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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, what I'd like
us to do is answer some

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questions I've posed here in
a really long problem.

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So let me take us through it.

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So we're going to consider
a position vector that's

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described by x of
t, y of t, 0.

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I want to have it in three space
so I can take a cross

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product later.

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And then we're going to suppose
that it actually has

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constant length.

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And that when I look at the
acceleration vector at t, it's

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actually equal to a constant
times r of t where the

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constant is not 0.

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OK?

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So I can, these are the two
things I know about this

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position vector.

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It has constant length for all
t, it has constant length.

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And the acceleration is always
equal to some constant times

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the position.

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OK, that's what I'm
giving you.

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And now I want you to use
those things and vector

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differentiation to show that r
dot v is equal to 0, where v

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is the velocity.

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And then to show that r
cross v is constant.

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So you're going to have to
figure out, essentially, the

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thing you have to figure out
is what relationship do you

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want to differentiate to
show these two things.

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OK?

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That's the hard part
of this problem.

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And then, I went to see
if you can give an

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example of such an r.

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So if you can give an example of
a position vector that has

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these properties.

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And maybe if you're having a
hard time, the first thing for

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you to do might be to think
about this, and to see if you

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can figure out an example of
that, and then see kind of how

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things work together
in that example.

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That may actually
proved helpful.

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So why don't you work on this
problem, pause the video, and

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then when you're feeling good
about your answer, you can

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bring the video back up and
I'll show you how I do it.

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OK, welcome back.

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So there was a lot to do in this
problem, but let me just

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remind you what the
framework is.

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We have a position vector, and
we know two things about it.

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We know that it has constant
length and we know that the

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acceleration is always
equal to some

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constant times the position.

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I didn't give you the constant,
but we know it's

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always equal to some constant
times the position.

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And then we wanted to show
two things using vector

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differentiation.

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We wanted to show that
r dot v was 0.

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And we wanted to show that
r cross v was constant.

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And then we want to talk about
what is an example.

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So let's start off and see if we
can figure out how to show

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that r dot v is equal to 0.

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And, you know, as you're
thinking about this problem,

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something that you want to
remember as you're thinking

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about this is, well, what are
the things that I know?

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I know that r dot r is constant,
so I'm going to

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write that down.

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r dot r--

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I'm not going to say is equal
to c, because that's a

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different constant--

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so I'll just, let me
just call it c1.

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OK?

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That's a different constant
than my c, maybe.

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OK, I know that r dot r is some
constant, and I want to

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show something about
r dot v. Right?

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So if I'm looking at this
and I say, well, I

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know this thing here.

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So it's the only dot product
relationship that I have.

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Because right now, I know a
relationship between a and r,

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and I know r has constant
length.

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Since that's all I gave you,
if I want to look at a dot

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product, this is the
relationship I know.

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The constant length thing.

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And so I know I somehow have to
use this one to figure out

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something

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about r dot v. OK.

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Well, what's the point that
we should realize?

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What is v?

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v is d dt of r.

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So I could take the derivative,
if I could take

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the derivative of just one of
these things, then I would get

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r dot v down here.

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Right?

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If I took d dt of just one of
these, I would get the r dot

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v. And d dt of this is 0.

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But I can't do that, right?

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Because if I take d dt of this
whole thing, I'm going to end

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up having to differentiate this
r once and leave this

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alone, and I'm going to have
to differentiate this r and

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leave this alone.

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But if you heard what I was
just saying, maybe you see

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that that's actually still
going to be OK.

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So let's look at what happens.

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I mean, this is all really I
have to work with, so I'm

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going to explore.

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Let's look at what happens when
I take d dt of r dot r.

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And I'm going to start leaving
off the hats here, because I'm

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going to leave them off
somewhere, so we'll just leave

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them off now, and then I
won't leave some off

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and put some on.

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So from here on out, I'm just
going to write r, v, and a

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without that hats, but
they're vectors.

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OK.

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So d dt of r dot r, well, I have
to take d dt of r, and

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then I dot that with r, and then
I take r dotted with--

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sorry--

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d dt of r.

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Right?

