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JOEL LEWIS: Hi.

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Welcome back to recitation.

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In lecture, you've been
learning about line integrals

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of vector fields.

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And I have one problem
here that requires

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you to do two line integrals of
vector fields on that subject.

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So OK.

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So in both problems, F is going
to be the vector field whose

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coordinates are x*y and
x squared plus y squared.

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And C is going to be the
arc of the parabola y

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equals x squared that
starts at the point (1, 1)

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and ends at the point (2, 4).

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All right?

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So what I'd like you to do
is to compute the integral

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over this curve C of F dot
dr in two different ways.

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So the first time,
I'd like you to use

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the natural parametrization x
equals t, y equals t squared.

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For me, at least, that's
the first parametrization

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that I think of when I
think about this curve.

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But I'd also like you
to do it again using

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a different
parametrization, using

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the parametrization x equals e
to the t, y equals e to the 2t.

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So this still parametrizes
the curve y equals x squared,

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since e to the t squared
equals e to the 2t.

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So why don't you pause the
video, have a go at computing

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this integral using these two
different parametrizations,

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come back, and we can
work it out together.

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So let's get started.

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We want to compute a line
integral of a vector field F.

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So we know that the integral
over our curve C of F

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dot dr-- well,
usually we write--

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F equals M, N. The two
components are M and N. So

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in this case, this is going to
be equal to the integral over C

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of M dx plus N dy.

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Now, in our case,
we know what F is.

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We know that F is
the vector field x*y,

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x squared plus y squared.

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So this is equal to the integral
over C of x*y dx plus x squared

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plus y squared dy.

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And now we just need to plug
in in our two different cases,

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into our parametrizations, and
this will turn into an integral

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that we can evaluate.

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So let's have a look-see.

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So in our first
parametrization over here,

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we have x equals t and
y equals t squared.

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And we want this parametrization
to go from the points (1, 1)

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to the point (2, 4).

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So that means that t-- x--
t is going from 1 to 2.

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OK.

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So I'm going to just write
equals, and continue on.

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So in this first part--
so t is our parameter,

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and it's going from 1 to 2.

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So x*t is going to be t times
t squared, and dx is dt.

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And then plus-- so x
squared plus y squared is

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t squared plus t to the fourth.

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And dy is times 2t dt,
since y is t squared.

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All right.

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So this is our integral.

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And now it's
straightforward to compute.

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This is a single variable
integral of a polynomial,

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so we might have some fraction
arithmetic in our future,

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but nothing horrible.

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So let's collect our dt's.

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So this is the
integral from 1 to 2.

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So here I've got 2 t cubed
plus 2 t to the fifth.

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And here I have t cubed.

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So that's 3 t cubed plus
2 t to the fifth dt.

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OK, and so now we
have to evaluate.

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So this is equal to--
let me just continue it.

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So 3 t cubed becomes 3
t to the fourth over 4.

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Plus 2 t to the
fifth becomes-- well,

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t to the sixth over 6--
2 t to the sixth over 6,

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so that's plus t to the
sixth over 3, between t

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equals 1 and 2.

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And let me just bring
that computation up here.

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So that's equal to 3 times
16 over 4, plus 64 over 3,

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minus 3/4 plus 1/3.

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OK, and now we
gotta work this out.

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So this is maybe 12, and
the thirds give me 21,

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so that's 33 minus 3/4.

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So that's 32 and 1/4, or
129 over 4, if you prefer.

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OK, so that's going to
be our answer there.

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So now let's go back
and do it again using

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this alternate parametrization.

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So in this alternate
parametrization,

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we had x equal e to the t,
and y equal e to the 2t.

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So in that case, that means
that dx-- so x is e to the t,

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y is e to the 2t, so our
integral that we want,

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let me rewrite it.

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Well, actually, let
me not rewrite it.

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Let me just go back here
and take a look at it.

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So we want the
integral of F dot dr,

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and we know that that's equal
to the integral over C of x*y dx

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plus x squared
plus y squared dy.

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So x is e to the t,
y is e to the 2t,

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so that means dx is
going to be e to the t dt

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and dy is going to
be 2 e to the 2t dt.

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So OK.

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So we'll just have to
plug these things in.

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So our integral-- maybe I'll
give it a name: What We Want.

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WWW.

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What We Want is equal
to the integral of,

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well we start plugging in.

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So x*y dx is e to the t
times e to the 2t times,

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we said dx is e to the t dt.

