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So, if you remember last time,
we looked at parametric

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equations -- -- as a way of
describing the motion of a point

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that moves in the plane or in
space as a function of time of

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your favorite parameter that
will tell you how far the motion

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has progressed.
And, I think we did it in

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detail the example of the
cycloid, which is the curve

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00:01:01,000 --> 00:01:09,000
traced by a point on a wheel
that's rolling on a flat

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surface.
So, we have this example where

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we have this wheel that's
rolling on the x-axis,

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and we have this point on the
wheel.

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And, as it moves around,
it traces a trajectory that

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moves more or less like this.
OK, so I'm trying a new color.

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Is this visible from the back?
So, no more blue.

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OK, so remember,
in general, we are trying to

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00:01:52,000 --> 00:01:58,000
find the position,
so, x of t, y of t,

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00:01:58,000 --> 00:02:09,000
maybe z of t if we are in space
-- -- of a moving point along a

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trajectory.
And, one way to think about

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this is in terms of the position
vector.

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00:02:25,000 --> 00:02:32,000
So, position vector is just the
vector whose components are

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coordinates of a point,
OK, so if you prefer,

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00:02:37,000 --> 00:02:43,000
that's the same thing as a
vector from the origin to the

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00:02:43,000 --> 00:02:50,000
moving point.
So, maybe our point is here, P.

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00:02:50,000 --> 00:03:02,000
So, this vector here -- This
vector here is vector OP.

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00:03:02,000 --> 00:03:12,000
And, that's also the position
vector r of t.

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00:03:12,000 --> 00:03:24,000
So, just to give you,
again, that example -- -- if I

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take the cycloid for a wheel of
radius 1,

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00:03:34,000 --> 00:03:41,000
and let's say that we are going
at unit speed so that the angle

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that we used as a parameter of
time is the same thing as time

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00:03:48,000 --> 00:03:53,000
when the position vector,
in this case,

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00:03:53,000 --> 00:04:00,000
we found to be,
just to make sure that they

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00:04:00,000 --> 00:04:07,000
have it right,
.

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00:04:07,000 --> 00:04:10,000
OK, that's a formula that you
should have in your notes from

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00:04:10,000 --> 00:04:13,000
last time, except we had theta
instead of t because we were

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using the angle.
But now I'm saying,

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we are moving at unit speed,
so time and angle are the same

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00:04:20,000 --> 00:04:24,000
thing.
So, now, what's interesting

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about this is we can analyze the
motion in more detail.

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OK, so, now that we know the
position of the point as a

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00:04:33,000 --> 00:04:37,000
function of time,
we can try to study how it

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00:04:37,000 --> 00:04:43,000
varies in particular things like
the speed and acceleration.

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00:04:43,000 --> 00:04:48,000
OK, so let's start with speed.
Well, in fact we can do better

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than speed.
Let's not start with speed.

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So, speed is a number.
It tells you how fast you are

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going along your trajectory.
I mean, if you're driving in a

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car, then it tells you how fast
you are going.

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00:05:01,000 --> 00:05:03,000
But, unless you have one of
these fancy cars with a GPS,

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00:05:03,000 --> 00:05:05,000
it doesn't tell you which
direction you're going.

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00:05:05,000 --> 00:05:08,000
And, that's useful information,
too, if you're trying to figure

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00:05:08,000 --> 00:05:10,000
out what your trajectory is.
So, in fact,

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there's two aspects to it.
One is how fast you are going,

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and the other is in what
direction you're going.

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That means actually we should
use a vector maybe to think

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about this.
And so, that's called the

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velocity vector.
And, the way we can get it,

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so, it's called usually V,
so, V here stands for velocity

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more than for vector.
And, you just get it by taking

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the derivative of a position
vector with respect to time.

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Now, it's our first time
writing this kind of thing with

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a vector.
So, the basic rule is you can

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take the derivative of a vector
quantity just by taking the

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derivatives of each component.
OK, so that's just dx/dt,

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00:06:06,000 --> 00:06:17,000
dy/dt, and if you have z
component, dz/dt.

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00:06:17,000 --> 00:06:32,000
So, let me -- OK,
so -- OK, so let's see what it

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is for the cycloid.
So, an example of a cycloid,

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well, so what do we get when we
take the derivatives of this

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formula there?
Well, so, the derivative of t

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is 1- cos(t).
The derivative of 1 is 0.

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00:07:08,000 --> 00:07:12,000
The derivative of -cos(t) is
sin(t).

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Very good.
OK, that's at least one thing

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you should remember from single
variable calculus.

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Hopefully you remember even
more than that.

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OK, so that's the velocity
vector.

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00:07:27,000 --> 00:07:31,000
It tells us at any time how
fast we are going,

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and in what direction.
So, for example, observe.

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00:07:37,000 --> 00:07:40,000
Remember last time at the end
of class we were trying to

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figure out what exactly happens
near the bottom point,

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when we have this motion that
seems to stop and go backwards.

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And, we answered that one way.
But, let's try to understand it

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in terms of velocity.
What if I plug t equals 0 in

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here?
Then, 1- cos(t) is 0,

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00:07:57,000 --> 00:08:01,000
sin(t) is 0.
The velocity is 0.

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00:08:01,000 --> 00:08:05,000
So, at the time,at that
particular time,

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our point is actually not
moving.

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00:08:08,000 --> 00:08:11,000
Of course, it's been moving
just before, and it starts

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00:08:11,000 --> 00:08:14,000
moving just afterwards.
It's just the instant,

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at that particular instant,
the speed is zero.

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So, that's especially maybe a
counterintuitive thing,

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but something is moving.
And at that time,

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it's actually stopped.
Now, let's see,

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so that's the vector.
And, it's useful.

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00:08:36,000 --> 00:08:39,000
But, if you want just the usual
speed as a number,

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then, what will you do?
Well, you will just take

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exactly the magnitude of this
vector.

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So, speed, which is the scalar
quantity is going to be just the

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magnitude of the vector,
V.