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Well, what do I get here? d
dt of r we said was v, so

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that's v dot r.

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And what do I get here?

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This is r dot--

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there's d dt of r again--
so I get r dot v.

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Now the great thing about the
dot product is that if I

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switch the order of these two,
it's still the same thing.

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So I can just write this as--

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well, I'll switch the
order of this one.

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So they both look like r dot v
plus r dot v, and that means I

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get 2 r dot v. Right?

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So d dt of r dot r is actually
2 of r dotted with v-- the

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position dotted with
the velocity.

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Now why is this going
to help me?

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Because what do I know about
this quantity r dot r?

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I know it's constant, right?

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So what is d dt of a constant?

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d dt of a constant is 0.

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So I actually started off
with knowing this was 0.

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So if I go through the whole
chain, I see 0 is equal to--

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let me put the 0 down
here again--

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0 is equal to 2 r dot
v, and so r dot v I

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know is equal to 0.

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What does that mean
geometrically?

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That means r and v
are orthogonal.

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And where is v going to sit?

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Well, if I come back over to how
I described r, r is in the

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xy plane, right?

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The z component is 0.

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So if I differentiate--

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if I take d dt of r--

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I'm going to have the derivative
of x to the

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function of t.

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And then whatever y is as a
function of t, I take that

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derivative.

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And this is still 0.

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So v is going to sit in the xy
plane, and based on what we

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know so far, we know it's
actually orthogonal to r.

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So they make a 90-degree
angle at all times t.

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OK, and how did we
do that again?

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I just want to remind you,
we knew one dot product

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relationship, that was r
dot r was a constant.

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So we differentiated that and
tried to see what happened.

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And the main point at the end of
it, was that when I had a v

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dot r, I could rewrite it as an
r dot v, and so I just end

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up with two of something that
I want to know about.

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So that's the main idea
of the first part.

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Now the second part was I
asked you to figure out

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something over here.

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We wanted to know that r
cross v is constant.

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OK?

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And t, it's always the same.

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All right?

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So let's think about if I want
to show that for every t

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something is constant,
I could show--

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actually, I've sort of
seen it already--

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I could show that its
derivative is 0.

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OK.

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So if r cross v is constant--
or If its derivative is 0, I

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should say--

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then r cross v was constant.

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Right?

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If its derivative is 0 for all
t, then r cross v is constant.

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Right?

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So that's really what we
want to exploit here.

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So let's look at the idea.

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So we want to-- again, let
me write it down--

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show r cross v is constant.

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And the strategy we're going
to use is to do this we're

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going to show that d dt of
r cross v is equal to 0.

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Right?

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If we can show that, then
this means that for

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all t it's the same.

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It's not changing in t.

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Right?

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So for all t, r cross v is going
to be the same, so r

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cross v is going
to be constant.

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So the difference between the
two problems was in the first

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problem you didn't quite know
maybe what expression you

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needed to differentiate
to find what you

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were looking for.

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Here, we know what we need to
differentiate, but we have to

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make sure we understand--

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to show this is constant--

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when we differentiate,
we should get 0.

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OK?

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So that's sort of a slightly
different type of problem.

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And I'm asking you maybe from
the other side here.

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So let's now, let's just
see what we get on

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the left-hand side.

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So what's d dt of r cross v?

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Well, d dt of r is v, right?

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So we get v cross v for
the first term.

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So take d dt of r, we get v.
We leave this v alone, and

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then we add to that, r cross--

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d dt of v, what's d dt of v--

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that's a.

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Right?

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Now, let's take a
look at this.

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Well, v cross v, v is
pointing in the same

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direction as itself.

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So when you try and take a cross
product of that, you

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know that the length of your
vector should be the area of

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the parallelogram formed
by these two vectors.

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But v is pointing in the same
direction as itself, so

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there's no area there.

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That's a geometric
interpretation of why this

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thing should be 0.

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Another reason is that remember
that your v cross v

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is going to include a sine theta
term, where theta is the

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angle between the two
vectors, right?

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That's another formula you have.
And so when you look at

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the angle between this vector
and itself, it's 0.

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And sine 0 is 0.

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So this is, in fact,
0 in that part.

