CHRISTINE BREINER: Welcome back to recitation.
In this video, what I'd like us to do is answer some
questions I've posed here in a really long problem.
So let me take us through it.
So we're going to consider a position vector that's
described by x of t, y of t, 0.
I want to have it in three space so I can take a cross
product later.
And then we're going to suppose that it actually has
constant length.
And that when I look at the acceleration vector at t, it's
actually equal to a constant times r of t where the
constant is not 0.
OK?
So I can, these are the two things I know about this
position vector.
It has constant length for all t, it has constant length.
And the acceleration is always equal to some constant times
the position.
OK, that's what I'm giving you.
And now I want you to use those things and vector
differentiation to show that r dot v is equal to 0, where v
is the velocity.
And then to show that r cross v is constant.
So you're going to have to figure out, essentially, the
thing you have to figure out is what relationship do you
want to differentiate to show these two things.
OK?
That's the hard part of this problem.
And then, I went to see if you can give an
example of such an r.
So if you can give an example of a position vector that has
these properties.
And maybe if you're having a hard time, the first thing for
you to do might be to think about this, and to see if you
can figure out an example of that, and then see kind of how
things work together in that example.
That may actually proved helpful.
So why don't you work on this problem, pause the video, and
then when you're feeling good about your answer, you can
bring the video back up and I'll show you how I do it.
OK, welcome back.
So there was a lot to do in this problem, but let me just
remind you what the framework is.
We have a position vector, and we know two things about it.
We know that it has constant length and we know that the
acceleration is always equal to some
constant times the position.
I didn't give you the constant, but we know it's
always equal to some constant times the position.
And then we wanted to show two things using vector
differentiation.
We wanted to show that r dot v was 0.
And we wanted to show that r cross v was constant.
And then we want to talk about what is an example.
So let's start off and see if we can figure out how to show
that r dot v is equal to 0.
And, you know, as you're thinking about this problem,
something that you want to remember as you're thinking
about this is, well, what are the things that I know?
I know that r dot r is constant, so I'm going to
write that down.
r dot r--
I'm not going to say is equal to c, because that's a
different constant--
so I'll just, let me just call it c1.
OK?
That's a different constant than my c, maybe.
OK, I know that r dot r is some constant, and I want to
show something about r dot v. Right?
So if I'm looking at this and I say, well, I
know this thing here.
So it's the only dot product relationship that I have.
Because right now, I know a relationship between a and r,
and I know r has constant length.
Since that's all I gave you, if I want to look at a dot
product, this is the relationship I know.
The constant length thing.
And so I know I somehow have to use this one to figure out
something
about r dot v. OK.
Well, what's the point that we should realize?
What is v?
v is d dt of r.
So I could take the derivative, if I could take
the derivative of just one of these things, then I would get
r dot v down here.
Right?
If I took d dt of just one of these, I would get the r dot
v. And d dt of this is 0.
But I can't do that, right?
Because if I take d dt of this whole thing, I'm going to end
up having to differentiate this r once and leave this
alone, and I'm going to have to differentiate this r and
leave this alone.
But if you heard what I was just saying, maybe you see
that that's actually still going to be OK.
So let's look at what happens.
I mean, this is all really I have to work with, so I'm
going to explore.
Let's look at what happens when I take d dt of r dot r.
And I'm going to start leaving off the hats here, because I'm
going to leave them off somewhere, so we'll just leave
them off now, and then I won't leave some off
and put some on.
So from here on out, I'm just going to write r, v, and a
without that hats, but they're vectors.
OK.
So d dt of r dot r, well, I have to take d dt of r, and
then I dot that with r, and then I take r dotted with--
sorry--
d dt of r.
Right?
Well, what do I get here? d dt of r we said was v, so
that's v dot r.
And what do I get here?
This is r dot--
there's d dt of r again-- so I get r dot v.
Now the great thing about the dot product is that if I
switch the order of these two, it's still the same thing.
So I can just write this as--
well, I'll switch the order of this one.
So they both look like r dot v plus r dot v, and that means I
get 2 r dot v. Right?
So d dt of r dot r is actually 2 of r dotted with v-- the
position dotted with the velocity.
Now why is this going to help me?
Because what do I know about this quantity r dot r?
I know it's constant, right?
So what is d dt of a constant?
d dt of a constant is 0.
So I actually started off with knowing this was 0.
So if I go through the whole chain, I see 0 is equal to--
let me put the 0 down here again--
0 is equal to 2 r dot v, and so r dot v I
know is equal to 0.
What does that mean geometrically?
That means r and v are orthogonal.
And where is v going to sit?
Well, if I come back over to how I described r, r is in the
xy plane, right?
The z component is 0.
So if I differentiate--
if I take d dt of r--
I'm going to have the derivative of x to the
function of t.
And then whatever y is as a function of t, I take that
derivative.
And this is still 0.
So v is going to sit in the xy plane, and based on what we
know so far, we know it's actually orthogonal to r.
So they make a 90-degree angle at all times t.
OK, and how did we do that again?
I just want to remind you, we knew one dot product
relationship, that was r dot r was a constant.
So we differentiated that and tried to see what happened.
And the main point at the end of it, was that when I had a v
dot r, I could rewrite it as an r dot v, and so I just end
up with two of something that I want to know about.
So that's the main idea of the first part.
Now the second part was I asked you to figure out
something over here.
We wanted to know that r cross v is constant.
