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 of what that

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curve looks like.

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You know, it's curving upwards
not quite as fast as a

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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 lengths of this
curve, come back, and we

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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

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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

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length, first you have to
compute the little piece of

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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, and THEN hopefully

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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
ds equals the square

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root of dx squared
plus dy squared.

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So this always reminds me of the
Pythagorean theorem, so I

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find it easy to remember.

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And then from here, it's not
very hard to get the other

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form, which 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 plus dy

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dx squared dx.

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

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the form that you can actually
use to integrate it, to

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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, y prime dy

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dx is just 3/2 x to the 1/2,
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--

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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 of x

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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 to x, so we need

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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 from 0 to 4 of

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square root of 1
plus 9/4 x dx.

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Now, this curve has the property
that this is an

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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, so

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I'm going to do an extra
substitution, and then it'll

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be really easy.

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

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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, computing their

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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 to choose a one that

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will have a value that we
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, and

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that's nice.

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So we can we can actually
write down what

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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

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dx, and since I want to
substitute it the other way, I

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could write that as
dx equals 4/9 du.

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And I also need to change
bounds, so when x equals 0,

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that goes to u, I put the 0
here, u is equal to 1 when x

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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 that

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I'm interested in is the
integral from 1 to 10 of 4/9

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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, so I'm going

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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

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3/2, is just 1.

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OK.

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So now if you wanted to,
you know, get a decimal

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approximation for this number,
you could put this into a

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calculator.

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You can also kind of eyeball
what this is, because 10, the

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square root of 10 is just a
little bigger than 3, so this

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is, you know, bigger than 27,
so this is bigger than 26.

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So this whole thing is probably
about 8 or a little

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bit larger.

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Probably going to be a little
bit larger than

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8, 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, I'm sure

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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 of arc

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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 that we

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can actually evaluate the
resulting integral in a nice

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closed form.

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So I'll stop there.