WEBVTT

00:00:00.000 --> 00:00:09.400
CHRISTINE BREINER: Welcome
back to recitation.

00:00:09.400 --> 00:00:11.960
In this video, I'd like
us to do the following two

00:00:11.960 --> 00:00:16.660
problems, both related to
the same position vector.

00:00:16.660 --> 00:00:19.950
So we're starting off with
a position vector defined

00:00:19.950 --> 00:00:24.750
as r of t is equal to 1 minus
2 t squared i plus t squared

00:00:24.750 --> 00:00:28.870
j plus negative 2
plus 2 t squared k.

00:00:28.870 --> 00:00:30.660
So that's our position
vector, and I'd

00:00:30.660 --> 00:00:32.760
like us to do the
following two things.

00:00:32.760 --> 00:00:35.360
And you'll notice this
problem is pretty much

00:00:35.360 --> 00:00:36.830
just a computational problem.

00:00:36.830 --> 00:00:38.400
We're going to make
sure that we know

00:00:38.400 --> 00:00:41.980
what these things I'm
about to talk about are,

00:00:41.980 --> 00:00:44.710
how you define them, and how
you get from the position vector

00:00:44.710 --> 00:00:45.830
to each of these things.

00:00:45.830 --> 00:00:50.050
So we want to compute
the velocity, the speed,

00:00:50.050 --> 00:00:55.250
the acceleration, and find the
unit tangent vector for r of t.

00:00:55.250 --> 00:00:56.886
And then, the
second part, we want

00:00:56.886 --> 00:01:00.750
to compute the arc length of the
trajectory from t equals 0 to t

00:01:00.750 --> 00:01:02.155
equals 2.

00:01:02.155 --> 00:01:04.030
So I'll give you a moment
to do that problem.

00:01:04.030 --> 00:01:06.154
Why don't you pause the
video, work on the problem.

00:01:06.154 --> 00:01:08.850
When you're ready to check your
work, bring the video back up

00:01:08.850 --> 00:01:10.100
and I'll show you how I do it.

00:01:18.930 --> 00:01:20.120
OK, welcome back.

00:01:20.120 --> 00:01:22.700
Well, hopefully, you felt
comfortable with answering

00:01:22.700 --> 00:01:23.850
these questions.

00:01:23.850 --> 00:01:26.650
So now I will answer them and
you can compare your answers

00:01:26.650 --> 00:01:27.200
with mine.

00:01:27.200 --> 00:01:31.180
So let me start off with part a.

00:01:31.180 --> 00:01:34.010
Part a, the first
thing we're going to do

00:01:34.010 --> 00:01:35.850
is find the velocity.

00:01:35.850 --> 00:01:39.560
So velocity is
really-- all we need

00:01:39.560 --> 00:01:42.770
to do is take the derivative
of the position vector

00:01:42.770 --> 00:01:43.640
with respect to t.

00:01:43.640 --> 00:01:46.260
So I'm just going to
take r prime of t.

00:01:46.260 --> 00:01:49.770
And now I'm going to write
it in the shorthand notation

00:01:49.770 --> 00:01:53.770
that you've seen, with the
brackets to denote that it's

00:01:53.770 --> 00:01:56.750
not a point, but it's a vector.

00:01:56.750 --> 00:01:59.780
So this is what
you've seen to denote

00:01:59.780 --> 00:02:01.380
a vector rather than a point.

00:02:01.380 --> 00:02:04.030
So the derivative with respect
to t of the first component

00:02:04.030 --> 00:02:07.100
is just negative 4t.

00:02:07.100 --> 00:02:09.550
The derivative with respect
to t of the second component

00:02:09.550 --> 00:02:12.950
is just 2t, because
we had t squared,

00:02:12.950 --> 00:02:15.450
so when we take its
derivative, we just get 2t.

00:02:15.450 --> 00:02:19.490
And the third component was
negative 2 plus 2 t squared,

00:02:19.490 --> 00:02:22.560
so when I take its
derivative, I get a 4t,

00:02:22.560 --> 00:02:24.882
so that is actually v of t, OK?

00:02:24.882 --> 00:02:26.340
And then the next
thing I asked you

00:02:26.340 --> 00:02:29.710
to do is determine the speed,
and the speed, of course,

00:02:29.710 --> 00:02:33.540
is just the length of the
velocity vector, right?

00:02:33.540 --> 00:02:39.434
So we just need to find the
length of v. Now, to do that,

00:02:39.434 --> 00:02:41.100
to remind ourselves
what we do for that,

00:02:41.100 --> 00:02:42.724
we actually take the
inner product of v

00:02:42.724 --> 00:02:45.640
with itself, the dot
product of v with itself,

00:02:45.640 --> 00:02:49.900
and then we take the
square root of that.

