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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, I'd like us to
consider the following problem.

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The first part is I'd like
to know, for what values of b

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is this vector field
F conservative?

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And F is defined as y*i plus
the quantity x plus b*y*z j plus

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the quantity y
squared plus 1, k.

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So as you can see,
the only thing

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we're allowed to
manipulate in this problem

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is b. b will be
some real number.

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And I want to know, what real
numbers can I put in for b so

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that this vector
field is conservative.

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The second part
of this problem is

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for each b-value you
determined from one,

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find a potential function.

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So fix the b-value for one of
the ones that is acceptable,

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based on number one, and then
find the potential function.

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And then the third part
says that you should explain

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why F dot dr is exact,
and this is obviously

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for the b-values
determined from one.

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So the second and third part
are once you know the b-values.

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And you're only going to
use those b-values that

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make F conservative, because
that's the place where we can

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talk about finding a
potential function,

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and that's where we can talk
about F dot dr being exact,

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are exactly those values.

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OK.

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So why don't you pause the
video, work on these three

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problems, and then when you're
feeling good about them,

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bring the video back up,
I'll show you what I did.

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OK, welcome back.

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Again, we're interested in doing
three things with this vector

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field.

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The first thing we
want to do is to find

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the values of b that make this
vector field conservative.

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So I will start with that part.

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And as we know from
the lecture, the thing

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I ultimately need to do is I
need to find the curl of F. OK,

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so the curl of F is going
to measure how far F

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is from being conservative.

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So if the curl of
F is 0, I'm going

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to have F being conservative.

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So that's really what I'm
interested in doing first.

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So I'm actually just going
to rewrite what the curl of F

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actually is.

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So I'm going to let F-- I'm
going to denote in our usual

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way by P, Q, R.
And in this case,

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that's specifically y comma x
plus b*y*z comma y squared plus

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1.

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OK, that's my P, Q, R. So
P is y, Q is x plus b*y*z,

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and R is y squared plus 1.

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And now the curl of F-- which
was found in the lecture,

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so I'm not going to show
you again how to find it,

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I'm just going to write the
formula for it-- is exactly

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the following vector.

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It's the derivative of the R-th
component with respect to y,

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minus the Q-th component
with respect to z.

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That's the i-value.

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The j-value is P sub
z minus R sub x, j.

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The k-th value is Q
sub x minus P sub y, k.

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OK, so there are
three components here.

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And let's just start figuring
out what these values are,

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and then we'll see what kind
of restrictions we have on b.

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So let's start doing this.

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So again, this is P,
this is Q, and this

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is R. So R sub y is the
derivative of this component

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with respect to y.

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That's just 2y.

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Q sub z is the derivative
of this with respect to z.

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Well, this is 0,
and this is b*y.

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OK.

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Let's look at the rest first.

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P sub z: the derivative of
this with respect to z is 0.

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The derivative of R
with respect to x is 0.

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So that doesn't have
any b's in it at all.

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And then Q sub x minus P sub y.

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Q is the middle one.

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Q sub x is 1.

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P is the first one.

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P sub y is 1.

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OK, so what do we get here?

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I should have written
equals there, maybe.

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OK, so the j-th component is 0.

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And the k-th component is 0.

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So all I'm left with
is 2y minus b*y, i.

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And if I want F to
be conservative,

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this quantity has
to be 0, so I see

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there's only one b-value
that's going to work,

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and that is b is equal to 2.

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OK?

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So I know in part one,
the answer to the question

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is just for b equals
2, is F conservative.

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That was, maybe, poorly phrased.

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F is conservative
only when b is 2.

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OK.

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And that's because the curl
of F is 0 only when b is 2.

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All right.

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So now we can move on
to the second part.

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And the second part is for
this particular value of b,

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find a potential function.

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And our strategy
for that is going

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to be one of the
methods from lecture.

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And it's going to be the
method from lecture that

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in three dimensions is
much easier than the other.

