I've been requested to do a ladder rate of change problem.
And I figure I would do one, because it really is the
classic rate of change problem.
And it's part of any complete calculus education.
So what is the ladder rate of change problem?
So let's say I have the ground.
And I will do the ground in brown.
Let me see.
Where am I going to draw my wall?
This is the ground.
That's too thin.
Let's say this is the ground.
A thicker ground.
That's the ground, and then I have a yellow wall.
This is my yellow wall.
And then leaning against that yellow wall and the
ground, I have a ladder.
This is the ladder.
I'm looking at it from the side.
I'm just going to give a side view of the ladder.
This is the wall, this is the ground, that's a ladder
leaning against the wall.
And let's say that my ladder is 10 meters long.
And let's say that for whatever reason I've kind of greased up
this wall and the floor here.
And so a normal ladder would just kind of stay where it is,
but especially because this floor is all greased up,
the bottom of this ladder is moving to the left.
So let me draw that.
Let me draw an arrow.
So the bottom of this ladder is moving to the left.
And let's say it's moving to the left at
4 meters per second.
So let's think about a couple of things; what's happening.
When this point right here is moving to the left at 4 meters
per second, as long as we have some height here-- you know, at
some point this whole thing is going to fall down.
The ladder, for whatever arbitrary reason, will just
be sliding along the ground.
Maybe someone's pulling it or whatever.
But before the latter completely falls to the ground,
while this is being pulled out this way, this point right here
is going to be coming down.
Right?
Think about it.
In fact, that's probably what's causing it to
slide to the left.
And actually it'll do it at an accelerating rate.
And we're not going to worry all about that.
We're just going to do the classic ladder problem.
But I just want to give you the intuition that if you were to
put a roller here and put a roller there, that that's
what a ladder would do.
It would slide down.
And when it slides down, this point would go to the base
of the floor right here.
Right?
This point would go down.
And this point would go out to the left, until the ladder is
in something like this position.
Until it is flat on the ground.
I'm just doing it right below the line.
Right?
At that point it would either stop, but if someone's pulling
it then it could continue to go at 4 meters per second.
Whatever.
So hopefully you're satisfied with that.
So my question to you is, what is the rate of change of this
point-- of the kind of top of the ladder-- what is its rate
of change downwards when the left point of the ladder right
here is going to the left at 4 meters per second, and this
distance right here-- let me do it in a different
color-- is 8 meters?
Actually, let me do the mirror image of that, just because I
don't want to use negative numbers and all of that.
I just realized that.
So let me just draw it really messy real quick on this side.
So that's the ground, that's the wall.
I'll do the ladder in a different color still.
That's the ladder.
It's 10 meters.
This point right here is moving to the right at
4 meters per second.
Actually, I'm going to do the horizontal component as x, and
I wanted this to be a positive x-value instead of a
negative x-value.
So that's why I did that.
We're 8 meters away, so it's positive x.
Right?
If we take this 0, 0.
And what I want to figure out is the rate of change of this
point of the ladder, the top point of the ladder.
How fast is that moving down?
Right?
So how do we think about this?
Well let's call this distance right here x, because it is.
It's kind of the horizontal displacement.
And let's call this distance up here y.
Right?
So at any given point in time, how does x, y and 10 relate?
Well, this is a right angle, so we can use the
Pythagorean theorem.
So we know that x squared plus y squared is equal to what?
It's equal to the hypotenuse squared.
It's equal to 10 squared, which is equal to 100.
So this right here describes, if you've watched the video on
implicit functions or implicit differentiation, this
is kind of an implicit function of x and y.
Right?
Or an implicit function of x, really.
So what can we do now?
Well, we're dealing with rates of change with respect to time.
Right?
This is 4 meters per second.
This is the rate of change, or this 4 meters per second.
This is the rate of change at which x is changing
with respect to time.
Right?
That's dx/dt.
How much does x move in a very small amount of time?
And what are we looking for?
We're looking for how fast is this y-value moving downward.
