## TTU Math2450 Calculus3 Sec 13.3

• 0:00 - 0:02
• 0:02 - 0:05
MAGDALENA TODA: Welcome
to our review of 13.1.
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How many of you didn't
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I have your exam, and yours.
• 0:10 - 0:11
And you have to wait.
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I don't have it with me.
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I have it in my office.
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If you have questions
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why don't you go ahead and
email me right after class.
• 0:23 - 0:31
Chapter 13 is a very
physical chapter.
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It has a lot to do with
mechanical engineering,
• 0:34 - 0:37
with mechanics,
physics, electricity.
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You're going to see things,
weird things like work.
• 0:45 - 0:47
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Do you remember the definition?
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So we define the work
as a path integral
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along the regular curve.
• 0:55 - 0:59
And by regular curve-- I'm
sorry if I'm repeating myself,
• 0:59 - 1:02
but this is part of the
deal-- R is the position
• 1:02 - 1:10
vector in R3 that is class C1.
• 1:10 - 1:15
That means differentiable and
derivatives are continuous.
• 1:15 - 1:19
Plus you are not
allowed to stop.
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So no matter how drunk,
the bug has to keep flying,
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and not even for a
fraction of a second is he
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or she allowed to
have velocity 0.
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At no point I want
to have velocity 0.
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And that's the position vector.
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And then you have some force
field acting on you-- no,
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acting on the particle
at every moment.
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So you have an F that is
acting at location xy.
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Maybe if you are in space,
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where x is a function of
t, y is a functional of t,
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z is a function of t,
which is the same as saying
• 2:07 - 2:12
that R of t, which is the given
position vector, is x of t
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y of t.
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Let me put angular bracket,
although I hate them,
• 2:16 - 2:20
because you like angular
brackets for vectors.
• 2:20 - 2:23
F is also a nice function.
• 2:23 - 2:24
How nice?
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We discussed a
little bit last time.
• 2:27 - 2:29
It really doesn't
have to be continuous.
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The book assumes it continues.
• 2:31 - 2:33
It has to be
integrable, so maybe it
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could be piecewise continuous.
• 2:36 - 2:43
So I had nice enough, was
it continues piecewise.
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• 2:46 - 2:51
And we define the work as
being the path integral over c.
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I keep repeating, because
that's going to be on the final
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as well.
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So all the notions
that are important
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should be given enough
attention in this class.
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Hi.
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So do you guys remember
how we denoted F?
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F was, in general, three
components in our F1, F2, F3.
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They are functions of
the position vector,
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or the position xyz.
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And the position is
a function of time.
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So all in all, after
you do all the work,
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keep in mind that when you
multiply with a dot product,
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the integral will give you what?
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A time integral?
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From a time T0 to a time
T1, you are here at time T0
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and you are here at time T1.
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piecewise, differentiable,
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you don't know what it is.
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But let's assume just a
very nice, smooth arc here.
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Of what?
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Of F1 times what is that?
x prime of t plus F2 times
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y prime of t, plus F3
times z prime of t dt.
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So keep in mind that
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And he was-- what was he?
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Was defined as the
velocity vector multiplied
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by the infinitesimal element dt.
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Say again, the
velocity vector prime
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was a vector in F3 quantified
by the infinitesimal element dt.
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So we reduce this Calc
3 notion path integral
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to a Calc 1 notion, which was a
simple integral from t0 to t1.
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And we've done a
lot of applications.
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What else have we done?
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We've done some
integral of this type
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over another curve, script c.
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I'm repeating mostly for Alex.
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You're caught in the process.
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And there are two or three
people who need an update.
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Maybe I have another
function of g and ds.
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And this is an integral that
in the end will depend on s.
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But s itself depends on t.
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So if I were to re-express
this in terms of d,
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how would I re-express
the whole thing?
• 5:35 - 5:41
g of s, of t, whatever that
is, then Mr. ds was what?
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STUDENT: s prime of t.
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MAGDALENA TODA: Right.
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So this was the-- that s
prime of t was the speed.
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The speed of the arc of a curve.
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So you have an R of
t and R3, a vector.
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And the speed was,
by definition,
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arc length element was
by definition integral
• 6:07 - 6:09
from 2t0 to t.
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Of the speed R prime
magnitude d tau.
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I'll have you put tau
because I'm Greek,
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and it's all Greek to me.
• 6:19 - 6:23
So the tau, some people call
the tau the dummy variable.
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I don't like to call it dumb.
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It's a very smart variable.
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It goes from t0 to t, so what
you have is a function of t.
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This guy is speed.
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So when you do that
here, ds becomes speed,
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R prime of t times dt.
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This was your old friend ds.
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And let me put it on top
of this guy with speed.
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Because he was so
important to you,
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• 6:56 - 7:04
So that was review of--
reviewing of 13.1 and 13.2
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There were some things in
13.3 that I pointed out
• 7:10 - 7:12
to you are important.
• 7:12 - 7:17
13.3 was independence of path.
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Everybody write, magic-- no.
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Magic section.
• 7:21 - 7:22
No, have to be serious.
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So that's independence of path
of certain type of integrals,
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of some integrals.
• 7:35 - 7:38
And an integral like
that, a path integral
• 7:38 - 7:41
is independent of path.
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When would such an animal--
look at this pink animal,
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inside-- when would
this not depend
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on the path you are taking
between two given points?
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So I can move on another
arc and another arc
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and another regular arc, and
all sorts of regular arcs.
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It doesn't matter
which path I'm taking--
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STUDENT: If that
force is conservative.
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MAGDALENA TODA: If the
force is conservative.
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Excellent, Alex.
• 8:07 - 8:13
And what did it mean for a
force to be conservative?
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How many of you
know-- it's no shame.
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Just raise hands.
• 8:17 - 8:20
If you forgot what it is,
• 8:20 - 8:23
But if you remember what it
means for a force F force
• 8:23 - 8:27
field-- may the force be
with you-- be conservative,
• 8:27 - 8:30
then what do you do?
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Say F is conservative
by definition.
• 8:33 - 8:39
• 8:39 - 8:50
When, if and only, F
there is a so-called--
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STUDENT: Scalar.
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MAGDALENA TODA: --potential.
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Scalar potential, thank you.
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I'll fix that.
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A scalar potential function f.
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didn't want to put this.
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Because a few
people told me they
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the symbolistics.
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This means "there exists."
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OK, smooth potential
such that-- at least
• 9:22 - 9:27
is differential [INAUDIBLE]
1 such that the nabla of f--
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what the heck is that?
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f will be the given F.
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And we saw all sorts of wizards
here, like, Harry Potter,
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[INAUDIBLE] well,
there are many,
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Alex, Erin, many,
many-- Matthew.
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So what did they do?
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They guessed the
scalar potential.
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there are 10 of them.
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It's a whole school
of Harry Potter.
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How do they find the little f?
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Through witchcraft.
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No.
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Normally you should--
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STUDENT: I've actually
done it through witchcraft.
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Tell you that?
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MAGDALENA TODA: You did.
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I think you can do it
through witchcraft.
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But practically everybody
has the ability to guess.
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Why do we have the ability
to guess and check?
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Because our brain does
the integration for you.
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Whether you tell your
brain to stop or not,
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sees is kind of function--
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and now I'm gonna
• 10:30 - 10:32
on a little harder one.
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I didn't want to do an
R2 value vector function.
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Let me go to R3.
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But I know that you have
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So let's say somebody
gave you a force field
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that is yz i plus xzj plus xyk.
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And you're going to jump and
say this is a piece of cake.
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I can see the scalar potential
and just wave my magic wand,
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and I get it.
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STUDENT: [INAUDIBLE]
• 11:08 - 11:09
MAGDALENA TODA: Oh my god, yes.
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Guys, you saw it fast.
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OK, I should be proud of you.
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And I am proud of you.
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where the students couldn't
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see any of the scalar
potentials that I gave them,
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that I asked them to guess.
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How did you deal with it?
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You integrate this
with respect to F?
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In the back of
• 11:30 - 11:32
And then you guessed
one, and then you
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said, OK so should be xyz.
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Does it verify my
other two conditions?
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And you say, oh yeah, it does.
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Because of I prime with respect
to y, I have exactly xz.
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If I prime with respect to c I
have exactly xy, so I got it.
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And even if somebody
said xyz plus 7,
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they would still be right.
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In the end you can have
any xyz plus a constant.
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In general it's not
so easy to guess.
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But there are lots of examples
of conservative forces where
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you simply cannot see the scalar
potential or cannot deduce it
• 12:07 - 12:10
like in a few seconds.
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Expect something easy,
though, like that,
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something that you can see.
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Let's see an example.
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Assume this is your force field
acting on a particle that's
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moving on a curving space.
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And it's stubborn and it
decides to move on a helix,
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because it's a-- I don't
know what kind of particle
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would move on a
helix, but suppose
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a lot of particles, just a
little train or a drunken bug
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or something.
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And you were moving
on another helix.
• 12:45 - 12:52
Now suppose that helix will
be R of t equals cosine t
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sine t and t where you
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What do I have at 0?
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The point 1, 0, 0.
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That's the point,
let's call it A.
• 13:09 - 13:13
And let's call this B. I
don't know what I want to do.
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I'll just do a
complete rotation,
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just to make my life easier.
• 13:19 - 13:23
And this is B. And that
will be A at t equals 0
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and B equals 2 pi.
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• 13:33 - 13:35
So what will this be at B?
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STUDENT: 1, 0, 2 pi.
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MAGDALENA TODA: 1, 0, and 2 pi.
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So you perform a complete
rotation and come back.
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conservative, you are lucky.
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Because you know the theorem
that says in that case
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the work integral will
be independent of path.
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And due to the theorem in-- what
section was that again-- 13.3,
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independence of path,
you know that this
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is going to be-- let me rewrite
it one more time with gradient
• 14:12 - 14:16
of f instead of big F.
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And this will become what,
f of the q-- not the q.
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In the book it's f of q minus
f of q. f of B minus f of A,
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right?
• 14:25 - 14:28
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What does this mean?
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You have to measure
the-- to evaluate
• 14:33 - 14:39
the coordinates of
this function xyz
• 14:39 - 14:50
where t equals 2 pi minus
xyz where t equals what?
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0.
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And now I have to be
careful, because I
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have to evaluate them.
• 14:58 - 15:07
So when t is 0 I have x
is 1, y is 0, and t is 0.
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In the end it doesn't matter.
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I can get 0-- I
can get 0 for this
• 15:13 - 15:15
and get 0 for that as well.
• 15:15 - 15:21
So when this is 2 pi I get
x equals 1, y equals 0,
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and t equals 2 pi.
• 15:23 - 15:27
So in the end, both products
are 0 and I got a 0.
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So although the [INAUDIBLE]
works very hard-- I mean,
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works hard in our perception
to get from a point
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to another-- the work is 0.
• 15:39 - 15:39
Why?
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Because it's a vector
value thing inside.
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And there are some
annihilations going on.
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So that reminds me
of another example.
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So we are done
with this example.
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Let's go back to our washer.
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I was just doing
laundry last night
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and I was thinking of
the washer example.
• 16:01 - 16:05
And I thought of a small
variation of the washer
• 16:05 - 16:09
example, just assuming that
I would give you a pop quiz.
• 16:09 - 16:12
And I'm not giving you
a pop quiz right now.
• 16:12 - 16:15
But if I gave you
a pop quiz now,
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two, the washer.
• 16:21 - 16:24
• 16:24 - 16:28
It is performing
a circular motion,
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and I want to know
the work performed
• 16:30 - 16:36
by the centrifugal force
between various points.
• 16:36 - 16:48
So have the circular motion,
the centrifugal force.
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This is the
centrifugal, I'm sorry.
• 16:51 - 16:54
I'll take the centrifugal force.
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And that was last
time we discussed
• 16:56 - 17:05
that, that was extending
• 17:05 - 17:07
the vector value position.
• 17:07 - 17:12
So you have that in
every point, xi plus yj.
• 17:12 - 17:15
And you want F to
be able xi plus yj.
• 17:15 - 17:20
But it points outside
from the point
• 17:20 - 17:23
on the circular trajectory.
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out what you performed
• 17:31 - 17:39
by F in one full rotation.
• 17:39 - 17:44
• 17:44 - 17:49
We gave the equation of motion,
being cosine t y sine t,
• 17:49 - 17:52
if you remember from last time.
• 17:52 - 17:57
And then W2, let's
say, is performed by F
• 17:57 - 18:02
from t equals 0 to t equals pi.
• 18:02 - 18:03
I want that as well.
• 18:03 - 18:13
And W2 performed by F from
t-- that makes t0 to t
• 18:13 - 18:20
equals pi-- t equals 0
to t equals pi over 4.
• 18:20 - 18:22
These are all very
easy questions,
• 18:22 - 18:25
and you should be able to
• 18:25 - 18:28
Now, let me tell you something.
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We are in plane, not in space.
• 18:30 - 18:31
But it doesn't matter.
• 18:31 - 18:35
would be 0, piece of cake.
• 18:35 - 18:40
well-trained that when
• 18:40 - 18:42
you look at the force
field like that,
• 18:42 - 18:44
• 18:44 - 18:45
is it conservative?
• 18:45 - 18:48
• 18:48 - 18:51
And it is conservative.
• 18:51 - 18:54
And that means little f is what?
• 18:54 - 18:57
• 18:57 - 19:00
Nitish said that yesterday.
• 19:00 - 19:01
Why did you go there?
• 19:01 - 19:03
You want to sleep today?
• 19:03 - 19:06
I'm just teasing you.
• 19:06 - 19:08
I got so comfortable with
you sitting in the front row.
• 19:08 - 19:10
STUDENT: I took his spot.
• 19:10 - 19:12
STUDENT: She doesn't like
you sitting over here.
• 19:12 - 19:13
MAGDALENA TODA: It's OK.
• 19:13 - 19:14
It's fine.
• 19:14 - 19:17
I still give him credit
for what he said last time.
• 19:17 - 19:20
So do you guys remember,
• 19:20 - 19:23
x squared plus y squared over
2, and he found the scalar
• 19:23 - 19:27
potential through witchcraft
in about a second and a half?
• 19:27 - 19:28
OK.
• 19:28 - 19:31
We are gonna conclude something.
• 19:31 - 19:36
Do you remember that I found the
• 19:36 - 19:39
I got W to be 0.
• 19:39 - 19:45
But if I were to find another
explanation why the work would
• 19:45 - 19:50
be 0 in this case, it
would have been 0 anyway
• 19:50 - 19:53
for any force field.
• 19:53 - 19:58
Even if I took the F
to be something else.
• 19:58 - 20:03
Assume that F would be
G. Really wild, crazy,
• 20:03 - 20:07
but still differentiable
vector value function.
• 20:07 - 20:09
G differential.
• 20:09 - 20:15
Would the work that we
want be the same for G?
• 20:15 - 20:16
STUDENT: Yeah.
• 20:16 - 20:17
MAGDALENA TODA: Why?
• 20:17 - 20:18
STUDENT: Because of
displacement scenario.
• 20:18 - 20:21
MAGDALENA TODA: Since
it's conservative,
• 20:21 - 20:23
you have a closed loop.
• 20:23 - 20:26
So the closed loop
will say, thick F
• 20:26 - 20:30
at that terminal point minus
thick F at the initial point.
• 20:30 - 20:34
But if a loop motion,
• 20:34 - 20:35
is the initial point.
• 20:35 - 20:36
Duh.
• 20:36 - 20:41
So you have the
same point, the P
• 20:41 - 20:44
equals qe if it's
a closed curve.
• 20:44 - 20:48
So for a closed curve--
we also call that a loop.
• 20:48 - 20:50
would have been too easy
• 20:50 - 20:55
and you would have gotten a
• 20:55 - 20:58
So any closed curve
is called a loop.
• 20:58 - 21:02
conservative-- attention,
• 21:02 - 21:05
you might have examples
like that in the exams--
• 21:05 - 21:09
then it doesn't matter
who little f is,
• 21:09 - 21:13
if p equals q you get 0 anyway.
• 21:13 - 21:16
But the reason why I
said you would get 0
• 21:16 - 21:21
on the example of last time
was a slightly different one.
• 21:21 - 21:24
What does the engineer
say to himself?
• 21:24 - 21:26
STUDENT: Force is perpendicular.
• 21:26 - 21:26
MAGDALENA TODA: Yeah.
• 21:26 - 21:27
Very good.
• 21:27 - 21:29
Whenever the force
is perpendicular
• 21:29 - 21:33
to the trajectory, I'm going
to get 0 for the force.
• 21:33 - 21:37
Because at every
moment the dot product
• 21:37 - 21:41
between the force and the
displacement direction,
• 21:41 - 21:45
which would be like dR, the
tangent to the displacement,
• 21:45 - 21:47
would be [INAUDIBLE].
• 21:47 - 21:50
And cosine of [INAUDIBLE] is 0.
• 21:50 - 21:51
Duh.
• 21:51 - 21:53
So that's another reason.
• 21:53 - 22:00
Reason of last time
was F perpendicular
• 22:00 - 22:05
to the R prime
direction, R prime
• 22:05 - 22:11
being the velocity-- look,
when I'm moving in a circle,
• 22:11 - 22:13
this is the force.