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00:09:01,000 --> 00:09:09,000
OK, so, in this case,
while it would be square root

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00:09:09,000 --> 00:09:18,000
of (1- cost)^2 sin^2(t),
and if you expand that,

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you will get,
let me take a bit more space,

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it's going to be square root of
1 - 2cos(t) cos^2(t) sin^2(t).

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It seems to simplify a little
bit because we have cos^2 plus

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sin^2.
That's 1.

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So, it's going to be the square
root of 2 - 2cos(t).

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So, at this point,
if I was going to ask you,

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00:09:52,000 --> 00:09:55,000
when is the speed the smallest
or the largest?

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You could answer based on that.
See, at t equals 0,

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well, that turns out to be
zero.

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00:10:01,000 --> 00:10:04,000
The point is not moving.
At t equals pi,

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00:10:04,000 --> 00:10:07,000
that ends up being the square
root of 2 plus 2,

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00:10:07,000 --> 00:10:09,000
which is 4.
So, that's 2.

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And, that's when you're truly
at the top of the arch,

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00:10:12,000 --> 00:10:15,000
and that's when the point is
moving the fastest.

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00:10:15,000 --> 00:10:18,000
In fact, they are spending
twice as fast as the wheel

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because the wheel is moving to
the right at unit speed,

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00:10:20,000 --> 00:10:24,000
and the wheel is also rotating.
So, it's moving to the right

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and unit speed relative to the
center so that the two effects

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add up, and give you a speed of
2.

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00:10:32,000 --> 00:10:36,000
Anyway, that's a formula we can
get.

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OK, now, what about
acceleration?

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So, here I should warn you that
there is a serious discrepancy

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between the usual intuitive
notion of acceleration,

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00:10:58,000 --> 00:11:02,000
the one that you are aware of
when you drive a car and the one

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00:11:02,000 --> 00:11:05,000
that we will be using.
So, you might think

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acceleration is just the
directive of speed.

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00:11:08,000 --> 00:11:13,000
If my car goes 55 miles an hour
on the highway and it's going a

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00:11:13,000 --> 00:11:15,000
constant speed,
it's not accelerating.

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00:11:15,000 --> 00:11:18,000
But, let's say that I'm taking
a really tight turn.

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Then, I'm going to feel
something.

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There is some force being
exerted.

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00:11:21,000 --> 00:11:24,000
And, in fact,
there is a sideways

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acceleration at that point even
though the speed is not

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changing.
So, the definition will take

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effect.
The acceleration is,

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00:11:34,000 --> 00:11:40,000
as a vector,
and the acceleration vector is

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just the derivative of a
velocity vector.

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00:11:47,000 --> 00:11:51,000
So, even if the speed is
constant, that means,

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even if a length of the
velocity vector stays the same,

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the velocity vector can still
rotate.

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00:11:59,000 --> 00:12:03,000
And, as it rotates,
it uses acceleration.

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00:12:03,000 --> 00:12:07,000
OK, and so this is the notion
of acceleration that's relevant

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00:12:07,000 --> 00:12:13,000
to physics when you find F=ma;
that's the (a) that you have in

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mind here.
It's a vector.

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00:12:17,000 --> 00:12:19,000
Of course, if you are moving in
a straight line,

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00:12:19,000 --> 00:12:20,000
then the two notions are the
same.

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00:12:20,000 --> 00:12:23,000
I mean, acceleration is also
going to be along the line,

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00:12:23,000 --> 00:12:25,000
and it's going to has to do
with the derivative of speed.

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But, in general,
that's not quite the same.

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So, for example,
let's look at the cycloid.

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If we take the example of the
cycloid, well,

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00:12:40,000 --> 00:12:44,000
what's the derivative of one
minus cos(t)?

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00:12:44,000 --> 00:12:52,000
It's sin(t).
And, what's the derivative of

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00:12:52,000 --> 00:12:55,000
sin(t)?
cos(t), OK.

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So, the acceleration vector is
.

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00:13:04,000 --> 00:13:09,000
So, in particular,
let's look at what happens at

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00:13:09,000 --> 00:13:13,000
time t equals zero when the
point is not moving.

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00:13:13,000 --> 00:13:20,000
Well, the acceleration vector
there will be zero from one.

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00:13:20,000 --> 00:13:28,000
So, what that means is that if
I look at my trajectory at this

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00:13:28,000 --> 00:13:35,000
point, that the acceleration
vector is pointing in that

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00:13:35,000 --> 00:13:39,000
direction.
It's the unit vector in the

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00:13:39,000 --> 00:13:43,000
vertical direction.
So, my point is not moving at

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that particular time.
But, it's accelerating up.

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So, that means that actually as
it comes down,

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first it's slowing down.
Then it stops here,

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00:13:53,000 --> 00:13:56,000
and then it reverses going back
up.

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00:13:56,000 --> 00:14:01,000
OK, so that's another way to
understand what we were saying

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00:14:01,000 --> 00:14:06,000
last time that the trajectory at
that point has a vertical

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00:14:06,000 --> 00:14:11,000
tendency because that's the
direction in which the motion is

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00:14:11,000 --> 00:14:16,000
going to occur just before and
just after time zero.

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00:14:16,000 --> 00:14:30,000
OK, any questions about that?
No.

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00:14:30,000 --> 00:14:36,000
OK, so I should insist maybe on
one thing,

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00:14:36,000 --> 00:14:41,000
which is that,
so, we can differentiate

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00:14:41,000 --> 00:14:46,000
vectors just component by
component,

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00:14:46,000 --> 00:14:50,000
OK, and we can differentiate
vector expressions according to

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00:14:50,000 --> 00:14:54,000
certain rules that we'll see in
a moment.

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One thing that we cannot do,
it's not true that the length

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00:15:02,000 --> 00:15:12,000
of dr dt, which is the speed,
is equal to the length of dt.

