WEBVTT
00:00:07.605 --> 00:00:08.230
JOEL LEWIS: Hi.
00:00:08.230 --> 00:00:09.355
Welcome back to recitation.
00:00:09.355 --> 00:00:11.563
In lecture, you've been
learning about line integrals
00:00:11.563 --> 00:00:12.560
of vector fields.
00:00:12.560 --> 00:00:16.435
And I have one problem
here that requires
00:00:16.435 --> 00:00:22.080
you to do two line integrals of
vector fields on that subject.
00:00:22.080 --> 00:00:22.630
So OK.
00:00:22.630 --> 00:00:28.490
So in both problems, F is going
to be the vector field whose
00:00:28.490 --> 00:00:32.070
coordinates are x*y and
x squared plus y squared.
00:00:32.070 --> 00:00:35.260
And C is going to be the
arc of the parabola y
00:00:35.260 --> 00:00:39.210
equals x squared that
starts at the point (1, 1)
00:00:39.210 --> 00:00:41.670
and ends at the point (2, 4).
00:00:41.670 --> 00:00:42.350
All right?
00:00:42.350 --> 00:00:45.280
So what I'd like you to do
is to compute the integral
00:00:45.280 --> 00:00:49.520
over this curve C of F dot
dr in two different ways.
00:00:49.520 --> 00:00:51.950
So the first time,
I'd like you to use
00:00:51.950 --> 00:00:56.570
the natural parametrization x
equals t, y equals t squared.
00:00:56.570 --> 00:00:58.770
For me, at least, that's
the first parametrization
00:00:58.770 --> 00:01:01.810
that I think of when I
think about this curve.
00:01:01.810 --> 00:01:03.620
But I'd also like you
to do it again using
00:01:03.620 --> 00:01:05.350
a different
parametrization, using
00:01:05.350 --> 00:01:10.090
the parametrization x equals e
to the t, y equals e to the 2t.
00:01:10.090 --> 00:01:14.490
So this still parametrizes
the curve y equals x squared,
00:01:14.490 --> 00:01:16.830
since e to the t squared
equals e to the 2t.
00:01:19.410 --> 00:01:24.040
So why don't you pause the
video, have a go at computing
00:01:24.040 --> 00:01:27.729
this integral using these two
different parametrizations,
00:01:27.729 --> 00:01:29.520
come back, and we can
work it out together.
00:01:37.650 --> 00:01:39.290
So let's get started.
00:01:39.290 --> 00:01:44.740
We want to compute a line
integral of a vector field F.
00:01:44.740 --> 00:01:51.965
So we know that the integral
over our curve C of F
00:01:51.965 --> 00:01:54.120
dot dr-- well,
usually we write--
00:01:54.120 --> 00:01:58.720
F equals M, N. The two
components are M and N. So
00:01:58.720 --> 00:02:02.400
in this case, this is going to
be equal to the integral over C
00:02:02.400 --> 00:02:08.270
of M dx plus N dy.
00:02:08.270 --> 00:02:10.700
Now, in our case,
we know what F is.
00:02:10.700 --> 00:02:13.075
We know that F is
the vector field x*y,
00:02:13.075 --> 00:02:14.770
x squared plus y squared.
00:02:14.770 --> 00:02:24.680
So this is equal to the integral
over C of x*y dx plus x squared
00:02:24.680 --> 00:02:28.400
plus y squared dy.
00:02:28.400 --> 00:02:31.720
And now we just need to plug
in in our two different cases,
00:02:31.720 --> 00:02:35.080
into our parametrizations, and
this will turn into an integral
00:02:35.080 --> 00:02:36.500
that we can evaluate.
00:02:36.500 --> 00:02:40.050
So let's have a look-see.
00:02:40.050 --> 00:02:43.590
So in our first
parametrization over here,
00:02:43.590 --> 00:02:49.970
we have x equals t and
y equals t squared.
00:02:49.970 --> 00:02:53.380
And we want this parametrization
to go from the points (1, 1)
00:02:53.380 --> 00:02:55.240
to the point (2, 4).
00:02:55.240 --> 00:03:00.161
So that means that t-- x--
t is going from 1 to 2.
00:03:00.161 --> 00:03:00.660
OK.
00:03:08.680 --> 00:03:12.290
So I'm going to just write
equals, and continue on.
00:03:12.290 --> 00:03:15.940
So in this first part--
so t is our parameter,
00:03:15.940 --> 00:03:18.320
and it's going from 1 to 2.
00:03:18.320 --> 00:03:28.400
So x*t is going to be t times
t squared, and dx is dt.
00:03:28.400 --> 00:03:33.170
And then plus-- so x
squared plus y squared is
00:03:33.170 --> 00:03:36.000
t squared plus t to the fourth.
