WEBVTT
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CHRISTINE BREINER: Welcome
back to recitation.
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In this video, I'd like us to
see how geometric methods can
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help us understand the flux of
a vector field across a curve.
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So in particular,
what we're going to do
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is try and use geometric
methods to compute
00:00:20.155 --> 00:00:24.940
the flux of four different
vector fields F across curve C.
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So I've labeled them
for later purposes.
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I've labeled them F_1,
F_2, F_3, and F_4,
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and they are as follows.
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F sub 1 is the
vector field that is
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a scalar function
of only the radius,
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times the vector x comma y.
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And the curve I'm
interested in in this part
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is the unit circle.
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The second one, part b, is the
vector field that is g of r--
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again, where g is
a scalar function
00:00:53.530 --> 00:00:56.020
and r is the radius,
so it depends only
00:00:56.020 --> 00:00:59.410
on the radius-- times
the vector minus y, x.
00:00:59.410 --> 00:01:02.470
And again the C will
be the unit circle.
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The third and fourth ones,
we use a different C,
00:01:05.300 --> 00:01:06.720
but they will be the same there.
00:01:06.720 --> 00:01:08.479
I'll point that out.
00:01:08.479 --> 00:01:10.020
So in the third one,
the vector field
00:01:10.020 --> 00:01:12.970
is 3 times the vector [1, 1].
00:01:12.970 --> 00:01:15.870
And C in this case will be
the segment connecting (0, 0)
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to (1, 1).
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So it's a piece of
the line y equals x.
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And then in part d, F will be
3 times the vector [-1, 1].
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And C is, again, this segment
from (0, 0) to (1, 1).
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So again, what I'd like you
to do is rather than trying
00:01:31.380 --> 00:01:34.550
to parametrize the curve and
do the entire calculation,
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I'd like to see you
try and understand
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the relationship between
each of these vector fields
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F and the normal to the
curve that they're on,
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and see if you can
figure out the flux based
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on that relationship.
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So why don't you pause the
video, give that a shot,
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and then when you're
ready to see how I did it,
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you can bring the video back up.
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OK, welcome back.
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Again, I'm going to try and
use my geometric intuition
00:02:06.110 --> 00:02:09.140
to understand what the flux is
for each of these four vector
00:02:09.140 --> 00:02:11.020
fields along these four curves.
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So what I'd like to do is, if
I'm going to try and understand
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geometrically what's
happening, it's
00:02:15.590 --> 00:02:17.086
always good to draw a picture.
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So I'm going to draw a picture
that I'll use for a and b,
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and then I'll draw
another picture later
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that I'll use for c and d.
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So notice again in a, my
curve is the unit circle.
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And I have a somewhat
explicit understanding
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of what the vector fields are.
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So I'm going to
draw the unit circle
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and see if I can figure
out where F_1 and F_2 are.
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It's not a perfect unit
circle, but it looks sort of
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like the unit circle.
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So this is going to
be my unit circle.
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And what I want
to point out first
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is that I was not
trying to trick you,
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but just to help you
notice, that in a and b,
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they both depended on
this radial function.
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But the radius on the
unit circle is fixed at 1.
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So in both of these
vector fields,
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it will simply be g of 1.
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So that will be
a constant value.
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So all that's giving me is some
scalar multiple of whatever
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the length of this one is.
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So it's this direction
times this scalar g of 1.
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One other thing I
want to point out
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is that both of these
vectors, F_1 and F_2,
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if I ignore the g part, and
I look at just the x comma
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y and the minus y comma x, is
these parts have unit length.
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And you should be able to see
that right away, because x
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and y are on the unit circle.
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And so the length of
the vector whose tail
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is at the origin and head is
on the edge of the unit circle
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has length 1.
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And then you can easily see
that this vector and this vector
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have the same length, because
their individual components are
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the same absolute value.
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We already understand a few
things about F_1 and F_2
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right away.
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That all the length
is coming from this g,
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and g is fixed all the
way around the unit circle
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at g of 1.
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Now let's figure out what the
flux is for these two things.