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00:15:12,000 --> 00:15:18,000
OK, this is completely false.
And, they are really not the

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same.
So, if you have to

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00:15:19,000 --> 00:15:24,000
differentiate the length of a
vector, but basically you are in

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00:15:24,000 --> 00:15:25,000
trouble.
If you really,

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00:15:25,000 --> 00:15:27,000
really want to do it,
well, the length of the vector

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00:15:27,000 --> 00:15:30,000
is the square root of the sums
of the squares of the

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00:15:30,000 --> 00:15:32,000
components,
and from that you can use the

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00:15:32,000 --> 00:15:34,000
formula for the derivative of
the square root,

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00:15:34,000 --> 00:15:36,000
and the chain rule,
and various other things.

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And, you can get there.
But, it will not be a very nice

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00:15:39,000 --> 00:15:42,000
expression.
There is no simple formula for

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00:15:42,000 --> 00:15:44,000
this kind of thing.
Fortunately,

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00:15:44,000 --> 00:15:48,000
we almost never have to compute
this kind of thing because,

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00:15:48,000 --> 00:15:51,000
after all, it's not a very
relevant quantity.

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00:15:51,000 --> 00:15:53,000
What's more relevant might be
this one.

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This is actually the speed.
This one, I don't know what it

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means.
OK.

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00:16:10,000 --> 00:16:14,000
So, let's continue our
exploration.

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00:16:14,000 --> 00:16:20,000
So, the next concept that I
want to define is that of arc

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length.
So, arc length is just the

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00:16:23,000 --> 00:16:26,000
distance that you have traveled
along the curve,

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00:16:26,000 --> 00:16:27,000
OK?
So, if you are in a car,

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00:16:27,000 --> 00:16:30,000
you know, it has mileage
counter that tells you how far

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00:16:30,000 --> 00:16:33,000
you've gone, how much fuel
you've used if it's a fancy car.

206
00:16:33,000 --> 00:16:37,000
And, what it does is it
actually integrates the speed of

207
00:16:37,000 --> 00:16:41,000
the time to give you the arc
length along the trajectory of

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00:16:41,000 --> 00:16:45,000
the car.
So, the usual notation that we

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00:16:45,000 --> 00:16:51,000
will have is (s) for arc length.
I'm not quite sure how you get

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00:16:51,000 --> 00:16:57,000
an (s) out of this,
but it's the usual notation.

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00:16:57,000 --> 00:17:14,000
OK, so, (s) is for distance
traveled along the trajectory.

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00:17:14,000 --> 00:17:16,000
And, so that makes sense,
of course, we need to fix a

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00:17:16,000 --> 00:17:19,000
reference point.
Maybe on the cycloid,

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00:17:19,000 --> 00:17:22,000
we'd say it's a distance
starting on the origin.

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00:17:22,000 --> 00:17:25,000
In general, maybe you would say
you start at time,

216
00:17:25,000 --> 00:17:28,000
t equals zero.
But, it's a convention.

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00:17:28,000 --> 00:17:31,000
If you knew in advance,
you could have,

218
00:17:31,000 --> 00:17:35,000
actually, your car's mileage
counter to count backwards from

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00:17:35,000 --> 00:17:38,000
the point where the car will die
and start walking.

220
00:17:38,000 --> 00:17:41,000
I mean, that would be
sneaky-freaky,

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00:17:41,000 --> 00:17:45,000
but you could have a negative
arc length that gets closer and

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00:17:45,000 --> 00:17:48,000
closer to zero,
and gets to zero at the end of

223
00:17:48,000 --> 00:17:51,000
a trajectory,
or anything you want.

224
00:17:51,000 --> 00:17:53,000
I mean, arc length could be
positive or negative.

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00:17:53,000 --> 00:17:56,000
Typically it's negative what
you are before the reference

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00:17:56,000 --> 00:18:01,000
point, and positive afterwards.
So, now, how does it relate to

227
00:18:01,000 --> 00:18:08,000
the things we've seen there?
Well, so in particular,

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00:18:08,000 --> 00:18:16,000
how do you relate arc length
and time?

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00:18:16,000 --> 00:18:22,000
Well, so, there's a simple
relation, which is that the rate

230
00:18:22,000 --> 00:18:26,000
of change of arc length versus
time,

231
00:18:26,000 --> 00:18:30,000
well, that's going to be the
speed at which you are moving,

232
00:18:30,000 --> 00:18:38,000
OK, because the speed as a
scalar quantity tells you how

233
00:18:38,000 --> 00:18:44,000
much distance you're covering
per unit time.

234
00:18:44,000 --> 00:18:47,000
OK, and in fact,
to be completely honest,

235
00:18:47,000 --> 00:18:51,000
I should put an absolute value
here because there is examples

236
00:18:51,000 --> 00:18:55,000
of curves maybe where your
motion is going back and forth

237
00:18:55,000 --> 00:18:59,000
along the same curve.
And then, you don't want to

238
00:18:59,000 --> 00:19:01,000
keep counting arc length all the
time.

239
00:19:01,000 --> 00:19:04,000
Actually, maybe you want to say
that the arc length increases

240
00:19:04,000 --> 00:19:05,000
and then decreases along the
curve.

241
00:19:05,000 --> 00:19:08,000
I mean, you get to choose how
you count it.

242
00:19:08,000 --> 00:19:10,000
But, in this case,
if you are moving back and

243
00:19:10,000 --> 00:19:12,000
forth, it would make more sense
to have the arc length first

244
00:19:12,000 --> 00:19:18,000
increase,
then decrease,

245
00:19:18,000 --> 00:19:26,000
increase again,
and so on.

246
00:19:26,000 --> 00:19:34,000
So -- So if you want to know
really what the arc length is,

247
00:19:34,000 --> 00:19:41,000
then basically the only way to
do it is to integrate speed

248
00:19:41,000 --> 00:19:45,000
versus time.
So, if you wanted to know how

249
00:19:45,000 --> 00:19:49,000
long an arch of cycloid is,
you have this nice-looking

250
00:19:49,000 --> 00:19:51,000
curve;
how long is it?