• 22:13 - 22:15
And I'm moving.
• 22:15 - 22:18
This is my velocity, is
the tangent to the circle.
• 22:18 - 22:22
And the velocity and the normal
are always perpendicular,
• 22:22 - 22:23
at every point.
• 22:23 - 22:24
That's why I have 0.
• 22:24 - 22:27
• 22:27 - 22:32
So note that even if I
didn't take a close look,
• 22:32 - 22:36
be from 0 to pi?
• 22:36 - 22:39
Still?
• 22:39 - 22:43
0 because of that.
• 22:43 - 22:44
0.
• 22:44 - 22:47
How about from 0 to pi over 4?
• 22:47 - 22:50
Still 0.
• 22:50 - 22:52
And of course if somebody
would not believe them,
• 22:52 - 22:55
if somebody would not
understand the theory,
• 22:55 - 22:58
they would do the work and
they would get to the answer
• 22:58 - 23:01
and say, oh my
god, yeah, I got 0.
• 23:01 - 23:03
All right?
• 23:03 - 23:04
OK.
• 23:04 - 23:10
Now, what if somebody--
and I want to spray this.
• 23:10 - 23:12
erase the board
• 23:12 - 23:15
and move onto example
three or whatever?
• 23:15 - 23:16
Yes?
• 23:16 - 23:17
OK.
• 23:17 - 23:19
All right.
• 23:19 - 23:22
STUDENT: Could you say
non-conservative force?
• 23:22 - 23:24
MAGDALENA TODA: Yeah,
that's what I-- exactly.
• 23:24 - 23:26
• 23:26 - 23:32
You are gonna guess my mind.
• 23:32 - 23:44
• 23:44 - 23:47
And I'm going to
pick a nasty one.
• 23:47 - 23:50
And since I'm doing
review anyway,
• 23:50 - 23:51
you may have one like that.
• 23:51 - 23:55
And you may have both one that
involves a conservative force
• 23:55 - 24:00
field and one that does not
involve a conservative force
• 24:00 - 24:01
field.
• 24:01 - 24:07
And we can ask you, find us the
work belong to different path.
• 24:07 - 24:12
And I've done this
type of example before.
• 24:12 - 24:15
Let's take F of
x and y in plane.
• 24:15 - 24:29
In our two I take xyi
plus x squared y of j.
• 24:29 - 24:35
And the problem would
involve my favorite picture,
• 24:35 - 24:39
y equals x squared and y
equals x, our two paths.
• 24:39 - 24:41
One is the straight path.
• 24:41 - 24:44
One is the [INAUDIBLE] path.
• 24:44 - 24:46
They go from 0,
0 to 1, 1 anyway.
• 24:46 - 24:49
• 24:49 - 24:58
find W1 along path one
• 24:58 - 25:01
and W2 along path two.
• 25:01 - 25:06
And of course,
example three, if this
• 25:06 - 25:10
were conservative
you would say, oh,
• 25:10 - 25:12
it doesn't matter
what path I'm taking,
• 25:12 - 25:15
I'm still getting
• 25:15 - 25:17
But is this conservative?
• 25:17 - 25:18
STUDENT: No.
• 25:18 - 25:20
Because you said it wasn't.
• 25:20 - 25:21
MAGDALENA TODA: Very good.
• 25:21 - 25:23
So how do you know?
• 25:23 - 25:26
That's one test
when you are in two.
• 25:26 - 25:31
There is the magic test that
says-- let's say this is M,
• 25:31 - 25:37
and let's say this is N. You
would have to check if M sub--
• 25:37 - 25:37
STUDENT: y.
• 25:37 - 25:38
MAGDALENA TODA: y.
• 25:38 - 25:39
Very good.
• 25:39 - 25:40
I'm proud of you.
• 25:40 - 25:42
3350, by the way.
• 25:42 - 25:44
Is equal to N sub x.
• 25:44 - 25:47
M sub y is x.
• 25:47 - 25:49
N sub x is 2xy.
• 25:49 - 25:52
They are not equal.
• 25:52 - 25:55
So that's me crying that I have
to do the work twice and get--
• 25:55 - 25:58
probably I'll get two
different examples.
• 25:58 - 26:01
• 26:01 - 26:04
• 26:04 - 26:09
how many of you opened the
book at section 13.2, 13.3.
• 26:09 - 26:12
it, any of them?
• 26:12 - 26:13
• 26:13 - 26:15
MAGDALENA TODA: Oh, good.
• 26:15 - 26:20
There is another criteria for
a force to be conservative.
• 26:20 - 26:22
If you are, it's piece of cake.
• 26:22 - 26:23
You do that, right?
• 26:23 - 26:24
MAGDALENA TODA: Yes, sir?
• 26:24 - 26:26
STUDENT: Curl has frequency 0.
• 26:26 - 26:27
MAGDALENA TODA: The curl
criteria, excellent.
• 26:27 - 26:29
The curl has to be zero.
• 26:29 - 26:39
So if F in R 3 is
conservative, then you'll
• 26:39 - 26:40
get different order curve.
• 26:40 - 26:42
Curl F is 0.
• 26:42 - 26:45
Now let's check what
the heck was curl.
• 26:45 - 26:48
You see, mathematics
is not a bunch
• 26:48 - 26:52
of these joint discussions
like other sciences.
• 26:52 - 26:55
In mathematics, if you don't
know a section or you skipped
• 26:55 - 26:58
it, you are sick, you
have a date that day,
• 26:58 - 27:03
you didn't study, then it's
all over because you cannot
• 27:03 - 27:07
understand how to work out the
problems and materials if you
• 27:07 - 27:08
skip the section.
• 27:08 - 27:12
Curl was the one
where we learned
• 27:12 - 27:16
that we used the determinant.
• 27:16 - 27:17
That's the easiest story.
• 27:17 - 27:21
It came with a t-shirt,
but that t-shirt really
• 27:21 - 27:26
doesn't help because
it's easier to,
• 27:26 - 27:28
the formula,
• 27:28 - 27:31
you set out the determinant.
• 27:31 - 27:34
So you have the operator
derivative with respect
• 27:34 - 27:41
to x, y z followed by what?
• 27:41 - 27:43
F1, F2, F3.
• 27:43 - 27:47
asking you if you did it
• 27:47 - 27:52
for this F, what is
the third component?
• 27:52 - 27:53
STUDENT: The 0.
• 27:53 - 27:55
MAGDALENA TODA: The
0, so this guy is 0.
• 27:55 - 28:00
This guy is X squared
Y, and this guy is xy.
• 28:00 - 28:02
And it should be
a piece of cake,
• 28:02 - 28:04
but I want to do
it one more time.
• 28:04 - 28:09
I times the minor derivative
of 0 with respect to y
• 28:09 - 28:12
is 0 minus derivative
of x squared
• 28:12 - 28:15
y respect to 0, all
right, plus j minus
• 28:15 - 28:18
j because I'm alternating.
• 28:18 - 28:20
You've known enough
• 28:20 - 28:23
to know why I'm expanding
along the first row.
• 28:23 - 28:26
I have a minus, all
right, then the x
• 28:26 - 28:34
of 0, 0 derivative of xy
respect to the 0 plus k times
• 28:34 - 28:38
the minor corresponding
to k derivative 2xy.
• 28:38 - 28:45
• 28:45 - 28:46
Oh, and the derivative--
• 28:46 - 28:49
• 28:49 - 28:53
STUDENT: Yeah, this
is the n equals 0.
• 28:53 - 28:54
MAGDALENA TODA: Oh,
yeah, that's the one
• 28:54 - 28:59
where it's not a because
that's not conservative.
• 28:59 - 29:00
So what do you get.
• 29:00 - 29:05
You get 2xy minus x, right?
• 29:05 - 29:07
But I don't know how to
write it better than that.
• 29:07 - 29:08
Well, it doesn't matter.
• 29:08 - 29:10
Leave it like that.
• 29:10 - 29:18
So this would be 0 if it only
if x would be 0, but otherwise y
• 29:18 - 29:19
was 1/2.
• 29:19 - 29:23
But in general, it
is not a 0, good.
• 29:23 - 29:30
So F is not
conservative, and then we
• 29:30 - 29:32
can say goodbye
to the whole thing
• 29:32 - 29:40
here and move on to
computing the works.
• 29:40 - 29:42
What is the only
way we can do that?
• 29:42 - 29:46
By parameterizing
the first path,
• 29:46 - 29:49
but I didn't say which
one is the first path.
• 29:49 - 29:53
This is the first path, so
x of t equals t, and y of t
• 29:53 - 29:55
equals t is your
parameterization.
• 29:55 - 30:04
The simplest one, and then
W1 will be integral of-- I'm
• 30:04 - 30:10
too lazy to write down x of t,
y of t, but this is what it is.
• 30:10 - 30:15
Times x prime of
t plus x squared
• 30:15 - 30:22
y times y prime of t dt where--
• 30:22 - 30:31
STUDENT: Isn't that just
xy dx y-- never mind.
• 30:31 - 30:34
MAGDALENA TODA: This is F2.
• 30:34 - 30:37
And this is x prime,
and this is y prime
• 30:37 - 30:40
because this thing is
just-- I have no idea.
• 30:40 - 30:42
STUDENT: Right,
• 30:42 - 30:47
is that not the same as just
F1 dx because we're going
• 30:47 - 30:50
to do a chain rule anyway.
• 30:50 - 30:54
MAGDALENA TODA: If I put
the x, I cannot put this.
• 30:54 - 30:57
OK, this times that is dx.
• 30:57 - 31:00
This guy times this guy is dx.
• 31:00 - 31:01
STUDENT: But then
you can't use your
• 31:01 - 31:04
MAGDALENA TODA: Then I
cannot use the t's then.
• 31:04 - 31:06
STUDENT: All right, there we go.
• 31:06 - 31:11
MAGDALENA TODA: All right, so
I have integral from 0 to 1 t,
• 31:11 - 31:15
t times 1 t squared.
• 31:15 - 31:18
If I make a mistake, that would
be a silly algebra mistake
• 31:18 - 31:19
[INAUDIBLE].
• 31:19 - 31:22
• 31:22 - 31:24
All right, class.
• 31:24 - 31:34
t cubed times 1dt,
how much is this?
• 31:34 - 31:38
t cubed over 3 plus t
to the fourth over 4.
• 31:38 - 31:39
STUDENT: It's just 2-- oh, no.
• 31:39 - 31:45
• 31:45 - 31:48
MAGDALENA TODA: Very good.
• 31:48 - 31:52
Do not expect that we kill you
with computations on the exams,
• 31:52 - 31:55
but that's not what we want.
• 31:55 - 31:58
We want to test if you have
the basic understanding of what
• 31:58 - 32:03
this is all about, not to
kill you with, OK, that.
• 32:03 - 32:05
I'm not going to say that
in front of the cameras,
• 32:05 - 32:07
but everybody knows that.
• 32:07 - 32:08
There are professors
who would like
• 32:08 - 32:10
to kill you with computations.
• 32:10 - 32:12
Now, we're living in
a different world.
• 32:12 - 32:16
If I gave you a long
polynomial sausage here
• 32:16 - 32:18
work with it, that
• 32:18 - 32:21
doesn't mean that I'm smart
because MATLAB can do it.
• 32:21 - 32:25
Mathematica, you get some
very nice simplifications
• 32:25 - 32:28
over there, so I'm
just trying to see
• 32:28 - 32:36
if rather than being able
to compute with no error,
• 32:36 - 32:41
you are having the basic
understanding of the concept.
• 32:41 - 32:45
And the rest can been done
by the mathematical software,
• 32:45 - 32:50
mathematicians are using.
• 32:50 - 32:53
years ago, I think
• 32:53 - 32:58
I knew colleagues at all the
ranks in academia who would not
• 32:58 - 33:02
touch Mathematica or
MATLAB or Maple say
• 33:02 - 33:05
that's like tool from
evil or something,
• 33:05 - 33:08
but now everybody uses.
• 33:08 - 33:11
Engineers use mostly
MATLAB as I told you.
• 33:11 - 33:16
Mathematicians use both
MATLAB and Mathematica.
• 33:16 - 33:19
Some of them use Maple,
especially the ones who
• 33:19 - 33:23
have demos for K-12
level teachers,
• 33:23 - 33:26
but MATLAB is a wonderful
tool, very pretty powerful
• 33:26 - 33:28
in many ways.
• 33:28 - 33:31
If you are doing any kind
of linear algebra project--
• 33:31 - 33:34
I noticed three or four of you
are taking linear algebra-- you
• 33:34 - 33:39
can always rely on MATLAB being
the best of all of the above.
• 33:39 - 33:40
OK, W2.
• 33:40 - 33:43
• 33:43 - 33:50
For W2, I have a parabola, and
it's, again, a piece of cake.
• 33:50 - 33:55
X prime will be 1,
y prime will be 2t.
• 33:55 - 33:56
When I write down
the whole thing,
• 33:56 - 33:59
I have to pay a little
bit of attention
• 33:59 - 34:02
when I substitute
especially when I'm
• 34:02 - 34:05
taking an exam under pressure.
• 34:05 - 34:09
• 34:09 - 34:14
x squared is t
squared, y is t squared
• 34:14 - 34:17
times y prime, which is 2t.
• 34:17 - 34:20
So now this is x prime.
• 34:20 - 34:21
This is y prime.
• 34:21 - 34:25
Let me change colors.
• 34:25 - 34:27
All politicians change colors.
• 34:27 - 34:29
But I'm not a
politician, but I'm
• 34:29 - 34:34
thinking it's useful for you
to see who everybody was.
• 34:34 - 34:39
This is the F1 in terms of t.
• 34:39 - 34:46
That's the idea of what that
is, and this is F2 in terms of t
• 34:46 - 34:48
as well.
• 34:48 - 34:50
• 34:50 - 34:54
Absolutely, I'm going to
• 34:54 - 34:57
Is it obviously to everybody
I'm going to get another answer?
• 34:57 - 34:58
STUDENT: Yeah.
• 34:58 - 35:01
MAGDALENA TODA: So I don't
have to put the t's here,
• 35:01 - 35:04
but I thought it was
sort of neat to see
• 35:04 - 35:06
that t goes from 0 to 1.
• 35:06 - 35:09
And what do I get?
• 35:09 - 35:16
This whole lot of them is t
cubed plus 2 t to the fifth.
• 35:16 - 35:19
• 35:19 - 35:26
So when I do the integration,
I get t to the 4 over 4 plus--
• 35:26 - 35:27
shut up, Magdalena, get people--
• 35:27 - 35:30
• 35:30 - 35:31
STUDENT: [INAUDIBLE].
• 35:31 - 35:32
MAGDALENA TODA: Very good.
• 35:32 - 35:36
Yeah, he's done
the simplification.
• 35:36 - 35:38
STUDENT: You get
the same values.
• 35:38 - 35:40
• 35:40 - 35:45
Plug in 1, you get 7/12 again.
• 35:45 - 35:48
MAGDALENA TODA: So I'm
asking you-- OK, what was it?
• 35:48 - 35:57
Solve 0, 1-- so I'm
• 35:57 - 36:01
think we get the same value?
• 36:01 - 36:03
Because the force
is not conservative,
• 36:03 - 36:07
and I went on another path.
• 36:07 - 36:10
I went on one path, and
I went on another path.
• 36:10 - 36:16
And look, obviously my
expression was different.
• 36:16 - 36:19
It's like one of those
math games or UIL games.
• 36:19 - 36:21
And look at the algebra.
• 36:21 - 36:24
The polynomials are different.
• 36:24 - 36:26
What was my luck here?
• 36:26 - 36:27
I took 1.
• 36:27 - 36:28
STUDENT: Yeah.
• 36:28 - 36:29
MAGDALENA TODA: I
could have taken 2.
• 36:29 - 36:36
So if instead of 1, I would
have taken another number,
• 36:36 - 36:38
then the higher the power,
the bigger the number
• 36:38 - 36:39
would have been.
• 36:39 - 36:40
I could have taken 2--
• 36:40 - 36:42
STUDENT: You could
have taken negative 1,
• 36:42 - 36:44
and you still wouldn't
• 36:44 - 36:49
MAGDALENA TODA: Yeah, there
are many reasons why that is.
• 36:49 - 36:54
But anyway, know that when you
take 1, 1 to every power is 1.
• 36:54 - 36:55
And yeah, you were lucky.
• 36:55 - 36:58
But in general, keep in
mind that if the force is
• 36:58 - 37:02
conservative, in
general, in most examples
• 37:02 - 37:05
you're not going to get the
• 37:05 - 37:11
because it does depend on
the path you want to take.
• 37:11 - 37:18
I think I have reviewed quite
everything that I wanted.
• 37:18 - 37:27
• 37:27 - 37:30
to move forward.
• 37:30 - 37:33
• 37:33 - 37:42
So I'm saying we are done
with sections 13.1, 13.2,
• 37:42 - 37:49
and 13.3, which was my
favorite because it's not
• 37:49 - 37:51
just the integral of
the path that I like,
• 37:51 - 37:55
but it's the so-called
fundamental theorem of calculus
• 37:55 - 38:05
3, which says, fundamental
theorem of the path integral
• 38:05 - 38:12
saying that you have f of the
endpoint minus f of the origin,
• 38:12 - 38:14
where little f is
that scalar potential
• 38:14 - 38:17
as the linear function
was concerned.
