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

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

I start with this little piece of a circle.

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

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

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

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

Here's a very specific setup for this problem.

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

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

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

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

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

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

level rising? That's the related rates question.

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

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

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

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

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

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

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

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

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

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

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

centimeters. Now that's enough information now to

figure out the height of the cone, all right?

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

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

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

All right? The hypotenuse has length square root of

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

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

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

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

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

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

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

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

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

h. The trouble is that there's really two

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

the cone, I can use similar triangles, right?

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

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

triangle. And that means that this length here, the

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

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

I can simplify this expression just a little bit.

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

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

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

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

know how fast things are changing. So I differentiate.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

means that the current water height is 3 centimeters.

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

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

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

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

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

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

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

dt. So now I know how quickly the water height

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

centimeters per second at this particular moment.

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

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

of the cone. And then the formula for volume involved

both the radius and the height of the cone.

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

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

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

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

those pieces coming together.