251
00:19:51,000 --> 00:19:55,000
Well, you'd have to basically
integrate this quantity from t

252
00:19:55,000 --> 00:19:57,000
equals zero to 2 pi.

253
00:20:24,000 --> 00:20:28,000
And, to say the truth,
I don't really know how to

254
00:20:28,000 --> 00:20:31,000
integrate that.
So, we don't actually have a

255
00:20:31,000 --> 00:20:34,000
formula for the length at this
point.

256
00:20:34,000 --> 00:20:41,000
However, we'll see one later
using a cool trick,

257
00:20:41,000 --> 00:20:47,000
and multi-variable calculus.
So, for now,

258
00:20:47,000 --> 00:20:52,000
we'll just leave the formula
like that, and we don't know how

259
00:20:52,000 --> 00:20:55,000
long it is.
Well, you can put that into

260
00:20:55,000 --> 00:20:57,000
your calculator and get the
numerical value.

261
00:20:57,000 --> 00:21:07,000
But, that's the best I can
offer.

262
00:21:07,000 --> 00:21:18,000
Now, another useful notion is
the unit vector to the

263
00:21:18,000 --> 00:21:25,000
trajectory.
So, the usual notation is T hat.

264
00:21:25,000 --> 00:21:28,000
It has a hat because it's a
unit vector, and T because it's

265
00:21:28,000 --> 00:21:32,000
tangent.
Now, how do we get this unit

266
00:21:32,000 --> 00:21:36,000
vector?
So, maybe I should have pointed

267
00:21:36,000 --> 00:21:40,000
out before that if you're moving
along some trajectory,

268
00:21:40,000 --> 00:21:43,000
say you're going in that
direction, then when you're at

269
00:21:43,000 --> 00:21:47,000
this point,
the velocity vector is going to

270
00:21:47,000 --> 00:21:53,000
be tangential to the trajectory.
It tells you the direction of

271
00:21:53,000 --> 00:21:57,000
motion in particular.
So, if you want a unit vector

272
00:21:57,000 --> 00:22:02,000
that goes in the same direction,
all you have to do is rescale

273
00:22:02,000 --> 00:22:05,000
it, so, at its length becomes
one.

274
00:22:05,000 --> 00:22:10,000
So, it's v divided by a
magnitude of v.

275
00:22:28,000 --> 00:22:33,000
So, it seems like now we have a
lot of different things that

276
00:22:33,000 --> 00:22:40,000
should be related in some way.
So, let's see what we can say.

277
00:22:40,000 --> 00:22:50,000
Well, we can say that dr by dt,
so, that's the velocity vector,

278
00:22:50,000 --> 00:22:59,000
that's the same thing as if I
use the chain rule dr/ds times

279
00:22:59,000 --> 00:23:06,000
ds/dt.
OK, so, let's think about this

280
00:23:06,000 --> 00:23:11,000
things.
So, this guy here we've just

281
00:23:11,000 --> 00:23:17,000
seen.
That's the same as the speed,

282
00:23:17,000 --> 00:23:21,000
OK?
So, this one here should be v

283
00:23:21,000 --> 00:23:28,000
divided by its length.
So, that means this actually

284
00:23:28,000 --> 00:23:34,000
should be the unit vector.
OK, so, let me rewrite that.

285
00:23:34,000 --> 00:23:40,000
It's T ds/dt.
So, maybe if I actually stated

286
00:23:40,000 --> 00:23:43,000
directly that way,
see, I'm just saying the

287
00:23:43,000 --> 00:23:46,000
velocity vector has a length and
a direction.

288
00:23:46,000 --> 00:23:51,000
The length is the speed.
The direction is tangent to the

289
00:23:51,000 --> 00:23:51,000
trajectory.

290
00:24:19,000 --> 00:24:25,000
So, the speed is ds/dt,
and the vector is T hat.

291
00:24:25,000 --> 00:24:33,000
And, that's how we get this.
So, let's try just to see why

292
00:24:33,000 --> 00:24:37,000
dr/ds should be T.
Well, let's think of dr/ds.

293
00:24:37,000 --> 00:24:42,000
dr/ds means position vector r
means you have the origin,

294
00:24:42,000 --> 00:24:47,000
which is somewhere out there,
and the vector r is here.

295
00:24:47,000 --> 00:24:51,000
So, dr/ds means we move by a
small amount,

296
00:24:51,000 --> 00:24:56,000
delta s along the trajectory a
certain distance delta s.

297
00:24:56,000 --> 00:25:00,000
And, we look at how the
position vector changes.

298
00:25:00,000 --> 00:25:08,000
Well, we'll have a small change.
Let me call that vector delta r

299
00:25:08,000 --> 00:25:13,000
corresponding to the size,
corresponding to the length

300
00:25:13,000 --> 00:25:17,000
delta s.
And now, delta r should be

301
00:25:17,000 --> 00:25:25,000
essentially roughly equal to,
well, its direction will be

302
00:25:25,000 --> 00:25:30,000
tangent to the trajectory.
If I take a small enough

303
00:25:30,000 --> 00:25:33,000
interval,
then the direction will be

304
00:25:33,000 --> 00:25:37,000
almost tensioned to the
trajectory times the length of

305
00:25:37,000 --> 00:25:41,000
it will be delta s,
the distance that I have

306
00:25:41,000 --> 00:25:45,000
traveled.
OK, sorry, maybe I should

307
00:25:45,000 --> 00:25:50,000
explain that on a separate
board.

308
00:25:50,000 --> 00:25:56,000
OK, so, let's say that we have
that amount of time,

309
00:25:56,000 --> 00:26:00,000
delta t.
So, let's zoom into that curve.

