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So, if you remember last time, we looked at parametric
equations -- -- as a way of describing the motion of a point
that moves in the plane or in space as a function of time of
your favorite parameter that will tell you how far the motion
has progressed. And, I think we did it in
detail the example of the cycloid, which is the curve
traced by a point on a wheel that's rolling on a flat
surface. So, we have this example where
we have this wheel that's rolling on the x-axis,
and we have this point on the wheel.
And, as it moves around, it traces a trajectory that
moves more or less like this. OK, so I'm trying a new color.
Is this visible from the back? So, no more blue.
OK, so remember, in general, we are trying to
find the position, so, x of t, y of t,
maybe z of t if we are in space -- -- of a moving point along a
trajectory. And, one way to think about
this is in terms of the position vector.
So, position vector is just the vector whose components are
coordinates of a point, OK, so if you prefer,
that's the same thing as a vector from the origin to the
moving point. So, maybe our point is here, P.
So, this vector here -- This vector here is vector OP.
And, that's also the position vector r of t.
So, just to give you, again, that example -- -- if I
take the cycloid for a wheel of radius 1,
and let's say that we are going at unit speed so that the angle
that we used as a parameter of time is the same thing as time
when the position vector, in this case,
we found to be, just to make sure that they
have it right, .
OK, that's a formula that you should have in your notes from
last time, except we had theta instead of t because we were
using the angle. But now I'm saying,
we are moving at unit speed, so time and angle are the same
thing. So, now, what's interesting
about this is we can analyze the motion in more detail.
OK, so, now that we know the position of the point as a
function of time, we can try to study how it
varies in particular things like the speed and acceleration.
OK, so let's start with speed. Well, in fact we can do better
than speed. Let's not start with speed.
So, speed is a number. It tells you how fast you are
going along your trajectory. I mean, if you're driving in a
car, then it tells you how fast you are going.
But, unless you have one of these fancy cars with a GPS,
it doesn't tell you which direction you're going.
And, that's useful information, too, if you're trying to figure
out what your trajectory is. So, in fact,
there's two aspects to it. One is how fast you are going,
and the other is in what direction you're going.
That means actually we should use a vector maybe to think
about this. And so, that's called the
velocity vector. And, the way we can get it,
so, it's called usually V, so, V here stands for velocity
more than for vector. And, you just get it by taking
the derivative of a position vector with respect to time.
Now, it's our first time writing this kind of thing with
a vector. So, the basic rule is you can
take the derivative of a vector quantity just by taking the
derivatives of each component. OK, so that's just dx/dt,
dy/dt, and if you have z component, dz/dt.
So, let me -- OK, so -- OK, so let's see what it
is for the cycloid. So, an example of a cycloid,
well, so what do we get when we take the derivatives of this
formula there? Well, so, the derivative of t
is 1- cos(t). The derivative of 1 is 0.
The derivative of -cos(t) is sin(t).
Very good. OK, that's at least one thing
you should remember from single variable calculus.
Hopefully you remember even more than that.
OK, so that's the velocity vector.
It tells us at any time how fast we are going,
and in what direction. So, for example, observe.
Remember last time at the end of class we were trying to
figure out what exactly happens near the bottom point,
when we have this motion that seems to stop and go backwards.
And, we answered that one way. But, let's try to understand it
in terms of velocity. What if I plug t equals 0 in
here? Then, 1- cos(t) is 0,
sin(t) is 0. The velocity is 0.
So, at the time,at that particular time,
our point is actually not moving.
Of course, it's been moving just before, and it starts
moving just afterwards. It's just the instant,
at that particular instant, the speed is zero.
So, that's especially maybe a counterintuitive thing,
but something is moving. And at that time,
it's actually stopped. Now, let's see,
so that's the vector. And, it's useful.
But, if you want just the usual speed as a number,
then, what will you do? Well, you will just take
exactly the magnitude of this vector.
So, speed, which is the scalar quantity is going to be just the
magnitude of the vector, V.
OK, so, in this case, while it would be square root
of (1- cost)^2 sin^2(t), and if you expand that,
you will get, let me take a bit more space,
it's going to be square root of 1 - 2cos(t) cos^2(t) sin^2(t).
It seems to simplify a little bit because we have cos^2 plus
sin^2. That's 1.
So, it's going to be the square root of 2 - 2cos(t).
So, at this point, if I was going to ask you,
when is the speed the smallest or the largest?
You could answer based on that. See, at t equals 0,
well, that turns out to be zero.