• 38:17 - 38:24
I'm going to call it the
fundamental theorem of path
• 38:24 - 38:26
integral.
• 38:26 - 38:29
Last time I told you the
fundamental theorem of calculus
• 38:29 - 38:32
• 38:32 - 38:35
We refer to that in Calc 1.
• 38:35 - 38:39
But this one is the fundamental
theorem of path integral.
• 38:39 - 38:43
Remember it because at
least one problem out of 15
• 38:43 - 38:45
or something on the
final, and there are not
• 38:45 - 38:46
going to be very many.
• 38:46 - 38:49
It's going to ask you to
know that result. This is
• 38:49 - 38:52
an important theorem.
• 38:52 - 38:56
And another important theorem
that is starting right now
• 38:56 - 38:58
is Green's theorem.
• 38:58 - 39:03
Green's theorem is
a magic result. I
• 39:03 - 39:05
have a t-shirt with it.
• 39:05 - 39:06
I didn't bring it today.
• 39:06 - 39:08
Maybe I'm going to bring
it next time First,
• 39:08 - 39:12
I want you to see
the result, and then
• 39:12 - 39:15
I'll bring the t-shirt
to the exam, so OK.
• 39:15 - 39:18
• 39:18 - 39:25
Assume that you have a
soup called Jordan curve.
• 39:25 - 39:28
• 39:28 - 39:32
You see, mathematicians don't
• 39:32 - 39:34
by their names.
• 39:34 - 39:37
We are crazy people, but
we don't have a big ego.
• 39:37 - 39:42
We would not say a theorem
of myself or whatever.
• 39:42 - 39:45
We never give our names to that.
• 39:45 - 39:51
But all through calculus you
saw all sorts of results.
• 39:51 - 39:57
Like you see the Jordan
curve is a terminology,
• 39:57 - 40:00
but then you see
everywhere the Linus rule.
• 40:00 - 40:02
Did Linus get to
call it his own rule?
• 40:02 - 40:06
No, but Euler's
number, these are
• 40:06 - 40:09
things that were
discovered, and in honor
• 40:09 - 40:12
of that particular
mathematician,
• 40:12 - 40:13
we call them names.
• 40:13 - 40:16
We call them the name
of the mathematician.
• 40:16 - 40:18
• 40:18 - 40:23
Out of curiosity for
0.5 extra credit points,
• 40:23 - 40:25
find out who Jordan was.
• 40:25 - 40:33
Jordan curve is a closed
curve that, in general,
• 40:33 - 40:35
could be piecewise continuous.
• 40:35 - 40:41
• 40:41 - 40:43
So you have a closed
loop over here.
• 40:43 - 40:50
So in general, I could
have something like that
• 40:50 - 40:54
that does not enclose.
• 40:54 - 40:57
That encloses a
domain without holes.
• 40:57 - 41:04
• 41:04 - 41:08
Holes are functions
of the same thing.
• 41:08 - 41:10
STUDENT: So doesn't it
need to be continuous?
• 41:10 - 41:12
MAGDALENA TODA:
No, I said it is.
• 41:12 - 41:14
STUDENT: You said, piecewise.
• 41:14 - 41:15
MAGDALENA TODA: Ah, piecewise.
• 41:15 - 41:16
This is piecewise.
• 41:16 - 41:18
STUDENT: Oh, so it's piecewise.
• 41:18 - 41:18
OK.
• 41:18 - 41:20
MAGDALENA TODA: So you
have a bunch of arcs.
• 41:20 - 41:23
Finitely many, let's
• 41:23 - 41:26
Finitely many arcs,
they have corners,
• 41:26 - 41:30
but you can see define the
integral along such a path.
• 41:30 - 41:34
• 41:34 - 41:38
Oh, and also for another
0.5 extra credit,
• 41:38 - 41:40
find out who Mr.
Green was because he
• 41:40 - 41:43
has several theorems that
are through mathematics
• 41:43 - 41:46
and free mechanics and
variation calculus.
• 41:46 - 41:51
There are several identities
that are called Greens.
• 41:51 - 41:52
There is this famous
Green's theorem,
• 41:52 - 41:55
but there are Green's
first identity,
• 41:55 - 41:58
Green's second identity,
and all sorts of things.
• 41:58 - 42:02
And find out who Mr.
Green was, and as a total,
• 42:02 - 42:04
you have 1 point extra credit.
• 42:04 - 42:09
And you can turn in a regular
essay like a two-page thing.
• 42:09 - 42:13
You want biography of these
mathematicians if you want,
• 42:13 - 42:16
just a few paragraphs.
• 42:16 - 42:19
So what does Green's theorem do?
• 42:19 - 42:26
Green's theorem is
a remarkable result
• 42:26 - 42:31
to the double integral.
• 42:31 - 42:38
It's a remarkable
and famous result.
• 42:38 - 42:48
integral on the closed
• 42:48 - 43:07
curve to a double integral
over the domain enclosed.
• 43:07 - 43:10
I can see the domain
inside, but you
• 43:10 - 43:15
have to understand it's
enclosed by the curve.
• 43:15 - 43:21
• 43:21 - 43:24
All right, and assume
that you have--
• 43:24 - 43:36
M and N are C1 functions of
x and y, what does it mean?
• 43:36 - 43:38
M is a function of xy.
• 43:38 - 43:40
N is a function of xy in plane.
• 43:40 - 43:43
Both of them are differentiable
with continuous derivative.
• 43:43 - 43:47
• 43:47 - 43:48
They are differentiable.
• 43:48 - 43:50
You can take the
partial derivatives,
• 43:50 - 43:52
and all the partial
derivatives are continuous.
• 43:52 - 43:55
That's what we mean
by being C1 functions.
• 43:55 - 43:59
And there the magic
happens, so let me show you
• 43:59 - 44:02
where the magic happens.
• 44:02 - 44:06
This in the box,
the path integral
• 44:06 - 44:21
over c of M dx plus Ndy is
equal to the double integral
• 44:21 - 44:22
over the domain enclosed.
• 44:22 - 44:24
OK, this is the c.
• 44:24 - 44:27
On the boundary you
go counterclockwise
• 44:27 - 44:29
like any respectable
mathematician
• 44:29 - 44:34
would go in a trigonometric
sense, just counterclockwise.
• 44:34 - 44:37
And this is the domain
being closed by c.
• 44:37 - 44:40
• 44:40 - 44:44
And you put here the
integral, which is magic.
• 44:44 - 44:46
This is easy to
remember for you.
• 44:46 - 44:48
This is not easy to
remember unless I
• 44:48 - 44:50
take the t-shirt to
the exam, and you
• 44:50 - 44:52
cheat by looking at my t-shirt.
• 44:52 - 44:55
No, by the time of the
exam, I promised you
• 44:55 - 44:59
you are going to have at least
one week, seven days or more,
• 44:59 - 45:03
10-day period in which
we will study samples,
• 45:03 - 45:06
various samples of old finals.
• 45:06 - 45:08
I'm going to go ahead and
send you some by email.
• 45:08 - 45:11
Do you mind?
• 45:11 - 45:14
In the next week
after this week, we
• 45:14 - 45:15
are going to start reviewing.
• 45:15 - 45:20
And by dA I mean dxdy, the
usual area limit in Cartesian
• 45:20 - 45:24
coordinates the way you
are used to it the most.
• 45:24 - 45:27
• 45:27 - 45:30
And then, Alex is looking
at it and said, well,
• 45:30 - 45:32
then I tell her that
the most elegant way
• 45:32 - 45:35
is to put it with dxdy.
• 45:35 - 45:38
This is what we call a
one form in mathematics.
• 45:38 - 45:40
What is a one form.
• 45:40 - 45:44
It is a linear combination of
this infinitesimal elements
• 45:44 - 45:47
dxdy in plane with some
scalar functions of x
• 45:47 - 45:49
and y in front of her.
• 45:49 - 45:51
OK, so what do we do?
• 45:51 - 45:53
We integrate the one form.
• 45:53 - 45:57
The book doesn't talk about one
forms because the is actually
• 45:57 - 46:01
written for the average
student, the average freshman
• 46:01 - 46:05
or the average
sophomore, but I think
• 46:05 - 46:08
we have an exposure to
the notion of one form,
• 46:08 - 46:11
so I can get a little bit
more elegant and more rigorous
• 46:11 - 46:12
in my speech.
• 46:12 - 46:16
student, you most likely
• 46:16 - 46:18
would know this is a one form.
• 46:18 - 46:22
That's actually the
definition of a one form.
• 46:22 - 46:23
And you'll say, what is this?
• 46:23 - 46:27
This is actually two
form, but you are
• 46:27 - 46:28
going to say, wait a minute.
• 46:28 - 46:31
I have a scalar
function, whatever
• 46:31 - 46:34
that is, from the integration
in front of the dxdy
• 46:34 - 46:40
you want but you never said
that dxdy is a two form.
• 46:40 - 46:45
Actually, I did, and I
didn't call it a two form.
• 46:45 - 46:47
Do you remember that
I introduced to you
• 46:47 - 46:50
some magic wedge product?
• 46:50 - 46:54
• 46:54 - 46:58
And we said, this is
a tiny displacement.
• 46:58 - 47:00
Dx infinitesimal is small.
• 47:00 - 47:02
Imagine how much the
video we'll there
• 47:02 - 47:05
is an infinitesimal
displacement dx
• 47:05 - 47:08
and an infinitesimal
displacement dy,
• 47:08 - 47:11
and you have some
sort of a sign area.
• 47:11 - 47:15
So we said, we don't
just take dxdy,
• 47:15 - 47:19
but we take a product
between dxdy with a wedge,
• 47:19 - 47:22
meaning that if I
change the order,
• 47:22 - 47:24
I'm going to have minus dy here.
• 47:24 - 47:29
This is typical exterior
derivative theory-- exterior
• 47:29 - 47:31
derivative theory.
• 47:31 - 47:35
And it's a theory that
starts more or less
• 47:35 - 47:36
• 47:36 - 47:40
And many people get their
master's degree in math
• 47:40 - 47:43
and never get to see it, and
I pity them, but this life.
• 47:43 - 47:47
On the other hand, when
you have dx, which dx--
• 47:47 - 47:51
the area between dx and dx is 0.
• 47:51 - 47:53
So we're all very happy
I get rid of those.
• 47:53 - 47:56
When I have the sign
between the displacement,
• 47:56 - 47:58
dy and itself is 0.
• 47:58 - 48:00
So these are the
basic properties
• 48:00 - 48:05
that we started
• 48:05 - 48:08
I want to show you what happens.
• 48:08 - 48:16
I'm going to-- yeah,
I'm going to erase here.
• 48:16 - 48:23
• 48:23 - 48:26
I'm going to show
you later I'm going
• 48:26 - 48:33
to prove this theorem to you
later using these tricks that I
• 48:33 - 48:35
just showed you here.
• 48:35 - 48:54
I will provide proof
to this formula, OK?
• 48:54 - 48:58
And let's take a look at
that, and we say, well,
• 48:58 - 49:00
can I memorize that by
the time of the final?
• 49:00 - 49:02
Yes, you can.
• 49:02 - 49:13
this, it can actually
• 49:13 - 49:19
didn't think would be possible.
• 49:19 - 49:21
For example, example
1, and I say,
• 49:21 - 49:26
that would be one of
the most basic ones.
• 49:26 - 49:39
Find the geometric meaning of
the integral over a c where
• 49:39 - 49:40
c is a closed loop.
• 49:40 - 49:42
OK, c is a loop.
• 49:42 - 49:47
Piecewise define Jordan
curve-- Jordan curve.
• 49:47 - 49:50
And I integrate out
of something weird.
• 49:50 - 49:51
And you say, oh, my God.
• 49:51 - 49:52
Look at her.
• 49:52 - 49:59
She picked some weird function
where the path from the dx
• 49:59 - 50:06
is M, and the path in front of
dy is N, the M and N functions.
• 50:06 - 50:08
Why would pick like that?
• 50:08 - 50:11
You wouldn't know yet, but
if you apply Green's theorem,
• 50:11 - 50:14
assuming you believe
it's true, you
• 50:14 - 50:18
have double integral over the
domain enclosed by this loop.
• 50:18 - 50:25
The loop is enclosing
this domain of what?
• 50:25 - 50:32
Now, I'm trying to shut up,
and I'm want you to talk.
• 50:32 - 50:36
What am I going to
write over here?
• 50:36 - 50:37
STUDENT: 1 plus 1.
• 50:37 - 50:41
MAGDALENA TODA: 1 plus
1, how fun is that?
• 50:41 - 50:46
Y minus 1, 1 plus 1 equals
2 last time I checked,
• 50:46 - 50:49
and this is dA.
• 50:49 - 50:53
And what do you think
this animal would be?
• 50:53 - 50:56
The cast of 2 always can escape.
• 50:56 - 51:01
So if we don't want
it, just kick it out.
• 51:01 - 51:04
What is the remaining
double integral for d of DA?
• 51:04 - 51:07
You have seen this guy all
through the Calculus 3 course.
• 51:07 - 51:10
You're tired of it.
• 51:10 - 51:14
You said, I cannot wait for
this semester to be over
• 51:14 - 51:19
because this is the double
integral of 1dA over d.
• 51:19 - 51:22
What in the world is that?
• 51:22 - 51:24
That is the--
• 51:24 - 51:25
STUDENT: --area.
• 51:25 - 51:27
MAGDALENA TODA: Area, very good.
• 51:27 - 51:31
This is the area of the
domain d inside the curve.
• 51:31 - 51:35
• 51:35 - 51:39
So you have discovered
something beautiful
• 51:39 - 51:47
that the area of the domain
enclosed by a Jordan curve
• 51:47 - 51:51
equals 1/2 because you pull
the two out in front here,
• 51:51 - 51:56
it's going to be 1/2 of the path
integrals over the boundary.
• 51:56 - 51:59
This is called
boundary of a domain.
• 51:59 - 52:00
c is the boundary of the domain.
• 52:00 - 52:05
• 52:05 - 52:07
Some mathematicians--
I don't know
• 52:07 - 52:11
how far you want to go with your
education, but in a few years
• 52:11 - 52:13
you might become
• 52:13 - 52:19
And even some engineers use this
notation boundary of d, del d.
• 52:19 - 52:22
That means the boundaries,
the frontier of a domain.
• 52:22 - 52:24
The fence of a ranch.
• 52:24 - 52:27
That is the del d, but
don't tell the rancher
• 52:27 - 52:31
because he will take his gun
out and shoot you thinking
• 52:31 - 52:34
that you are off the hook or
you are after something weird.
• 52:34 - 52:38
So that's the boundary
of the domain.
• 52:38 - 52:42
And then you have
minus ydx plus xdy.
• 52:42 - 52:46
• 52:46 - 52:48
MAGDALENA TODA: We discover
something beautiful.
• 52:48 - 52:50
Something important.
• 52:50 - 52:53
with this exercise--
• 52:53 - 52:59
one which I could even--
I could even call a lemma.
• 52:59 - 53:04
Lemma is not quite a
theorem, because it's based--
• 53:04 - 53:05
could be based on a theorem.
• 53:05 - 53:09
It's a little result that can
be proved in just a few lines
• 53:09 - 53:12
without being something
sophisticated based
• 53:12 - 53:16
on something you
knew from before.
• 53:16 - 53:20
So this is called a lemma.
• 53:20 - 53:26
When you have a sophisticated
area to compute--
• 53:26 - 53:30
or even can you prove-- if you
believe in Green's theorem,
• 53:30 - 53:33
can you prove that the
area inside the circle
• 53:33 - 53:35
is pi r squared?
• 53:35 - 53:40
Can you prove that the
area inside of an ellipse
• 53:40 - 53:41
is-- I don't know what.
• 53:41 - 53:44
Do you know the area
inside of an ellipse?
• 53:44 - 53:48
Nobody taught me in school.
• 53:48 - 53:51
I don't know why
it's so beautiful.
• 53:51 - 53:55
I learned what an ellipse
• 53:55 - 53:59
in high school and again
a review as a freshman
• 53:59 - 54:01
analytic geometry.
• 54:01 - 54:03
So we've seen conics again--
• 54:03 - 54:05
STUDENT: I think we did
• 54:05 - 54:07
We might have seen it.
• 54:07 - 54:08
MAGDALENA TODA:
But nobody told me
• 54:08 - 54:10
like-- I give you an ellipse.
• 54:10 - 54:12
Compute the area inside.
• 54:12 - 54:13
• 54:13 - 54:15
And I didn't know
the formula until I
• 54:15 - 54:18
became an assistant professor.
• 54:18 - 54:19
I was already in my thirties.
• 54:19 - 54:24
That's a shame to see that
thing for the first time OK.
• 54:24 - 54:28
So let's see if we believe
this lemma, and the Green's
• 54:28 - 54:29
theorem of course.
• 54:29 - 54:32
But let's apply the
lemma, primarily
• 54:32 - 54:33
from the Green's theorem.