310
00:26:00,000 --> 00:26:12,000
So, we have r at time t.
We have r at time t plus delta

311
00:26:12,000 --> 00:26:17,000
t.
This vector here I will call

312
00:26:17,000 --> 00:26:23,000
delta r.
The length of this vector is

313
00:26:23,000 --> 00:26:28,000
delta s.
And, the direction is

314
00:26:28,000 --> 00:26:36,000
essentially that of a vector.
OK, so, delta s over delta t,

315
00:26:36,000 --> 00:26:43,000
that's the distance traveled
divided by the time.

316
00:26:43,000 --> 00:26:46,000
That's going to be close to the
speed.

317
00:26:46,000 --> 00:26:57,000
And, delta r is approximately T
times delta s.

318
00:26:57,000 --> 00:27:04,000
So, now if I divide both sides
by delta t, I get this.

319
00:27:04,000 --> 00:27:07,000
And, if I take the limit as
delta t turns to zero,

320
00:27:07,000 --> 00:27:10,000
then I get the same formula
with the derivatives and with an

321
00:27:10,000 --> 00:27:13,000
equality.
It's an approximation.

322
00:27:13,000 --> 00:27:15,000
The approximation becomes
better and better if I go to

323
00:27:15,000 --> 00:27:16,000
smaller intervals.

324
00:27:38,000 --> 00:27:44,000
OK, are there any questions
about this?

325
00:27:44,000 --> 00:27:59,000
Yes?
Yes, that's correct.

326
00:27:59,000 --> 00:28:01,000
OK, so let's be more careful,
actually.

327
00:28:01,000 --> 00:28:12,000
So, you're asking about whether
the delta r is actually strictly

328
00:28:12,000 --> 00:28:16,000
tangent to the curve.
Is that -- That's correct.

329
00:28:16,000 --> 00:28:20,000
Actually, delta r is not
strictly tangent to anything.

330
00:28:20,000 --> 00:28:23,000
So, maybe I should draw another
picture.

331
00:28:23,000 --> 00:28:29,000
If I'm going from here to here,
then delta r is going to be

332
00:28:29,000 --> 00:28:36,000
this arc inside the curve while
the vector will be going in this

333
00:28:36,000 --> 00:28:39,000
direction, OK?
So, they are not strictly

334
00:28:39,000 --> 00:28:41,000
parallel to each other.
That's why it's only

335
00:28:41,000 --> 00:28:44,000
approximately equal.
Similarly, this distance,

336
00:28:44,000 --> 00:28:48,000
the length of delta r is not
exactly the length along the

337
00:28:48,000 --> 00:28:50,000
curve.
It's actually a bit shorter.

338
00:28:50,000 --> 00:28:53,000
But, if we imagine a smaller
and smaller portion of the

339
00:28:53,000 --> 00:28:56,000
curve,
then this effect of the curve

340
00:28:56,000 --> 00:29:00,000
being a curve and not a straight
line becomes more and more

341
00:29:00,000 --> 00:29:02,000
negligible.
If you zoom into the curve

342
00:29:02,000 --> 00:29:04,000
sufficiently,
then it looks more and more

343
00:29:04,000 --> 00:29:07,000
like a straight line.
And then, what I said becomes

344
00:29:07,000 --> 00:29:18,000
true in the limit.
OK? Any other questions?

345
00:29:18,000 --> 00:29:35,000
No? OK.
So, what happens next?

346
00:29:35,000 --> 00:29:39,000
OK, so let me show you a nice
example of why we might want to

347
00:29:39,000 --> 00:29:43,000
use vectors to study parametric
curves because,

348
00:29:43,000 --> 00:29:46,000
after all, a lot of what's here
you can just do in coordinates.

349
00:29:46,000 --> 00:29:48,000
And, we don't really need
vectors.

350
00:29:48,000 --> 00:29:51,000
Well, and truly,
vectors being a language,

351
00:29:51,000 --> 00:29:54,000
you never strictly need it,
but it's useful to have a

352
00:29:54,000 --> 00:30:02,000
notion of vectors.
So, I want to tell you a bit

353
00:30:02,000 --> 00:30:14,000
about Kepler's second law of
celestial mechanics.

354
00:30:14,000 --> 00:30:20,000
So, that goes back to 1609.
So, that's not exactly recent

355
00:30:20,000 --> 00:30:24,000
news, OK?
But, still I think it's a very

356
00:30:24,000 --> 00:30:29,000
interesting example of why you
might want to use vector methods

357
00:30:29,000 --> 00:30:33,000
to analyze motions.
So, what happened back then was

358
00:30:33,000 --> 00:30:39,000
Kepler was trying to observe the
motion of planets in the sky,

359
00:30:39,000 --> 00:30:42,000
and trying to come up with
general explanations of how they

360
00:30:42,000 --> 00:30:44,000
move.
Before him, people were saying,

361
00:30:44,000 --> 00:30:46,000
well, they cannot move in a
circle.

362
00:30:46,000 --> 00:30:48,000
But maybe it's more complicated
than that.

363
00:30:48,000 --> 00:30:51,000
We need to add smaller circular
motions on top of each other,

364
00:30:51,000 --> 00:30:53,000
and so on.
They have more and more

365
00:30:53,000 --> 00:30:56,000
complicated theories.
And then Kepler came with these

366
00:30:56,000 --> 00:31:00,000
laws that said basically that
planets move in an ellipse

367
00:31:00,000 --> 00:31:03,000
around the sun,
and that they move in a very

368
00:31:03,000 --> 00:31:07,000
specific way along that ellipse.
So, there's actually three

369
00:31:07,000 --> 00:31:11,000
laws, but let me just tell you
about the second one that has a

370
00:31:11,000 --> 00:31:17,000
very nice vector interpretation.
So, what Kepler's second law

371
00:31:17,000 --> 00:31:24,000
says is that the motion of
planets is, first of all,

372
00:31:24,000 --> 00:31:36,000
they move in a plane.
And second, the area swept out

373
00:31:36,000 --> 00:31:51,000
by the line from the sun to the
planet is swept at constant

374
00:31:51,000 --> 00:31:57,000
time.
Sorry, is swept at constant

375
00:31:57,000 --> 00:32:04,000
rate.
From the sun to the planet,

376
00:32:04,000 --> 00:32:14,000
it is swept out by the line at
a constant rate.