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So if this is 0, then
we get what we want.

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Well, I only gave you one other
bit of information in

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the problem.

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And if you remember, it was that
a is always equal to a

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constant times r.

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So I can rewrite this right-hand
side as r cross a

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constant times r.

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And because of properties of
these cross products, I can

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pull out the constant.

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Or I can actually, I guess I
don't need to pull it out to

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talk about it, but it's nicer
if I pull it out.

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And look at what I have here.

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I have the exact same situation
as v cross v. I

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mean, this is still pointing
in the same direction.

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Constant times r and r still
point in the same direction as

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00:10:15,270 --> 00:10:16,790
if I were to compare r and r.

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00:10:16,790 --> 00:10:19,220
So I didn't have to pull out the
constant, but then right

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00:10:19,220 --> 00:10:22,370
here it's very easy to see
that this is also 0.

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00:10:22,370 --> 00:10:25,120
So I had 0 plus this,
so I get 0.

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00:10:25,120 --> 00:10:27,640
So, I've shown through
this process--

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00:10:27,640 --> 00:10:30,070
maybe I should have written
equal signs here--

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00:10:30,070 --> 00:10:34,720
that d dt of r cross v is
actually equal to 0.

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00:10:34,720 --> 00:10:38,870
And so you see that this cross
product between r and v--

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00:10:38,870 --> 00:10:41,640
which we know are orthogonal
sitting in the xy plane--

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00:10:41,640 --> 00:10:45,100
that it's always the same, it's
always the same vector.

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00:10:45,100 --> 00:10:48,160
And now I asked you to give an
example, and maybe you thought

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00:10:48,160 --> 00:10:50,605
of an example first, and then
thought of how it worked.

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00:10:50,605 --> 00:10:58,100

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00:10:58,100 --> 00:10:58,964
And so the easiest example is
if you let r of t equal

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00:10:58,964 --> 00:10:59,068
cosine t sine t.

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00:10:59,068 --> 00:11:00,318
So the easiest example--

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00:11:00,318 --> 00:11:01,930

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00:11:01,930 --> 00:11:04,400
there are others, obviously--

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00:11:04,400 --> 00:11:07,480
is if you let r of t equal
cosine t, sine t comma 0.

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00:11:07,480 --> 00:11:11,920
Sorry, I was thinking about
it in three space, right?

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00:11:11,920 --> 00:11:14,150
And in fact, if I were
to scale this,

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00:11:14,150 --> 00:11:15,100
it would still work.

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00:11:15,100 --> 00:11:18,060
I could put any constant
in front.

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00:11:18,060 --> 00:11:18,160
This carves out--

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00:11:18,160 --> 00:11:20,890
I let t go between 0
and 2 pi, or even

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minus infinity to infinity--

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00:11:22,080 --> 00:11:27,380
I'm just carving out that the
position vector is always on

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00:11:27,380 --> 00:11:28,950
the unit circle on this case.

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00:11:28,950 --> 00:11:29,590
Right?

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00:11:29,590 --> 00:11:32,220
If I put a constant in front,
it's on another circle.

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00:11:32,220 --> 00:11:36,580
So for whatever values of t I'm
letting myself vary over,

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00:11:36,580 --> 00:11:41,160
all the vectors are going to lie
on some part of a circle.

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00:11:41,160 --> 00:11:41,790
OK?

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00:11:41,790 --> 00:11:44,600
And so this is maybe
the easy example.

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00:11:44,600 --> 00:11:46,660
Maybe you want to calculate,
just to give yourself some

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00:11:46,660 --> 00:11:50,870
practice, what v of t is and
what a of t actually is.

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00:11:50,870 --> 00:11:53,800
What these two quantities
actually are, and then look at

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00:11:53,800 --> 00:11:54,380
what happens.

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00:11:54,380 --> 00:12:00,660
What happens with r and v, and
see why r dot v is equal to 0.

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00:12:00,660 --> 00:12:00,772
What you should see--

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00:12:00,772 --> 00:12:03,600
I'll try and give you a picture
of it geometrically.