• 54:33 - 54:37
Can we actually prove
that the area of the disk
• 54:37 - 54:41
is pi r squared and the
area of the ellipse--
• 54:41 - 54:43
inside the ellipse
will be god knows what.
• 54:43 - 54:47
And we will discover
that by ourselves.
• 54:47 - 54:49
I think that's the
beauty of mathematics.
• 54:49 - 54:54
Because every now and then
even if you discover things
• 54:54 - 54:56
that people have known
for hundreds of years,
• 54:56 - 54:58
it still gives you
the satisfaction
• 54:58 - 55:02
that you did something by
yourself-- all on yourself.
• 55:02 - 55:06
Like, when you feel
build a helicopter or you
• 55:06 - 55:08
build a table.
• 55:08 - 55:10
There are many more
beautiful tables
• 55:10 - 55:12
that were built before
you, but still it's
• 55:12 - 55:15
a lot of satisfaction that
you do all by yourself.
• 55:15 - 55:17
It's the same with mathematics.
• 55:17 - 55:23
So let's see what we can
do now for the first time.
• 55:23 - 55:25
Not for the first time.
• 55:25 - 55:28
We do it in other ways.
• 55:28 - 55:38
Can you prove using the lemma
or Green's theorem-- which
• 55:38 - 55:43
is the same thing-- either
one-- that the area of the disk
• 55:43 - 55:48
of radius r-- this is the r.
• 55:48 - 55:52
r is pi r squared.
• 55:52 - 55:56
• 55:56 - 55:58
I hope so.
• 55:58 - 56:00
And the answer is, I hope so.
• 56:00 - 56:01
And that's all.
• 56:01 - 56:04
• 56:04 - 56:09
Area of the disk of radius r.
• 56:09 - 56:10
Oh my god.
• 56:10 - 56:13
So you go, well.
• 56:13 - 56:19
If I knew the parameterization
of that boundary C,
• 56:19 - 56:21
it would be a piece of cake.
• 56:21 - 56:26
Because I would just-- I know
how to do a path integral now.
• 56:26 - 56:28
I've learned in the
previous sections,
• 56:28 - 56:30
so this should be easy.
• 56:30 - 56:33
Can we do that?
• 56:33 - 56:33
So let's see.
• 56:33 - 56:36
• 56:36 - 56:38
Without computing
the double integral,
• 56:38 - 56:41
because I can always do
that with polar coordinates.
• 56:41 - 56:43
And we are going to do that.
• 56:43 - 56:48
• 56:48 - 56:50
Let's do that as
well, as practice.
• 56:50 - 56:54
Because so you
review for the exam.
• 56:54 - 56:57
• 56:57 - 57:01
But another way to
do it would be what?
• 57:01 - 57:07
1/2 integral over the circle.
• 57:07 - 57:14
And how do I parametrize a
• 57:14 - 57:15
Who tells me?
• 57:15 - 57:18
x of t will be--
that was Chapter 10.
• 57:18 - 57:21
Everything is a
circle in mathematics.
• 57:21 - 57:22
STUDENT: r cosine t.
• 57:22 - 57:23
MAGDALENA TODA: r cosine t.
• 57:23 - 57:24
Excellent.
• 57:24 - 57:26
y of t is?
• 57:26 - 57:27
STUDENT: r sine t.
• 57:27 - 57:29
MAGDALENA TODA: r sine t.
• 57:29 - 57:32
So, finally I'm going to
go ahead and use this one.
• 57:32 - 57:37
And I'm going to say, well,
minus y to be plugged in.
• 57:37 - 57:40
• 57:40 - 57:43
This is minus y.
• 57:43 - 57:45
Multiply by dx.
• 57:45 - 57:48
Well, you say, wait a
minute. dx with respect.
• 57:48 - 57:49
What is dx?
• 57:49 - 57:52
dx is just x prime dt.
• 57:52 - 57:55
Dy is just y prime dt.
• 57:55 - 57:56
And t goes out.
• 57:56 - 57:58
It's banished.
• 57:58 - 58:00
No, he's the most important guy.
• 58:00 - 58:03
So t goes from something
to something else.
• 58:03 - 58:05
We will see that later.
• 58:05 - 58:07
What is x prime dt?
• 58:07 - 58:13
X prime is minus r sine
theta-- sine t, Magdalena.
• 58:13 - 58:16
Minus r sine t.
• 58:16 - 58:18
That was x prime.
• 58:18 - 58:19
Change the color.
• 58:19 - 58:23
Give people some
variation in their life.
• 58:23 - 58:32
Plus r cosine t,
because this x--
• 58:32 - 58:33
STUDENT: [INAUDIBLE].
• 58:33 - 58:39
• 58:39 - 58:46
MAGDALENA TODA: --times
the y, which is r cosine t.
• 58:46 - 58:49
So it suddenly became beautiful.
• 58:49 - 58:52
It looks-- first it looks ugly,
but now it became beautiful.
• 58:52 - 58:53
Why?
• 58:53 - 58:54
How come it became beautiful?
• 58:54 - 58:57
STUDENT: Because you got sine
squared plus cosine square.
• 58:57 - 58:58
MAGDALENA TODA:
Because I got a plus.
• 58:58 - 59:02
If you pay attention, plus sine
squared plus cosine squared.
• 59:02 - 59:05
So I have, what is sine
squared plus cosine squared?
• 59:05 - 59:08
I heard that our
students in trig--
• 59:08 - 59:12
Poly told me-- who still don't
know that this is the most
• 59:12 - 59:14
important thing you
learn in trigonometry--
• 59:14 - 59:15
is Pythagorean theorem.
• 59:15 - 59:16
Right?
• 59:16 - 59:24
So you have 1/2 integral
• 59:24 - 59:27
STUDENT: r squared--
• 59:27 - 59:28
MAGDALENA TODA:
r-- no, I'm lazy.
• 59:28 - 59:31
I'm going slow-- r.
• 59:31 - 59:33
dt.
• 59:33 - 59:35
T from what to what?
• 59:35 - 59:37
From 0 times 0.
• 59:37 - 59:40
I'm starting whatever
I want, actually.
• 59:40 - 59:44
I go counterclockwise
I'm into pi.
• 59:44 - 59:46
STUDENT: Why is
that not r squared?
• 59:46 - 59:48
It should be r squared.
• 59:48 - 59:49
MAGDALENA TODA: I'm sorry, guys.
• 59:49 - 59:50
I'm sorry.
• 59:50 - 59:54
I don't know what
I am-- r squared.
• 59:54 - 59:57
1/2 r squared times 2 pi.
• 59:57 - 60:01
• 60:01 - 60:03
So we have pi r squared.
• 60:03 - 60:07
And if you did not
tell me it's r squared,
• 60:07 - 60:10
we wouldn't have
• 60:10 - 60:10
That's good.
• 60:10 - 60:14
• 60:14 - 60:16
What's the other way to do it?
• 60:16 - 60:19
If a problem on
• 60:19 - 60:22
you prove in two different
ways that the rubber
• 60:22 - 60:25
disk is pi r squared using
Calc 3, or whatever--
• 60:25 - 60:26
STUDENT: Would require--
• 60:26 - 60:28
MAGDALENA TODA: The
double integral, right?
• 60:28 - 60:29
Right?
• 60:29 - 60:31
STUDENT: Could have done
Cartesian coordinates as well.
• 60:31 - 60:33
If that counts as a second way.
• 60:33 - 60:34
MAGDALENA TODA: Yeah.
• 60:34 - 60:35
You can-- OK.
• 60:35 - 60:36
What could this be?
• 60:36 - 60:37
Oh my god.
• 60:37 - 60:42
This would be minus 1
to 1 minus square root
• 60:42 - 60:46
1 minus x squared to square
root 1 minus x squared.
• 60:46 - 60:47
Am i right guys?
• 60:47 - 60:47
STUDENT: Yep.
• 60:47 - 60:49
MAGDALENA TODA: 1 dy dx.
• 60:49 - 60:51
Of course it's a pain.
• 60:51 - 60:54
STUDENT: You could double that
and set the bottoms both equal
• 60:54 - 60:55
to 0.
• 60:55 - 60:56
MAGDALENA TODA: Right.
• 60:56 - 61:01
So we can do by symmetry--
• 61:01 - 61:03
STUDENT: Yeah.
• 61:03 - 61:05
MAGDALENA TODA: I'm--
shall I erase or leave it.
• 61:05 - 61:08
Are you understand
what Alex is saying?
• 61:08 - 61:12
This is 2i is the integral
that you will get.
• 61:12 - 61:14
STUDENT: Just write
it next to it--
• 61:14 - 61:15
MAGDALENA TODA: I tell
you four times, you
• 61:15 - 61:16
see, Alex, because you have--
• 61:16 - 61:17
STUDENT: Oh, yeah.
• 61:17 - 61:19
MAGDALENA TODA: --symmetry
with respect to the x-axis,
• 61:19 - 61:21
and symmetry with
respect to y-axis.
• 61:21 - 61:27
And you can take 0
to 1 and 0 to that.
• 61:27 - 61:29
And you have x from 0 to 1.
• 61:29 - 61:34
You have y from 0 to stop.
• 61:34 - 61:35
Square root of 1 minus x square.
• 61:35 - 61:37
Like the strips.
• 61:37 - 61:42
And you have 4
times that A1, which
• 61:42 - 61:45
would be the area of
• 61:45 - 61:46
You can do that, too.
• 61:46 - 61:47
It's easier.
• 61:47 - 61:50
But the best way to do that is
not in Cartesian coordinates.
• 61:50 - 61:53
The best way is to do
it in polar coordinates.
• 61:53 - 61:57
Always remember
• 61:57 - 62:01
So if you have
Jacobian r-- erase.
• 62:01 - 62:03
Let's put r here again.
• 62:03 - 62:08
And then dr d theta.
• 62:08 - 62:10
But now you say, wait
a minute, Magdalena.
• 62:10 - 62:12
You said r is fixed.
• 62:12 - 62:12
Yes.
• 62:12 - 62:14
And that's why I
need to learn Greek,
• 62:14 - 62:16
because it's all Greek to me.
• 62:16 - 62:19
rho as a variable.
• 62:19 - 62:24
And I say, rho is
between 0 and r.
• 62:24 - 62:25
r is fixed.
• 62:25 - 62:27
That's my [INAUDIBLE].
• 62:27 - 62:32
Big r is not usually written
as a variable from 0 to some.
• 62:32 - 62:33
I cannot use that.
• 62:33 - 62:37
So I have to us a Greek letter,
whether I like it or not.
• 62:37 - 62:40
And theta is from 0 to 2 pi.
• 62:40 - 62:42
And I still get the same thing.
• 62:42 - 62:47
I get r-- rho squared
over 2 between 0 and r.
• 62:47 - 62:49
And I have 2 pi.
• 62:49 - 62:53
And in the end that means
pi r squared, and I'm back.
• 62:53 - 62:56
And you say, wait,
this is Example 4.
• 62:56 - 62:57
Whatever example.
• 62:57 - 62:59
Is it Example 4, 5?
• 62:59 - 63:01
You say, this is
a piece of cake.
• 63:01 - 63:06
I have two methods showing
me that area of the disk
• 63:06 - 63:07
is so pi r squared.
• 63:07 - 63:08
It's so trivial.
• 63:08 - 63:12
Yeah, then let's move
on and do the ellipse.
• 63:12 - 63:15
Or we could have been
smart and done the ellipse
• 63:15 - 63:17
from the beginning.
• 63:17 - 63:19
And then the circular
disk would have
• 63:19 - 63:23
been just a trivial, particular
example of the ellipse.
• 63:23 - 63:25
But let's do the ellipse
with this magic formula
• 63:25 - 63:27
that I just taught you.
• 63:27 - 63:30
• 63:30 - 63:34
In the finals-- I'm going to
send you a bunch of finals.
• 63:34 - 63:37
You're going to be
amused, because you're
• 63:37 - 63:38
going to look at
them and you say,
• 63:38 - 63:42
regardless of the year and
semester when the final was
• 63:42 - 63:44
given for Calc 3,
there was always
• 63:44 - 63:49
one of the problems at the
end using direct application
• 63:49 - 63:51
of Green's theorem.
• 63:51 - 63:53
So Green's theorem
is an obsession,
• 63:53 - 63:55
and not only at Tech.
• 63:55 - 63:59
I was looking UT Austin,
A&M, other schools--
• 63:59 - 64:06
California Berkley-- all the
Calc 3 courses on the final
• 64:06 - 64:11
have at least one application--
direct application
• 64:11 - 64:12
applying principal.
• 64:12 - 64:13
OK.
• 64:13 - 64:17
• 64:17 - 64:19
So what did I say?
• 64:19 - 64:22
I said that we have
to draw an ellipse.
• 64:22 - 64:25
How do we draw an ellipse
without making it up?
• 64:25 - 64:27
That's the question.
• 64:27 - 64:29
STUDENT: Draw a circle.
• 64:29 - 64:30
MAGDALENA TODA: Draw a circle.
• 64:30 - 64:32
• 64:32 - 64:34
OK.
• 64:34 - 64:35
All right.
• 64:35 - 64:40
And guys this
• 64:40 - 64:43
So I'm doing what I can.
• 64:43 - 64:46
• 64:46 - 64:49
I should have tried
more coffee today,
• 64:49 - 64:52
because I'm getting
insecure and very shaky.
• 64:52 - 64:53
OK.
• 64:53 - 64:58
So I have the ellipse
in standard form
• 64:58 - 65:02
of center O, x squared over
x squared plus y squared
• 65:02 - 65:05
over B squared equals 1.
• 65:05 - 65:08
And now you are going to
me who is A and who is B?
• 65:08 - 65:09
What are they called?
• 65:09 - 65:10
Semi--
• 65:10 - 65:11
STUDENT: Semiotics.
• 65:11 - 65:12
MAGDALENA TODA: Semiotics.
• 65:12 - 65:15
A and B. Good.
• 65:15 - 65:19
Find the area.
• 65:19 - 65:22
I don't like-- OK.
• 65:22 - 65:27
Let's put B inside, and let's
put C outside the boundary.
• 65:27 - 65:43
So area of the ellipse domain
D will be-- by the lemma-- 1/2
• 65:43 - 65:46
integral over C.
• 65:46 - 65:47
This is C. Is not f.
• 65:47 - 65:48
Don't confuse it.
• 65:48 - 65:51
It is my beautiful
script C. I've
• 65:51 - 65:52
tried to use it many times.
• 65:52 - 65:55
Going to be minus y dx plus xdy.
• 65:55 - 65:58
• 65:58 - 65:59
Again, why was that?
• 65:59 - 66:05
Because we said this
is M and this is N,
• 66:05 - 66:09
and Green's theorem will give
you double integral of N sub x
• 66:09 - 66:11
minus M sub y.
• 66:11 - 66:14
So you have 1 minus
minus 1, which is 2.
• 66:14 - 66:16
And 2 knocked that out.
• 66:16 - 66:16
OK.
• 66:16 - 66:19
That's how we prove it.
• 66:19 - 66:20
OK.
• 66:20 - 66:24
Problem is that I do not the
parametrization of the ellipse.
• 66:24 - 66:28
And if somebody doesn't help me,
I'm going to be in big trouble.
• 66:28 - 66:33
• 66:33 - 66:34
And I'll start
cursing and I'm not
• 66:34 - 66:37
allowed to curse in
front of the classroom.
• 66:37 - 66:41
But you can help me on
that, because this reminds
• 66:41 - 66:46
you of a famous Greek identity.
• 66:46 - 66:49
The fundamental trig identity.
• 66:49 - 66:52
If this would be cosine
squared of theta,
• 66:52 - 66:55
and this would be sine squared
of theta, as two animals,
• 66:55 - 66:57
their sum would be 1.
• 66:57 - 67:01
And whenever you have sums
of sum squared thingies,
• 67:01 - 67:04
then you have to think trig.
• 67:04 - 67:07
So, what would be
good as a parameter?
• 67:07 - 67:07
OK.
• 67:07 - 67:10
What would be good
as a parametrization
• 67:10 - 67:12
to make this come true?
• 67:12 - 67:15
STUDENT: You have the cosine
of theta would equal x over x.
• 67:15 - 67:16
MAGDALENA TODA: Uh-huh.
• 67:16 - 67:18
So then x would be A times--
• 67:18 - 67:19
STUDENT: The cosine of theta.
• 67:19 - 67:22
MAGDALENA TODA:
Do you like theta?
• 67:22 - 67:24
You don't, because
you're not Greek.
• 67:24 - 67:25
That's the problem.
• 67:25 - 67:27
If you were Greek,
you would like it.
• 67:27 - 67:29
is not here anymore.
• 67:29 - 67:31
Greek from Cypress.
• 67:31 - 67:38
And he could claim that the
most important-- most important
• 67:38 - 67:40
alphabet is the
Greek one, and that's
• 67:40 - 67:44
why the mathematicians
• 67:44 - 67:45
OK?
• 67:45 - 67:47
B sine t.
• 67:47 - 67:48
How do you check?
• 67:48 - 67:49
You always think, OK.
• 67:49 - 67:51
This over that is cosine.
• 67:51 - 67:53
This over this is sine.