377
00:32:14,000 --> 00:32:23,000
OK, so that's an interesting
law because it tells you,

378
00:32:23,000 --> 00:32:27,000
once you know what the orbit of
the planet looks like,

379
00:32:27,000 --> 00:32:30,000
it tells you how fast it's
going to move on that orbit.

380
00:33:09,000 --> 00:33:19,000
OK, so let me explain again.
So, this law says maybe the

381
00:33:19,000 --> 00:33:27,000
sun, let's put the sun here at
the origin, and let's have a

382
00:33:27,000 --> 00:33:34,000
planet.
Well, the planet orbits around

383
00:33:34,000 --> 00:33:41,000
the sun -- -- in some
trajectory.

384
00:33:41,000 --> 00:33:45,000
So, this is supposed to be
light blue.

385
00:33:45,000 --> 00:33:49,000
Can you see that it's different
from white?

386
00:33:49,000 --> 00:33:51,000
No?
OK, me neither.

387
00:33:51,000 --> 00:33:53,000
[LAUGHTER]
OK, it doesn't really matter.

388
00:33:53,000 --> 00:33:55,000
So, the planet moves on its
orbit.

389
00:33:55,000 --> 00:34:00,000
And, if you wait for a certain
time, then a bit later it would

390
00:34:00,000 --> 00:34:04,000
be here, and then here,
and so on.

391
00:34:04,000 --> 00:34:09,000
Then, you can look at the
amount of area inside this

392
00:34:09,000 --> 00:34:12,000
triangular wedge.
And, the claim is that the

393
00:34:12,000 --> 00:34:16,000
amount of area in here is
proportional to the time

394
00:34:16,000 --> 00:34:18,000
elapsed.
So, in particular,

395
00:34:18,000 --> 00:34:21,000
if a planet is closer to the
sun, then it has to go faster.

396
00:34:21,000 --> 00:34:25,000
And, if it's farther away from
the sun, then it has to go

397
00:34:25,000 --> 00:34:28,000
slower so that the area remains
proportional to time.

398
00:34:28,000 --> 00:34:32,000
So, it's a very sophisticated
prediction.

399
00:34:32,000 --> 00:34:36,000
And, I think the way he came to
it was really just by using a

400
00:34:36,000 --> 00:34:39,000
lot of observations,
and trying to measure what was

401
00:34:39,000 --> 00:34:44,000
true that wasn't true.
But, let's try to see how we

402
00:34:44,000 --> 00:34:49,000
can understand that in terms of
all we know today about

403
00:34:49,000 --> 00:34:52,000
mechanics.
So, in fact,

404
00:34:52,000 --> 00:34:56,000
what happens is that Newton,
so Newton was quite a bit

405
00:34:56,000 --> 00:35:04,000
later.
That was the late 17th century

406
00:35:04,000 --> 00:35:13,000
instead of the beginning of the
17th century.

407
00:35:13,000 --> 00:35:30,000
So, he was able to explain this
using his laws for gravitational

408
00:35:30,000 --> 00:35:36,000
attraction.
And, you'll see that if we

409
00:35:36,000 --> 00:35:41,000
reformulate Kepler's Law in
terms of vectors,

410
00:35:41,000 --> 00:35:43,000
and if we work a bit with these
vectors,

411
00:35:43,000 --> 00:35:46,000
we are going to end up with
something that's actually

412
00:35:46,000 --> 00:35:49,000
completely obvious to us now.
At the time,

413
00:35:49,000 --> 00:35:52,000
it was very far from obvious,
but to us now to completely

414
00:35:52,000 --> 00:35:59,000
obvious.
So, let's try to see,

415
00:35:59,000 --> 00:36:15,000
what does Kepler's law say in
terms of vectors?

416
00:36:15,000 --> 00:36:24,000
OK, so, let's think of what
kinds of vectors we might want

417
00:36:24,000 --> 00:36:31,000
to have in here.
Well, it might be good to think

418
00:36:31,000 --> 00:36:38,000
of, maybe, the position vector,
and maybe its variation.

419
00:36:38,000 --> 00:36:46,000
So, if we wait a certain amount
of time, we'll have a vector,

420
00:36:46,000 --> 00:36:53,000
delta r, which is the change in
position vector a various

421
00:36:53,000 --> 00:36:59,000
interval of time.
OK, so let's start with the

422
00:36:59,000 --> 00:37:02,000
first step.
What's the most complicated

423
00:37:02,000 --> 00:37:05,000
thing in here?
It's this area swept out by the

424
00:37:05,000 --> 00:37:08,000
line.
How do we express that area in

425
00:37:08,000 --> 00:37:12,000
terms of vectors?
Well, I've almost given the

426
00:37:12,000 --> 00:37:14,000
answer by drawing this picture,
right?

427
00:37:14,000 --> 00:37:18,000
If I take a sufficiently small
amount of time,

428
00:37:18,000 --> 00:37:22,000
this shaded part looks like a
triangle.

429
00:37:22,000 --> 00:37:25,000
So, we have to find the area of
the triangle.

430
00:37:25,000 --> 00:37:27,000
Well, we know how to do that
now.

431
00:37:27,000 --> 00:37:34,000
So, the area is approximately
equal to one half of the area of

432
00:37:34,000 --> 00:37:40,000
a parallelogram that I could
form from these vectors.

433
00:37:40,000 --> 00:37:46,000
And, the area of a
parallelogram is given by the

434
00:37:46,000 --> 00:37:52,000
magnitude of a cross product.
OK, so, I should say,

435
00:37:52,000 --> 00:37:56,000
this is the area swept in time
delta t.

436
00:37:56,000 --> 00:38:00,000
You should think of delta t as
relatively small.