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So if you notice first, F
sub 1 is the vector [x, y]
00:04:10.220 --> 00:04:12.190
times the scalar g of 1.
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So if I come over here,
I want to point out--
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I'll do this part
in white first--
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that if I'm at this point--
this is the point (x, y)--
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the vector [x, y] is
equal to this vector.
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So if I think about putting
that at this point--
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I'm going to draw
it here-- I get
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something that looks like this.
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So this is F sub 1, probably.
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I'm going to have to make
one comment about that.
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But notice, this should look
like it's all in one direction.
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So this is the vector [x, y].
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If I slide it so
its tail is here,
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it's again in the
same direction,
00:04:51.460 --> 00:04:54.800
and now I've just
scaled it by g of 1.
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Now, this is assuming,
obviously, that g of 1
00:04:56.920 --> 00:04:57.471
is positive.
00:04:57.471 --> 00:04:59.720
So we're going to assume
that throughout this problem.
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I'll mention what happens
when g is negative at the end.
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So this is my vector F sub 1.
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Let's think about what is
the normal to this curve.
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The normal to this
curve, actually
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at each point on
the circle, points
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in exactly the same
direction as F sub 1.
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Because if I'm parametrizing
in this direction,
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the normal-- I'll
draw one down here
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so we can see what it looks
like-- the normal-- actually,
00:05:21.850 --> 00:05:24.290
let me come from there--
the normal is going
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to look something like this.
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OK.
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So it would be connecting
from the origin
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to that point on
the circle, and keep
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going out in that direction.
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That's the normal direction.
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So F sub 1, if g is
a positive function,
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it points in exactly the
same direction as the normal.
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If g is a negative function,
it points in exactly
00:05:41.330 --> 00:05:42.460
the opposite direction.
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So F_1 would be
flipped exactly around
00:05:44.590 --> 00:05:46.350
if g was a negative function.
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So if I want to compute the
flux for part a-- I'll do it
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down here-- if I want to
compute the flux, remember,
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I'm taking the integral
along the curve
00:05:54.110 --> 00:05:58.020
of F dotted with n ds.
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Well, F dotted
with n is constant.
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And that's the main point that's
going to make this easier.
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At each point-- I guess I should
say F_1-- F_1 dotted with n
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is always equal to the length of
F_1 times the length of n times
00:06:12.740 --> 00:06:15.090
cosine of the
angle between them.
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With a very quick
calculation, you
00:06:16.520 --> 00:06:19.990
can see that winds
up being g of 1.
00:06:19.990 --> 00:06:22.980
So the only reason I don't have
to worry about absolute value,
00:06:22.980 --> 00:06:25.575
is if g is positive, I'm
pointing in the same direction.
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If g is negative, I'm pointing
in the opposite direction.
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And so the cosine theta is
minus 1 instead of plus 1.
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You might want to check
that for yourself,
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but this is just g of one.
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So that's a constant.
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So this is actually
equal to g of 1 times
00:06:39.970 --> 00:06:42.820
the integral over C of ds.
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Now, what is this?
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If I integrate this, I should
pick up exactly the length
00:06:46.590 --> 00:06:47.800
of the curve.
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OK.
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Because this is the
derivative of arc length,
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so when I integrate
this, I get arc length.
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But it's a unit circle,
so the arc length
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is just the circumference
of the unit circle.
00:06:56.960 --> 00:07:03.500
So that's 2*pi times g of 1.
00:07:03.500 --> 00:07:04.312
OK.
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And that's all you get.
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That's it.
00:07:06.560 --> 00:07:08.685
So we didn't actually have
to parametrize anything.
00:07:08.685 --> 00:07:11.250
We just had to understand
F_1 relating to the normal.
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So I didn't draw the normal
here, but if I take this normal
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and I spin around
to here, the normal
00:07:16.150 --> 00:07:18.880
is in the same
direction as F sub 1.
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So now let's look at F sub 2.
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And let's do this first
by pointing something out
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about the relationship
between F sub 2 and F sub 1.
00:07:27.750 --> 00:07:31.230
Notice that if I take F sub
1 and I dot it with F sub 2,
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I get 0.
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Right?