The point is not moving. At t equals pi,
that ends up being the square root of 2 plus 2,
which is 4. So, that's 2.
And, that's when you're truly at the top of the arch,
and that's when the point is moving the fastest.
In fact, they are spending twice as fast as the wheel
because the wheel is moving to the right at unit speed,
and the wheel is also rotating. So, it's moving to the right
and unit speed relative to the center so that the two effects
add up, and give you a speed of 2.
Anyway, that's a formula we can get.
OK, now, what about acceleration?
So, here I should warn you that there is a serious discrepancy
between the usual intuitive notion of acceleration,
the one that you are aware of when you drive a car and the one
that we will be using. So, you might think
acceleration is just the directive of speed.
If my car goes 55 miles an hour on the highway and it's going a
constant speed, it's not accelerating.
But, let's say that I'm taking a really tight turn.
Then, I'm going to feel something.
There is some force being exerted.
And, in fact, there is a sideways
acceleration at that point even though the speed is not
changing. So, the definition will take
effect. The acceleration is,
as a vector, and the acceleration vector is
just the derivative of a velocity vector.
So, even if the speed is constant, that means,
even if a length of the velocity vector stays the same,
the velocity vector can still rotate.
And, as it rotates, it uses acceleration.
OK, and so this is the notion of acceleration that's relevant
to physics when you find F=ma; that's the (a) that you have in
mind here. It's a vector.
Of course, if you are moving in a straight line,
then the two notions are the same.
I mean, acceleration is also going to be along the line,
and it's going to has to do with the derivative of speed.
But, in general, that's not quite the same.
So, for example, let's look at the cycloid.
If we take the example of the cycloid, well,
what's the derivative of one minus cos(t)?
It's sin(t). And, what's the derivative of
sin(t)? cos(t), OK.
So, the acceleration vector is .
So, in particular, let's look at what happens at
time t equals zero when the point is not moving.
Well, the acceleration vector there will be zero from one.
So, what that means is that if I look at my trajectory at this
point, that the acceleration vector is pointing in that
direction. It's the unit vector in the
vertical direction. So, my point is not moving at
that particular time. But, it's accelerating up.
So, that means that actually as it comes down,
first it's slowing down. Then it stops here,
and then it reverses going back up.
OK, so that's another way to understand what we were saying
last time that the trajectory at that point has a vertical
tendency because that's the direction in which the motion is
going to occur just before and just after time zero.
OK, any questions about that? No.
OK, so I should insist maybe on one thing,
which is that, so, we can differentiate
vectors just component by component,
OK, and we can differentiate vector expressions according to
certain rules that we'll see in a moment.
One thing that we cannot do, it's not true that the length
of dr dt, which is the speed, is equal to the length of dt.
OK, this is completely false. And, they are really not the
same. So, if you have to
differentiate the length of a vector, but basically you are in
trouble. If you really,
really want to do it, well, the length of the vector
is the square root of the sums of the squares of the
components, and from that you can use the
formula for the derivative of the square root,
and the chain rule, and various other things.
And, you can get there. But, it will not be a very nice
expression. There is no simple formula for
this kind of thing. Fortunately,
we almost never have to compute this kind of thing because,
after all, it's not a very relevant quantity.
What's more relevant might be this one.
This is actually the speed. This one, I don't know what it
means. OK.
So, let's continue our exploration.
So, the next concept that I want to define is that of arc
length. So, arc length is just the
distance that you have traveled along the curve,
OK? So, if you are in a car,
you know, it has mileage counter that tells you how far
you've gone, how much fuel you've used if it's a fancy car.
And, what it does is it actually integrates the speed of
the time to give you the arc length along the trajectory of
the car. So, the usual notation that we
will have is (s) for arc length. I'm not quite sure how you get
an (s) out of this, but it's the usual notation.
OK, so, (s) is for distance traveled along the trajectory.
And, so that makes sense, of course, we need to fix a
reference point. Maybe on the cycloid,
we'd say it's a distance starting on the origin.
In general, maybe you would say you start at time,
t equals zero. But, it's a convention.
If you knew in advance, you could have,
actually, your car's mileage counter to count backwards from
the point where the car will die and start walking.
I mean, that would be sneaky-freaky,
but you could have a negative arc length that gets closer and
closer to zero, and gets to zero at the end of
a trajectory, or anything you want.
I mean, arc length could be positive or negative.
Typically it's negative what you are before the reference
point, and positive afterwards. So, now, how does it relate to
the things we've seen there? Well, so in particular,
how do you relate arc length and time?
