we know, oh, I guess, actually, we
need a few more things.
Take it back.
So now we need, we're after, as I said, a particular moment
in time, and at our particular moment in time, so I haven't
used this one given, yet, that 5 feet of the wall--
oh, sorry--
5 feet of the ladder projects over the wall.
So OK, so at our moment, so at the key moment, we have that d
is equal to 5.
OK, so that means 20 minus d is equal to 15.
What else does that mean?
So that means that at that moment, x in this right
triangle is the third leg in a right triangle with hypotenuse
15 and one leg 12.
So that's one of your three, four, five triangles, but
blown up a little bit.
So we have at that moment, that x is equal to 9.
And OK, so at that moment, to figure out y, we can use this
other relationship that we have. So 20 over 15 is equal
to y over 12, so y is equal to 20 over 15 times 12.
So that's 4 over 3, so that's 16.
So at our key moment in time, these are the values that
we're going to end up plugging in.
And also, dx dt is equal to 5.
x is being increased.
Increased?
Yeah x is, right, so as the foot gets pulled away, x is
getting bigger.
So x is being increased at 5 units per second.
All right, so these are all the things we're
going to plug in.
All right, good.
So now we are after dy dt at this moment. and we don't
have, and we know dx dt.
We don't have a direct relationship between y and x,
but do have a direct relationship between y and d.
So if we got dd dt-- sorry-- dd dt, then we could get dy
dt, and because of this relationship,
we can get dd dt.
All right.
So let's start off by doing that.
So OK, so this is an identity.
It holds always so we can differentiate it.
And we're going to differentiate it
with respect to t.
x and d are functions of t.
So OK.
So differentiating this identity I get
2x times dx dt plus--
well, OK, derivative of 144 with respect to t is 0--
is equal to--
I guess I could expand this out, but it's easier just to
use the chain rule right away-- so this is equal to 2
times 20 minus d times the derivative of 20 minus d with
respect to t, which is minus dd dt.
All right.
I beg the gods of math notation to
forgive me for dd dt.
But, OK.
So, allright, so now, this is always true.
And what we want is the value of dd dt at our particular
moment in time.
So at our particular moment in time we have, well let's
see,so OK, we have x, we have the dx dt, and we have d.
So we can just plug all those things in.
So, at our moment, let me see how I'm going to do this.
I'll do it over here.
At our moment, so 2x is 18 times dx dt is 5 equals 2
times 20 minus d is 15 times minus dd dt.
And dd dt is what we're after.
So OK.
So divide 2 into this is 9.
45 divided by 15 is 3, so dd dt is equal to minus 3 at the
moment that we're interested in.
All right.
Great.
So now we have dd dt, and so now we can go back to the
second relationship we have, the one that relates d and y,
and we can do the same thing here.
We can take a derivative, use the chain rule, use, and get a
relationship between dy dt and d and dd dt.
And then we'll be able to plug in all these values we have
for our particular moment.
So OK, so I think I'm just going to
differentiate this straight.
There are some simplifications I could do first, but it'll
work fine if we just do it straight.
So OK, so I need to compute the derivative.
This is an identity.
I can take its derivative, so I differentiate the left hand
side with respect to t, so I get 20 over-- so this is a
minus first power-- so I get minus 20 over 20 minus d
quantity squared times--
I need the derivative of the bottom-- is minus dd dt.
And on the right hand side I get dy dt over 12.
So I have this relationship between d and y, I take its
derivative, now I have a relationship that involves dy
dt, which is the thing that I'm after.
So, whew.
So now, I'm back at this plugging in stage.
So at our moment in time, I know what d is, and I just
figured out what dd dt is, so I can just plug those values
in to figure out what dy dt is.
So OK, so let's do that.
So at this moment in time we have--
so let me just multiply through by the 12 and that
will let us solve-- so we have dy dt is equal to 12 times,
well, so it's minus 20 over 15 squared, times minus, minus 3.
Times 3.
So this is--
all right.
Arithmetic.
Not my favorite thing in the world.
So we've got two 3's here.
We're going to end up with a 5 in the denominator, and then
we've got 4 times minus 4, so that's minus 16 over 5.
So.
All right.
So that's the answer.
Let's remember what we've done.
So minus 16 over 5, that's a negative number.
Why is it a negative number?
Well, as we pull this, the bottom of this ladder here
away from the wall, the top of the ladder is going to fall
downwards, it's going to get closer to the ground.
And y is the vertical distance between the ground and the top
of the ladder.
So that's shrinking; y is shrinking and should have a
negative derivative.
OK.
Negative derivative, minus 16 over 5 feet per second.
So it would take, if this instantaneous rate held up,
which it won't in this situation--
if it held up it would take 5 seconds to get down, so it's
not falling very quickly.
OK.
That seems about right.
Anything else?
I guess we computed that dd dt was also negative.
Same thing here-- when you pull this way, this length is
going to shrink, so d is getting smaller, so it has
negative derivative.
OK.
So at this moment in time, the top of the ladder is falling--
the thing we're interested in-- at exactly 16 over 5 feet
per second.
And we're all set.