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PROFESSOR: Hi everybody.

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Welcome back to recitation.

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In lecture you did a bunch
of examples of related rates

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problems.

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So I have a couple more
for you to do today.

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So here we've got--
OK, so we've got

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air being blown into
a spherical balloon

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at a rate of 1,000 cubic
centimeters per second.

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So the question is,
how fast is the radius

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growing when the radius
is equal to 8 centimeters?

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And then OK, so I've got
a second question, which

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is, how fast is the surface
area growing at that same time?

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So why don't you take
a few minutes, work

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this one out for
yourself, come back,

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and we'll work it out together.

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All right.

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Welcome back.

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So this question, like all
related rates questions,

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has the property that
the calculus is typically

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very straightforward,
but that there's

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some geometric or
algebraic setup.

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So in this case, it's
straight up geometry.

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So we have a balloon, we
know it's a perfect sphere,

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we know how fast the
volume is changing.

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So OK, but we need to know how
fast the radius is changing.

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So in order to do that
we need to figure out

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a relationship between
the radius and the volume.

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And then we can just do
implicit differentiation

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like we've been doing.

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So for example, OK,
so for a sphere--

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the setup is not so bad
for this first one--

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so we know that for a sphere the
volume is equal to 4/3 pi times

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the radius cubed.

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So that's the fundamental
relationship between the volume

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and radius of a sphere, and
it's true for every sphere

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everywhere in Euclidean space.

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And we're given
also, that the volume

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is changing at a constant rate
of 1,000 centimeters cubed

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per second.

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So dV/dt is just
given to be 1,000.

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You know, leave off the
units at this point.

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So the question
is, what is dr/dt?

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That's what we're
trying to figure out.

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We're trying to figure
out how fast the radius is

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changing at the moment
when the radius is

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equal to 8 centimeters.

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So how can we do that?

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Well, this fundamental
relationship, it's an identity.

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It always holds.

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So that means we can
differentiate it.

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So if we take the
derivative of this identity,

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well, V on the left
just becomes dV/dt.

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And on the right we want to
do implicit differentiation.

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So here r is changing
with respect to time.

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r is a function of t.

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Well, OK.

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So 4/3 pi is a constant.

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So that when we differentiate
nothing happens.

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4/3 pi.

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And so now we differentiate
r cubed with respect to t,

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so that gives us 3 r
squared times dr/dt.

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So that's just the chain
rule in action there.

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And now what we
want is this dr/dt.

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Right?

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That's the thing that
we're looking for, is

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how fast the radius is growing.

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So that's dr/dt.

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And we want it at the
moment when r is equal to 8.

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So when r is equal to 8,
this implies that-- well, OK,

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so dV/dt is 1,000 always.

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And it implies that--
OK, so it's equal to,

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1,000 is equal to 4/3 pi times
3 times 8 squared times dr/dt.

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So at this moment that
we're interested in,

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we have this equation
to solve, for dr/dt,

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and this is a nice,
simple equation to solve.

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You just divide
through by everything

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on the right-hand
side other than dr/dt.

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So this implies that dr/dt
is equal to-- well, OK,

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so I have to divide
1,000 by all this stuff.

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I think it works out to
something like 125 over 32*pi.

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All right.

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So that's the exact value.

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Maybe you're interested
in sort of knowing

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about how large this is.

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So 32*pi is pretty close to 100,
so this is about 1.2 something.

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So, all right.

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So there we go.

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So that answers
the first question.

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At that moment the radius is
growing at a rate of 125 over

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32*pi centimeters per second.

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OK.

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So now how about the second
question that we've got here?

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What about the surface area?

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So again, we know how fast
now the radius is changing

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and we know how fast
the volume is changing.

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So in order to figure out
how fast the surface area is

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changing, we need something
that relates the surface

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area to either the
volume or the radius.

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Now the relationship between
surface area and volume

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is something that we could
sort of work out if we had to,

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but it's a lot easier to
write down the surface

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area in terms of the radius.

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So let's do that.

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So we have, I'm going to
use the letter S to denote

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surface area of a sphere.

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So again, it's a general
identity, you know,

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a geometric fact that the
surface area of a sphere is

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equal to 4*pi times
the radius squared.

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And this is always true.

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And now the thing that
we want is the rate

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of change of the surface area.

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So the rate of change
is the derivative.

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So we want to compute the
derivative here, dS/dt.

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So OK, so we just do it.

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So dS/dt is equal to--
well, 4*pi hangs around.

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And again, we
differentiate r squared.

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r is a function of t, so we
have to use the chain rule here.

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So this is times 2r times dr/dt.

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So this is an identity.

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So this is true always.

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And now we want to know, at
this particular moment in time,

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when r is equal to
8, what is ds/dt?

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And in order figure that
out, well OK, we just

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have to be able to plug
in for everything else.

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So when r equals 8--
all right, well luckily,

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you know, if we were just
starting this problem

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from scratch here,
we'd have a problem.

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Which is we wouldn't
know what dr/dt was.

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But luckily, we've already
figured it out, right?

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In the first part
of the problem.

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So we know that when
r is equal to 8,

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dr/dt is equal to
125 over 32*pi.

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Did I copy that right?

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Yes, I did.

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OK.

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So OK, so in this case, the
equation we have to solve

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is just completely
straightforward.

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We just plug in the values and
it's already solved for us.

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So that's nice.

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So that we get that ds/dt at
that moment is equal to-- well,

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it's 4*pi times 2 times 8 is
the radius times 125 over 32*pi.

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Oh boy, and all right.

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So we can work this out
if we want, I guess.

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That's 32*pi to cancel,
so that's equal to 250.

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And I guess the units there
better be centimeters squared

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per second.

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So at this moment in
time the surface area

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is growing by 250 centimeters
squared per second.

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So that's all we had to do.

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So we're all set.