OK?
And t, it's always the same.
All right?
So let's think about if I want to show that for every t
something is constant, I could show--
actually, I've sort of seen it already--
I could show that its derivative is 0.
OK.
So if r cross v is constant-- or If its derivative is 0, I
should say--
then r cross v was constant.
Right?
If its derivative is 0 for all t, then r cross v is constant.
Right?
So that's really what we want to exploit here.
So let's look at the idea.
So we want to-- again, let me write it down--
show r cross v is constant.
And the strategy we're going to use is to do this we're
going to show that d dt of r cross v is equal to 0.
Right?
If we can show that, then this means that for
all t it's the same.
It's not changing in t.
Right?
So for all t, r cross v is going to be the same, so r
cross v is going to be constant.
So the difference between the two problems was in the first
problem you didn't quite know maybe what expression you
needed to differentiate to find what you
were looking for.
Here, we know what we need to differentiate, but we have to
make sure we understand--
to show this is constant--
when we differentiate, we should get 0.
OK?
So that's sort of a slightly different type of problem.
And I'm asking you maybe from the other side here.
So let's now, let's just see what we get on
the left-hand side.
So what's d dt of r cross v?
Well, d dt of r is v, right?
So we get v cross v for the first term.
So take d dt of r, we get v. We leave this v alone, and
then we add to that, r cross--
d dt of v, what's d dt of v--
that's a.
Right?
Now, let's take a look at this.
Well, v cross v, v is pointing in the same
direction as itself.
So when you try and take a cross product of that, you
know that the length of your vector should be the area of
the parallelogram formed by these two vectors.
But v is pointing in the same direction as itself, so
there's no area there.
That's a geometric interpretation of why this
thing should be 0.
Another reason is that remember that your v cross v
is going to include a sine theta term, where theta is the
angle between the two vectors, right?
That's another formula you have. And so when you look at
the angle between this vector and itself, it's 0.
And sine 0 is 0.
So this is, in fact, 0 in that part.
So if this is 0, then we get what we want.
Well, I only gave you one other bit of information in
the problem.
And if you remember, it was that a is always equal to a
constant times r.
So I can rewrite this right-hand side as r cross a
constant times r.
And because of properties of these cross products, I can
pull out the constant.
Or I can actually, I guess I don't need to pull it out to
talk about it, but it's nicer if I pull it out.
And look at what I have here.
I have the exact same situation as v cross v. I
mean, this is still pointing in the same direction.
Constant times r and r still point in the same direction as
if I were to compare r and r.
So I didn't have to pull out the constant, but then right
here it's very easy to see that this is also 0.
So I had 0 plus this, so I get 0.
So, I've shown through this process--
maybe I should have written equal signs here--
that d dt of r cross v is actually equal to 0.
And so you see that this cross product between r and v--
which we know are orthogonal sitting in the xy plane--
that it's always the same, it's always the same vector.
And now I asked you to give an example, and maybe you thought
of an example first, and then thought of how it worked.
And so the easiest example is if you let r of t equal
cosine t sine t.
So the easiest example--
there are others, obviously--
is if you let r of t equal cosine t, sine t comma 0.
Sorry, I was thinking about it in three space, right?
And in fact, if I were to scale this,
it would still work.
I could put any constant in front.
This carves out--
I let t go between 0 and 2 pi, or even
minus infinity to infinity--
I'm just carving out that the position vector is always on
the unit circle on this case.
Right?
If I put a constant in front, it's on another circle.
So for whatever values of t I'm letting myself vary over,
all the vectors are going to lie on some part of a circle.
OK?
And so this is maybe the easy example.
Maybe you want to calculate, just to give yourself some
practice, what v of t is and what a of t actually is.
What these two quantities actually are, and then look at
what happens.
What happens with r and v, and see why r dot v is equal to 0.
What you should see--
I'll try and give you a picture of it geometrically.
What you should see is that, you know, if this is--
we'll see if I can effectively do a two-dimensional drawing
in three space-- if this is the r I'm looking at, that v
has to be coming in this way-- out this way.
So there's r and there's v. And this angle-- because I'm
trying to squash what was a circle--
I guess I'll look from above first. My picture--
from above-- is looking like something like this.
There's r and there's v, and that's a right angle.
Right?
So if I look from coming down on to the xy plane, here's a
position vector, here's its velocity, they're orthogonal.
And as I move all the way around the circle, that
position vector and the velocity are going to keep
that relationship.
When I look at this--
I'm trying to insert in the z-axis here.
When I look at this, if I look at what is the cross product
of these two vectors, well it's always going to point in
the z-direction.
It's going to point straight in the z-direction from here.
Because it's orthogonal to both of these, it's going to
point straight in the z-direction.
I know r is constant length.
I can then see v is constant length from the
example I have here.
And as I rotate, I'm going to get that
this is constant length.
OK?
So this is where the picture of what we were actually
describing vary much more generally in the
first part of this.
So the main point that I want us to see in this problem is
that when we want to find information about
relationships between r, v, and a-- this position and
velocity and acceleration--
what we can do is differentiate
these vector fields.
We saw an example when you were looking at I think
Kepler's second law you saw this in lecture.
But I just want to show you that you can use this if you
don't necessarily know explicitly things about a
position vector.
You can still find out things about its relationship if
you're given some information.
You don't actually have to have a formula to find out
some information about these relationships.
So I think that's where I'll stop.