Right?
So we're looking for dy/dt.
That's essentially what the question is asking for.
How fast is y changing as a function of time when x is
changing at a function of time at 4 meters per second,
and the value of x is equal to 8 meters.
So I know that seems very complicated, but essentially
all we do is we take the derivative of both sides of
this equation with respect to time.
So let's do that.
So let's take the derivative with respect to time of x
squared plus y squared.
Equals-- and we have to do it to both sides of this equation.
100.
So what's the derivative with respect to time of x squared?
Well, it's the chain rule.
It's the derivative with respect to x of this
entire expression.
So that's 2x times the derivative of x with
respect to time.
That's just the chain rule.
Plus the derivative of this expression with respect to y.
So that's 2y.
Times the derivative of y with respect to time.
And if you found this to be a little confusing you might
want to watch the video that we actually just did on
implicit differentiation.
It gives a little bit more intuition why this really
just is the chain rule.
But hopefully you get this a little bit.
And then how much does 100 change with respect to time?
Well, this 100 value is never going to change, because,
frankly, the length of the ladder is never
going to change.
Right?
That's where that 100 comes from.
It's 10 squared.
So the rate of 100 with respect to time, well it never
changes so it's equal to 0.
Well what do we know?
We know x.
Right?
We know x is what?
x is, at this point in time, 8 meters.
We know the change of x with respect to time is
4 meters per second.
Do we know y?
Well, I think we do.
When x is 8, what is y?
Well, we could just use this formula right here.
We have 8 squared plus y squared is equal to 100.
So 64 plus y squared is equal to 100.
Or y squared is equal to 36.
So y is plus or minus 6, but we know that we're only dealing
with positive values-- at least in this example right here.
So y is equal to 6.
So we know x, we know the rate of change of x versus time.
That was given in the problem-- 4 meters per second.
We know y.
So then we can solve for the rate of change of y
with respect to time.
So let's do that.
Let's just substitute the values.
So what's x?
We're saying, well what is the rate of change of y with
respect to time when x is 8?
So we get 2 times 8-- that's x.
Times the change in-- so this is 8.
What's the change in x with respect to time at
this moment in time?
It's 4 meters per second.
That's 4.
So 2 times 8 times 4.
Plus 2y.
We just figured out that y is 6.
So plus 2 times 6.
Times-- and this is what we're solving for.
dy/dt is equal to 0.
So what is this?
2 times 8 is 16.
Times 4 is 64.
Right?
So 64 plus 12 dy/dt.
And let me erase some stuff.
I tend to try to crowd too much into one window,
making things confusing.
Actually, we've actually already drawn that.
So let me erase my-- woops.
I think that is good enough.
OK.
So then we can simplify up here.
That equals 0.
Subtract 64 from both sides, we get 12 dy/dt
is equal to minus 64.
And then you get the change in y with respect to time is
equal to minus 64 over 12.
And what's the greatest common factor?
I think it's a 4.
So that equals minus 16 over 3.
Right?
Or you could say that's 5 and 1/3, or 5.3.
And it's going to be meters per second.
That's the units.
Let me write it that way.
Is equal to minus 5.33-- I mean, you could whatever--
meters per second.
And there we have it.
And then you might say, hey, why is there that minus there?
Well it makes sense, because while the x-value, it's rate
of change with respect to time, it's going in the
positive direction.
Right?
At least the way I drew it here.
The x is going to the right, so it's change is positive.
But what's happening to the y-value?
The y-value, you could say its velocity is downward.
So that's where that negative value comes from, because
it has a negative velocity.
I mean really this is kind of just the x and y velocities
of those different points.
And you know, you can think about it a little bit that at
least at that point in time while the x-value is increasing
at 4 meters per second, the y-value is moving even
faster downward.
So that's something to think about.
But you might want to experiment with it at different
points in time, because it's going to be dependent on where
we are and kind of what does the right triangle look like.
But anyway, that is the classic rate of change problem--
the ladder problem.
I will see you in the next video.