00:02:49.900 --> 00:02:52.070
So let's look at what
the dot product will be.

00:02:52.070 --> 00:02:54.610
Let me find the
squared thing first,

00:02:54.610 --> 00:02:57.830
and then I will take
the square root.

00:02:57.830 --> 00:03:00.780
So v dotted with v,
the first component

00:03:00.780 --> 00:03:03.440
I'm going to have negative
4t quantity squared,

00:03:03.440 --> 00:03:06.760
so that's going to
be 16 t squared.

00:03:06.760 --> 00:03:08.370
And then the second
component is going

00:03:08.370 --> 00:03:10.960
to be 2t quantity
squared, so I'm

00:03:10.960 --> 00:03:13.737
going to have plus 4 t squared.

00:03:13.737 --> 00:03:15.320
And the third component
is going to be

00:03:15.320 --> 00:03:17.910
another-- it's going to
be 4t quantity squared,

00:03:17.910 --> 00:03:21.720
so I get another 16 t squared.

00:03:21.720 --> 00:03:27.980
So when I add those up, I
believe I get 36 t squared?

00:03:27.980 --> 00:03:28.960
Yes, good.

00:03:31.500 --> 00:03:34.310
And so then, I just have to take
the square root of both sides

00:03:34.310 --> 00:03:37.570
to get what the speed
actually is instead

00:03:37.570 --> 00:03:39.130
of the square of the speed.

00:03:41.730 --> 00:03:46.090
So I get 6t, OK?

00:03:46.090 --> 00:03:50.080
So that's the velocity;
that's the speed.

00:03:50.080 --> 00:03:51.580
Now I need to find
the acceleration

00:03:51.580 --> 00:03:54.100
and I need to find the
unit tangent vector.

00:03:54.100 --> 00:03:55.250
OK, so let me see.

00:03:55.250 --> 00:03:58.040
I will come over here.

00:03:58.040 --> 00:04:00.760
Let me step off here and I
will find the acceleration

00:04:00.760 --> 00:04:02.200
and the unit tangent vector.

00:04:02.200 --> 00:04:04.110
So the acceleration,
if you remember,

00:04:04.110 --> 00:04:06.110
the acceleration is
actually just the derivative

00:04:06.110 --> 00:04:08.380
of the velocity
with respect to t.

00:04:08.380 --> 00:04:11.290
So the acceleration
is going to be

00:04:11.290 --> 00:04:14.250
the derivative of negative
4t is just negative 4.

00:04:14.250 --> 00:04:18.330
The derivative of 2t is just
2, and the derivative of 4t

00:04:18.330 --> 00:04:20.660
is just 4, all
with respect to t.

00:04:20.660 --> 00:04:21.750
OK.

00:04:21.750 --> 00:04:26.420
So the acceleration vector is
equal to negative 4 comma 2

00:04:26.420 --> 00:04:27.860
comma 4, so you
see this actually

00:04:27.860 --> 00:04:30.040
has constant acceleration.

00:04:30.040 --> 00:04:31.540
So at any point,
your acceleration

00:04:31.540 --> 00:04:33.940
is always this value,
so it's not surprising

00:04:33.940 --> 00:04:36.984
that our velocity is
increasing, and actually, it's

00:04:36.984 --> 00:04:39.150
increasing-- you'll notice,
each of these components

00:04:39.150 --> 00:04:40.230
is constant.

00:04:40.230 --> 00:04:42.800
The velocity, each of
the components is linear,

00:04:42.800 --> 00:04:45.900
and if we went back, we
look at the position vector,

00:04:45.900 --> 00:04:47.950
each of those
components is quadratic.

00:04:47.950 --> 00:04:49.780
And this is exactly
what you expect

00:04:49.780 --> 00:04:52.730
from just your
understanding of derivatives

00:04:52.730 --> 00:04:54.810
in single-variable calculus.

00:04:54.810 --> 00:04:57.669
If you start off with a constant
and you find an antiderivative,

00:04:57.669 --> 00:05:00.210
it's going to be linear, and
you find another antiderivative,

00:05:00.210 --> 00:05:02.520
you're going to
have a quadratic,

00:05:02.520 --> 00:05:04.680
so we shouldn't be
surprised by any of that.

00:05:04.680 --> 00:05:06.937
Now we have one more
thing to do with Part a,

00:05:06.937 --> 00:05:08.770
and that is to find the
unit tangent vector.

00:05:08.770 --> 00:05:11.180
And that's fairly easy,
because all we have to do

00:05:11.180 --> 00:05:15.350
is-- if you notice, we
have the velocity vector

00:05:15.350 --> 00:05:17.686
and we have its length.

