CHRISTINE BREINER: Welcome back to recitation.
In this video, I'd like us to consider
the following problem.
The first part is I'd like to know for what values of b is
this vector field F conservative.
And F is defined as y i plus the quantity x plus byz j,
plus the quantity y squared plus 1, k.
So as you can see, the only thing we're allowed to
manipulate in this problem is b. b will be some real number.
And I want to know what real numbers can I put in for b so
that this vector field is conservative.
The second part of this problem is for each b-value
you determined from one, find a potential function.
So fix the b-value for one of the ones that is acceptable,
based on number one, and then find the potential function.
And then the third part says that you should explain why F
dot dr is exact, and this is obviously for the b values
determined from one.
So the second and third part are once
you know the b-values.
And you're only going to use those b-values that make F
conservative, because that's the place where we can talk
about finding a potential function, and where we can
talk about F dot dr being exact are
exactly those values.
OK.
So why don't you pause the video, work on these three
problems, and then when you're feeling good about them, bring
the video back up and I'll show you what I did.
OK, welcome back.
Again, we're interested in doing three things with this
vector field.
The first thing we want to do is to find the values of b
that make this vector field conservative.
So I will start with that part.
And as we know from the lecture, the thing I
ultimately need to do is I need to find the curl of F.
OK, so the curl of F is going to measure how far F is from
being conservative.
So if the curl of F is 0, I'm going to have F being
conservative.
So that's really what I'm interested in doing first.
So I'm actually just going to rewrite what the curl of F
actually is.
So I'm going to let F--
I'm going to denote in our usual way by P, Q, R--
in this case, be specifically y comma x plus byz comma y
squared plus 1.
OK, that's my P, Q, R. So P is y, Q is x plus byz, and R is y
squared plus 1.
And now the curl of F--
which was found in the lecture, so I'm not going to
show you again how to find it, I'm just going to write the
formula for it-- is exactly the following vector.
It's the derivative of the R component with respect to y,
minus the Q component with respect to z.
That's the i-value.
The j-value is P sub z minus R sub x, j.
The k-value is Q sub x minus P sub y, k.
OK, so there are three components here.
And let's just start figuring out with these values are, and
then we'll see what kind of restrictions we have on b.
So let's start doing this.
So again, this is P, this is Q, and this is R. So R sub y
is the derivative of this component with respect to y.
That's just 2y.
Q sub z is the derivative of this with respect to z.
Well, this is 0, and this is by.
OK.
Let's look at the rest first. P sub z: the derivative of
this with respect to z is 0.
The derivative of R with respect to x is 0.
So that doesn't have any b's in it at all.
And then Q sub x minus P sub y.
Q is the middle one.
Q sub x is 1.
P is the first one.
P sub y is 1.
OK, so what do we get here?
I should have written equals there, maybe.
OK, so the j component is 0.
And the k component is 0.
So all I'm left with is 2y minus by, i.
And if I want F to be conservative, this quantity
has to be 0, so I see there's only one b-value that's going
to work, and that is b is equal to 2.
OK?
So I know in part one, the answer to the question is just
for b equals 2 is F conservative.
That was, maybe, poorly phrased.
F is conservative only when b is 2.
OK.
And that's because the curl of F is 0 only when b is 2.
All right.
So now we can move on to the second part.
And the second part is for this particular value of b,
find a potential function.
And our strategy for that is going to be one of the methods
from lecture.
And it's going to be the method from lecture that in
three dimensions is much easier than the other.
So the one method in lecture that's easy in three
dimensions is where you start at the origin, and you
integrate F dot dr along a curve that's
made up of line segments.
So this strategy I've done before in two dimensions in
one of the problems in recitation.
Now, we'll see it in three dimensions.
So what we're going to do is we're going to integrate along
a certain curve F dot dr. And this curve is going to go from
the origin to (x1, y1, z1).
And that will give us f of (x1, y1, z1).
OK.
So this is a sort of general strategy, and now we'll talk
about it specifically.
This will actually be f of (x1, y1, z1) plus a constant,
but we'll deal with that part right at the end.
OK, so C in this case is going to be made up of three curves.
And I'm going to draw them in a picture and then we're going
to describe them.
So I'm going to start at (0, 0, 0).
My first curve will go out to x1 comma 0, 0, and that's
going to be the curve C1.
Oops.
I want that to go the other way.
That way.
OK, C1 is going to go from the origin to x1 comma 0, 0.
So the y- and z-values are going to be 0 and 0 all the
way along, and the x-value is going to change.
My next one--
I'm going to make it long so I can have
enough room to write--
that's going to be my C2.
And that's going to be x1 comma y1 comma 0.
So in the end, what I've done is I've taken my x-value, I've
kept it fixed all the way along here, but I'm varying
the y-value out to y1.
And then the last one is going to go straight up.
Right there.
And so it's going to be with the x-value and y-value fixed.
And at the end, I will be at x1, y1, z1.
And this is C3.
And this one, I'm going to move in this direction.
Those are my three curves.
And I want to point out that in order to understand how to
simplify this problem, I'm going to have to remind myself
what F dot dr is.
OK.
So F dot dr is P dx plus Q dy plus R dz.
Right?
That's what F dot dr is.