00:03:36.000 --> 00:03:41.720
And dy is times 2t dt,
since y is t squared.
00:03:41.720 --> 00:03:42.430
All right.
00:03:42.430 --> 00:03:43.694
So this is our integral.
00:03:43.694 --> 00:03:45.360
And now it's
straightforward to compute.
00:03:45.360 --> 00:03:50.680
This is a single variable
integral of a polynomial,
00:03:50.680 --> 00:03:54.290
so we might have some fraction
arithmetic in our future,
00:03:54.290 --> 00:03:55.610
but nothing horrible.
00:03:55.610 --> 00:03:59.210
So let's collect our dt's.
00:03:59.210 --> 00:04:01.520
So this is the
integral from 1 to 2.
00:04:01.520 --> 00:04:05.900
So here I've got 2 t cubed
plus 2 t to the fifth.
00:04:05.900 --> 00:04:07.960
And here I have t cubed.
00:04:07.960 --> 00:04:19.020
So that's 3 t cubed plus
2 t to the fifth dt.
00:04:19.020 --> 00:04:20.970
OK, and so now we
have to evaluate.
00:04:20.970 --> 00:04:23.480
So this is equal to--
let me just continue it.
00:04:23.480 --> 00:04:33.630
So 3 t cubed becomes 3
t to the fourth over 4.
00:04:33.630 --> 00:04:36.730
Plus 2 t to the
fifth becomes-- well,
00:04:36.730 --> 00:04:39.180
t to the sixth over 6--
2 t to the sixth over 6,
00:04:39.180 --> 00:04:45.430
so that's plus t to the
sixth over 3, between t
00:04:45.430 --> 00:04:47.420
equals 1 and 2.
00:04:47.420 --> 00:04:51.940
And let me just bring
that computation up here.
00:04:51.940 --> 00:05:05.630
So that's equal to 3 times
16 over 4, plus 64 over 3,
00:05:05.630 --> 00:05:12.200
minus 3/4 plus 1/3.
00:05:12.200 --> 00:05:14.600
OK, and now we
gotta work this out.
00:05:14.600 --> 00:05:18.530
So this is maybe 12, and
the thirds give me 21,
00:05:18.530 --> 00:05:21.010
so that's 33 minus 3/4.
00:05:21.010 --> 00:05:31.030
So that's 32 and 1/4, or
129 over 4, if you prefer.
00:05:31.030 --> 00:05:35.300
OK, so that's going to
be our answer there.
00:05:35.300 --> 00:05:37.880
So now let's go back
and do it again using
00:05:37.880 --> 00:05:40.300
this alternate parametrization.
00:05:40.300 --> 00:05:42.300
So in this alternate
parametrization,
00:05:42.300 --> 00:05:47.710
we had x equal e to the t,
and y equal e to the 2t.
00:05:47.710 --> 00:05:53.240
So in that case, that means
that dx-- so x is e to the t,
00:05:53.240 --> 00:05:55.740
y is e to the 2t, so our
integral that we want,
00:05:55.740 --> 00:05:58.547
let me rewrite it.
00:05:58.547 --> 00:06:00.130
Well, actually, let
me not rewrite it.
00:06:00.130 --> 00:06:02.220
Let me just go back here
and take a look at it.
00:06:02.220 --> 00:06:04.670
So we want the
integral of F dot dr,
00:06:04.670 --> 00:06:08.620
and we know that that's equal
to the integral over C of x*y dx
00:06:08.620 --> 00:06:11.280
plus x squared
plus y squared dy.
00:06:11.280 --> 00:06:14.670
So x is e to the t,
y is e to the 2t,
00:06:14.670 --> 00:06:18.870
so that means dx is
going to be e to the t dt
00:06:18.870 --> 00:06:23.650
and dy is going to
be 2 e to the 2t dt.
00:06:23.650 --> 00:06:24.150
So OK.
00:06:24.150 --> 00:06:26.070
So we'll just have to
plug these things in.
00:06:26.070 --> 00:06:29.630
So our integral-- maybe I'll
give it a name: What We Want.
00:06:35.400 --> 00:06:36.950
WWW.
00:06:36.950 --> 00:06:40.950
What We Want is equal
to the integral of,
00:06:40.950 --> 00:06:42.050
well we start plugging in.
00:06:42.050 --> 00:06:49.420
So x*y dx is e to the t
times e to the 2t times,
00:06:49.420 --> 00:06:53.460
we said dx is e to the t dt.
00:06:53.460 --> 00:06:56.960
Plus, now x squared
plus y squared,
00:06:56.960 --> 00:07:01.665
so that's e to the 2t--
that's x squared-- plus e
00:07:01.665 --> 00:07:10.920
to the 4t-- that's y squared--
times 2 d to the 2t dt.