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Plus, now x squared
plus y squared,

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so that's e to the 2t--
that's x squared-- plus e

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to the 4t-- that's y squared--
times 2 d to the 2t dt.

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That's dy.

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Whew.

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OK.

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But this integral is
missing something, right?

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Because it needs to be
a definite integral.

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Right now, I've written
down an indefinite integral.

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So that's a problem.

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So we need to think, what
are the bounds on this curve?

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What are we integrating
from and to?

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So let's go all the way back and
look at the problem over here.

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So this is the
parametrization of the curve,

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and we need it to connect
the points (1, 1) and (2, 4).

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So we need to know
which value of t

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puts this parametrization
at the point (1, 1),

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and which value of t puts
it at the point (2, 4).

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Well, I think the easiest way
to do that is to solve this,

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and we have t equals
natural log of x.

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So natural log of 1 is
0, and natural log of 2

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is natural log of 2.

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So OK.

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So we want t to go from
0 to natural log of 2.

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To go from this
point to that one.

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OK, so 0 to natural log of 2.

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Let's remember that, and come
back over here and put that in.

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So we want this
integral that we're

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trying to compute using this
alternate parametrization--

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we've already written down what
the integrand is-- but it's

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going to be the
integral, we just said,

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from 0 to the natural log of 2.

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OK.

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Great.

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So now let's take this integral
and let's start simplifying it.

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So let's see.

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This is an e to the t times
e to the 2t times e to the t,

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so that's e to the 4t dt.

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And here we have-- that looks
like a 2 e to the 4t plus 2 e

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to the 6t dt.

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So if we combine like terms,
we get the integral from 0

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to natural log of 2.

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So we had an e to
the 4t and a 2 e

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to the 4t, so that's 3 e to
the 4t plus 2 e to the 6t dt.

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OK, 3 e to the 4t
plus 2 e to the 6t.

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OK.

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And this is also a
different-looking integral

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than the one we had before,
but it's not a difficult one

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to evaluate, because it's just
e to a constant times t dt.

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Pretty straightforward
stuff, I think.

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So what does it
actually give us?

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Let me come up here
and write it up here.

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So this is equal to-- all
right, integrate e to the 4t,

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and you get e to the 4t over
4, so this is 3/4 e to the 4t,

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plus 2/6-- so that's
plus 1/3 e to the 6t,

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and we're evaluating
that between 0 and ln 2.

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And I will leave it
to you to confirm

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that what you get
when you do this

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is you get exactly 129 over 4
when you plug in these values

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and take the difference.

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So one thing to notice
here is that doing it

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with the second parametrization,
we got 129 over 4.

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Now, if you look
back just right here,

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you see that when we did it
with the first parametrization,

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we also got 129 over 4.

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Now this is a general
phenomenon, right?

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The parametrization
doesn't change

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the value of the line integral.

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So when you have a line
integral of a vector field,

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you can parametrize the curve
that you're integrating over

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in a bunch of different ways.

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I mean, in fact, infinitely
many different ways.

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There are all sorts
of different choices,

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and they all give
the same answer.

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They have to.

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So no matter what
parametrization you choose,

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you get the same number
out for the line integral

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over the same curve of
the same vector field.

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All right.

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So what that means
for you is that when

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you're given just a
curve and a vector field

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and you get to choose
your parametrization,

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you should always just choose
whatever the nicest one is.

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Because it doesn't
matter which one

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you choose, and
there's no reason

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ever to choose anything
harder than necessary.

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All right.

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So now, I want to quickly
recap what we did.

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So we had a curve
and a vector field F,

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and we were asked to compute
the line integral of F dot dr

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and we did it two
different ways.

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So in both cases,
we used the fact

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that F dot dr is M dx plus N dy.

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We wrote that out using the
known values of M and N,

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and using the known
components of F.

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And then we plugged
in for both cases.

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In this case, we used that
first parametrization.

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And then over to
my left, we used

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this second parametrization
that I gave you.

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And in both cases, that
reduced it to an integral

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in terms of a single variable t
that was the parameter that we

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could compute.

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And the answer came out the
same both times, as it had to.

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So one thing you
can notice is if you

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compare this with the
preceding lecture video,

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you'll see that this curve
and vector field, they

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connect the same points and
it's the same vector field,

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and these answers
both differ from both

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of the answers in the
preceding recitation video.

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So this is further illustration
that different curves can

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give you different values, but
if you take the same curve,

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different
parametrizations always

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have to give you same values
for that line integral

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of the same vector field.

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I'll end there.