Well, so, there's a simple relation, which is that the rate
of change of arc length versus time,
well, that's going to be the speed at which you are moving,
OK, because the speed as a scalar quantity tells you how
much distance you're covering per unit time.
OK, and in fact, to be completely honest,
I should put an absolute value here because there is examples
of curves maybe where your motion is going back and forth
along the same curve. And then, you don't want to
keep counting arc length all the time.
Actually, maybe you want to say that the arc length increases
and then decreases along the curve.
I mean, you get to choose how you count it.
But, in this case, if you are moving back and
forth, it would make more sense to have the arc length first
increase, then decrease,
increase again, and so on.
So -- So if you want to know really what the arc length is,
then basically the only way to do it is to integrate speed
versus time. So, if you wanted to know how
long an arch of cycloid is, you have this nice-looking
curve; how long is it?
Well, you'd have to basically integrate this quantity from t
equals zero to 2 pi.
And, to say the truth, I don't really know how to
integrate that. So, we don't actually have a
formula for the length at this point.
However, we'll see one later using a cool trick,
and multi-variable calculus. So, for now,
we'll just leave the formula like that, and we don't know how
long it is. Well, you can put that into
your calculator and get the numerical value.
But, that's the best I can offer.
Now, another useful notion is the unit vector to the
trajectory. So, the usual notation is T hat.
It has a hat because it's a unit vector, and T because it's
tangent. Now, how do we get this unit
vector? So, maybe I should have pointed
out before that if you're moving along some trajectory,
say you're going in that direction, then when you're at
this point, the velocity vector is going to
be tangential to the trajectory. It tells you the direction of
motion in particular. So, if you want a unit vector
that goes in the same direction, all you have to do is rescale
it, so, at its length becomes one.
So, it's v divided by a magnitude of v.
So, it seems like now we have a lot of different things that
should be related in some way. So, let's see what we can say.
Well, we can say that dr by dt, so, that's the velocity vector,
that's the same thing as if I use the chain rule dr/ds times
ds/dt. OK, so, let's think about this
things. So, this guy here we've just
seen. That's the same as the speed,
OK? So, this one here should be v
divided by its length. So, that means this actually
should be the unit vector. OK, so, let me rewrite that.
It's T ds/dt. So, maybe if I actually stated
directly that way, see, I'm just saying the
velocity vector has a length and a direction.
The length is the speed. The direction is tangent to the
trajectory.
So, the speed is ds/dt, and the vector is T hat.
And, that's how we get this. So, let's try just to see why
dr/ds should be T. Well, let's think of dr/ds.
dr/ds means position vector r means you have the origin,
which is somewhere out there, and the vector r is here.
So, dr/ds means we move by a small amount,
delta s along the trajectory a certain distance delta s.
And, we look at how the position vector changes.
Well, we'll have a small change. Let me call that vector delta r
corresponding to the size, corresponding to the length
delta s. And now, delta r should be
essentially roughly equal to, well, its direction will be
tangent to the trajectory. If I take a small enough
interval, then the direction will be
almost tensioned to the trajectory times the length of
it will be delta s, the distance that I have
traveled. OK, sorry, maybe I should
explain that on a separate board.
OK, so, let's say that we have that amount of time,
delta t. So, let's zoom into that curve.
So, we have r at time t. We have r at time t plus delta
t. This vector here I will call
delta r. The length of this vector is
delta s. And, the direction is
essentially that of a vector. OK, so, delta s over delta t,
that's the distance traveled divided by the time.
That's going to be close to the speed.
And, delta r is approximately T times delta s.
So, now if I divide both sides by delta t, I get this.
And, if I take the limit as delta t turns to zero,
then I get the same formula with the derivatives and with an
equality. It's an approximation.
The approximation becomes better and better if I go to
smaller intervals.
OK, are there any questions about this?
Yes? Yes, that's correct.
OK, so let's be more careful, actually.
So, you're asking about whether the delta r is actually strictly
tangent to the curve. Is that -- That's correct.
Actually, delta r is not strictly tangent to anything.
So, maybe I should draw another picture.
If I'm going from here to here, then delta r is going to be
this arc inside the curve while the vector will be going in this
direction, OK? So, they are not strictly
parallel to each other. That's why it's only
approximately equal. Similarly, this distance,
the length of delta r is not exactly the length along the
curve. It's actually a bit shorter.
