WEBVTT

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Hi.

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Welcome back to recitation.

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Today we're going to do a
nice little problem involving

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computing the arc
length of a curve.

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So in particular, consider the
curve given by the equation y

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equals x to the 3/2.

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So I have here a kind
of mediocre sketch

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of what that curve looks like.

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You know, it's curving
upwards not quite

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as fast as a parabola would.

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So I'm interested in the piece
of that curve for x between 0

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and 4, which I've drawn here.

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So why don't you take a
minute, pause the video,

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compute the arc length
of this curve, come back,

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and we can compute it together.

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All right.

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Welcome back.

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Hopefully you had some luck
computing this arc length here.

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So let's set about doing it.

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So I'm sure you remember that
in order to compute arc length,

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first you have to compute the
little piece of arc length ds.

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And we have a couple of
different formulas for that.

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So then after that,
you get an integral,

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and then hopefully it's an
integral you can compute.

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So let's remember what ds is.

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So there are a couple of
different ways to remember it.

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One way, that I
like, is to write

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ds equals the square root of
dx squared plus dy squared.

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So this always reminds me
of the Pythagorean theorem,

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so I find it easy to remember.

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And then from
here, it's not very

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hard to get the
other form, which

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is, you can divide through
by a dx squared inside

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and multiply by dx outside.

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So you can also write this
as the square root of 1

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plus dy/dx squared dx.

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And when you write it in
this form-- it's, you know,

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this is the form
that you can actually

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use to integrate it, to actually
compute the value in question.

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So in our case, we have y
as a function of x, right?

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So we just have
to compete dy/dx.

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So y is x to the 3/2, so
dy/dx is easy to compute,

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y prime, dy/dx is
just 3/2 x to the 1/2,

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or 3/2 square root of x.

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So ds, then-- well, we just
have to plug it in there.

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So that means ds is equal to
the square root of 1 plus-- OK.

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So now you have to square this.

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Well, 3/2 squared is just
9/4, and the square root

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of x squared is x.

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So this is 9/4 x dx.

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So this is the thing that
we want to integrate.

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And now you need
bounds of integration.

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So in our case, this is dx.

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We want to integrate
with respect

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to x, so we need bounds on x.

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And luckily we have them.

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We have 0 less than or equal
to x, less than or equal to 4,

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the bounds that we want.

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So the arc length in
question is the integral

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from 0 to 4 of square
root of 1 plus 9/4 x dx.

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Now, this curve has
the property that this

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is an integral we actually
know how to compute.

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Right?

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There's a-- well, OK.

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So I always lose track of
my constants when I do this,

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so I'm going to do an
extra substitution,

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and then it'll be really easy.

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But you know, this is an
integral-- many of you

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can probably do this
one in your heads,

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basically, at this point.

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This is unusual.

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Even most polynomials
that you write down,

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computing their arc
length is really hard.

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You get nasty things popping up.

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So, you know, I
sort of conspired

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to choose a one that
will have a value that we

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can integrate by hand.

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You don't need to resort to
any sort of numerical method.

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But it happens, in this
case, that that did happen,

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and that's nice.

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So we can we can actually write
down what this arc length is.

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So I'm going to do
the substitution, u

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equals 1 plus 9/4 x.

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So with this substitution, I
get that du is equal to 9/4 dx,

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and since I want to
substitute it the other way,

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I could write that
as dx equals 4/9 du.

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And I also need
to change bounds,

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so when x equals 0, that
goes to u, I put the 0 here,

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u is equal to 1 when
x is equal to 4.

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So I put 4 in here.

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That goes to u equals
10, and so, OK.

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With those substitutions,
I get that the arc length

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that I'm interested in is
the integral from 1 to 10

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of 4/9 times the
square root of u du.

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OK.

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And so now this is,
you know, really easy.

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So this is u to the 1/2,
so I integrate that,

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so I'm going to get u to
the 3/2 divided by 3/2.

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So this is 4/9 times u to the
3/2 divided by 3/2 between u

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equals 1 and u equals 10.

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OK.

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So I can divide here, so this
becomes 8/27 is the constant.

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So this is 8 over 27 times 10
to the 3/2 minus 1 to the 3/2,

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is just 1.

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OK.

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So now if you
wanted to, you know,

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get a decimal approximation
for this number,

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you could put this
into a calculator.

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You can also kind
of eyeball what

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this is, because 10,
the square root of 10

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is just a little bigger than
3, so this is, you know,

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bigger than 27, so
this is bigger than 26.

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So this whole thing is probably
about 8 or a little bit larger.

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Probably going to be a
little bit larger than 8,

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would be my guess.

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So that's, you know,
just rough eyeballing.

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Since you're all sitting
in front of a computer,

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I'm sure you can get a more
precise estimate on your own.

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But there we go.

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So very much just applying the
sort of straightforward tools

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that we've developed for
computing arc lengths.

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You know, using our formulas
for the little element

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of arc length, for the
differential of arc length.

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Computing a derivative,
plugging it in.

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And it happens, in this
case, that we got something

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that we can actually
evaluate the resulting

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integral in a nice closed form.

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So I'll stop there.