• 67:53 - 67:54
I square them.
• 67:54 - 67:56
I get exactly that
and I get a 1.
• 67:56 - 67:57
Good.
• 67:57 - 67:58
I'm in good shape.
• 67:58 - 68:01
I know that this
implicit equation--
• 68:01 - 68:05
this is an implicit
equation-- happens if and only
• 68:05 - 68:11
if I have this system of
the parametrization with t
• 68:11 - 68:17
between-- anything I want,
including the basic 0 to 2
• 68:17 - 68:18
pi interval.
• 68:18 - 68:22
And then if I were to move
all around for time real t
• 68:22 - 68:26
I would wind around that the
circle infinitely many times.
• 68:26 - 68:29
Between time equals
minus infinity--
• 68:29 - 68:33
that nobody remembers-- and
time equals plus infinity--
• 68:33 - 68:36
that nobody will
ever get to know.
• 68:36 - 68:39
So those are the values of it.
• 68:39 - 68:41
All the real values, actually.
• 68:41 - 68:45
I only needed from 0 to 2
pi to wind one time around.
• 68:45 - 68:47
And this is the idea.
• 68:47 - 68:49
I wind one time around.
• 68:49 - 68:51
Now people-- you're going
to see mathematicians
• 68:51 - 68:53
are not the greatest people.
• 68:53 - 69:01
I've seen engineers and
physicists use a lot this sign.
• 69:01 - 69:02
Do you know what this means?
• 69:02 - 69:04
STUDENT: It means
one full revolution.
• 69:04 - 69:07
MAGDALENA TODA: It
means a full revolution.
• 69:07 - 69:10
You're going to have
a loop-- loops, that's
• 69:10 - 69:11
whatever you want.
• 69:11 - 69:13
Here and goes counterclockwise.
• 69:13 - 69:16
And they put this
little sign showing
• 69:16 - 69:22
I'm going counterclockwise on
a closed curved, or a loop.
• 69:22 - 69:23
All right.
• 69:23 - 69:24
Don't think they are crazy.
• 69:24 - 69:27
This was used in lots
of scientific papers
• 69:27 - 69:31
in math, physics, and
engineering, and so on.
• 69:31 - 69:31
OK.
• 69:31 - 69:35
• 69:35 - 69:37
Let's do it then.
• 69:37 - 69:38
Can we do it by ourselves?
• 69:38 - 69:39
I think so.
• 69:39 - 69:40
That's see.
• 69:40 - 69:42
1/2 is 1.
• 69:42 - 69:45
And I don't like
the pink marker.
• 69:45 - 69:47
Integral log.
• 69:47 - 69:52
Time from 0 to 2 pi
should be measured.
• 69:52 - 69:56
y minus B sine t.
• 69:56 - 70:02
• 70:02 - 70:04
dx-- what tells me that?
• 70:04 - 70:07
STUDENT: B minus--
• 70:07 - 70:07
• 70:07 - 70:08
MAGDALENA TODA: Very good.
• 70:08 - 70:10
Minus A sine t.
• 70:10 - 70:11
How hard is that?
• 70:11 - 70:16
It's a piece of cake Plus x--
• 70:16 - 70:18
STUDENT: A cosine.
• 70:18 - 70:19
MAGDALENA TODA: Very good.
• 70:19 - 70:22
A cosine t.
• 70:22 - 70:23
TImes--
• 70:23 - 70:26
STUDENT: B cosine t.
• 70:26 - 70:29
MAGDALENA TODA: --B cosine t.
• 70:29 - 70:30
And dt.
• 70:30 - 70:33
And this thing-- look at it.
• 70:33 - 70:33
It's huge.
• 70:33 - 70:36
It looks huge, but it's
so beautiful, because--
• 70:36 - 70:37
STUDENT: AB.
• 70:37 - 70:37
MAGDALENA TODA: AB.
• 70:37 - 70:39
Why is it AB?
• 70:39 - 70:44
It's AB because sine squared
plus cosine squared inside
• 70:44 - 70:46
becomes 1.
• 70:46 - 70:50
And I have plus AB,
plus AB, AB out.
• 70:50 - 70:52
Kick out the AB.
• 70:52 - 70:57
Kick out the A and
the B and you get
• 70:57 - 71:02
something beautiful-- sine
squared t plus cosine squared
• 71:02 - 71:03
• 71:03 - 71:05
And he says, I'm 1.
• 71:05 - 71:08
Look how beautiful
life is for you.
• 71:08 - 71:09
Finally, we proved it.
• 71:09 - 71:11
What did we prove?
• 71:11 - 71:12
We are almost there.
• 71:12 - 71:13
We got a 1/2.
• 71:13 - 71:16
• 71:16 - 71:19
A constant value kick out, AB.
• 71:19 - 71:22
• 71:22 - 71:22
STUDENT: Times 2 pi.
• 71:22 - 71:24
MAGDALENA TODA: Times 2 pi.
• 71:24 - 71:27
• 71:27 - 71:28
Good.
• 71:28 - 71:30
2 goes away.
• 71:30 - 71:33
And we got a magic thing that
nobody taught us in school,
• 71:33 - 71:35
because they were mean.
• 71:35 - 71:37
They really didn't want
us to learn too much.
• 71:37 - 71:39
That's the thingy.
• 71:39 - 71:41
AB pi.
• 71:41 - 71:45
AB pi is what we were
hoping for, because, look.
• 71:45 - 71:48
I mean it's almost
too good to be true.
• 71:48 - 71:54
Well, it's a disk of radius
r, A and B are equal.
• 71:54 - 71:56
And they are the
• 71:56 - 71:59
And that's why we
have pi r squared
• 71:59 - 72:01
as a particular
example of the disk
• 72:01 - 72:05
of the area of this ellipse.
• 72:05 - 72:08
When I saw it the first
time, I was like, well,
• 72:08 - 72:12
I'm glad that I lived to be
30 or something to learn this.
• 72:12 - 72:18
to me in K-12 or in college.
• 72:18 - 72:23
And I was a completing-- I was
a PhD and I didn't know it.
• 72:23 - 72:25
And then I said, oh,
that's why-- pi AB.
• 72:25 - 72:27
Yes, OK.
• 72:27 - 72:28
All right.
• 72:28 - 72:32
So it's so easy to
understand once you-- well.
• 72:32 - 72:33
Once you learn the section.
• 72:33 - 72:35
If you don't learn
the section you
• 72:35 - 72:39
will not be able to understand.
• 72:39 - 72:39
OK.
• 72:39 - 72:40
All right.
• 72:40 - 72:42
I'm going to go
• 72:42 - 72:45
And I'll show you
an example that
• 72:45 - 72:50
was popping up like an obsession
with the numbers changed
• 72:50 - 72:53
in most of the final exams
that happen in the last three
• 72:53 - 72:59
years, regardless of
who wrote the exam.
• 72:59 - 73:05
Because this problem really
matches the learning outcomes,
• 73:05 - 73:09
any good university
• 73:09 - 73:11
around the world.
• 73:11 - 73:12
So you'll say, wow.
• 73:12 - 73:13
It's so easy.
• 73:13 - 73:17
I could not believe it
that-- how easy it is.
• 73:17 - 73:26
But once you see it, you
will-- you'll say, wow.
• 73:26 - 73:27
It's easy.
• 73:27 - 73:35
• 73:35 - 73:35
OK.
• 73:35 - 73:41
• 73:41 - 73:44
[CHATTER]
• 73:44 - 73:46
Let's try this one.
• 73:46 - 73:49
You have a circle.
• 73:49 - 73:58
and the circle will be
• 73:58 - 74:04
and origin 0 of 4, 9, 0, and 0.
• 74:04 - 74:09
• 74:09 - 74:17
And I'm going to
write-- I'm going
• 74:17 - 74:20
to give you-- first I'm going
to give you a very simple one.
• 74:20 - 74:31
• 74:31 - 74:38
Compute in the
simplest possible way.
• 74:38 - 74:42
If you don't want to
parametrize the circle--
• 74:42 - 74:43
you can always
parametrize the circle.
• 74:43 - 74:44
Right?
• 74:44 - 74:45
But you don't want to.
• 74:45 - 74:49
You want to do it the
fastest possible way
• 74:49 - 74:51
without parameterizing
the circle.
• 74:51 - 74:54
Without writing down
what I'm writing down.
• 74:54 - 74:55
You are in a hurry.
• 74:55 - 74:59
You have 20-- 15 minutes
• 74:59 - 75:01
And you're looking
at me and say, I
• 75:01 - 75:02
hope I get an A in this final.
• 75:02 - 75:06
So what do you have to
remember when you look at that?
• 75:06 - 75:10
• 75:10 - 75:15
M and M. M and M.
No, M and N. OK.
• 75:15 - 75:19
And you have to remember
that you are over a circle
• 75:19 - 75:20
so you have a closed loop.
• 75:20 - 75:22
And that's a Jordan curve.
• 75:22 - 75:24
That's enclosing a disk.
• 75:24 - 75:28
So you have a relationship
between the path
• 75:28 - 75:35
integral along the C and the
area along the D-- over D.
• 75:35 - 75:36
Which is of what?
• 75:36 - 75:39
Is N sub x minus M sub y.
• 75:39 - 75:41
So let me write it
in this form, which
• 75:41 - 75:46
is the same thing my students
mostly prefer to write it as.
• 75:46 - 75:49
N sub x minus M sub y.
• 75:49 - 75:52
The t-shirt I have
has it written
• 75:52 - 75:57
like that, because it was
bought from nerdytshirt.com
• 75:57 - 76:01
And it was especially
created to impress nerds.
• 76:01 - 76:04
And of course if you
look at the del notation
• 76:04 - 76:07
that gives you that kind
of snobbish attitude
• 76:07 - 76:12
that you aren't a scientist.
• 76:12 - 76:12
OK.
• 76:12 - 76:16
So what is this
going to be then?
• 76:16 - 76:19
Double integral over d.
• 76:19 - 76:22
And sub x is up
here so it gave 5.
• 76:22 - 76:25
And sub y is a piece of cake.
• 76:25 - 76:37
3 dx dy equals 2 out times
the area of the disk, which
• 76:37 - 76:38
is something you know.
• 76:38 - 76:41
And I'm not going to ask you
to prove that all over again.
• 76:41 - 76:43
So you have to say 2.
• 76:43 - 76:47
I know the area of the
disk-- pi r squared.
• 76:47 - 76:48
• 76:48 - 76:49
And you leave the room.
• 76:49 - 76:50
And that's it.
• 76:50 - 76:52
It's almost too
easy to believe it,
• 76:52 - 76:58
but it was always there in
the simplest possible way.
• 76:58 - 77:03
And now I'm wondering, if I
were to give you something hard,
• 77:03 - 77:08
because-- you know my theory
that when you practice
• 77:08 - 77:12
at something in
the classroom you
• 77:12 - 77:17
have to be working on harder
things in the classroom
• 77:17 - 77:19
to do better in the exam.
• 77:19 - 77:23
So let me cook up
something ugly for you.
• 77:23 - 77:26
The same kind of disk.
• 77:26 - 77:28
And I'm changing the functions.
• 77:28 - 77:33
And I'll make it
more complicated.
• 77:33 - 77:36
• 77:36 - 77:40
Let's see how you
perform on this one.
• 77:40 - 77:47
• 77:47 - 77:50
We avoided that one,
probably, on finals
• 77:50 - 77:52
because I think the
majority of students
• 77:52 - 77:58
wouldn't have understood what
theorem they needed to apply.
• 77:58 - 78:00
It looks a little bit scary.
• 78:00 - 78:02
But let's say that I've
given you the hint,
• 78:02 - 78:05
apply Greens theorem
on the same path
• 78:05 - 78:10
integral, which is a circle
of origin 0 and radius r.
• 78:10 - 78:14
I now draw counterclockwise.
• 78:14 - 78:18
You apply Green's theorem and
you say, I know how to do this,
• 78:18 - 78:21
because now I know the theorem.
• 78:21 - 78:27
This is M. This is N. And I--
my t-shirt did not say M and N.
• 78:27 - 78:31
It said P and Q. Do you
want to put P and Q?
• 78:31 - 78:32
I put P and Q.
• 78:32 - 78:35
So I can-- I can have this
like it is on my t-shirt.
• 78:35 - 78:39
So this is going
to be P sub x-- no.
• 78:39 - 78:40
Q sub x.
• 78:40 - 78:41
Sorry.
• 78:41 - 78:45
M and N. So the second
one with respect to x.
• 78:45 - 78:49
The one that sticks to the y
is prime root respect to x.
• 78:49 - 78:54
The one that sticks to dx is
prime root with respect to y.
• 78:54 - 78:57
And I think one
time-- the one time
• 78:57 - 79:01
when that my friend and
colleague wrote that,
• 79:01 - 79:03
he did it differently.
• 79:03 - 79:06
He wrote something
like, just-- I'll
• 79:06 - 79:10
put-- I don't remember what.
• 79:10 - 79:11
He put this one.
• 79:11 - 79:14
• 79:14 - 79:17
Then the student
was used to dx/dy
• 79:17 - 79:19
and got completely confused.
• 79:19 - 79:26
So pay attention to
what you are saying.
• 79:26 - 79:29
Most of us write it
in x and y first.
• 79:29 - 79:33
And we can see that the
derivative with respect
• 79:33 - 79:39
to x of q, because that is
the one next to be the y.
• 79:39 - 79:42
When he gave it to me
like that, he messed up
• 79:42 - 79:45
everybody's notations.
• 79:45 - 79:46
No.
• 79:46 - 79:47
Good students steal data.
• 79:47 - 79:50
So you guys have to
put it in standard form
• 79:50 - 79:53
and pay attention to
what you are doing.
• 79:53 - 79:54
All right.
• 79:54 - 79:57
So that one form can
be swapped by people
• 79:57 - 79:59
who try to play games.
• 79:59 - 80:03
• 80:03 - 80:08
Now in this one-- So you
have q sub x minus b sub y.
• 80:08 - 80:16
You have 3x squared minus
minus, or just plus, 3y squared.
• 80:16 - 80:16
Good.
• 80:16 - 80:17
Wonderful.
• 80:17 - 80:20
Am I happy, do you
think I'm happy?
• 80:20 - 80:23
Why would I be so happy?
• 80:23 - 80:26
Why is this a happy thing?
• 80:26 - 80:28
something more wild.
• 80:28 - 80:28
I don't.
• 80:28 - 80:30
I'm happy I don't.
• 80:30 - 80:32
Why am I so happy?
• 80:32 - 80:34
Let's see.
• 80:34 - 80:40
3 out over the disk.
• 80:40 - 80:42
Is this ringing a bell?
• 80:42 - 80:49
• 80:49 - 80:50
Yeah.
• 80:50 - 80:53
It's r squared if I
do this in former.
• 80:53 - 80:59
So if I do this in former,
its going to be rdr, d theta.
• 80:59 - 81:01
So life is not as
hard as you believe.
• 81:01 - 81:04
It can look like
a harder problem,
• 81:04 - 81:06
but in reality, it's not really.
• 81:06 - 81:12
So I have 3 times-- now, I
have r squared, I have r cubed.
• 81:12 - 81:17
r cubed dr d theta, r between.
• 81:17 - 81:21
• 81:21 - 81:31
r was between 0 and big
R. Theta will always
• 81:31 - 81:34
be between 0 and 2 pi.
• 81:34 - 81:42
So, I want you, without
• 81:42 - 81:44
and tell me what you got.
• 81:44 - 81:47
STUDENT: Just say it?
• 81:47 - 81:48
MAGDALENA TODA: Yep.
• 81:48 - 81:53
STUDENT: 3/2, pi
r to the fourth.
• 81:53 - 81:54
MAGDALENA TODA: So
how did you do that?
• 81:54 - 81:57
You said, r to the
4 over 4, coming
• 81:57 - 82:00
from integration times the 2 pi,
coming from integration times
• 82:00 - 82:02
3.
• 82:02 - 82:05
Are you guys with me?
• 82:05 - 82:09
Is everybody with me on this?
• 82:09 - 82:12
OK so, we will simplify
• 82:12 - 82:17
What regard is the
• 82:17 - 82:18
STUDENT: How did he
solve that integral
• 82:18 - 82:20
without switching the poles?
• 82:20 - 82:26
• 82:26 - 82:29
MAGDALENA TODA: It would
have been a killer.
• 82:29 - 82:31
Let me write it out.
• 82:31 - 82:32
[LAUGHTER]
• 82:32 - 82:35
Because you want to
write it out, of course.
• 82:35 - 82:41
OK, 3 integral, integral
x squared plus y
• 82:41 - 82:45
squared, dy/dx, just to make
my life a little bit funnier,
• 82:45 - 82:50
and then y between minus
square root-- you're
• 82:50 - 82:52
looking for trouble, huh?
• 82:52 - 83:00
Y squared minus x squared to
r squared minus s squared.
• 83:00 - 83:01
And again, you could
do what you just
• 83:01 - 83:05
said, split into four integrals
over four different domains,
• 83:05 - 83:07
or two up and down.
• 83:07 - 83:12
And minus r and are
you guys with me?
• 83:12 - 83:15
And then, when you go
and integrate that,
• 83:15 - 83:23
you integrate with respect
to y-- [INAUDIBLE].
