[music].
>> I'm going to blow up this balloon.
Let's suppose that I'm blowing air into
that balloon at a constant rate, so the
balloon's volume is changing at a constant
rate.
How quickly is the radius of the balloon
changing?
We can model the balloon as a sphere.
And recall the volume of a sphere in terms
of its radius r.
The volume of a sphere of radius r is 4
3rds pi r cubed.
So how fast does the radius change as I
blow the balloon up?
So I've got the original equation for the
volume of a sphere of radius r.
I'm going to differentiate both sides with
respect to time.
So the derivative of v, with respect to
time, is I'm going to write dv dt.
And the derivative of the other side, with
respect to time, we've gotta think a
little bit more about.
Now 4 3rds and pi are just constants.
So I can pull them out of the derivative.
But I've still gotta figure out what the
time derivative of r cubed is.
To figure out the time derivative of r
cubed, I'm regarding r, remember, as a
function of t.
So I use the chain rule, and the
derivative of cubing is 3 times the inside
function squared times the derivative of
the inside function, so dr dt.
So dv dt is 4 3rds pi times 3 times r
squared times dr dt.
Let's solve for dr dt, the rate of change
in the balloon's radius.
All right, so I'm going to solve for dr
dt.
Well, first thing I can do is, I've got a
dividing by 3, and a multiplying by 3.
So I don't need to do both of those
operations.
So this is 4 pi r squared times dr dt.
The other side is just dv dt.
So I'll divide both sides by 4 pi r
squared.
And I'll find out that, dr dt is dv dt
divided by 4 pi r squared.
Can we really make use of this equation?
I, does this help us at all?
Yeah, this makes sense.
We started with the original equation for
the volume of a sphere in terms of its
radius.
And I differentitated this equation.
And then solved for dr dt.
And what this equation is telling me is
that the change in radius, which is what
dr dt is measuring, the rate of change in
the size of the balloon, well, it depends
on how quickly the volume's changing,
which is dv dt.
But it also depends upon the radius,
right?
This is this r squared in the denominator.
In particular, a really big balloon
doesn't get a whole lot bigger for a given
change in volume, compared to a little
tiny balloon, which gets quite a bit
bigger for the same change in volume.
It all has to do with that r squared in
the denominator.
When the balloon's radius is very small, a
given change in volume results in quite a
dramatic change in the radius.
But once the radius is large, the r
squared in the denominator means that dr
dt is then much smaller for the same
change in volume.
Another breath, and yeah, I mean, the
thing's still getting bigger, but much
less dramatically.
.