And so what I'm interested in, I'm going to integrate each of
these things along C1, C2, C3.
But let's notice what happens along certain
numbers of these curves.
If we come back over here.
On C1, y is fixed and z as fixed.
So dy and dz are both 0.
So on C1, I only have to integrate P. OK, so I'm going
to keep track of that.
On C1, which C1 is parametrized by x--
0 to x--
I only need to worry about the P. P of x, 0, 0, dx.
This is my C1 component, and there's nothing else there,
because these two are both 0.
Right?
Now let's consider what happens on C2.
If I look here.
On C2, the x-value is fixed and the z-value is fixed. x is
fixed at x1 and z is fixed at 0.
And so dx and dz are both 0, because x
and z are not changing.
So there's only a dy component.
So on C2, which is parametrized in y--
from 0 to y1--
I'm only interested in Q at x1 comma y comma 0, dy.
Again, this component is 0 on C2 because dx is 0.
And this component is 0 on C2 because dz is 0.
And this component--
I'm evaluating it--
x is fixed at x1, z is fixed at 0, and the y is varying
from 0 to y1.
And then there's one more component, and I'm going to
write it below, and then we'll do the rest over here.
And the third component is the C3 component.
Now, not surprisingly--
if I come back over here--
because x and y are fixed all along the C3 component, the
only thing that's changing is z.
So dx and dy are 0, so I'm only worried
about the dz part.
OK.
So again, as happened before, I only had P in the first one
and Q in the second one, and now I have R only
in the third one.
And it's parametrized in z, from 0 to z1.
That's what z varies over.
And it's going to be R at x1 comma y1 comma z dz.
Because the x's are fixed at x1, the y is fixed at y1, but
z is varying from 0 to z1.
All right.
So I have these three parts, and now I just have to fill
them in with the vector field that I have. I want to find
what P is at (x, 0, 0), what Q is at (x1, y1, 0), and what R
is at (x1, y1, z).
And then integrate.
So I have two steps left.
One is plugging in and one is evaluating.
So let me remind us what P, Q, and R actually are, and then
we'll see what we get.
Let me write it again.
Maybe here.
P, Q, R was equal to y comma x plus 2yz--
I'll put it here so you don't have to look and I don't have
to look-- and then y squared plus 1.
OK.
So P at x comma 0 comma 0.
Well, if I plug in 0 for y, P is 0.
So P at (x, 0, 0) is equal to 0.
So I get nothing to integrate in the first part.
That's nice.
OK.
Now, what is Q at x1 comma y comma 0?
Well, that would be an x1 here.
0 for y makes this term go away.
So it's just equal to x1.
Right?
And then R at x1 comma y1 comma z is y1 squared plus 1.
So now I'm going to substitute these in to what I'm
integrating.
So in the first one, there's nothing there.
Let me just write it right here.
The Q is going to be the integral from
0 to y1 of x1 dy.
And the R part is going to be the integral from 0 to z1 of
y1 squared plus 1 dz.
OK, so the P part was disappeared.
This is the Q part evaluated where I
needed it to be evaluated.
It's just x1 dy.
And the R part evaluated at (x1, y1, z) is just y1
squared plus 1.
And so I integrate that in z.
So if I integrate this in y, all I get is x1 y
evaluated 0 and y1.
So here I just get an x1 y1.
Right?
And then here, if I integrate this in z, I just get a z--
and so I evaluate that at z1 and 0--
I just get z1 times y1 squared plus 1.
OK.
So this is actually my potential function.
And so let me write it formally.
I should actually say, this is my final answer.
Right?
I was integrating.
This is actually what I get.
And so what I was trying to find-- if you remember.
I'm going to come back here and just mention it again.
What I was doing was I was integrating along a curve, F
dot dr, to give me f of (x1, y1, z1).
Right?
So now I've found it.
The only thing I said is we also have to allow for there
to be a constant.
OK.
So the potential function is actually exactly this function
plus a constant.
OK.
So this is f of (x1, y1, z1).
And since I don't have much room above, I'll
just write it below.
So that's my potential function for this vector
field, capital F, when it is conservative.
So when b is equal to 2.
OK, and there was one last part to this question.
Right?
So if we come all the way back over, we're
reminded of one last part.
It was explain why F dot dr is exact for the b values
determined from number one.
And the reason is exactly because of
the following thing.
F is conservative based on the fact that b is 2.
And so when we talk about when F dot dr is exact, the
simplest case is capital F is conservative, and I'm on a
simply connected domain.
OK.
And if you notice, capital F is defined for all values x,
y, z, and is differentiable for all values x, y, z.
So F is defined and differentiable
everywhere on R3.
R3 is simply connected.
So we have a conservative vector field on a simply
connected region.
And that's what it means.
That's one way that we have of knowing F dot dr is exact.
And so that actually answers the
third part of the question.
So again, let me just remind you what we did.
We started with a vector field F, we found values for b that
made that vector field conservative, and then we used
one of the techniques in class to find a potential function
for that value of b.
So there were a number of steps involved, but
ultimately, again, it's the same type of problem you've
seen before when F was a vector field in two
dimensions.
So it shouldn't be feeling too different from some of the
stuff you saw earlier.
OK, I think that's where I'll stop.