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DAVID JORDAN: Hello , welcome
back to recitation.

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The problem I'd like to work on
with you now is a long one.

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So it's going to be practice
computing line integrals.

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So to begin with, we have this
function of two variables.

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f is x to the fifth
plus 3xy cubed.

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And we have this--

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C is the upper semi-circle going
from (1, 0) to (-1, 0).

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So it's this upper semi-circle
here that we often consider.

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And so the first thing that we
want to do is to just compute

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the gradient capital F
to be the gradient of

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this function f.

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And then parts b through d,
we're going to compute this

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line integral of this vector
field f along this curve C.

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We're going to compute it
in three different ways.

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So first of all, we're going to
compute it directly, just

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using the definition.

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And then in Part c, we're going
to compute it using the

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path independence of line
integrals and we're going to

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replace the path C with
a simpler path.

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And then finally in Part d,
we're going to use the

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fundamental theorem
of line integrals.

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Now when you do Part b,
what I want you to do

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is set up the integral.

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You're going to get a very
complicated integral that I

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wouldn't want to
try to compute.

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So just set up the integral
completely and then go ahead

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and move on to Part c and d.

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So why don't you pause the video
and work on that, and

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we'll check back in
a few minutes and

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we'll solve it together.

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Welcome back.

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I hope you had some luck working
these problems. So

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let's do the easy one first,
computing the gradient.

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So for the gradient we just
take the two partial

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derivatives.

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So we get 5x to the fourth
plus 3y cubed.

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That's the partial derivative in
the x-direction, and in the

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y-direction, we just
get 9xy squared.

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So now for Part b, we're
asked to compute

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this integral directly.

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So we have to recall
what it means.

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So first of all, if we go back
over here, we have this curve

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C. And we need to give a
parameterization for it, and

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so we're going to introduce
a parameterization r of a

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variable t, and we're going
to use that to do our

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computations.

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So let's set r of t--

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so this is our usual circle
that we're used to working

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with, so we're just going
to take the usual

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parameterization,
cos t and sin t.

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And what's important is that the
range is going to be from

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t equals 0 to t equals pi.

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It's t equals pi because we
don't want to go all the way

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around the circle.

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We just want to go halfway
around until we get to

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negative 1.

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So if that is r of t, then
we can compute the

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differential dr of t.

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And so it's going to be just
taking the derivative.

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So we have negative sin
t and cos t dt.

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And so now we can just
write out this

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line integral directly.

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So the integral over C of
F dot dr just becomes--

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so we have the integral
from t equals 0 to pi.

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Those are the ranges
for our curve.

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And now we're going to take the
dot product of F, which

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was (5x to the fourth plus
3y cubed, 9xy squared).

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We're just going to dot this
with our dr vector, which is

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(-sin t, cos t).

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All together we have dt.

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And so now, notice that here
we've got the variables x and

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y, and here we've got the
variables t, but because of

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our parameterization, we
actually know that, for

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instance, x is cos
t and y is sin t.

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So we can write this all out.

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So we have 5 cos to the fourth
t plus 3 sin cubed t.

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So that's this guy written
out in terms of t.

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And then we multiply it
by a negative sin t.

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And then to that we add
the other components.

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So we have plus a 9.

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So we have cos t coming from the
x and another cos t here.

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So we have cos squared t, and
we have a sin squared t dt.

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OK, so that's what it means to
compute this line integral

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directly, and it's not
something that I

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look forward to doing.

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So let's see if we can use path
independence to make our

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lives a little bit simpler.

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So that's going to be c.

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So what we want to do is we want
to replace our original

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curve C with any other curve
that has the same starting

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point and the same
ending point.

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And the curve that I would like
to use is just a straight

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line connecting them.

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There's lots of different
choices that you could do, but

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to me this one seems
the most natural.

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So let's give that a try.

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So let's let r of t be
the curve (-t,0).

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Negative t because we want it
to run moving to the left.

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And then our range is just going
to be from minus 1 to 1.

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So when t is minus 1, then we
get minus the negative 1 and

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it starts at 1.

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And when t is 1 it goes
to negative 1.

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And notice that it goes along
the y equals 0 axis.

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So now we can do the same
computation that we did before

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but we can use this curve.

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So the thing that I want to
emphasize is that if we're

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computing a line integral
of a gradient function--

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so of a function which
is conservative--

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then we can use any line and
we can use any path that

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connects the two end points.

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We can replace our path.

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And so that's what we did.

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We replaced C1 with C2.

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So now this becomes much
easier in two ways.

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So we'll see.

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So our range now is just t
goes from minus 1 to 1.

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And so dr here is
just (-1, 0).

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That's dr. And there's a dt.

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And let's see.

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So now F, we had this value
for F, but notice that the

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y-coordinate is always
0 along this curve.

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So the y-coordinate being 0
means that we just have 5t to

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the fourth and then 0 here.

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That's it, because we set y
to be 0 along this curve.

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So all together this is a very
nice integral to do.

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So just taking this dot product,
all we have is minus

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5t to the fourth dt.

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That's simplified greatly.

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And we just have minus t to the
fifth from 1 to minus 1.

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And so we get simply minus 2.

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So that was a much, much more
straightforward integral to do

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than the one we started with.

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Now finally, in d, we're
suggested to use the

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fundamental theorem
of line integrals.

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So let's remember
what that says.

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That says that if we have any
curve and the line integral

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that we're taking--

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if we know that we're taking the
line integral not of any

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vector field, but of a vector
field which is already the

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gradient of f, then that tells
us that this is simply f of

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the endpoint minus f of the
starting point of our curve.

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So really we don't need to
do any integral at all.

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And so let's see.

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So recall that f was x to the
fifth plus 3xy cubed.

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And so the endpoint--

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so we just need to take
f of (-1, 0) and

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subtract f of (1, 0).

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And so plugging this all in
together we get minus 1 minus

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a positive 1.

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All together we get minus 2.

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And, of course, this does agree
with what we did when we

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computed using the
line integrals.