If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:7:43

AP Calc: CHA‑3 (EU), CHA‑3.D (LO), CHA‑3.D.1 (EK), CHA‑3.D.2 (EK)

So let's say that we've
got a pool of water and I drop a rock into the
middle of that pool of water. And a little while later,
a little wave, a ripple has formed that is moving
radially outward from where I dropped the rock. So let's see how
well I can draw that. So it's moving
radially outwards. So that is the
ripple that is formed from me dropping the
rock into the water. So it's a circle centered
at where the rock initially hit the water. And let's say right at this
moment the radius of the circle is equal to 3 centimeters. And we also know that
the radius is increasing at a rate of 1
centimeter per second. So radius growing at rate
of 1 centimeter per second. So given this, right
now our circle, our ripple circle has a
radius of 3 centimeters. And we know that the
radius is growing at 1 centimeter per second. Given that, at what rate
is the area growing? At what rate is area
of circle growing? Interesting. So let's think
about what we know and then what we
don't know, what we're trying to figure out. So if we call this radius
r, we know that right now r is equal to 3 centimeters. We also know the
rate at which r is changing with respect to time. We also know this information
right over here. dr dt, the rate at which the radius is
changing with respect to time, is 1 centimeter per second. Now what do we
need to figure out? Well, they say at what rate is
the area of the circle growing? So we need to figure
out at what rate is the area of
the circle-- where a is the area of the circle--
at what rate is this growing? This is what we
need to figure out. So what might be
useful here is if we can come up with a relationship
between the area of the circle and the radius of the
circle and maybe take the derivative with
respect to time. And we'll have to use a
little bit of the chain rule to do that. So what is the relationship
at any given point in time between the
area of the circle and the radius of the circle? Well, this is
elementary geometry. The area of a circle
is going to be equal to pi times the radius
of the circle squared. Now what we want
to do is figure out the rate at which the area is
changing with respect to time. So why don't we
take the derivative of both sides of this
with respect to time? And let me give myself a
little more real estate. Actually, let me just
rewrite what I just had. So pi r squared. Area is equal to pi r squared. I'm going to take the
derivative of both sides of this with respect to time. So the derivative
with respect to time. I'm not taking the
derivative with respect to r, I'm taking the derivative
with respect to time. So on the left-hand
side right over here, I'm going to have the
derivative of our area. Actually, let me just write
it in that green color. I'm going to have the
derivative of our area with respect to time
on the left-hand side. And on the right-hand
side, what do I have? Well, if I'm taking the
derivative of a constant times something, I can take
the constant out. So let me just do that. Pi times the derivative with
respect to time of r squared. And to make it a little bit
clearer what I'm about to do, why I'm using the
chain rule, we're assuming that r is
a function of time. If r wasn't a
function of time then area wouldn't be a
function of time. So instead of just writing
r, let me make it explicit that it is a function of time. I'll write r of t. So it's r of t,
which we're squaring. And we want to find the
derivative of this with respect to time. And here we just have
to apply the chain rule. We're taking the derivative of
something squared with respect to that something. So the derivative of that
something squared with respect to the something is going
to be 2 times that something to the first power. Let me make it clear. This is the derivative of r of t
squared with respect to r of t. The derivative of something
squared with respect to that something. If it was a derivative of x
squared with respect to x, we'd have 2x. If it was the derivative of r
of t squared with respect to r of t, it's 2r of t. But this doesn't get us just
the derivative with respect to time. This is just the derivative
with respect to r of t. The derivative at which this
changes with respect to time, we have to multiply this
times the rate at which r of t changes with respect to time. So the rate at which r of t
changes with respect to time? Well, we could just
write that as dr dt. These are equivalent
expressions. And of course, we
have our pi out front. And I just want to
emphasize this is just the chain rule right over here. The derivative of something
squared with respect to time is going to be the derivative
of the something squared with respect to the something. So that's 2 times the
something, times the derivative of that something
with respect to time. I can't emphasize enough. What we did right over here,
this is the chain rule. That is the chain rule. So we're left with
pi times this is equal to the derivative of
our area with respect to time. Now let me rewrite
all this again just so it cleans
up a little bit. So we have the derivative
of our area with respect to time is equal to
pi times-- actually let me put that 2 out front. Is equal to 2 times
pi times-- I can now switch back to just
calling this r. We know that r is
a function of t. So I'll just write 2
pi times r times dr dt. Actually, let me
make the r in blue. 2 pi r dr dt. Now what do we know? We know what r is. We know that r, at this
moment right in time, is 3 centimeters. Right now r is 3 centimeters. We know dr dt right now is
1 centimeter per second. We know this is 1
centimeter per second. So what's da dt
going to be equal to? Well, it's going to be equal
to-- do that same green-- 2 pi times 3 times 3
times 1 times-- that's purple-- times 1
centimeter per second. And let's make sure we
get the units right. So we have a centimeter
times a centimeter. So it's going to be centimeters. That's too dark of a color. It's going to be square
centimeters, centimeters times centimeters, square
centimeter per second, which is the exact units we
need for a change in area. So we have da dt is equal to
this. da, the rate at which area is changing with respect
to time, is equal to 6 pi. So it's going to be a little
bit over 18 centimeters squared per second. Right at that moment. Yep, 3 times 2 pi. So 6 pi centimeters
squared per second is how fast the
area is changing. And we are done.