437
00:38:00,000 --> 00:38:05,000
I mean, the scale of a planet
that might still be a few days,

438
00:38:05,000 --> 00:38:09,000
but small compared to the other
old trajectory.

439
00:38:09,000 --> 00:38:16,000
So, let's remember that the
amount by which we moved,

440
00:38:16,000 --> 00:38:20,000
delta r,
is approximately equal to v

441
00:38:20,000 --> 00:38:25,000
times delta t,
OK, and just using the

442
00:38:25,000 --> 00:38:36,000
definition of a velocity vector.
So, let's use that.

443
00:38:36,000 --> 00:38:43,000
Sorry, so it's approximately
equal to r cross v magnitude

444
00:38:43,000 --> 00:38:48,000
times delta t.
I can take out the delta t,

445
00:38:48,000 --> 00:38:52,000
which is scalar.
So, now, what does it mean to

446
00:38:52,000 --> 00:38:55,000
say that area is swept at a
constant rate?

447
00:38:55,000 --> 00:39:00,000
It means this thing is
proportional to delta t.

448
00:39:00,000 --> 00:39:05,000
So, that means,
so, the law says,

449
00:39:05,000 --> 00:39:15,000
in fact, that the length of
this cross product r cross v

450
00:39:15,000 --> 00:39:25,000
equals a constant.
OK, r cross v has constant

451
00:39:25,000 --> 00:39:31,000
length.
Any questions about that?

452
00:39:31,000 --> 00:39:37,000
No? Yes?
Yes, let me try to explain that

453
00:39:37,000 --> 00:39:40,000
again.
So, what I'm claiming is that

454
00:39:40,000 --> 00:39:46,000
the length of the cross products
r cross v measures the rate at

455
00:39:46,000 --> 00:39:50,000
which area is swept by the
position vector.

456
00:39:50,000 --> 00:39:52,000
I should say,
with a vector of one half of

457
00:39:52,000 --> 00:39:55,000
this length is the rate at which
area is swept.

458
00:39:55,000 --> 00:39:58,000
How do we see that?
Well, let's take a small time

459
00:39:58,000 --> 00:40:01,000
interval, delta t.
In time, delta t,

460
00:40:01,000 --> 00:40:05,000
our planet moves by v delta t,
OK?

461
00:40:05,000 --> 00:40:08,000
So, if it moves by v delta t,
it means that this triangle up

462
00:40:08,000 --> 00:40:12,000
there has two sides.
One is the position vector,

463
00:40:12,000 --> 00:40:14,000
r.
The other one is v delta t.

464
00:40:14,000 --> 00:40:18,000
So, its area is given by one
half of the magnitude of a cross

465
00:40:18,000 --> 00:40:21,000
product.
That's the formula we've seen

466
00:40:21,000 --> 00:40:24,000
for the area of a triangle in
space.

467
00:40:24,000 --> 00:40:28,000
So, the area is one half of the
cross product,

468
00:40:28,000 --> 00:40:33,000
r, and v delta t,
magnitude of the cross product.

469
00:40:33,000 --> 00:40:37,000
So, to say that the rate at
which area is swept is constant

470
00:40:37,000 --> 00:40:39,000
means that these two are
proportional.

471
00:40:39,000 --> 00:40:42,000
Area divided by delta t is
constant at our time.

472
00:40:42,000 --> 00:40:51,000
And so, this is constant.
OK, now, what about the other

473
00:40:51,000 --> 00:40:58,000
half of the law?
Well, it says that the motion

474
00:40:58,000 --> 00:41:04,000
is in a plane,
and so we have a plane in which

475
00:41:04,000 --> 00:41:09,000
the motion takes place.
And, it contains,

476
00:41:09,000 --> 00:41:12,000
also, the sun.
And, it contains the

477
00:41:12,000 --> 00:41:16,000
trajectory.
So, let's think about that

478
00:41:16,000 --> 00:41:20,000
plane.
Well, I claim that the position

479
00:41:20,000 --> 00:41:25,000
vector is in the plane.
OK, that's what we are saying.

480
00:41:25,000 --> 00:41:28,000
But, there is another vector
that I know it is in the plane.

481
00:41:28,000 --> 00:41:32,000
You could say the position
vector at another time,

482
00:41:32,000 --> 00:41:34,000
or at any time,
but in fact,

483
00:41:34,000 --> 00:41:40,000
what's also true is that the
velocity vector is in the plane.

484
00:41:40,000 --> 00:41:44,000
OK, if I'm moving in the plane,
then position and velocity are

485
00:41:44,000 --> 00:41:50,000
in there.
So, the plane of motion

486
00:41:50,000 --> 00:41:59,000
contains r and v.
So, what's the direction of the

487
00:41:59,000 --> 00:42:08,000
cross product r cross v?
Well, it's the direction that's

488
00:42:08,000 --> 00:42:19,000
perpendicular to this plane.
So, it's normal to the plane of

489
00:42:19,000 --> 00:42:24,000
motion.
And, that means, now,

490
00:42:24,000 --> 00:42:28,000
that actually we've put the two
statements in there into a

491
00:42:28,000 --> 00:42:33,000
single form because we are
saying r cross v has constant

492
00:42:33,000 --> 00:42:37,000
length and constant direction.
In fact, in general,

493
00:42:37,000 --> 00:42:40,000
maybe I should say something
about this.

494
00:42:40,000 --> 00:42:42,000
So, if you just look at the
position vector,

495
00:42:42,000 --> 00:42:45,000
and the velocity vector for any
motion at any given time,

496
00:42:45,000 --> 00:42:48,000
then together,
they determine some plane.

497
00:42:48,000 --> 00:42:51,000
And, that's the plane that
contains the origin,

498
00:42:51,000 --> 00:42:54,000
the point, and the velocity
vector.

499
00:42:54,000 --> 00:42:56,000
If you want,
it's the plane in which the

500
00:42:56,000 --> 00:42:59,000
motion seems to be going at the
given time.