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So the one method in lecture
that's easy in three dimensions

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is where you start
at the origin,

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and you integrate F dot
dr along a curve that's

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made up of line segments.

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So this strategy I've done
before in two dimensions,

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in one of the problems
in recitation.

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Now, we'll see it
in three dimensions.

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So what we're going
to do is we're

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going to integrate along
a certain curve F dot dr.

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And this curve is going
to go from the origin to

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(x_1, y_1, z_1).

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And that will give us
f of (x_1, y_1, z_1).

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OK.

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So this is a sort
of general strategy,

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and now we'll talk
about it specifically.

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This will actually be f of
(x_1, y_1, z_1) plus a constant,

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but we'll deal with that
part right at the end.

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OK, so C in this case is going
to be made up of three curves.

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And I'm going to draw
them, in a picture,

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and then we're going
to describe them.

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So I'm going to
start at (0, 0, 0).

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My first curve will go
out to x_1 comma 0, 0,

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and that's going to
be the curve C_1.

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Oops.

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I want that to go the other way.

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That way.

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OK, C_1 is going to go from
the origin to x_1 comma 0, 0.

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So the y- and z-values
are going to be

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0 and 0 all the way along, and
the x-value is going to change.

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My next one-- I'm
going to make it long

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so I can have enough
room to write-- that's

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going to be my C_2.

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And that's going to be
x_1 comma y_1 comma 0.

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So in the end, what I've done
is I've taken my x-value,

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I've kept it fixed all
the way along here,

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but I'm varying the
y-value out to y_1.

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And then the last one is
going to go straight up.

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Right there.

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And so it's going to be with
the x-value and y-value fixed.

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And at the end, I will
be at x_1, y_1, z_1.

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And this is C3.

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So those are my three
curves-- And this one,

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I'm going to move
in this direction.

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Those are my three curves.

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And I want to point
out that in order

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to understand how to
simplify this problem,

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I'm going to have to remind
myself what F dot dr is.

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OK.

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So F dot dr is P*dx
plus Q*dy plus R*dz.

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Right?

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That's what F dot dr is.

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And so what I'm
interested in, I'm

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going to integrate each of these
things along C_1, C_2, C_3.

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But let's notice what
happens along certain numbers

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of these curves.

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If we come back over here.

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On C_1, y is fixed
and z as fixed.

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So dy and dz are both 0.

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So on C1, I only have
to integrate P. OK,

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so I'm going to
keep track of that.

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On C_1-- which,
C_1 is parametrized

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by x, 0 to x-- I
only need to worry

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about the P. P of x, 0, 0, dx.

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This is my C_1 component, and
there's nothing else there,

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because these two are both 0.

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Right?

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Now let's consider
what happens on C_2.

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If I look here.

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On C_2, the x-value is fixed
and the z-value is fixed.

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x is fixed at x_1
and z is fixed at 0.

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And so dx and dz are both
0, because x and z are not

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changing.

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So there's only a dy component.

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So on C_2, which is parametrized
in y-- from 0 to y_1--

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I'm only interested in Q
at x_1 comma y comma 0, dy.

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Again, this component is
0 on C_2 because dx is 0.

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And this component is 0
on C_2 because dz is 0.

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And this component--
I'm evaluating it--

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x is fixed at x_1,
z is fixed at 0,

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and the y is varying
from 0 to y1.

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And then there's
one more component,

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and I'm going to write
it below, and then

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we'll do the rest over here.

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And the third component
is the C_3 component.

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Now, not surprisingly--
if I come back over here--

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because x and y are fixed
all along the C3 component,

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the only thing
that's changing is z.

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So dx and dy are 0, so I'm
only worried about the dz part.

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OK.

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So again, as happened
before, I only

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had P in the first one
and Q in the second one,

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and now I have R,
only, in the third one.

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And it's parametrized
in z, from 0 to z_1.

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That's what z varies over.

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And it's going to be R at
x_1 comma y_1 comma z dz.