00:07:32.900 --> 00:07:38.150
Because ignoring even the scalar
part, I get x times minus y,
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plus y times x,
which gives me 0.
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If the scalars come along
for the ride, I still get 0.
00:07:44.640 --> 00:07:47.810
So F sub 1 and F sub
2 are orthogonal.
00:07:47.810 --> 00:07:51.144
And in fact-- you can
do this for yourself,
00:07:51.144 --> 00:07:52.810
but if I come over
and draw the picture,
00:07:52.810 --> 00:07:55.660
F sub 2 is going to be F
sub 1 rotated by 90 degrees.
00:07:58.560 --> 00:07:59.620
Something like this.
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This is my F sub 2.
00:08:01.590 --> 00:08:06.270
Again, if g of 1 was negative,
F sub 1 would be this direction,
00:08:06.270 --> 00:08:08.960
and F sub 2 would
then be around here.
00:08:08.960 --> 00:08:11.670
But ultimately, it's not
going to matter in this case
00:08:11.670 --> 00:08:13.750
whether g is positive or
negative, because notice
00:08:13.750 --> 00:08:14.860
what happens.
00:08:14.860 --> 00:08:19.130
If I want to integrate F sub
2 dotted with the normal,
00:08:19.130 --> 00:08:21.930
notice the normal is in the
same direction as F sub 1,
00:08:21.930 --> 00:08:24.400
so F sub 2 dotted
with the normal is 0.
00:08:24.400 --> 00:08:27.500
So if I'm going to integrate the
function 0 all along the curve,
00:08:27.500 --> 00:08:33.410
I shouldn't be surprised that
my answer to part b is 0.
00:08:33.410 --> 00:08:35.190
So there was even
less work in part b.
00:08:35.190 --> 00:08:37.450
Because I immediately
had that F sub
00:08:37.450 --> 00:08:40.199
2 is really in the direction
of the tangent to the curve.
00:08:40.199 --> 00:08:42.490
And so I have something in
the direction of the tangent
00:08:42.490 --> 00:08:44.370
dotted with something
in the direction
00:08:44.370 --> 00:08:47.930
of the normal-- in fact,
the normal-- so I get 0.
00:08:47.930 --> 00:08:48.430
All right.
00:08:48.430 --> 00:08:51.239
So now I'm going to draw
a picture for c and d,
00:08:51.239 --> 00:08:52.780
and then we're going
to use that one.
00:08:55.960 --> 00:08:58.070
And I have to make sure
I come on this side.
00:08:58.070 --> 00:08:59.630
Sorry about that.
00:08:59.630 --> 00:09:00.795
So here is (0, 0).
00:09:04.110 --> 00:09:06.930
And here is (1, 1).
00:09:06.930 --> 00:09:07.930
Actually, you know what?
00:09:07.930 --> 00:09:09.020
I'm going to make
it a little longer.
00:09:09.020 --> 00:09:10.160
I might need more room.
00:09:10.160 --> 00:09:14.970
So (1, 1) I'll make
a little further up.
00:09:14.970 --> 00:09:15.960
OK?
00:09:15.960 --> 00:09:18.192
(0, 0) and (1, 1),
and that's my curve.
00:09:18.192 --> 00:09:19.260
There's (1, 1).
00:09:19.260 --> 00:09:20.240
OK.
00:09:20.240 --> 00:09:23.190
Now, if I parametrize
it in this direction,
00:09:23.190 --> 00:09:26.360
then I can draw my normal.
00:09:26.360 --> 00:09:28.760
Because it's a line
segment, my normal
00:09:28.760 --> 00:09:31.030
is constant in its
length and direction.
00:09:31.030 --> 00:09:39.040
So at any given point, it's
exactly equal to this vector,
00:09:39.040 --> 00:09:40.350
up to the right scaling.
00:09:40.350 --> 00:09:45.740
And it should be
something like [1, -1]
00:09:45.740 --> 00:09:47.722
divided by square root 2.
00:09:47.722 --> 00:09:48.430
That's my normal.
00:09:48.430 --> 00:09:50.930
If you want it precisely,
that's what it is.