00:07:10.920 --> 00:07:12.891
That's dy.
00:07:12.891 --> 00:07:13.390
Whew.
00:07:13.390 --> 00:07:13.974
OK.
00:07:13.974 --> 00:07:15.890
But this integral is
missing something, right?
00:07:15.890 --> 00:07:18.434
Because it needs to be
a definite integral.
00:07:18.434 --> 00:07:20.600
Right now, I've written
down an indefinite integral.
00:07:20.600 --> 00:07:21.730
So that's a problem.
00:07:21.730 --> 00:07:26.190
So we need to think, what
are the bounds on this curve?
00:07:26.190 --> 00:07:28.150
What are we integrating
from and to?
00:07:28.150 --> 00:07:33.310
So let's go all the way back and
look at the problem over here.
00:07:33.310 --> 00:07:36.460
So this is the
parametrization of the curve,
00:07:36.460 --> 00:07:41.250
and we need it to connect
the points (1, 1) and (2, 4).
00:07:41.250 --> 00:07:43.650
So we need to know
which value of t
00:07:43.650 --> 00:07:47.680
puts this parametrization
at the point (1, 1),
00:07:47.680 --> 00:07:50.590
and which value of t puts
it at the point (2, 4).
00:07:50.590 --> 00:07:53.270
Well, I think the easiest way
to do that is to solve this,
00:07:53.270 --> 00:07:57.200
and we have t equals
natural log of x.
00:07:57.200 --> 00:08:02.260
So natural log of 1 is
0, and natural log of 2
00:08:02.260 --> 00:08:03.530
is natural log of 2.
00:08:03.530 --> 00:08:04.030
So OK.
00:08:04.030 --> 00:08:07.350
So we want t to go from
0 to natural log of 2.
00:08:07.350 --> 00:08:09.880
To go from this
point to that one.
00:08:09.880 --> 00:08:11.580
OK, so 0 to natural log of 2.
00:08:11.580 --> 00:08:14.770
Let's remember that, and come
back over here and put that in.
00:08:14.770 --> 00:08:16.780
So we want this
integral that we're
00:08:16.780 --> 00:08:20.550
trying to compute using this
alternate parametrization--
00:08:20.550 --> 00:08:23.150
we've already written down what
the integrand is-- but it's
00:08:23.150 --> 00:08:25.090
going to be the
integral, we just said,
00:08:25.090 --> 00:08:27.840
from 0 to the natural log of 2.
00:08:30.570 --> 00:08:31.231
OK.
00:08:31.231 --> 00:08:31.730
Great.
00:08:31.730 --> 00:08:36.220
So now let's take this integral
and let's start simplifying it.
00:08:36.220 --> 00:08:37.540
So let's see.
00:08:37.540 --> 00:08:39.990
This is an e to the t times
e to the 2t times e to the t,
00:08:39.990 --> 00:08:43.820
so that's e to the 4t dt.
00:08:43.820 --> 00:08:48.620
And here we have-- that looks
like a 2 e to the 4t plus 2 e
00:08:48.620 --> 00:08:50.680
to the 6t dt.
00:08:50.680 --> 00:08:58.190
So if we combine like terms,
we get the integral from 0
00:08:58.190 --> 00:08:59.470
to natural log of 2.
00:08:59.470 --> 00:09:01.130
So we had an e to
the 4t and a 2 e
00:09:01.130 --> 00:09:12.710
to the 4t, so that's 3 e to
the 4t plus 2 e to the 6t dt.
00:09:12.710 --> 00:09:15.570
OK, 3 e to the 4t
plus 2 e to the 6t.
00:09:15.570 --> 00:09:16.080
OK.
00:09:16.080 --> 00:09:19.517
And this is also a
different-looking integral
00:09:19.517 --> 00:09:21.850
than the one we had before,
but it's not a difficult one
00:09:21.850 --> 00:09:26.380
to evaluate, because it's just
e to a constant times t dt.
00:09:26.380 --> 00:09:29.346
Pretty straightforward
stuff, I think.
00:09:29.346 --> 00:09:30.720
So what does it
actually give us?
00:09:30.720 --> 00:09:33.370
Let me come up here
and write it up here.
00:09:33.370 --> 00:09:37.720
So this is equal to-- all
right, integrate e to the 4t,
00:09:37.720 --> 00:09:45.040
and you get e to the 4t over
4, so this is 3/4 e to the 4t,
00:09:45.040 --> 00:09:51.680
plus 2/6-- so that's
plus 1/3 e to the 6t,
00:09:51.680 --> 00:09:55.630
and we're evaluating
that between 0 and ln 2.