But, if we imagine a smaller and smaller portion of the
curve, then this effect of the curve
being a curve and not a straight line becomes more and more
negligible. If you zoom into the curve
sufficiently, then it looks more and more
like a straight line. And then, what I said becomes
true in the limit. OK? Any other questions?
No? OK. So, what happens next?
OK, so let me show you a nice example of why we might want to
use vectors to study parametric curves because,
after all, a lot of what's here you can just do in coordinates.
And, we don't really need vectors.
Well, and truly, vectors being a language,
you never strictly need it, but it's useful to have a
notion of vectors. So, I want to tell you a bit
about Kepler's second law of celestial mechanics.
So, that goes back to 1609. So, that's not exactly recent
news, OK? But, still I think it's a very
interesting example of why you might want to use vector methods
to analyze motions. So, what happened back then was
Kepler was trying to observe the motion of planets in the sky,
and trying to come up with general explanations of how they
move. Before him, people were saying,
well, they cannot move in a circle.
But maybe it's more complicated than that.
We need to add smaller circular motions on top of each other,
and so on. They have more and more
complicated theories. And then Kepler came with these
laws that said basically that planets move in an ellipse
around the sun, and that they move in a very
specific way along that ellipse. So, there's actually three
laws, but let me just tell you about the second one that has a
very nice vector interpretation. So, what Kepler's second law
says is that the motion of planets is, first of all,
they move in a plane. And second, the area swept out
by the line from the sun to the planet is swept at constant
time. Sorry, is swept at constant
rate. From the sun to the planet,
it is swept out by the line at a constant rate.
OK, so that's an interesting law because it tells you,
once you know what the orbit of the planet looks like,
it tells you how fast it's going to move on that orbit.
OK, so let me explain again. So, this law says maybe the
sun, let's put the sun here at the origin, and let's have a
planet. Well, the planet orbits around
the sun -- -- in some trajectory.
So, this is supposed to be light blue.
Can you see that it's different from white?
No? OK, me neither.
[LAUGHTER] OK, it doesn't really matter.
So, the planet moves on its orbit.
And, if you wait for a certain time, then a bit later it would
be here, and then here, and so on.
Then, you can look at the amount of area inside this
triangular wedge. And, the claim is that the
amount of area in here is proportional to the time
elapsed. So, in particular,
if a planet is closer to the sun, then it has to go faster.
And, if it's farther away from the sun, then it has to go
slower so that the area remains proportional to time.
So, it's a very sophisticated prediction.
And, I think the way he came to it was really just by using a
lot of observations, and trying to measure what was
true that wasn't true. But, let's try to see how we
can understand that in terms of all we know today about
mechanics. So, in fact,
what happens is that Newton, so Newton was quite a bit
later. That was the late 17th century
instead of the beginning of the 17th century.
So, he was able to explain this using his laws for gravitational
attraction. And, you'll see that if we
reformulate Kepler's Law in terms of vectors,
and if we work a bit with these vectors,
we are going to end up with something that's actually
completely obvious to us now. At the time,
it was very far from obvious, but to us now to completely
obvious. So, let's try to see,
what does Kepler's law say in terms of vectors?
OK, so, let's think of what kinds of vectors we might want
to have in here. Well, it might be good to think
of, maybe, the position vector, and maybe its variation.
So, if we wait a certain amount of time, we'll have a vector,
delta r, which is the change in position vector a various
interval of time. OK, so let's start with the
first step. What's the most complicated
thing in here? It's this area swept out by the
line. How do we express that area in
terms of vectors? Well, I've almost given the
answer by drawing this picture, right?
If I take a sufficiently small amount of time,
this shaded part looks like a triangle.
So, we have to find the area of the triangle.
Well, we know how to do that now.
So, the area is approximately equal to one half of the area of
a parallelogram that I could form from these vectors.
And, the area of a parallelogram is given by the
magnitude of a cross product. OK, so, I should say,
this is the area swept in time delta t.
You should think of delta t as relatively small.
I mean, the scale of a planet that might still be a few days,
but small compared to the other old trajectory.
So, let's remember that the amount by which we moved,
delta r, is approximately equal to v
times delta t, OK, and just using the
definition of a velocity vector. So, let's use that.
Sorry, so it's approximately equal to r cross v magnitude
times delta t. I can take out the delta t,
which is scalar. So, now, what does it mean to
say that area is swept at a constant rate?
It means this thing is proportional to delta t.
So, that means, so, the law says,
in fact, that the length of this cross product r cross v
equals a constant. OK, r cross v has constant
length. Any questions about that?