00:09:50.930 --> 00:09:53.520
You don't actually need it to
solve this problem, though.
00:09:53.520 --> 00:09:56.030
But that's what it is if
you want it precisely.
00:09:56.030 --> 00:09:59.440
Now let's look at what
F_3 is, and then we'll
00:09:59.440 --> 00:10:00.780
look at what F_4 actually is.
00:10:00.780 --> 00:10:03.450
So let's look at F sub 3.
00:10:03.450 --> 00:10:08.960
F sub 3 was the
vector 3 times [1, 1].
00:10:08.960 --> 00:10:11.390
So if we come back
to our picture,
00:10:11.390 --> 00:10:16.990
at any point on this curve, if
I go in the [1, 1] direction,
00:10:16.990 --> 00:10:19.860
I stay parallel to this curve.
00:10:22.329 --> 00:10:23.870
I don't want to draw
the whole thing,
00:10:23.870 --> 00:10:25.620
because it would take
up the entire curve.
00:10:25.620 --> 00:10:27.420
It's longer than
the curve itself.
00:10:27.420 --> 00:10:32.520
But F sub 3, at any given
point, points in this direction.
00:10:32.520 --> 00:10:37.920
So this is F sub 3, but not
as long as it actually is.
00:10:37.920 --> 00:10:40.500
But only the direction is
going to matter in this case.
00:10:40.500 --> 00:10:43.000
So F sub 3 is pointing
in this direction.
00:10:43.000 --> 00:10:45.809
So if I want to compute the
flux, I dot it with the normal.
00:10:45.809 --> 00:10:46.850
But look at what happens.
00:10:46.850 --> 00:10:50.580
The normal is orthogonal
to F sub 3 at every point.
00:10:50.580 --> 00:10:52.760
And so F sub 3 dotted
with the normal is 0.
00:10:52.760 --> 00:10:57.150
And so again, for exactly
the same reason as part b,
00:10:57.150 --> 00:10:59.960
in part c, the flux is 0.
00:10:59.960 --> 00:11:03.260
So again, it's exactly the
same, that the vector field
00:11:03.260 --> 00:11:05.700
I was looking at and I wanted
to compute the flux for,
00:11:05.700 --> 00:11:08.460
is actually tangent to
the curve at every point.
00:11:08.460 --> 00:11:11.190
And so when I dot
it to the normal
00:11:11.190 --> 00:11:13.870
to the curve at
every point, I get 0.
00:11:13.870 --> 00:11:16.420
And so computing the flux is 0.
00:11:16.420 --> 00:11:20.460
So now, I have one more to
do, and that one is part d.
00:11:20.460 --> 00:11:23.010
And in this case, F
sub 4-- let me just
00:11:23.010 --> 00:11:28.600
remind you-- is 3 times
the vector [-1, 1].
00:11:28.600 --> 00:11:30.760
And so if we go back
to our picture here,
00:11:30.760 --> 00:11:32.830
F sub 4, if I compare
it to the normal,
00:11:32.830 --> 00:11:36.226
in fact, what I
get is very long.
00:11:36.226 --> 00:11:37.850
This is probably not
quite long enough.
00:11:37.850 --> 00:11:43.400
But that's at least the
direction of F sub 4.
00:11:43.400 --> 00:11:46.580
And so F sub 4 is exactly
the opposite direction
00:11:46.580 --> 00:11:48.350
to the normal.
00:11:48.350 --> 00:11:54.290
So if I want to compute the flux
of F sub 4 along this curve,
00:11:54.290 --> 00:11:57.100
all I have to understand is F
sub 4 dotted with the normal
00:11:57.100 --> 00:11:59.120
and the length of the curve.
00:11:59.120 --> 00:12:02.600
This is exactly the same
type of solution as part a.
00:12:02.600 --> 00:12:03.540
So let's notice this.
00:12:03.540 --> 00:12:06.690
First, F sub 4, the
length is 3 root 2.
00:12:06.690 --> 00:12:08.610
You can compute
that pretty quickly.
00:12:08.610 --> 00:12:10.870
The length of n is just 1.
