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[music].
>> Let's suppose that I have a cone of a

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certain size.
Suppose, to make this very concrete, that

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I start with this little piece of a
circle.

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This circle has radius square root of 90
centimeters, and the length of this arc

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here is a bit more than 18 centimeters,
specifically, it's 6 pi centimeters.

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I could cut this thing out, and then fold
it up to make a cone.

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Now let's fill that cone with water.
How fast then, is the water level rising?

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Here's a very specific setup for this
problem.

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All right, given this cone, here's the
story.

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Water is pouring into that cone at the
constant rate of 1 milliliter per second,

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and remember that a milliliter of water is
1 cubic centimeter of water.

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Right now, there is pi milliliters of
water in the cone.

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So the current volume of water is pi
milliliters of water in the cone.

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More water is pouring in.
The question is how fast is the water

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level rising?
That's the related rates question.

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I know how fast the water is coming in.
I know the change in volume.

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I want to know how quickly the water
height is changing.

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We can set this up as a related rates
problem using the same four step process.

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Draw a picture, write down an equation,
differentiate that equation, and evaluate

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the resulting derivative.
Now first I'm going to draw a diagram of

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the cone.
I built the cone out of this little piece

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of a circle, a circle of radius square
root 90 centimeters.

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And this arc length is 6 pi centimeters.
Now that's enough information to figure

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out the shape of the resulting cone.
Since the curved part here is 6 pi

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centimeters, that tells me that the
circumference of this top circle of the

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cone is 6 pi centimeters.
That tells me the radius of this is 3

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centimeters.
Now that's enough information now to

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figure out the height of the cone, all
right?

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I built the cone out of this piece of
circle, whose radius was square root of 90

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centimeters, and that ends up becoming
this length here, square root of 90.

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And I know this length is 3, and what I've
really got here is a right triangle.

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All right?
The hypotenuse has length square root of

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90, this leg has length 3, and that tells
me that this leg, which is the height of

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the cone, is 9 centimeters.
How does that diagram relate to the water

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level?
Of course, what I'm really interested in

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is, how much water is in the cone.
And the water has some height, which I'm

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calling h.
And the water itself forms a cone, with

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some radius r.
Now I need to write down an equation that

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relates all of these quantities.
I'm trying to understand the volume of

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water in the cone, and here's the formula
for the volume of a cone of radius r and

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height h.
The volume is one third pi r squared times

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h.
The trouble is that there's really two

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variables here, not just one.
Now if I think back to original shape of

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the cone, I can use similar triangles,
right?

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This big triangle is such that this length
is three times this length.

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And the triangle that I get here from the
water is similar to that original

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triangle.
And that means that this length here, the

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radius, must be 1 3rd the height.
That then lets me write down the equation

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for the volume of water, just in terms of
a single variable, h.

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I can simplify this expression just a
little bit.

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So I can combine the h squared and the h
to give me an h cubed.

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And the 1 3rd squared becomes a 1 9th,
which combines with the 1 3rd, to give me

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1 27th.
So the volume is 1 27th pi times the

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height cubed.
So now I've got an equation and I want to

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know how fast things are changing.
So I differentiate.

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So then I differentiate both sides here
with respect to time, thinking of v and h

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as a function of t.
Well, now dv dt is the time derivative of

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v, and this is the time derivative of
this, right?

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The 1 27th pi is just a constant multiple.
But I have to differentiate h cubed,

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thinking of h as a function of time.
And I do that with the chain rule, and the

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derivative of something cubed is 3 times
the inside function squared times the

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derivative of the inside, which is dh dt.
To finish this problem off, remember what

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I'm trying to do.
I'm trying to figure out dh dt.

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I know v at this particular moment, and I
know dv dt.

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That's enough information for me to figure
out dh dt.

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Let's see how.
Well here's what I'm initially told.

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I'm told that there's pi milliliter of
water currently in the cone, and I'm told

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that more water's pouring in at a rate of
1 milliliter per second.

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Now, I'd like to know the current water
height, and I'm claiming it's 3

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centimeters.
I know that, because I've got this

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formula, right?
I know that the volume is 1 27th pi times

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the current water height cubed.
And in this particular case, if the volume

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is pi milliliters of water, then I can
solve for h, all right?

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I divide both sides by pi and multiply by
27 and I find that 27 is h cubed, and that

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means that the current water height is 3
centimeters.

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Now once I know that, I can take this fact
and the fact that I'm told dv dt, and plug

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those in to the formula that relates dv dt
and dh dt.

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Now I'm interested in dh dt, but I know
that dv dt is 1.

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And I know h, in this particular moment,
is 3 centimeters.

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Now all I've gotta do is just solve this
equation for dh dt.

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And that's not so bad, right?
The 3 times 3 squared that's a 27, which

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cancels this 1 over 27 and then I divide
both sides by pi and 1 over pi then is dh

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dt.
So now I know how quickly the water height

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is changing.
It's changing at the rate of 1 over pi

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centimeters per second at this particular
moment.

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All told, this is a great example of
related rates.

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I mean, there was all this geometry that I
had to do to figure out the original shape

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of the cone.
And then the formula for volume involved

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both the radius and the height of the
cone.

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And I had to do similar triangles to write
down a single equation for the volume of

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the water, in terms of the water height.
And then I differentiated that, and then I

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went back and plugged in the original
values that the statement gave me to

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figure out dh dt.
So this is really a great example of all

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those pieces coming together.