Welcome back.
I'm now going to actually do a couple more rates of change
problems because I think the first one I did was probably a
little bit more complicated than I wanted, and the hard
part was actually the geometry and not the calculus.
So let's just say I drop a rock into a pond
and it has a ripple.
And the ripple spreads out-- it'll probably have multiple
ripples, but let's just say we focus on one ripple.
And the ripple is spreading out.
So let me draw.
This is where I dropped the rock, and then this is
a ripple in the pond.
It would be an actual circle, which isn't what
I drew, but close enough.
So let me say that-- let's call r.
r is not for ripple.
r is the distance that the ripple is from the center
from where it started from the get go.
And it's not by chance that that's also for radius.
So let's say that the ripple is moving out at
2 meters per second.
So the rate at which r is changing with respect to t
is 2 meters per second.
And what I want to know is what is the rate at which the area
that's kind of included by this ripple-- this entire area-- how
fast is that changing with respect to time.
When the ripple is let's say 3 meters from the center.
When the ripple is equal to 3 meters.
Well do we know any relation between the area of the circle
essentially, and the distance of the ripple, or the radius?
Well yes we do.
Area is equal to pi r squared.
So we want to figure out the rate at which area changes
with respect to time.
So what does a chain rule tell us?
The chain rule tells us that the rate at which a changes
with respect to t is equal to the rate at which a changes
with respect to r times the rate at which r changes
with respect to t.
Well we already know this piece, right?
We already know the rate at which r is changing
with respect to time.
That's right here.
All we have to do to figure out the rate at which a is changing
with respect to time is we have to figure out the rate at which
a changes with respect to r.
Well that's just the derivative.
The derivative with respect to r.
The derivative with respect to r-- do it of both sides-- of a
equals the derivative with respect to r of pi r squared.
So we get da/dr is equal to-- what's the derivative
with respect to r here?
Well that's easy.
Just 2 times pi r to the 1.
That's pretty interesting by itself that the rate at which
the derivative of the area-- formula really-- is what?
This formula by itself looks interesting, right?
That's a formula for the circumference of a circle.
So the derivative of an area of a circle with respect to the
radius is a circumference.
That is mildly fascinating.
Something even more interesting is to figure out what the
antiderivative is of the area and compare that to the
volume of a sphere.
You can look it up on Wikipedia or something.
Or even compare that to the surface area of the sphere, and
just keep picking derivatives and integrals and they'll be
some pretty interesting relationships.
But anyway.
Back to the problem.
Not to go on too far of a tangent.
So we figured out the rate at which 8 changes
with respect to r.
So going back to our original thing, we now know that the
rate at which a changes with respect to t is equal to the
rate at which a changes with respect to r.
Well that's just right here, right?
The rate at which a changes with respect to r.
Well that's just 2 pi r times the rate at which r is
changing with respect to t.
Well the rate at which r changes with respect to
t is just right here.
2 meters per second, we figured that out.
So 2 pi r times 2-- well I won't include the units
because that might confuse you right now.
So the rate at at which a is changing with respect
to t is 2 pi r times 2.
Well are we done yet?
Well no, because we haven't substituted yet for r.
So what is r?
Well we know. r is three meters, so da/dt is equal to
2 times pi times 3 times 2.
So what is that?
2 times 3 is 6 times 2 is 12.
12 pi and it's meters squared per second, because it's the
rate at which area-- square meters-- is changing
with respect to time.
And if we multiply out the units we would have gotten
the same thing right here.
So in the next video i'm going to do a slightly harder rate of
change problem just so that you see that it doesn't only apply
to kind of purely geometric things.
It applies to pretty much anything that you can
find a relationship between two values.
See you in the next module.