00:12:10.870 --> 00:12:11.482
It's a normal.
00:12:11.482 --> 00:12:12.940
That's why this
stuff didn't really
00:12:12.940 --> 00:12:14.950
matter what exactly it was.
00:12:14.950 --> 00:12:17.710
It's good to know what
direction it points in.
00:12:17.710 --> 00:12:20.040
So F sub 4 dotted
with n is exactly
00:12:20.040 --> 00:12:25.340
3 root 2 times cosine of the
angle between n and F sub 4.
00:12:25.340 --> 00:12:26.350
The angle is pi.
00:12:26.350 --> 00:12:29.610
You notice they differ
by 180 degrees, right?
00:12:29.610 --> 00:12:33.080
So it's cosine pi,
which is minus 1.
00:12:33.080 --> 00:12:37.160
So what I do in part d, is I'm
integrating along the curve
00:12:37.160 --> 00:12:42.240
the constant negative
3 square root 2 ds.
00:12:42.240 --> 00:12:46.690
This is exactly F sub 4
dotted with the normal.
00:12:46.690 --> 00:12:51.030
And so as before, this is
going to equal to negative 3
00:12:51.030 --> 00:12:52.870
root 2 times the arc length.
00:12:52.870 --> 00:12:55.110
Because the integral
over the curve
00:12:55.110 --> 00:12:57.590
ds is going to be
the arc length.
00:12:57.590 --> 00:13:00.480
And the arc length
is very easy to see.
00:13:00.480 --> 00:13:02.470
You've gone over 1 and up 1.
00:13:02.470 --> 00:13:02.970
Right?
00:13:02.970 --> 00:13:06.050
So Pythagorean theorem,
understanding right triangles,
00:13:06.050 --> 00:13:10.160
however you want to do it, the
length of this curve is root 2.
00:13:10.160 --> 00:13:16.380
So this works out to be negative
3 root 2 root 2, which is just
00:13:16.380 --> 00:13:19.210
negative 6.
00:13:19.210 --> 00:13:22.450
So let me just remind you,
there were four problems here.
00:13:22.450 --> 00:13:25.720
There are two sets of
problems, where in each case,
00:13:25.720 --> 00:13:27.560
you have one similar
to the other.
00:13:27.560 --> 00:13:30.490
So let me point this out
one more time, and just
00:13:30.490 --> 00:13:31.530
sort of step back.
00:13:31.530 --> 00:13:34.060
We had a circle, and
then in the next part
00:13:34.060 --> 00:13:35.140
we had a line segment.
00:13:35.140 --> 00:13:36.990
But in the circle,
one of the problems
00:13:36.990 --> 00:13:39.927
had the vector in the
direction of the normal,
00:13:39.927 --> 00:13:41.760
and you wanted to compute
the flux for that.
00:13:41.760 --> 00:13:44.580
And in the other, the vector
was tangent to the curve,
00:13:44.580 --> 00:13:46.205
and so it was orthogonal
to the normal.
00:13:46.205 --> 00:13:48.620
And you wanted to compute
the flux for that.
00:13:48.620 --> 00:13:51.020
Obviously, when you're
tangent to the curve
00:13:51.020 --> 00:13:53.800
and then orthogonal to
the normal, you get 0.
00:13:53.800 --> 00:13:56.370
And that was the
case for b and for c.
00:13:56.370 --> 00:14:00.390
When you're normal to the curve
and of constant length-- which
00:14:00.390 --> 00:14:04.490
was the case actually for both a
and d-- then all you have to do
00:14:04.490 --> 00:14:08.810
is find F dotted
with the normal,
00:14:08.810 --> 00:14:11.320
and then find the arc length,
and multiply them together.
00:14:11.320 --> 00:14:16.010
So that was the real strategy
we had to use for a and for d.
00:14:16.010 --> 00:14:19.390
So hopefully, this helps you
see how the geometric quantities
00:14:19.390 --> 00:14:22.380
are interacting to understand
the flux of a vector field
00:14:22.380 --> 00:14:23.200
across a curve.
00:14:23.200 --> 00:14:25.032
And that's where I'll stop.