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00:12:03,600 --> 00:12:09,310
What you should see is that,
you know, if this is--

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00:12:09,310 --> 00:12:12,450
we'll see if I can effectively
do a two-dimensional drawing

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00:12:12,450 --> 00:12:16,310
in three space-- if this is the
r I'm looking at, that v

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00:12:16,310 --> 00:12:19,370
has to be coming in this
way-- out this way.

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00:12:19,370 --> 00:12:21,980
So there's r and there's v. And
this angle-- because I'm

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00:12:21,980 --> 00:12:23,770
trying to squash what
was a circle--

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00:12:23,770 --> 00:12:27,280
I guess I'll look from above
first. My picture--

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00:12:27,280 --> 00:12:29,900
from above-- is looking like
something like this.

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00:12:29,900 --> 00:12:33,350
There's r and there's v, and
that's a right angle.

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00:12:33,350 --> 00:12:33,640
Right?

301
00:12:33,640 --> 00:12:37,880
So if I look from coming down
on to the xy plane, here's a

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00:12:37,880 --> 00:12:41,590
position vector, here's its
velocity, they're orthogonal.

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00:12:41,590 --> 00:12:43,660
And as I move all the way
around the circle, that

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00:12:43,660 --> 00:12:45,480
position vector and the velocity
are going to keep

305
00:12:45,480 --> 00:12:47,750
that relationship.

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00:12:47,750 --> 00:12:47,865
When I look at this--

307
00:12:47,865 --> 00:12:50,810
I'm trying to insert
in the z-axis here.

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00:12:50,810 --> 00:12:53,690
When I look at this, if I look
at what is the cross product

309
00:12:53,690 --> 00:12:56,000
of these two vectors, well it's
always going to point in

310
00:12:56,000 --> 00:12:56,790
the z-direction.

311
00:12:56,790 --> 00:12:59,130
It's going to point straight in
the z-direction from here.

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00:12:59,130 --> 00:13:01,740
Because it's orthogonal to both
of these, it's going to

313
00:13:01,740 --> 00:13:04,740
point straight in
the z-direction.

314
00:13:04,740 --> 00:13:06,910
I know r is constant length.

315
00:13:06,910 --> 00:13:08,850
I can then see v is constant
length from the

316
00:13:08,850 --> 00:13:09,750
example I have here.

317
00:13:09,750 --> 00:13:11,290
And as I rotate, I'm
going to get that

318
00:13:11,290 --> 00:13:13,030
this is constant length.

319
00:13:13,030 --> 00:13:13,350
OK?

320
00:13:13,350 --> 00:13:19,110
So this is where the picture
of what we were actually

321
00:13:19,110 --> 00:13:21,320
describing vary much more
generally in the

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00:13:21,320 --> 00:13:22,930
first part of this.

323
00:13:22,930 --> 00:13:26,200
So the main point that I want
us to see in this problem is

324
00:13:26,200 --> 00:13:28,980
that when we want to find
information about

325
00:13:28,980 --> 00:13:32,840
relationships between r, v,
and a-- this position and

326
00:13:32,840 --> 00:13:34,710
velocity and acceleration--

327
00:13:34,710 --> 00:13:36,660
what we can do is differentiate

328
00:13:36,660 --> 00:13:37,640
these vector fields.

329
00:13:37,640 --> 00:13:40,500
We saw an example when you
were looking at I think

330
00:13:40,500 --> 00:13:44,070
Kepler's second law you
saw this in lecture.

331
00:13:44,070 --> 00:13:47,380
But I just want to show you that
you can use this if you

332
00:13:47,380 --> 00:13:51,530
don't necessarily know
explicitly things about a

333
00:13:51,530 --> 00:13:52,330
position vector.

334
00:13:52,330 --> 00:13:56,340
You can still find out things
about its relationship if

335
00:13:56,340 --> 00:13:57,740
you're given some information.

336
00:13:57,740 --> 00:14:00,430
You don't actually have to have
a formula to find out

337
00:14:00,430 --> 00:14:02,910
some information about
these relationships.

338
00:14:02,910 --> 00:14:05,000
So I think that's
where I'll stop.

339
00:14:05,000 --> 00:14:05,646