00:05:17.686 --> 00:05:19.310
And so to find the
unit tangent vector,

00:05:19.310 --> 00:05:21.100
all we have to do
is take the velocity

00:05:21.100 --> 00:05:24.110
and divide it by its length,
and that will normalize it.

00:05:24.110 --> 00:05:28.140
That means that its length
will be one at that point,

00:05:28.140 --> 00:05:30.710
because you're taking a
vector, dividing by its length,

00:05:30.710 --> 00:05:33.760
so the length of the new vector
will have to be length one.

00:05:33.760 --> 00:05:36.670
So let me write that down.

00:05:36.670 --> 00:05:38.940
And actually, I guess the
point to remember here

00:05:38.940 --> 00:05:43.710
is that the velocity vector
is tangent to the path you're

00:05:43.710 --> 00:05:45.395
carving out, to the trajectory.

00:05:50.020 --> 00:05:52.440
OK, so this is a vector.

00:05:52.440 --> 00:05:54.220
This is a scalar.

00:05:54.220 --> 00:05:58.440
So I'm going to take
1 over 6t, and I'm

00:05:58.440 --> 00:06:06.940
going to multiply it
by negative 4t, 2t, 4t,

00:06:06.940 --> 00:06:09.360
and this gives me,
when I do my division,

00:06:09.360 --> 00:06:18.490
looks like it gives me a
negative 2/3, right, 1/3, 2/3.

00:06:21.880 --> 00:06:28.370
So that is the unit
tangent vector, OK?

00:06:28.370 --> 00:06:31.330
OK, and now, we have one
more point we want to make,

00:06:31.330 --> 00:06:34.449
and that is having
to do with the arc

00:06:34.449 --> 00:06:35.490
length of the trajectory.

00:06:35.490 --> 00:06:37.198
That was the second
part of this problem,

00:06:37.198 --> 00:06:39.370
was to find the arc length
of the trajectory from t

00:06:39.370 --> 00:06:41.320
equals 0 to t equals 2.

00:06:41.320 --> 00:06:44.770
So let me just draw
another line here.

00:06:44.770 --> 00:06:47.330
And what we want to do
there then is-- really

00:06:47.330 --> 00:06:56.610
what we want to do is we want
to integrate the speed, right?

00:06:56.610 --> 00:06:59.040
We want to integrate
the speed from 0 to 2.

00:06:59.040 --> 00:07:03.440
So this-- let me come over
here-- this absolute v,

00:07:03.440 --> 00:07:11.060
you might have also seen
it written as ds/dt, right?

00:07:11.060 --> 00:07:15.060
And so we want to
integrate this in dt--

00:07:15.060 --> 00:07:18.220
in t, sorry-- from 0 to 2.

00:07:18.220 --> 00:07:21.180
And so we come over here.

00:07:21.180 --> 00:07:27.200
We want to integrate
from 0 to 2, 6t dt.

00:07:27.200 --> 00:07:28.140
That's fairly easy.

00:07:28.140 --> 00:07:33.890
That's going to be 6 t squared
over 2, evaluated from 0 to 2.

00:07:33.890 --> 00:07:36.170
And so when I write
that down, I'm

00:07:36.170 --> 00:07:39.950
going to get 6 times 4
divided by 2, 24 divided by 2,

00:07:39.950 --> 00:07:42.020
I just get 12, and
the other term is 0.

00:07:42.020 --> 00:07:45.700
So the arc length
is-- of the trajectory

00:07:45.700 --> 00:07:50.310
from 0 to 2 is just 12 units.

00:07:50.310 --> 00:07:52.760
So this really was a purely
computational type of problem.

00:07:52.760 --> 00:07:54.910
All we were doing, if
you come back over here

00:07:54.910 --> 00:07:57.030
and you recall what
we were trying to do,

00:07:57.030 --> 00:08:00.050
is we started off with
a position vector.

00:08:00.050 --> 00:08:02.251
We just did a lot
of computation.

00:08:02.251 --> 00:08:04.376
We found the velocity, the
speed, the acceleration,

00:08:04.376 --> 00:08:06.507
the unit tangent
vector, and then

00:08:06.507 --> 00:08:08.840
we just wanted to find the
arc length of the trajectory.

00:08:08.840 --> 00:08:11.110
So this is all
very computational,

00:08:11.110 --> 00:08:12.760
but just to make
sure we understood

00:08:12.760 --> 00:08:14.850
what all the terms
meant and how they

00:08:14.850 --> 00:08:17.190
were related to one another.

00:08:17.190 --> 00:08:18.950
So I'll stop there.