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Because the x's are fixed at
x_1, the y is fixed at y_1,

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but z is varying from 0 to z_1.

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All right.

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So I have these
three parts, and now

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I just have to fill them in with
the vector field that I have.

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I want to find what
P is at (x, 0, 0),

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what Q is at (x_1, y_1, 0),
and what R is at (x_1, y_1, z).

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And then integrate.

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So I have two steps left.

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One is plugging in
and one is evaluating.

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So let me remind us what P, Q,
and R actually are, and then

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we'll see what we get.

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Let me write it again.

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Maybe here.

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[P, Q, R] was equal to
y comma x plus 2y*z--

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I'll put it here so you don't
have to look and I don't have

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to look-- and then
y squared plus 1.

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OK.

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So P at x comma 0 comma 0.

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Well, if I plug in
0 for y, P is 0.

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So P at (x, 0, 0) is equal to 0.

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So I get nothing to
integrate in the first part.

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That's nice.

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OK.

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Now, what is Q at
x_1 comma y comma 0?

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Well, that would be an x_1 here.

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0 for y makes this term go away.

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So it's just equal to x_1.

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Right?

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00:11:09,930 --> 00:11:18,660
And then R at x_1 comma y_1
comma z is y_1 squared plus 1.

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So now I'm going
to substitute these

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into what I'm integrating.

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So in the first one,
there's nothing there.

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Let me just write it right here.

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The Q is going to
be the integral

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from 0 to y_1 of x_1 dy.

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00:11:34,150 --> 00:11:39,160
And the R part is going to
be the integral from 0 to z_1

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of y_1 squared plus 1 dz.

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OK, so the P part
was disappeared.

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This is the Q part
evaluated where

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I needed it to be evaluated.

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It's just x_1 dy.

241
00:11:52,310 --> 00:11:55,200
And the R part evaluated
at (x_1, y_1, z)

242
00:11:55,200 --> 00:11:57,130
is just y_1 squared plus 1.

243
00:11:57,130 --> 00:11:59,500
And so I integrate that in z.

244
00:11:59,500 --> 00:12:03,910
So if I integrate this in y,
all I get is x_1*y evaluated 0

245
00:12:03,910 --> 00:12:04,720
and y1.

246
00:12:04,720 --> 00:12:08,070
So here I just get an x_1*y_1.

247
00:12:08,070 --> 00:12:09,510
Right?

248
00:12:09,510 --> 00:12:12,800
And then here, if I integrate
this in z, I just get a z--

249
00:12:12,800 --> 00:12:14,600
and so I evaluate
that at z_1 and 0--

250
00:12:14,600 --> 00:12:18,890
I just get z_1 times
y1 squared plus 1.

251
00:12:18,890 --> 00:12:19,400
OK.

252
00:12:19,400 --> 00:12:22,330
So this is actually
my potential function.

253
00:12:22,330 --> 00:12:23,932
And so let me write it formally.

254
00:12:23,932 --> 00:12:25,890
I should actually say,
this is my final answer.

255
00:12:25,890 --> 00:12:26,760
Right?

256
00:12:26,760 --> 00:12:27,920
I was integrating.

257
00:12:27,920 --> 00:12:29,730
This is actually what I get.

258
00:12:29,730 --> 00:12:31,180
And so what I was
trying to find,

259
00:12:31,180 --> 00:12:33,429
if you remember-- I'm going
to come back here and just

260
00:12:33,429 --> 00:12:34,190
mention it again.

261
00:12:34,190 --> 00:12:36,740
What I was doing was I was
integrating along a curve F dot

262
00:12:36,740 --> 00:12:39,210
dr, to give me f
of (x_1, y_1, z_1).

263
00:12:39,210 --> 00:12:40,010
Right?

264
00:12:40,010 --> 00:12:41,530
So now I've found it.