No? Yes? Yes, let me try to explain that
again. So, what I'm claiming is that
the length of the cross products r cross v measures the rate at
which area is swept by the position vector.
I should say, with a vector of one half of
this length is the rate at which area is swept.
How do we see that? Well, let's take a small time
interval, delta t. In time, delta t,
our planet moves by v delta t, OK?
So, if it moves by v delta t, it means that this triangle up
there has two sides. One is the position vector,
r. The other one is v delta t.
So, its area is given by one half of the magnitude of a cross
product. That's the formula we've seen
for the area of a triangle in space.
So, the area is one half of the cross product,
r, and v delta t, magnitude of the cross product.
So, to say that the rate at which area is swept is constant
means that these two are proportional.
Area divided by delta t is constant at our time.
And so, this is constant. OK, now, what about the other
half of the law? Well, it says that the motion
is in a plane, and so we have a plane in which
the motion takes place. And, it contains,
also, the sun. And, it contains the
trajectory. So, let's think about that
plane. Well, I claim that the position
vector is in the plane. OK, that's what we are saying.
But, there is another vector that I know it is in the plane.
You could say the position vector at another time,
or at any time, but in fact,
what's also true is that the velocity vector is in the plane.
OK, if I'm moving in the plane, then position and velocity are
in there. So, the plane of motion
contains r and v. So, what's the direction of the
cross product r cross v? Well, it's the direction that's
perpendicular to this plane. So, it's normal to the plane of
motion. And, that means, now,
that actually we've put the two statements in there into a
single form because we are saying r cross v has constant
length and constant direction. In fact, in general,
maybe I should say something about this.
So, if you just look at the position vector,
and the velocity vector for any motion at any given time,
then together, they determine some plane.
And, that's the plane that contains the origin,
the point, and the velocity vector.
If you want, it's the plane in which the
motion seems to be going at the given time.
Now, of course, if your motion is not in a
plane, then that plane will change.
It's, however, instant, if a plane in which
the motion is taking place at a given time.
And, to say that the motion actually stays in that plane
forever means that this guy will not change direction.
OK, so -- [LAUGHTER] [APPLAUSE]
OK, so, Kepler's second law is actually equivalent to saying
that r cross v equals a constant vector, OK?
That's what the law says. So, in terms of derivatives,
it means d by dt of r cross v is the zero vector.
OK, now, so there's an interesting thing to note,
which is that we can use the usual product rule for
derivatives with vector expressions,
with dot products or cross products.
There's only one catch, which is that when we
differentiate a cross product, we have to be careful that the
guy on the left stays on the left.
The guy on the right stays on the right.
OK, so, if you know that uv prime equals u prime v plus uv
prime, then you are safe. If you know it as u prime v
cross v prime u, then you are not safe.
OK, so it's the only thing to watch for.
So, product rule is OK for taking the derivative of a dot
product. There, you don't actually even
need to be very careful about all the things or the derivative
of a cross product. There you just need to be a
little bit more careful. OK, so, now that we know that,
we can write this as dr/dt cross v plus r cross dv/dt,
OK? Well, let's reformulate things
slightly. So, dr dt already has a name.
In fact, that's v. OK, that's what we call the
velocity vector. So, this is v cross v plus r
cross, what is dv/dt? That's the acceleration,
a, equals zero. OK, so now what's the next step?
Well, we know what v cross v is because, remember,
a vector cross itself is always zero, OK?
So, this is the same r cross a equals zero,
and that's the same as saying that the cross product of two
vectors is zero exactly when the parallelogram of the form has no
area. And, the way in which that
happens is if they are actually parallel to each other.
So, that means the acceleration is parallel to the position.
OK, so, in fact, what Kepler's second law says
is that the acceleration is parallel to the position vector.
And, since we know that acceleration is caused by a
force that's equivalent to the fact that the gravitational
force --
-- is parallel to the position vector, that means,
well, if you have the sun here at the origin,
and if you have your planets, well, the gravitational force
caused by the sun should go along this line.
In fact, the law doesn't even say whether it's going towards
the sun or away from the sun. Well, what we know now is that,
of course, the attraction is towards the sun.
But, Kepler's law would also be true, actually,
if things were going away. So, in particular,
say, electric force also has this property of being towards
the central charge. So, actually,
if you look at motion of charged particles in an electric
field caused by a point charged particle, it also satisfies
Kepler's law, satisfies the same law.
OK, that's the end for today, thanks.