• 83:23 - 83:25
Well he's right,
so you can get x
• 83:25 - 83:28
squared y plus y cubed over 3.
• 83:28 - 83:31
• 83:31 - 83:34
Between those points, minus 12.
• 83:34 - 83:36
And from that moment,
that would just
• 83:36 - 83:40
leave it and go for a walk.
• 83:40 - 83:43
I will not have the
patience to do this.
• 83:43 - 83:45
Just a second, Matthew.
• 83:45 - 83:47
For this kind of
stuff, of course
• 83:47 - 83:50
I could put this in Maple.
• 83:50 - 83:54
You know Maple has these
little interactive fields,
• 83:54 - 83:56
like little squares?
• 83:56 - 83:59
And you go inside there
• 83:59 - 84:03
And even if it looks very ugly,
Maple will spit you the answer.
• 84:03 - 84:06
If you know your
syntax and do it right,
• 84:06 - 84:08
even if you don't
switch to polar
• 84:08 - 84:10
coordinates or put
it in Cartesian.
• 84:10 - 84:13
Give it the right data, and
it's going to spit the answer.
• 84:13 - 84:14
Yes, Matthew?
• 84:14 - 84:16
STUDENT: I was
out of the room, I
• 84:16 - 84:19
was wondering why
it's now y cubed.
• 84:19 - 84:21
MAGDALENA TODA: Because if
you integrate with respect
• 84:21 - 84:24
to y first--
• 84:24 - 84:27
STUDENT: Because when I
walked out, it was negative y.
• 84:27 - 84:29
MAGDALENA TODA: If
I didn't put minus.
• 84:29 - 84:30
STUDENT: It's a new problem.
• 84:30 - 84:32
• 84:32 - 84:35
He walked out of the room
during the previous problem
• 84:35 - 84:36
and came back after this one.
• 84:36 - 84:38
And now he's confused.
• 84:38 - 84:40
MAGDALENA TODA: You don't
• 84:40 - 84:40
STUDENT: Oh.
• 84:40 - 84:41
No.
• 84:41 - 84:44
• 84:44 - 84:46
I like the polar coordinates.
• 84:46 - 84:48
MAGDALENA TODA: Let
• 84:48 - 84:50
before I talk any further.
• 84:50 - 84:53
I was about to put a plus here.
• 84:53 - 84:56
What would have been the problem
if I had put a plus here?
• 84:56 - 85:00
• 85:00 - 85:03
If I worked this out,
I would have gotten
• 85:03 - 85:06
x squared minus y squared.
• 85:06 - 85:08
Would that have been
the end of the world?
• 85:08 - 85:10
No.
• 85:10 - 85:16
But it would have complicated
my life a little bit more.
• 85:16 - 85:21
Let's do that one as well.
• 85:21 - 85:22
STUDENT: I was
just curious of how
• 85:22 - 85:25
you do any of these problems
when you can't switch to polar.
• 85:25 - 85:28
MAGDALENA TODA: Right, let's see
what-- because Actually, even
• 85:28 - 85:32
in this case, life is not so
hard, not as hard as you think.
• 85:32 - 85:35
The persistence in that matters.
• 85:35 - 85:38
You never give up on a
problem that freaks you out.
• 85:38 - 85:41
That's the definition
of a mathematician.
• 85:41 - 85:48
3x squared minus 3y
squared over dx/dy.
• 85:48 - 85:50
Do it slowly because
I'm not in a hurry.
• 85:50 - 85:56
We are almost done with 13.4.
• 85:56 - 85:57
This is OK, right?
• 85:57 - 85:59
Just the minus sign again?
• 85:59 - 86:01
STUDENT: Well not
the minus sign.
• 86:01 - 86:04
I was just wondering because
in the previous problem
• 86:04 - 86:07
you were doing the ellipse, you
started out with the equation
• 86:07 - 86:09
with the negative y--
• 86:09 - 86:11
MAGDALENA TODA:
For this one that's
• 86:11 - 86:15
just the limit that says that
this is the go double integral
• 86:15 - 86:19
of the area of the domain.
• 86:19 - 86:23
It's just a consequence--
or correlate if you want.
• 86:23 - 86:28
It's a consequence
of Green's theorem.
• 86:28 - 86:31
When you forget that consequence
of Green's theorem and we say
• 86:31 - 86:32
goodbye to that.
• 86:32 - 86:36
But while you were out,
this is Green's theorem.
• 86:36 - 86:39
The real Green's theorem,
the one that was a teacher.
• 86:39 - 86:41
There are several
Greens I can give you.
• 86:41 - 86:43
The famous Green
theorem is the one
• 86:43 - 86:47
I said when you have--
this is what we apply here.
• 86:47 - 86:51
The integral of M dx plus M dy.
• 86:51 - 87:01
You have a double integral of
M sub x minus M sub y over c.
• 87:01 - 87:03
• 87:03 - 87:08
So I'm assuming we would have
had this case of maybe me not
• 87:08 - 87:11
paying attention, or
being mean and not wanting
• 87:11 - 87:14
to give you a simple problem.
• 87:14 - 87:18
And what do you
do in such a case?
• 87:18 - 87:20
It's not obvious to
everybody, but you will see.
• 87:20 - 87:22
It's so pretty at some
point, if you know
• 87:22 - 87:24
how to get out of the mess.
• 87:24 - 87:27
• 87:27 - 87:31
I'm using polar coordinates.
• 87:31 - 87:36
So that's arc of sine, so I
have to go back to the basics.
• 87:36 - 87:40
If I go back to the
basics, ideas come to me.
• 87:40 - 87:42
Right?
• 87:42 - 87:46
So, OK.
• 87:46 - 87:52
r-- let's put dr d theta,
just to get rid of it,
• 87:52 - 87:54
because it's on my nerves.
• 87:54 - 88:00
This is 0 to 2 pi,
this is 0 to r.
• 88:00 - 88:03
And now, you say,
OK, in our mind,
• 88:03 - 88:07
because we are lazy
people, plug in those
• 88:07 - 88:11
and pull out what you can.
• 88:11 - 88:15
One 3 out equals for what?
• 88:15 - 88:18
Inside, you have r squared.
• 88:18 - 88:21
Do you agree?
• 88:21 - 88:30
expression, which is cosine
• 88:30 - 88:33
squared theta, minus
i squared theta.
• 88:33 - 88:35
And you're going to ask me why.
• 88:35 - 88:36
• 88:36 - 88:40
You just square these
and subtract them,
• 88:40 - 88:44
and see what in the world
you're going to get.
• 88:44 - 88:48
Because you get r squared
times cosine squared,
• 88:48 - 88:50
minus i squared.
• 88:50 - 88:51
I'm too lazy to write
down the argument.
• 88:51 - 88:53
But you know we
have trigonometry.
• 88:53 - 88:56
• 88:56 - 88:58
Yes, you see why it's
important for you
• 88:58 - 89:02
to learn trigonometry
when you are little.
• 89:02 - 89:06
You may be 50 or
60, in high school,
• 89:06 - 89:09
or you may be freshman year.
• 89:09 - 89:12
I don't care when, but you
have to learn that this is
• 89:12 - 89:14
the cosine of the double angle.
• 89:14 - 89:16
How many of you remember that?
• 89:16 - 89:19
Maybe you learned that?
• 89:19 - 89:19
Remember that?
• 89:19 - 89:21
OK.
• 89:21 - 89:26
I don't blame you at all
when you don't remember,
• 89:26 - 89:31
because since I've been
the main checker of finals
• 89:31 - 89:39
for the past five years--
it's a lot of finals.
• 89:39 - 89:41
Yeah, the i is there.
• 89:41 - 89:43
That's exactly what
I wanted to tell you,
• 89:43 - 89:46
that's why I left some room.
• 89:46 - 89:53
This data would be t.
• 89:53 - 89:56
The double angle formula did
not appear on many finals.
• 89:56 - 89:58
And I was thinking
it's a period.
• 89:58 - 90:00
instructors, generally they
• 90:00 - 90:07
say students have trouble
remembering or understanding
• 90:07 - 90:10
this later on, by
avoiding the issue,
• 90:10 - 90:14
you sort of bound to it for
the first time in Cal 2,
• 90:14 - 90:17
because there are any
geometric formulas.
• 90:17 - 90:22
And then, you bump again
inside it in Cal 3.
• 90:22 - 90:23
And it never leaves you.
• 90:23 - 90:28
So this, just knowing this
• 90:28 - 90:31
Let me put the r nicely here.
• 90:31 - 90:34
And now finally, we know
how to solve it, because I'm
• 90:34 - 90:35
going to go ahead and erase.
• 90:35 - 90:45
• 90:45 - 90:49
So why it is good for us is
that-- as Matthew observed
• 90:49 - 90:53
a few moments ago,
whenever you have
• 90:53 - 90:57
a product of a function, you
not only in a function in theta
• 90:57 - 91:01
because you can separate them
• 91:01 - 91:03
between the rhos.
• 91:03 - 91:04
In two different products.
• 91:04 - 91:06
So that's would be this theorem.
• 91:06 - 91:11
And you have 3 times-- the
part that depends only on r,
• 91:11 - 91:14
and the part that depends
only on theta, let's
• 91:14 - 91:15
put them separate.
• 91:15 - 91:22
We need theta, and
dr. And what do you
• 91:22 - 91:25
integrate when you integrate?
• 91:25 - 91:25
r cubed.
• 91:25 - 91:29
Attention, do not do rr.
• 91:29 - 91:31
From 0 to r.
• 91:31 - 91:32
OK?
• 91:32 - 91:34
STUDENT: And cosine theta?
• 91:34 - 91:40
MAGDALENA TODA: And then you
have a 0 to 2 pi, cosine 2.
• 91:40 - 91:42
now, let me give
you-- Let me tell you
• 91:42 - 91:45
what it is, because when
I was young, I was naive
• 91:45 - 91:48
and I always started with that.
• 91:48 - 91:52
part, the trig part in theta.
• 91:52 - 91:54
Because that becomes 0.
• 91:54 - 91:57
So no matter how
ugly this is, I've
• 91:57 - 92:00
playing games with us,
• 92:00 - 92:04
and they were giving us
some extremely ugly thing
• 92:04 - 92:06
that would take you forever
for you to integrate.
• 92:06 - 92:09
Or sometimes, it would have
been impossible to integrate.
• 92:09 - 92:12
But then, the whole
thing would have been 0
• 92:12 - 92:14
because when you
integrate cosine 2 theta,
• 92:14 - 92:17
it goes to sine theta.
• 92:17 - 92:21
Sine 2 theta at 2 pi and 0
are the same things, 0 minus 0
• 92:21 - 92:21
equals z.
• 92:21 - 92:24
• 92:24 - 92:27
I cannot tell you how many
• 92:27 - 92:29
play this game with us.
• 92:29 - 92:30
They give us something
that discouraged us.
• 92:30 - 92:34
No, it's not a piece of cake,
compared to what I have.
• 92:34 - 92:37
Some integral value
will go over two lines,
• 92:37 - 92:40
with a huge polynomial
or something.
• 92:40 - 92:44
But in the end, the integral
was 0 for such a result. Yes?
• 92:44 - 92:45
STUDENT: So I have a question.
• 92:45 - 92:50
Could we take that force and
prove that it was conservative?
• 92:50 - 92:55
MAGDALENA TODA: So now
that I'm questioning this,
• 92:55 - 93:00
I'm not questioning
you, but I-- is
• 93:00 - 93:06
the force, that is with you--
what is the original force
• 93:06 - 93:08
• 93:08 - 93:17
If I take y cubed i plus x cubed
j-- and you have to be careful.
• 93:17 - 93:19
Is this conservative?
• 93:19 - 93:23
• 93:23 - 93:25
STUDENT: Yeah.
• 93:25 - 93:28
MAGDALENA TODA: Really?
• 93:28 - 93:31
Why would we pick
a conservative?
• 93:31 - 93:34
STUDENT: Y squared plus
x squared over 2 is--
• 93:34 - 93:36
MAGDALENA TODA: Why is
it not conservative?
• 93:36 - 93:39
• 93:39 - 93:41
IT doesn't pass the hole test.
• 93:41 - 93:44
• 93:44 - 93:48
So p sub y is not
equal to q sub x.
• 93:48 - 93:52
If you primed this with respect
to y, you get that dy squared.
• 93:52 - 93:55
Prime this with this respect
to x, you get 3x squared.
• 93:55 - 93:57
So it's not concerned with him.
• 93:57 - 94:01
And still, I'm
getting-- it's a loop,
• 94:01 - 94:05
and I'm getting a 0, sort
of like I would expect it
• 94:05 - 94:08
I had any dependence of that.
• 94:08 - 94:09
What is the secret here?
• 94:09 - 94:14
STUDENT: That is conservative,
given a condition.
• 94:14 - 94:17
MAGDALENA TODA: Yes,
given a condition
• 94:17 - 94:21
that your x and y are moving
on the serpent's circle.
• 94:21 - 94:26
And that happens, because this
is a symmetric expression,
• 94:26 - 94:28
and x and y are
moving on a circle,
• 94:28 - 94:31
and one is the cosine theta
and one is sine theta.
• 94:31 - 94:35
So in the end, it
simplifies out.
• 94:35 - 94:40
But in general, if I would
have this kind of problem--
• 94:40 - 94:44
if somebody asked me is this
• 94:44 - 94:46
Let me give you a
few more examples.
• 94:46 - 94:58
• 94:58 - 95:15
One example that maybe will look
hard to most people is here.
• 95:15 - 95:37
• 95:37 - 95:50
The vector value function
given by f of x, y incline,
• 95:50 - 95:52
are two values.
• 95:52 - 95:55
No, I mean define two
values of [INAUDIBLE].
• 95:55 - 96:17
• 96:17 - 96:19
A typical exam problem.
• 96:19 - 96:23
And I saw it at
Texas A&M, as well.
• 96:23 - 96:28
So maybe some people like this
kind of a, b, c, d problem.
• 96:28 - 96:29
Is f conservative?
• 96:29 - 96:37
• 96:37 - 96:38
STUDENT: Yep
• 96:38 - 96:40
MAGDALENA TODA:
• 96:40 - 96:41
Good for you guys.
• 96:41 - 96:45
So if I gave you one that
has three components what
• 96:45 - 96:47
did you have to do?
• 96:47 - 96:50
Compute the curl.
• 96:50 - 96:52
You can, of course, compute
the curl also on this one
• 96:52 - 96:55
and have 0 for the
third component.
• 96:55 - 97:02
But the simplest thing
is to do f1 and f2.
• 97:02 - 97:07
f1 prime with respect to y
equals f2 prime with respect
• 97:07 - 97:08
to x.
• 97:08 - 97:13
So I'm going to
make a smile here.
• 97:13 - 97:16
And you realize that the authors
of such a problem, whether they
• 97:16 - 97:21
are at Tech or at Texas
A&M. They do that on purpose
• 97:21 - 97:28
so that you can use this
result to the next level.
• 97:28 - 98:04
And they're saying compute
the happy u over the curve
• 98:04 - 98:23
x cubed and y cubed equals 8 on
the path that connects points
• 98:23 - 98:29
2, 1 and 1, 2 in [INAUDIBLE].
• 98:29 - 98:39
• 98:39 - 98:46
Does this integral depend on f?
• 98:46 - 98:51
• 98:51 - 98:52
State why.
• 98:52 - 98:58
• 98:58 - 99:05
And you see, they don't tell
you find the scalar potential.
• 99:05 - 99:07
many of you will
• 99:07 - 99:09
be able to see it
because you have
• 99:09 - 99:14
good mathematical intuition,
and a computer process
• 99:14 - 99:17
planning in the background
over all the other processes.
• 99:17 - 99:19
We are very visual people.
• 99:19 - 99:22
If you realize that every time
just there with each other
• 99:22 - 99:26
through the classroom, there
are hundreds of distractions.
• 99:26 - 99:28
There's the screen,
there is somebody
• 99:28 - 99:31
who's next to you
who's sneezing,
• 99:31 - 99:35
all sorts of distractions.
• 99:35 - 99:38
unit can still
• 99:38 - 99:41
function, trying to
integrate and find the scalar
• 99:41 - 99:42
potential, which is a miracle.
• 99:42 - 99:46
I don't know how we managed
to do that after all.
• 99:46 - 99:50
If you don't manage to do that,
what do you have to set up?
• 99:50 - 99:55
You have to say, find is
there-- well, you know there is.
• 99:55 - 100:00
So you're not going to question
the existence of the scalar
• 100:00 - 100:04
potential You know it exists,
but you don't know what it is.
• 100:04 - 100:10
What is f such that f sub
x would be 6xy plus 1,
• 100:10 - 100:14
and m sub y will be 3x squared?
• 100:14 - 100:18
And normally, you would
have to integrate backwards.
• 100:18 - 100:21
Now, I'll give you 10 seconds.
• 100:21 - 100:25
If in 10 seconds, you don't
find me a scalar potential,
• 100:25 - 100:27
I'm going to make you
integrate backwards.
• 100:27 - 100:31
So this is finding the scalar
potential by integration.