265
00:12:41,530 --> 00:12:43,650
The only thing I said
is we also have to allow

266
00:12:43,650 --> 00:12:44,941
for there to be a constant.

267
00:12:44,941 --> 00:12:45,440
OK.

268
00:12:45,440 --> 00:12:50,000
So the potential function is
actually exactly this function

269
00:12:50,000 --> 00:12:52,990
plus a constant.

270
00:12:52,990 --> 00:12:53,950
OK.

271
00:12:53,950 --> 00:12:56,280
So this is f of (x_1, y_1, z_1).

272
00:12:56,280 --> 00:12:57,905
And since I don't
have much room above,

273
00:12:57,905 --> 00:12:59,373
I'll just write it below.

274
00:13:03,450 --> 00:13:05,330
This is f of x_1, y_1, z_1.

275
00:13:05,330 --> 00:13:10,810
So that's my potential function
for this vector field, capital

276
00:13:10,810 --> 00:13:12,920
F, when it is conservative.

277
00:13:12,920 --> 00:13:15,290
So when b is equal to 2.

278
00:13:15,290 --> 00:13:17,470
OK, and there was one last
part to this question.

279
00:13:17,470 --> 00:13:17,970
Right?

280
00:13:17,970 --> 00:13:19,920
So if we come all
the way back over,

281
00:13:19,920 --> 00:13:22,790
we're reminded of one last part.

282
00:13:22,790 --> 00:13:26,160
It was explain why F dot dr
is exact for the b values

283
00:13:26,160 --> 00:13:27,790
determined from number one.

284
00:13:27,790 --> 00:13:32,410
And the reason is exactly
because of the following thing.

285
00:13:32,410 --> 00:13:36,360
F is conservative based
on the fact that b is 2.

286
00:13:36,360 --> 00:13:42,240
And so when we talk about
when F dot dr is exact,

287
00:13:42,240 --> 00:13:47,080
the simplest case is
capital F is conservative,

288
00:13:47,080 --> 00:13:49,810
and I'm on a simply
connected domain.

289
00:13:49,810 --> 00:13:50,440
OK.

290
00:13:50,440 --> 00:13:57,320
And if you notice, capital F is
defined for all values x, y, z,

291
00:13:57,320 --> 00:13:59,900
and is differentiable for
all values of x, y, z.

292
00:13:59,900 --> 00:14:03,250
So F is defined and
differentiable everywhere

293
00:14:03,250 --> 00:14:04,550
on R^3.

294
00:14:04,550 --> 00:14:06,040
R^3 is simply connected.

295
00:14:06,040 --> 00:14:08,120
So we have a
conservative vector field

296
00:14:08,120 --> 00:14:09,650
on a simply connected region.

297
00:14:09,650 --> 00:14:10,700
And that's what it means.

298
00:14:10,700 --> 00:14:14,100
That's one way that we have
of knowing F dot dr is exact.

299
00:14:14,100 --> 00:14:17,910
And so that actually answers
the third part of the question.

300
00:14:17,910 --> 00:14:20,010
So again, let me just
remind you what we did.

301
00:14:20,010 --> 00:14:22,640
We started with
a vector field F,

302
00:14:22,640 --> 00:14:26,320
we found values for b that made
that vector field conservative,

303
00:14:26,320 --> 00:14:27,915
and then we used one
of the techniques

304
00:14:27,915 --> 00:14:31,950
in class to find a potential
function for that value of b.

305
00:14:31,950 --> 00:14:34,350
So there were a number
of steps involved,

306
00:14:34,350 --> 00:14:37,240
but ultimately, again, it's
the same type of problem

307
00:14:37,240 --> 00:14:41,750
you've seen before, when F was a
vector field in two dimensions.

308
00:14:41,750 --> 00:14:45,110
So it shouldn't be feeling
too different from some

309
00:14:45,110 --> 00:14:47,120
of the stuff you saw earlier.

310
00:14:47,120 --> 00:14:49,520
OK, I think that's
where I'll stop.