The following content is
provided under a Creative Commons license.
Your support will help MIT OpenCourseWare continue to offer
high quality educational resources for free.
To make a donation or to view additional materials from
hundreds of MIT courses, visit MIT OpenCourseWare at
ocw.mit.edu. 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.