• 100:31 - 100:34
The way you should, if
you weren't very smart.
• 100:34 - 100:38
But I think you're
smart enough to smell
• 100:38 - 100:42
the potential-- Very good.
• 100:42 - 100:44
But what if you don't?
• 100:44 - 100:46
• 100:46 - 100:50
So we had one or two
student who figured it out.
• 100:50 - 100:51
What if you don't?
• 100:51 - 100:55
If you don't, you can still do
perfectly fine on this problem.
• 100:55 - 101:01
Let's see how we do it
without seeing or guessing.
• 101:01 - 101:03
His brain was running
in the background.
• 101:03 - 101:05
He came up with the answer.
• 101:05 - 101:06
He's happy.
• 101:06 - 101:10
He can move on to
the next level.
• 101:10 - 101:12
STUDENT: Integrate both
sides with respect to r.
• 101:12 - 101:16
MAGDALENA TODA: Right, and
then mix and match them.
• 101:16 - 101:18
Make them in work.
• 101:18 - 101:21
So try to integrate
with respect to x.
• 101:21 - 101:26
6y-- or plus 1, I'm sorry guys.
• 101:26 - 101:29
And once you get it,
you're going to get--
• 101:29 - 101:32
STUDENT: 3x squared y plus x.
• 101:32 - 101:35
MAGDALENA TODA: And
plus a c of what?
• 101:35 - 101:38
And then take this fellow and
prime it with respect to y.
• 101:38 - 101:41
And you're going to
get-- it's not hard.
• 101:41 - 101:44
You're going to get dx
squared plus nothing,
• 101:44 - 101:50
plus c from the y, and it's
good because I gave you
• 101:50 - 101:51
a simple one.
• 101:51 - 101:54
So sometimes you can
have something here,
• 101:54 - 101:57
but in this case, it was just 0.
• 101:57 - 102:00
So c is kappa as a constant.
• 102:00 - 102:05
So instead of why we teach
found with a plus kappa here,
• 102:05 - 102:08
and it still does it.
• 102:08 - 102:13
So on such a problem,
I don't know,
• 102:13 - 102:18
but I think I would give equal
weights to it, B and C. Compute
• 102:18 - 102:21
the path integral
over the curve.
• 102:21 - 102:24
This is horrible, as
an increasing curve.
• 102:24 - 102:27
But I know that
there is a path that
• 102:27 - 102:29
connects the points 2, 1 and 1.
• 102:29 - 102:30
What I have to pay
attention to in my mind
• 102:30 - 102:33
is that these points
actually are on the curve.
• 102:33 - 102:36
And they are, because I
have 8 times 1 equals 8,
• 102:36 - 102:38
1 times 8 equals 8.
• 102:38 - 102:41
So while I was writing it,
I had to think a little bit
• 102:41 - 102:43
on the problem.
• 102:43 - 102:45
If you were to
draw-- well that's
• 102:45 - 102:48
for you have to find
out when you go home.
• 102:48 - 102:52
What do you think
this is going to be?
• 102:52 - 102:55
• 102:55 - 102:58
Actually, we have to
do it now, because it's
• 102:58 - 103:02
a lot simpler than
you think it is.
• 103:02 - 103:07
x and y will be positive,
I can also restrict that.
• 103:07 - 103:09
It looks horrible, but
it's actually much easier
• 103:09 - 103:10
than you think.
• 103:10 - 103:16
So how do I compute that path
integral that makes the points?
• 103:16 - 103:19
I'm going to have
fundamental there.
• 103:19 - 103:23
• 103:23 - 103:27
Which has f of x at q
minus f, with p, which
• 103:27 - 103:31
says that little f is here.
• 103:31 - 103:41
3x squared y plus
x at 2, 1 minus 3x
• 103:41 - 103:47
squared y plus x at 1, 2.
• 103:47 - 103:53
So all I have to do is
• 103:53 - 103:57
see what I'm actually doing?
• 103:57 - 103:58
It's funny.
• 103:58 - 104:02
Which one is the origin, and
which one is the endpoint?
• 104:02 - 104:04
The problem doesn't tell you.
• 104:04 - 104:07
It tells you only you are
connecting the two points.
• 104:07 - 104:10
But which one is the alpha,
and which one is the omega?
• 104:10 - 104:11
Where do you start?
• 104:11 - 104:13
You start here or
you start here?
• 104:13 - 104:16
• 104:16 - 104:17
OK.
• 104:17 - 104:19
Sort of arbitrary.
• 104:19 - 104:22
How do you handle this problem?
• 104:22 - 104:26
Depending on the direction--
pick one direction you move on
• 104:26 - 104:29
along the r, it's up to you.
• 104:29 - 104:32
And then you get an answer, and
if you change the direction,
• 104:32 - 104:34
what's going to happen
to the integral?
• 104:34 - 104:38
It's just change the
sign and that's all.
• 104:38 - 104:43
3 times 4, times 1, plus 2--
guys, keep an eye on my algebra
• 104:43 - 104:48
don't want to mess up.
• 104:48 - 104:50
Am I right, here?
• 104:50 - 104:50
STUDENT: Yes.
• 104:50 - 104:52
MAGDALENA TODA: So how much?
• 104:52 - 104:54
14, is it?
• 104:54 - 104:56
STUDENT: It's 7.
• 104:56 - 104:57
MAGDALENA TODA: Minus 7.
• 104:57 - 105:06
• 105:06 - 105:07
Good.
• 105:07 - 105:08
Wonderful.
• 105:08 - 105:12
So we know what to get,
and we know this does not
• 105:12 - 105:13
depend on the fact.
• 105:13 - 105:17
How much blah, blah,
blah does the instructor
• 105:17 - 105:21
expect for you to get full
credit on the problem?
• 105:21 - 105:22
STUDENT: Just enough to explain.
• 105:22 - 105:24
MAGDALENA TODA: Just
enough to explain.
• 105:24 - 105:30
About 2 lines or 1 line saying
you can say anything really.
• 105:30 - 105:34
You can say this is the theorem
that either shows independence
• 105:34 - 105:36
of that integral.
• 105:36 - 105:43
If the force F vector value
function is conservative,
• 105:43 - 105:47
then this is what
you have to write.
• 105:47 - 105:49
This doesn't depend
on the path c.
• 105:49 - 105:51
And you apply the
fundamental theorem
• 105:51 - 105:54
of path integrals for
the scalar potential.
• 105:54 - 105:58
And that scalar potential
depends on the endpoints
• 105:58 - 106:00
that you're taking.
• 106:00 - 106:02
And the value of
the work depends--
• 106:02 - 106:06
the work depends only on the
scalar potential and the two
• 106:06 - 106:08
points.
• 106:08 - 106:08
That's enough.
• 106:08 - 106:10
That's more than enough.
• 106:10 - 106:14
What if somebody's
not good with wording?
• 106:14 - 106:17
I'm not going to write
her all that explanation.
• 106:17 - 106:22
I'm just going to say whatever.
• 106:22 - 106:25
I'm going to give
her the theorem
• 106:25 - 106:27
in mathematical compressed way.
• 106:27 - 106:31
And I don't care if she
understands it or not.
• 106:31 - 106:35
Even if you write this
formula with not much wording,
• 106:35 - 106:37
I still give you credit.
• 106:37 - 106:39
But I would prefer
that you give me
• 106:39 - 106:42
some sort of-- some
sort of explanation.
• 106:42 - 106:43
Yes, sir.
• 106:43 - 106:44
STUDENT: You said answer was 0.
• 106:44 - 106:46
Then it would have
been path independent?
• 106:46 - 106:51
• 106:51 - 106:54
MAGDALENA TODA: No, the
answer would not be for sure 0
• 106:54 - 106:57
if it was a longer loop.
• 106:57 - 106:59
If it were a longer
closed curve,
• 106:59 - 107:04
that way where it
starts, it ends.
• 107:04 - 107:07
Even if I take a weekly
• 107:07 - 107:09
I still get 7, right?
• 107:09 - 107:11
That's the whole idea.
• 107:11 - 107:13
• 107:13 - 107:15
• 107:15 - 107:21
how do you find out?
• 107:21 - 107:26
Because I don't know how
many of you figured out
• 107:26 - 107:29
what kind of curve that is.
• 107:29 - 107:33
And it looks like an enemy
to you, but there is a catch.
• 107:33 - 107:39
It's an old friend of
yours and you don't see it.
• 107:39 - 107:40
So what is the curve?
• 107:40 - 107:41
What is the curve?
• 107:41 - 107:47
And what is this arc of a
curve between 2, 1 and 1, 2?
• 107:47 - 107:48
Can we find out what that is?
• 107:48 - 107:49
Of course, or cubic.
• 107:49 - 107:50
It's a fake cubic.
• 107:50 - 107:54
It's a fake cubic--
• 107:54 - 107:56
STUDENT: To function together?
• 107:56 - 107:58
MAGDALENA TODA: Let's
see what this is.
• 107:58 - 108:03
xy cubed minus 2 cubed equals 0.
• 108:03 - 108:06
well, our teachers--
• 108:06 - 108:14
I think our teachers teach us
when we were little that this,
• 108:14 - 108:17
if you divided by a
minus- I wasn't little.
• 108:17 - 108:19
I was in high school.
• 108:19 - 108:21
Well, 14-year-old.
• 108:21 - 108:22
STUDENT: A cubed.
• 108:22 - 108:23
STUDENT: A squared.
• 108:23 - 108:24
MAGDALENA TODA: A squared.
• 108:24 - 108:26
STUDENT: Minus 2AB.
• 108:26 - 108:27
Plus 2AB.
• 108:27 - 108:29
MAGDALENA TODA: Very good.
• 108:29 - 108:31
Plus AB, not 2AB.
• 108:31 - 108:32
STUDENT: Oh, darn.
• 108:32 - 108:34
MAGDALENA TODA: Plus B squared.
• 108:34 - 108:35
Suppose you don't believe.
• 108:35 - 108:37
That proves this.
• 108:37 - 108:38
Let's multiply.
• 108:38 - 108:42
A cubed plus A squared
B plus AB squared.
• 108:42 - 108:44
I'm done with the
first multiplication.
• 108:44 - 108:50
Minus BA squared minus
AB squared minus B cubed.
• 108:50 - 108:52
Do they cancel out?
• 108:52 - 108:54
Yes.
• 108:54 - 108:55
Good.
• 108:55 - 108:58
Cancel out.
• 108:58 - 109:00
And cancel out.
• 109:00 - 109:02
Out, poof.
• 109:02 - 109:03
We've proved it, why?
• 109:03 - 109:09
Because maybe some of you--
nobody gave it to proof before.
• 109:09 - 109:12
• 109:12 - 109:18
So as an application,
what is this?
• 109:18 - 109:18
There.
• 109:18 - 109:19
Who is A and who is B?
• 109:19 - 109:24
A is xy, B is 2.
• 109:24 - 109:34
So you have xy minus 2 times
all this fluffy guy, xy
• 109:34 - 109:42
squared plus 2xy plus--
• 109:42 - 109:45
STUDENT: 4.
• 109:45 - 109:45
MAGDALENA TODA: 4.
• 109:45 - 109:49
And I also said, because
I was sneaky, that's why.
• 109:49 - 109:55
or harder. xy is positive.
• 109:55 - 109:58
When I said xy was positive,
what was I intending?
• 109:58 - 110:03
I was intending for you to see
that this cannot be 0 ever.
• 110:03 - 110:08
So the only possible
for you to have 0 here
• 110:08 - 110:10
is when xy equals 2.
• 110:10 - 110:14
And xy equals 2 is a
much simpler curve.
• 110:14 - 110:18
And I want to know
if you realize
• 110:18 - 110:22
that this will have the points
2,1 and 1, 2 staring at you.
• 110:22 - 110:23
Have a nice day today.
• 110:23 - 110:25
Take care.
• 110:25 - 110:27
And good luck.
• 110:27 - 110:32
• 110:32 - 110:34
What is it?
• 110:34 - 110:35
STUDENT: [INAUDIBLE].
• 110:35 - 110:37
MAGDALENA TODA:
Some sort of animal.
• 110:37 - 110:38
It's a curve, a linear curve.
• 110:38 - 110:42
It's not a line.
• 110:42 - 110:43
What is it?
• 110:43 - 110:48
I was talking a little bit
• 110:48 - 110:50
• 110:50 - 110:53
Is this a conic?
• 110:53 - 110:53
Yeah.
• 110:53 - 110:55
What is a conic?
• 110:55 - 111:00
A conic is any kind of
curve that looks like this.
• 111:00 - 111:05
In general form--
oh my god, ABCD.
• 111:05 - 111:08
Now I got my ABC
plus f equals 0.
• 111:08 - 111:10
This is a conic in plane.
• 111:10 - 111:14
My conic is missing
everything else.
• 111:14 - 111:16
And B is 0.
• 111:16 - 111:19
And there is a way where
you-- I showed you how you
• 111:19 - 111:22
know what kind of conic it is.
• 111:22 - 111:28
A, A, B, B, C. A is
positive is-- no, A is 0,
• 111:28 - 111:32
B is-- it should be 2 here.
• 111:32 - 111:34
So you split this in half.
• 111:34 - 111:37
1/2, 1/2, and 0.
• 111:37 - 111:41
The determinant of this is
negative, the discriminant.
• 111:41 - 111:44
That's why we call it
• 111:44 - 111:45
So it cannot be an ellipse.
• 111:45 - 111:46
So what the heck is it?
• 111:46 - 111:47
STUDENT: [INAUDIBLE].
• 111:47 - 111:49
MAGDALENA TODA: Well, I'm silly.
• 111:49 - 111:50
I should have pulled out for y.
• 111:50 - 111:53
• 111:53 - 111:57
And I knew that it
goes down like 1/x.
• 111:57 - 112:01
But I'm asking you, why in
the world is that a conic?
• 112:01 - 112:02
Because you say, wait.
• 112:02 - 112:03
Wait a minute.
• 112:03 - 112:10
I know this curve since I was
five year old in kindergarten.
• 112:10 - 112:13
And this is the point 2, 1.
• 112:13 - 112:17
• 112:17 - 112:17
It's on it.
• 112:17 - 112:23
And there is a symmetric
• 112:23 - 112:25
1, 2.
• 112:25 - 112:27
And between the
two points, there
• 112:27 - 112:32
is just one arc of a curve.
• 112:32 - 112:34
And this is the path that
you are dragging some object
• 112:34 - 112:35
with force f.
• 112:35 - 112:38
You are computing
the work of a-- maybe
• 112:38 - 112:41
you're computing the work of
a neutron between those two
• 112:41 - 112:43
locations.
• 112:43 - 112:44
It's a--
• 112:44 - 112:45
STUDENT: Hyperbola?
• 112:45 - 112:46
MAGDALENA TODA: Hyperbola.
• 112:46 - 112:47
Why Nitish?
• 112:47 - 112:48
Yes, sir.
• 112:48 - 112:50
STUDENT: I was just
wondering, couldn't we
• 112:50 - 112:52
have gone to xy equals 2 plane?
• 112:52 - 112:53
STUDENT: Yeah, way quicker.
• 112:53 - 112:55
STUDENT: x cubed, y
cubed equals 2 cubed.
• 112:55 - 112:57
Then you'd just do both sides--
• 112:57 - 112:57
MAGDALENA TODA:
That's what I did.
• 112:57 - 112:58
STUDENT: The cubed root.
• 112:58 - 112:59
MAGDALENA TODA:
Didn't I do that?
• 112:59 - 113:03
No, because in
general, it's not--
• 113:03 - 113:07
you cannot say if and only
if xy equals 2 in general.
• 113:07 - 113:11
You have to write to
decompose the polynomial.
• 113:11 - 113:12
You were lucky
this was positive.
• 113:12 - 113:15
STUDENT: Well, because
we divided by x cubed.
• 113:15 - 113:17
We could have just
divided everything
• 113:17 - 113:19
by x cubed, and then taken
the cube root of both sides.
• 113:19 - 113:20
MAGDALENA TODA: He's
saying the same thing.
• 113:20 - 113:24
But in mathematics, we don't--
let me show you something.
• 113:24 - 113:25
STUDENT: It would
work for this case,
• 113:25 - 113:27
but not necessarily
for all cases?
• 113:27 - 113:28
MAGDALENA TODA: Yeah.
• 113:28 - 113:39
Let me show you some other
example where you just-- how
• 113:39 - 113:41
do you solve this equation?
• 113:41 - 113:46
By the way, a math
field test is coming.
• 113:46 - 113:48
No, only if you're a math major.
• 113:48 - 113:51
Sorry, junior or senior.
• 113:51 - 113:53
In one math field test,
you don't have to take it.
• 113:53 - 113:57
But some people who
• 113:57 - 114:01
if they take the math field
test, that replaces the GRE,
• 114:01 - 114:03
if the school agrees.
• 114:03 - 114:06
how many roots does it have
• 114:06 - 114:08
and what kind?
• 114:08 - 114:11
Two are imaginary
and one is real.
• 114:11 - 114:15
But everybody said
• 114:15 - 114:17
How can it have one root
if it's a cubic equation?
• 114:17 - 114:19
So one root.
• 114:19 - 114:21
x1 is 1.