00:09:55.630 --> 00:09:58.870
And I will leave it
to you to confirm
00:09:58.870 --> 00:10:00.310
that what you get
when you do this
00:10:00.310 --> 00:10:06.420
is you get exactly 129 over 4
when you plug in these values
00:10:06.420 --> 00:10:08.950
and take the difference.
00:10:08.950 --> 00:10:11.770
So one thing to notice
here is that doing it
00:10:11.770 --> 00:10:14.710
with the second parametrization,
we got 129 over 4.
00:10:14.710 --> 00:10:17.220
Now, if you look
back just right here,
00:10:17.220 --> 00:10:19.680
you see that when we did it
with the first parametrization,
00:10:19.680 --> 00:10:22.150
we also got 129 over 4.
00:10:22.150 --> 00:10:24.300
Now this is a general
phenomenon, right?
00:10:24.300 --> 00:10:26.020
The parametrization
doesn't change
00:10:26.020 --> 00:10:27.840
the value of the line integral.
00:10:27.840 --> 00:10:30.370
So when you have a line
integral of a vector field,
00:10:30.370 --> 00:10:33.742
you can parametrize the curve
that you're integrating over
00:10:33.742 --> 00:10:34.950
in a bunch of different ways.
00:10:34.950 --> 00:10:38.242
I mean, in fact, infinitely
many different ways.
00:10:38.242 --> 00:10:39.950
There are all sorts
of different choices,
00:10:39.950 --> 00:10:41.366
and they all give
the same answer.
00:10:41.366 --> 00:10:42.280
They have to.
00:10:42.280 --> 00:10:44.470
So no matter what
parametrization you choose,
00:10:44.470 --> 00:10:46.930
you get the same number
out for the line integral
00:10:46.930 --> 00:10:49.290
over the same curve of
the same vector field.
00:10:49.290 --> 00:10:49.790
All right.
00:10:49.790 --> 00:10:52.120
So what that means
for you is that when
00:10:52.120 --> 00:10:54.870
you're given just a
curve and a vector field
00:10:54.870 --> 00:10:56.860
and you get to choose
your parametrization,
00:10:56.860 --> 00:10:58.910
you should always just choose
whatever the nicest one is.
00:10:58.910 --> 00:10:59.930
Because it doesn't
matter which one
00:10:59.930 --> 00:11:01.430
you choose, and
there's no reason
00:11:01.430 --> 00:11:04.250
ever to choose anything
harder than necessary.
00:11:04.250 --> 00:11:05.280
All right.
00:11:05.280 --> 00:11:09.040
So now, I want to quickly
recap what we did.
00:11:09.040 --> 00:11:15.770
So we had a curve
and a vector field F,
00:11:15.770 --> 00:11:21.050
and we were asked to compute
the line integral of F dot dr
00:11:21.050 --> 00:11:23.050
and we did it two
different ways.
00:11:23.050 --> 00:11:25.250
So in both cases,
we used the fact
00:11:25.250 --> 00:11:30.080
that F dot dr is M dx plus N dy.
00:11:30.080 --> 00:11:33.010
We wrote that out using the
known values of M and N,
00:11:33.010 --> 00:11:35.900
and using the known
components of F.
00:11:35.900 --> 00:11:38.530
And then we plugged
in for both cases.
00:11:38.530 --> 00:11:41.600
In this case, we used that
first parametrization.
00:11:41.600 --> 00:11:44.650
And then over to
my left, we used
00:11:44.650 --> 00:11:47.610
this second parametrization
that I gave you.
00:11:47.610 --> 00:11:51.070
And in both cases, that
reduced it to an integral
00:11:51.070 --> 00:11:55.220
in terms of a single variable t
that was the parameter that we
00:11:55.220 --> 00:11:55.880
could compute.
00:11:55.880 --> 00:12:00.250
And the answer came out the
same both times, as it had to.
00:12:00.250 --> 00:12:02.330
So one thing you
can notice is if you
00:12:02.330 --> 00:12:04.810
compare this with the
preceding lecture video,
00:12:04.810 --> 00:12:08.190
you'll see that this curve
and vector field, they
00:12:08.190 --> 00:12:10.790
connect the same points and
it's the same vector field,
00:12:10.790 --> 00:12:12.760
and these answers
both differ from both
00:12:12.760 --> 00:12:16.620
of the answers in the
preceding recitation video.
00:12:16.620 --> 00:12:19.850
So this is further illustration
that different curves can
00:12:19.850 --> 00:12:22.730
give you different values, but
if you take the same curve,
00:12:22.730 --> 00:12:24.200
different
parametrizations always
00:12:24.200 --> 00:12:27.490
have to give you same values
for that line integral
00:12:27.490 --> 00:12:28.940
of the same vector field.
00:12:28.940 --> 00:12:30.630
I'll end there.