We've learned a lot about derivatives, and now we will
use them to solve something that is hopefully may
be kind of useful.
So let's just start with a review of the chain rule,
and I'm going to write in a different way.
So let's say I had the function f of g of x.
So what I'm going to do is I'm going actually write this in a
way that might be a bit foreign to you, but I think with a
little bit of explanation you'll realize that this is the
same thing as the chain rule that we all know and love.
I changed colors really for no reason.
Just to change colors.
Sometimes I do that.
I'll change colors again.
The derivative of f of g of x is equal to the rate at which
f changes with respect to g.
So that's df to dg times the rate at which g changes
with respect to x.
And you're saying Sal, this looks completely foreign to me.
And one, you could think about what I just said and
I think it'll make sense.
Or I could just rewrite this in the traditional
chain rule format.
And this isn't really the traditional.
What I wrote is the traditional, but the
way I've taught you.
So the way I taught you, what's the change of
f with respect to g?
Well that's just f prime of g of x times-- and what's the
rate at which g changes with respect to x?
Well that's just g prime of x.
So hopefully this make sense to you that these are just
two different ways of writing the chain rule.
I think what was this, this is either Lagrange or Leibniz's
notation, and this the other guy.
But I forget.
And this actually makes sense just from a fractions point of
view that this term and that term cancels out and you get,
you know that this equals that rate at which f changes
with respect x.
So with that out of the way, let's use this to solve
something kind of hopefully a little bit neat.
Let's say I have a cone.
Let's say it's one of those cups you have
at the water cooler.
That's the top of the cup.
And let's say at any point in the clone the ratio of the
radius of the cone to the height-- if this is the
height, then the radius is 1/2 of height.
At any point.
As you know, it's a fixed cone, these are lines on both sides.
So that ratio stays fixed.
So just to start off with some-- you maybe never
even learned this.
This will help us with the problem later on.
What's the volume of-- let's say I'm filling water
up to height h.
Let's say this is where the water line is.
But that's where the water is in this cone.
What's the volume assuming that the height is h.
Assuming that I've let's say 8 centimeters of water.
What is the volume of water that I put in the cone?
Well, if you don't know it, and I sometimes-- well actually
later when we do rotations of solid into integration modules
I can actually prove this to you-- but the volume of a cone
is equal to 1/3 base times height, where the base is
actually, you can kind of view it as the surface
area of the water.
And what's the base?
Well that's just equal to volume is equal to 1/3.
The base is equal to pi r squared, where r is the radius.
This is just kind of solid geometry review.
And what's the radius in this case?
It's 1/3 pi r squared times h.
And in this example I said that the radius
is 1/2 of the height.
So this equals 1/3 pi 1/2 the height-- I just replaced it
for r squared-- times h.
We haven't done any calculus yet.
This is just complicated-- not complicated really-- just a
little bit hairy solid geometry.
And if we simplify that we get volume is equal to-- let's see.
So we get 1/2 squared is 1/4 times 1/3 is 1/12 pi, and
then h squared times h, h to the third.
Very interesting, Now let's start doing some calculus.
I think this might blow your mind.
Let's say I am pouring water into this cup at a rate of 1
cubic centimeter per second.
And for you metric jocks, you'll know sometimes doctors
or nurses will say 1 cc, and that's also equal
to 1 millimeter.
Just for other frames of reference, but I like to stay
in centimeters cubed per second because we're going
to work with centimeters in multiple dimensions.
But let's say we're pouring 1 centimeter cubed per
second into this cup.
I want to know-- this is an interesting question-- how fast
is the water level rising.
How fast is h changing, the height of the water, at the
moment when h is equal to 2 centimeters.
So how do we do that?
This just give us a static picture of if we know the
height we can figure out the volume.
But if we figure it out, if we took the rate of change with
respect to time off of both sides of this equation,
something interesting might happen.
So let's take the derivative with respect time
on both sides.
So the derivative with respect to time of the volume.
Well that's just dv/dt.
And what's the derivative with respect to time of
this side of the equation?
Well it's the rate which v changes with respect to h.
This is v as a function of h, right?
So it's going to be dv as a function-- how fast does v
change with respect to h-- times how fast does h
change with respect to t.
This is just the chain rule up here.
I want you to think about this a little bit.
It might not seem obvious but all we're doing
is the chain rule.
It's a little confusing because I had no t's in this equation
before, and all of a sudden I'm picking a derivative
with respect to t.
But let's just say that h actually is a function of t,
which as you know it is.
So let's then just solve that.
So dv-- the derivative of v with respect to t-- is equal
to the derivative of v with respect to h.
Well that's easy.
The derivative of v with respect to h is just simple.
So 3 times 1/12, that's 3/12.
That's 1/4 pi, so we can just write pi over 4 h squared.
So that's this part.
And in this part is just still-- see, I can rewrite
in this new color dh/dt, times dh/dt.
So you're saying Sal, what have you now done.
I said the rate of which the volume is changing with respect
to time is equal to pi over 4 times the height squared times
the rate at which the height is changing with respect to time.
So can this solve the problem for us?
Well what do we know?
We know the rate at which the volume is changing
with respect to time.
1 centimeter cubed per second squared.
So we could say dv/dt is equal to 1.
I'm going to get rid of the units.
Most physics instructors would cringe.
But 1 centimeter cubed per second is equal to pi over 4.
This one is v the rate at which the volume is changing
with respect to time.
pi over 4 h squared.
We know what h is right now. h is 2, so it's
2 squared times 4.
We said the height is 2 right then, so 2 squared
is 4 times dh/dt.
So this cancels out, and we get 1 is equal to pi times dh/dt.
And we solve for dt/dt we get-- let me make sure
not to confuse you.
The rate at which the height of water is changing with respect
to time is just 1 over pi.
Fascinating.
Or 1 over pi centimeters per second.
So we can figure out what this number is, it's going to
be like 0.3 something.
So 0.3 something centimeters per second is the rate at which
the height of the water level is going to change as I put 1
centimeter cubed of volume into this cup per second.
I have probably confused you.
You may want to watch this again.
And I'll do a couple more videos with these rates of
change problems because these tend to confuse people, but
once you get the hang of it, I think you'll see that
they're not so bad.
I'll see you the next presentation.