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Hi.

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Welcome back to recitation.

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We've been talking
in class a little bit

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about parametric
equations and arc length.

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So let's do an example
of a problem where

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you compute an arc
length of a curve given

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by some parametric equations.

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So in particular, I have here
the parametric equations y

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equals t minus 1 over t, and
x equals t plus 1 over t,

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for 1 less than or equal to
t, less than or equal to 2.

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So those parametric equations
trace out some piece

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of a curve in the plane.

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And what I'd like you
to do is write down

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an integral whose value
is equal to the arc

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length of that curve.

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So the integral
you're going to get

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is going to be pretty
hard to evaluate.

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So I wouldn't recommend
you spend a whole lot

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of time trying to evaluate it.

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But just so we see that we,
you know, we can do this,

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and so we have the integral
whose value is the arc length.

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So take a few
minutes, work on that,

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pause the video, come back,
and we can work on it together.

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All right.

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So I hope you've had some
luck working this problem.

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Let's get started
on it together.

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So in order to compute the
arc length by an integral,

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we need to figure out what
the element of arc length

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is, that little piece ds.

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So this was true when
we had our curve given

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in rectangular coordinates,
and it's also true

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when our curve is given
in parametric coordinates.

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So in general, the little
element of arc length

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is given as ds equals the
square root of dx squared

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plus dy squared.

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Now, when our curve is given
in rectangular coordinates,

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usually what we do here is
we factor out a dx squared,

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and then we have dx by
dy, the derivative of y

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with respect to x.

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But here, we don't
have x or y given

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in terms of the other one.

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We have them both given in
terms of this parameter, t.

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So the parameter t is
the thing, is our sort

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of independent variable.

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It's the thing we're
going to want to end up

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integrating with respect to.

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So what we actually
want to do in this case,

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is we want to factor out
a dt from everywhere.

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So if you factor out a dt, what
this becomes-- or dt squared,

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I guess I should
say-- is that you

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get dx/dt squared plus dy/dt
squared, the whole thing square

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root, dt.

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OK.

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So this is just a little
algebraic manipulation

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involving differentials.

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And what this gives me is an
expression that will-- you

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know, x is a function of t. x
is given as a function of t,

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right here.

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So dx/dt is a function of t.

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And similarly, y is
given as a function of t,

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so dy/dt is a function of t.

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So this whole expression
is a function of t,

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so we have a function of t dt.

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So it's all set up
to be integrated.

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So in order to actually,
you know, compute

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with this formula, well, we just
take our y, and we take our x,

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and we take their derivatives.

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So from-- well, I guess
we could start with a y,

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since it's on top over there.

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So dy/dt.

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Well, you know.

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You just take a derivative.

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Right?

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So you have y is equal
to t minus 1 over t.

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And you take a
derivative of that.

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Derivative of t is just
1, derivative of 1 over t

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is minus 1 over t
squared, so this

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becomes-- but there's an
extra minus sign there,

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so this becomes 1
plus 1 over t squared.

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And similarly, dx by dt is equal
to 1 minus 1 over t squared.

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So you have these
two derivatives.

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So to compute arc
length, you plug them

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into this formula for the
element of arc length,

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and then you'll integrate.

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So OK.

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So we have to put them
into that formula.

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So what do we get?

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We have ds is equal to
the square root of, well,

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so dx/dt squared
plus dy/dt squared.

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So that's 1 minus 1
over t squared squared,

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plus 1 plus 1 over
t squared squared.

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dt outside.

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OK.

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And so now we, you know, maybe
we can simplify this a little.

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I would probably expand
out, at this point.

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So if you expand this out,
this is-- well, let's see.

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The first one is
just is 1 minus 2

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over t squared plus 1
over t to the fourth,

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plus-- that's the first
one we expanded out,

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and the second one,
we get 1 plus 2

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over t squared plus 1
over t to the fourth.

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All right.

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And if we put all this together,
say, over a common denominator,

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and simplify it a
little bit, we can

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see that this is equal to--
over t squared terms cancel out,

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and we end up with something
like square root of t

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to the fourth plus 1
over t squared, dt.

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OK.

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So this is all this ds that
we've been playing with.

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So the actual arc length is what
we get when we integrate ds.

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So we need appropriate bounds.

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So where do we get those bounds?

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Well, we go back
to the question,

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and here, the question
had to have told us.

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Right?

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If the question didn't tell
us, then it didn't actually

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describe a piece of curve,
or a finite piece of curve,

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really, is the point.

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So the question tells us that we
want this from t equals 1 to 2.

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So the arc length, we just
integrate our little element

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of arc length over the
appropriate interval.

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So that's t from 1 to 2
of the square root of t

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to the fourth plus
1 over t squared dt.

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All right.

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So as I said at
the beginning, this

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is not an integral
that's readily

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susceptible to the techniques
that we've learned.

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You know, if you
were interested,

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you could try some
numerical methods on it.

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I'm sure, you know, computer
algebra software can spit out,

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at worst, a good
numerical approximation,

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if you're curious.

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One more thing I want to
say about this question,

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is that we did all this without
ever trying to draw the curve,

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or think about what
the curve looks like.

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So it's not that hard to
get some sort of basic sense

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of the behavior of this curve.

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For example, when t
gets very, very large,

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we see that what happens is
that x and y, well, the 1 over t

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gets small, so x and y
get close to each other.

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But x is just a tiny,
tiny bit bigger.

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So for, as t gets
very, very big,

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you see that the
point on the curve

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is getting very close
to the line y equals x,

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but not actually touching it.

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And you can, you know,
you can try other things,

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as t gets close to 0
from the positive side,

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as t gets close to 0
from the negative side.

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So this has some sort of
asymptote, it seems like.

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There's some other
things you could check.

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You could try to show
that the x-value is always

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either-- well,
when t is positive,

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it's always at least 2,
or when t is negative,

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it's always at least
negative 2, and there's

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a sort of gap in between.

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You could analyze
its properties there.

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The other thing you could do,
is you could try and solve

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these equations to
eliminate t, and get

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just rectangular equations
in terms of x and y.

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So I'm not going
to do that for you.

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I'll give you a hint about
how I would go about it.

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Which is, you could start off--
so you want to eliminate t.

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So I think the nicest
way to do that is

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if you add these two equations.

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You'll get that the 1
over t's cancel out,

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and you get 2t equals x plus
y, or t equals x plus y over 2.

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And then you can take, and
you could substitute it back

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into one of these equations.

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And when you substitute
it back, then you'll

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have an equation that just
involves x and y, which,

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with a little more
simplification,

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takes a form that you should
already be familiar with.

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OK.

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And yes.

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Arc length.

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This is how we do arc
length in parametric form.

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Works out pretty nice,
except in this case,

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we get an integral we can't
evaluate at the very end.

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Too bad.

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All right.

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I'll leave it with that.