501
00:42:59,000 --> 00:43:01,000
Now, of course,
if your motion is not in a

502
00:43:01,000 --> 00:43:03,000
plane, then that plane will
change.

503
00:43:03,000 --> 00:43:06,000
It's, however,
instant, if a plane in which

504
00:43:06,000 --> 00:43:09,000
the motion is taking place at a
given time.

505
00:43:09,000 --> 00:43:13,000
And, to say that the motion
actually stays in that plane

506
00:43:13,000 --> 00:43:17,000
forever means that this guy will
not change direction.

507
00:43:17,000 --> 00:43:25,000
OK, so -- [LAUGHTER]
[APPLAUSE]

508
00:43:25,000 --> 00:43:42,000
OK, so, Kepler's second law is
actually equivalent to saying

509
00:43:42,000 --> 00:43:55,000
that r cross v equals a constant
vector, OK?

510
00:43:55,000 --> 00:44:04,000
That's what the law says.
So, in terms of derivatives,

511
00:44:04,000 --> 00:44:14,000
it means d by dt of r cross v
is the zero vector.

512
00:44:14,000 --> 00:44:20,000
OK, now, so there's an
interesting thing to note,

513
00:44:20,000 --> 00:44:23,000
which is that we can use the
usual product rule for

514
00:44:23,000 --> 00:44:26,000
derivatives with vector
expressions,

515
00:44:26,000 --> 00:44:28,000
with dot products or cross
products.

516
00:44:28,000 --> 00:44:30,000
There's only one catch,
which is that when we

517
00:44:30,000 --> 00:44:34,000
differentiate a cross product,
we have to be careful that the

518
00:44:34,000 --> 00:44:36,000
guy on the left stays on the
left.

519
00:44:36,000 --> 00:44:40,000
The guy on the right stays on
the right.

520
00:44:40,000 --> 00:44:44,000
OK, so, if you know that uv
prime equals u prime v plus uv

521
00:44:44,000 --> 00:44:47,000
prime, then you are safe.
If you know it as u prime v

522
00:44:47,000 --> 00:44:50,000
cross v prime u,
then you are not safe.

523
00:44:50,000 --> 00:44:52,000
OK, so it's the only thing to
watch for.

524
00:44:52,000 --> 00:45:05,000
So, product rule is OK for
taking the derivative of a dot

525
00:45:05,000 --> 00:45:10,000
product.
There, you don't actually even

526
00:45:10,000 --> 00:45:14,000
need to be very careful about
all the things or the derivative

527
00:45:14,000 --> 00:45:18,000
of a cross product.
There you just need to be a

528
00:45:18,000 --> 00:45:27,000
little bit more careful.
OK, so, now that we know that,

529
00:45:27,000 --> 00:45:39,000
we can write this as dr/dt
cross v plus r cross dv/dt,

530
00:45:39,000 --> 00:45:42,000
OK?
Well, let's reformulate things

531
00:45:42,000 --> 00:45:47,000
slightly.
So, dr dt already has a name.

532
00:45:47,000 --> 00:45:50,000
In fact, that's v.
OK, that's what we call the

533
00:45:50,000 --> 00:45:55,000
velocity vector.
So, this is v cross v plus r

534
00:45:55,000 --> 00:46:04,000
cross, what is dv/dt?
That's the acceleration,

535
00:46:04,000 --> 00:46:11,000
a, equals zero.
OK, so now what's the next step?

536
00:46:11,000 --> 00:46:15,000
Well, we know what v cross v is
because, remember,

537
00:46:15,000 --> 00:46:18,000
a vector cross itself is always
zero, OK?

538
00:46:18,000 --> 00:46:30,000
So, this is the same r cross a
equals zero,

539
00:46:30,000 --> 00:46:35,000
and that's the same as saying
that the cross product of two

540
00:46:35,000 --> 00:46:39,000
vectors is zero exactly when the
parallelogram of the form has no

541
00:46:39,000 --> 00:46:41,000
area.
And, the way in which that

542
00:46:41,000 --> 00:46:45,000
happens is if they are actually
parallel to each other.

543
00:46:45,000 --> 00:46:50,000
So, that means the acceleration
is parallel to the position.

544
00:46:50,000 --> 00:46:55,000
OK, so, in fact,
what Kepler's second law says

545
00:46:55,000 --> 00:47:02,000
is that the acceleration is
parallel to the position vector.

546
00:47:02,000 --> 00:47:05,000
And, since we know that
acceleration is caused by a

547
00:47:05,000 --> 00:47:08,000
force that's equivalent to the
fact that the gravitational

548
00:47:08,000 --> 00:47:08,000
force --

549
00:47:13,000 --> 00:47:18,000
-- is parallel to the position
vector, that means,

550
00:47:18,000 --> 00:47:22,000
well, if you have the sun here
at the origin,

551
00:47:22,000 --> 00:47:27,000
and if you have your planets,
well, the gravitational force

552
00:47:27,000 --> 00:47:32,000
caused by the sun should go
along this line.

553
00:47:32,000 --> 00:47:34,000
In fact, the law doesn't even
say whether it's going towards

554
00:47:34,000 --> 00:47:37,000
the sun or away from the sun.
Well, what we know now is that,

555
00:47:37,000 --> 00:47:39,000
of course, the attraction is
towards the sun.

556
00:47:39,000 --> 00:47:41,000
But, Kepler's law would also be
true, actually,

557
00:47:41,000 --> 00:47:44,000
if things were going away.
So, in particular,

558
00:47:44,000 --> 00:47:48,000
say, electric force also has
this property of being towards

559
00:47:48,000 --> 00:47:50,000
the central charge.
So, actually,

560
00:47:50,000 --> 00:47:54,000
if you look at motion of
charged particles in an electric

561
00:47:54,000 --> 00:47:58,000
field caused by a point charged
particle, it also satisfies

562
00:47:58,000 --> 00:48:01,000
Kepler's law,
satisfies the same law.

563
00:48:01,000 --> 00:48:03,000
OK, that's the end for today,
thanks.