[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.