• 114:21 - 114:23
The other two are imaginary.
• 114:23 - 114:24
This is the case in this also.
• 114:24 - 114:26
You have some imaginary roots.
• 114:26 - 114:31
So those roots
are funny, but you
• 114:31 - 114:36
would have to
solve this equation
• 114:36 - 114:42
because this is x minus 1
times x squared plus x plus 1.
• 114:42 - 114:46
So the roots are minus
1, plus minus square root
• 114:46 - 114:52
of b squared minus 4ac
over 2, which are minus 1
• 114:52 - 114:57
plus minus square
root of 3i over 2.
• 114:57 - 115:01
Do you guys know
how they are called?
• 115:01 - 115:06
You know them because in
some countries we learn them.
• 115:06 - 115:08
But do you know the notations?
• 115:08 - 115:09
STUDENT: What they call them?
• 115:09 - 115:10
MAGDALENA TODA: Yeah.
• 115:10 - 115:12
• 115:12 - 115:14
There is a Greek letter.
• 115:14 - 115:16
STUDENT: Iota.
• 115:16 - 115:17
MAGDALENA TODA: In
India, probably.
• 115:17 - 115:19
In my country, it was omega.
• 115:19 - 115:20
But I don't think--
• 115:20 - 115:21
STUDENT: In India, iota.
• 115:21 - 115:23
• 115:23 - 115:26
MAGDALENA TODA: But we call
them omega and omega squared.
• 115:26 - 115:29
Because one is the
square of the other.
• 115:29 - 115:30
They are, of course,
both imaginary.
• 115:30 - 115:36
And we call this the
cubic roots of unity.
• 115:36 - 115:39
• 115:39 - 115:42
You say Magdalena, why would
• 115:42 - 115:44
when everything is real?
• 115:44 - 115:44
OK.
• 115:44 - 115:48
It's real for the time being
while you are still with me.
• 115:48 - 115:50
The moment you're going
to say goodbye to me
• 115:50 - 115:55
and you know in 3350 your
life is going to change.
• 115:55 - 115:57
In that course,
• 115:57 - 116:03
to solve this equation just like
we asked all our 3350 students.
• 116:03 - 116:05
To our surprise,
the students don't
• 116:05 - 116:07
know what imaginary roots are.
• 116:07 - 116:08
Many, you know.
• 116:08 - 116:10
• 116:10 - 116:12
But the majority of
the students didn't
• 116:12 - 116:15
know how to get to
those imaginary numbers.
• 116:15 - 116:20
You're going to need to not
only use them, but also express
• 116:20 - 116:23
these in terms of trigonometry.
• 116:23 - 116:26
• 116:26 - 116:32
So just out of curiosity, since
I am already talking to you,
• 116:32 - 116:35
and since I've preparing you a
little bit for the differential
• 116:35 - 116:39
equations class where you
have lots of electric circuits
• 116:39 - 116:41
and applications
of trigonometry,
• 116:41 - 116:46
these imaginary numbers
can also be put-- they
• 116:46 - 116:51
are in general of the form
a plus ib. a plus minus ib.
• 116:51 - 116:55
And we agree that in
3350 you have to do that.
• 116:55 - 116:57
Out of curiosity,
is there anybody
• 116:57 - 117:03
who knows the trigonometric
form of these complex numbers?
• 117:03 - 117:06
STUDENT: Isn't it r e to the j--
• 117:06 - 117:10
• 117:10 - 117:14
MAGDALENA TODA: So you would
have exactly what he says here.
• 117:14 - 117:18
This number will
be-- if it's plus.
• 117:18 - 117:21
r e to the i theta.
• 117:21 - 117:26
He knows a little bit
more than most students.
• 117:26 - 117:34
And that is cosine
theta plus i sine theta.
• 117:34 - 117:37
Can you find me the
angle theta if I
• 117:37 - 117:43
want to write cosine theta
plus i sine theta or cosine
• 117:43 - 117:46
theta minus i sine theta?
• 117:46 - 117:50
Can you find me
the angle of theta?
• 117:50 - 117:51
Is it hard?
• 117:51 - 117:53
Is it easy?
• 117:53 - 117:55
What in the world is it?
• 117:55 - 118:00
• 118:00 - 118:01
Think like this.
• 118:01 - 118:04
We are done with this
example, but I'm just
• 118:04 - 118:08
saying some things that
• 118:08 - 118:12
If you want cosine
theta to be minus 1/2
• 118:12 - 118:21
and you want sine theta to be
root 3 over 2, which quadrant?
• 118:21 - 118:23
• 118:23 - 118:24
STUDENT: Second.
• 118:24 - 118:26
MAGDALENA TODA: The
• 118:26 - 118:27
Very good.
• 118:27 - 118:28
All right.
• 118:28 - 118:31
So think cosine.
• 118:31 - 118:36
If cosine would be a half and
sine would be root 3 over 2,
• 118:36 - 118:38
it would be in first quadrant.
• 118:38 - 118:40
And what angle would that be?
• 118:40 - 118:41
STUDENT: 60.
• 118:41 - 118:42
STUDENT: That's 60--
• 118:42 - 118:46
MAGDALENA TODA: 60 degrees,
which is pi over 3, right?
• 118:46 - 118:53
But pi over 3 is your
friend, so he's happy.
• 118:53 - 118:55
Well, he is there somewhere.
• 118:55 - 118:59
• 118:59 - 119:01
STUDENT: 120.
• 119:01 - 119:05
MAGDALENA TODA: Where you
are here, you are at what?
• 119:05 - 119:07
How much is 120-- very good.
• 119:07 - 119:09
How much is 120 pi?
• 119:09 - 119:11
STUDENT: 4 pi?
• 119:11 - 119:12
MAGDALENA TODA: No.
• 119:12 - 119:12
STUDENT: 2 pi over 3.
• 119:12 - 119:13
MAGDALENA TODA: 2 pi over 3.
• 119:13 - 119:14
Excellent.
• 119:14 - 119:17
So 2 pi over 3.
• 119:17 - 119:21
This would be if you
• 119:21 - 119:22
• 119:22 - 119:24
• 119:24 - 119:28
In degrees, that's 120 degrees.
• 119:28 - 119:39
So to conclude my detour
to introduction to 3350.
• 119:39 - 119:47
When they will ask you to solve
this equation, x cubed minus 1,
• 119:47 - 119:50
you have to tell them like that.
• 119:50 - 119:53
They will ask you to put
it in trigonometric form.
• 119:53 - 120:04
x1 is 1, x2 is cosine of 2 pi
over 3 plus i sine 2 pi over 3.
• 120:04 - 120:07
And the other one
is x3 equals cosine
• 120:07 - 120:16
of 2 pi over 3 minus
i sine of 2 pi over 3.
• 120:16 - 120:17
The last thing.
• 120:17 - 120:19
Because I should let you go.
• 120:19 - 120:20
There was no break.
• 120:20 - 120:23
• 120:23 - 120:26
We still have like 150 minutes.
• 120:26 - 120:29
I stole from you--
no, I stole really big
• 120:29 - 120:33
because we would have-- yeah,
we still have 15 minutes.
• 120:33 - 120:37
But the break was 10 minutes,
so I didn't give you a break.
• 120:37 - 120:40
What would this be if
you wanted to express it
• 120:40 - 120:43
in terms of another angle?
• 120:43 - 120:47
That's the last thing
• 120:47 - 120:49
STUDENT: [INAUDIBLE].
• 120:49 - 120:51
MAGDALENA TODA: Not minus.
• 120:51 - 120:53
Like cosine of an angle
plus i sine of an angle.
• 120:53 - 120:56
You would need to go to
• 120:56 - 120:58
• 120:58 - 120:58
STUDENT: 4.
• 120:58 - 121:00
MAGDALENA TODA:
You've said it before.
• 121:00 - 121:03
That would be 4 pi over 3.
• 121:03 - 121:05
And 4 pi over 3.
• 121:05 - 121:10
Keep in mind these things
with imaginary numbers because
• 121:10 - 121:14
in 3350, they will rely on
you knowing these things.
• 121:14 - 121:16
• 121:16 - 121:18
STUDENT: Then you apply
Euler's formula up there.
• 121:18 - 121:21
• 121:21 - 121:22
MAGDALENA TODA: Oh, yeah.
• 121:22 - 121:24
By the way, this is
called Euler's formula.
• 121:24 - 121:28
• 121:28 - 121:31
STUDENT: In middle
school, they teach you,
• 121:31 - 121:34
and they tell you when
discriminant is small,
• 121:34 - 121:36
there's no solutions.
• 121:36 - 121:37
MAGDALENA TODA: Yeah.
• 121:37 - 121:38
STUDENT: And you
go to [INAUDIBLE].
• 121:38 - 121:40
MAGDALENA TODA: When the
discriminant is less than 0,
• 121:40 - 121:42
there are no real solutions.
• 121:42 - 121:44
But you have in pairs
imaginary solutions.
• 121:44 - 121:46
They always come in pairs.
• 121:46 - 121:50
• 121:50 - 121:52
Do you want me to
show you probably
• 121:52 - 121:55
the most important problem
in 3350 in 2 minutes,
• 121:55 - 121:59
and then I'll let you go?
• 121:59 - 122:01
STUDENT: Sure.
• 122:01 - 122:08
MAGDALENA TODA: So somebody
gives you the equation
• 122:08 - 122:10
of the harmonic oscillator.
• 122:10 - 122:12
And you say, what
the heck is that?
• 122:12 - 122:17
You have a little spring
and you pull that spring.
• 122:17 - 122:19
And it's going to come back.
• 122:19 - 122:22
You displace it, it comes back.
• 122:22 - 122:24
It oscillates back and forth,
oscillates back and forth.
• 122:24 - 122:28
If you were to write the
solutions of the harmonic
• 122:28 - 122:30
oscillator in electric
circuits, there
• 122:30 - 122:31
would be oscillating functions.
• 122:31 - 122:37
So it has to do with sine and
cosine, so they must be trig.
• 122:37 - 122:39
If somebody gives
you this equation,
• 122:39 - 122:57
let's say ax squared-- y
double prime of x minus b.
• 122:57 - 122:59
Plus.
• 122:59 - 123:04
Equals to 0.
• 123:04 - 123:09
And here is a y equals 0.
• 123:09 - 123:12
Why would that
show up like that?
• 123:12 - 123:19
Well, Hooke's law tells
you that there is a force.
• 123:19 - 123:22
And there is a
force and the force
• 123:22 - 123:23
is mass times acceleration.
• 123:23 - 123:27
And acceleration is like this
type of second derivative
• 123:27 - 123:31
of the displacement.
• 123:31 - 123:38
And F and the displacement
are proportional,
• 123:38 - 123:41
when you write F
equals displacement,
• 123:41 - 123:44
let's call it y of x.
• 123:44 - 123:48
When you have y of x, x is time.
• 123:48 - 123:49
That's the displacement.
• 123:49 - 123:50
That's the force.
• 123:50 - 123:51
That's the k.
• 123:51 - 123:53
So you have a certain
Hooke's constant.
• 123:53 - 123:55
Hooke's law constant.
• 123:55 - 123:57
So when you write
this, Hooke's law
• 123:57 - 123:59
is going to become like that.
• 123:59 - 124:05
Mass times y double prime of
x equals-- this is the force.
• 124:05 - 124:07
k times y of x.
• 124:07 - 124:10
• 124:10 - 124:17
But it depends because
you can have plus minus.
• 124:17 - 124:19
So you can have plus or minus.
• 124:19 - 124:20
And these are
positive functions.
• 124:20 - 124:28
• 124:28 - 124:32
You have two equations
in that case.
• 124:32 - 124:40
One equation is the form y
double prime plus-- give me
• 124:40 - 124:40
a number.
• 124:40 - 124:43
Cy equals 0.
• 124:43 - 124:50
And the other one would be y
double prime minus cy equals 0.
• 124:50 - 124:51
All right.
• 124:51 - 124:55
Now, how hard is to
• 124:55 - 125:00
• 125:00 - 125:02
Can you guess the
solutions with naked eyes?
• 125:02 - 125:05
• 125:05 - 125:06
STUDENT: e to the x--
• 125:06 - 125:09
• 125:09 - 125:14
MAGDALENA TODA: So if you have--
you have e to the something.
• 125:14 - 125:18
If you didn't have a c, it
• 125:18 - 125:19
• 125:19 - 125:21
The c will act the
same in the end.
• 125:21 - 125:26
So here, what are the
possible solutions?
• 125:26 - 125:27
STUDENT: e to the--
• 125:27 - 125:29
MAGDALENA TODA: e to the t
is one of them. e to the x
• 125:29 - 125:33
is one of them, right?
• 125:33 - 125:36
So in the end, to
solve such a problem
• 125:36 - 125:37
they teach you the method.
• 125:37 - 125:40
You take the equation.
• 125:40 - 125:42
And for that, you associate
the so-called characteristic
• 125:42 - 125:44
equation.
• 125:44 - 125:47
For power 2, you put r squared.
• 125:47 - 125:51
Then you minus n for-- this
is how many times is it prime?
• 125:51 - 125:52
No times.
• 125:52 - 125:53
0 times.
• 125:53 - 125:55
So you put a 1.
• 125:55 - 125:58
If it's prime one times,
y prime is missing.
• 125:58 - 126:02
It's prime 1 time,
you would put minus r.
• 126:02 - 126:03
Equals 0.
• 126:03 - 126:07
And then you look at
the two roots of that.
• 126:07 - 126:08
And what are they?
• 126:08 - 126:09
Plus minus 1.
• 126:09 - 126:11
So r1 is 1, r2 is 2.
• 126:11 - 126:13
And there is a
theorem that says--
• 126:13 - 126:15
STUDENT: r2 is minus 1.
• 126:15 - 126:17
MAGDALENA TODA: r2 is minus 1.
• 126:17 - 126:20
Excuse me.
• 126:20 - 126:24
OK, there's a theorem that
says all the solutions
• 126:24 - 126:28
of this equation come as
linear combinations of e
• 126:28 - 126:31
to the r1t and e to the r2t.
• 126:31 - 126:33
So linear combination
means you can
• 126:33 - 126:39
take any number a and any
number b, or c1 and c2, anything
• 126:39 - 126:40
like that.
• 126:40 - 126:45
So all the solutions of
this will look like e
• 126:45 - 126:48
to the t with an a in
front plus e to the minus
• 126:48 - 126:50
t with a b in front.
• 126:50 - 126:53
Could you have seen
that with naked eye?
• 126:53 - 126:54
Well, yeah.
• 126:54 - 126:57
I mean, you are smart
and you guessed one.
• 126:57 - 126:59
An you said e to
the t satisfied.
• 126:59 - 127:02
Because if you put e to the
p and prime it as many times
• 127:02 - 127:05
as you want, you
still get e to the t.
• 127:05 - 127:06
So you get 0.
• 127:06 - 127:09
But nobody thought of-- or maybe
• 127:09 - 127:10
to the minus t.
• 127:10 - 127:11
STUDENT: Yeah.
• 127:11 - 127:12
through that one.
• 127:12 - 127:13
• 127:13 - 127:14
STUDENT: That's for a selection.
• 127:14 - 127:16
MAGDALENA TODA: So even if
you take e to the minus t,
• 127:16 - 127:17
• 127:17 - 127:20
And you get this thing.
• 127:20 - 127:24
All right, all the combinations
will satisfy the same equation
• 127:24 - 127:25
as well.
• 127:25 - 127:27
This is a superposition
principle.
• 127:27 - 127:29
With this, it was easy.
• 127:29 - 127:32
But this is the so-called
harmonic oscillator equation.
• 127:32 - 127:36
• 127:36 - 127:40
So either you have it simplified
y double prime plus y equals 0,
• 127:40 - 127:46
or you have some constant c.
• 127:46 - 127:49
Well, what do you
do in that case?
• 127:49 - 127:51
Let's assume you have 1.
• 127:51 - 127:54
Who can guess the solutions?
• 127:54 - 127:56
STUDENT: 0 and cosine--
• 127:56 - 127:58
MAGDALENA TODA: No, 0
is the trivial solution
• 127:58 - 127:59
and it's not going to count.
• 127:59 - 128:04
You can get it from the
combination of the--
• 128:04 - 128:05
STUDENT: y equals sine t.
• 128:05 - 128:07
MAGDALENA TODA:
Sine t is a solution
• 128:07 - 128:12
because sine t prime is cosine.
• 128:12 - 128:14
When you prime it
again, it's minus sine.
• 128:14 - 128:17
minus sine, you get 0.
• 128:17 - 128:19
So you just guessed
1 and you're right.
• 128:19 - 128:21
Make a face.
• 128:21 - 128:22
Do you see another one?
• 128:22 - 128:23
STUDENT: Cosine t.
• 128:23 - 128:24
MAGDALENA TODA: Cosine.
• 128:24 - 128:26
• 128:26 - 128:29
They are independent,
linear independent.
• 128:29 - 128:31
And so the multitude
of solutions
• 128:31 - 128:34
for that-- I taught you
a whole chapter in 3350.
• 128:34 - 128:37
Now you don't have
to take it anymore--
• 128:37 - 128:40
is going to be a equals sine t--
• 128:40 - 128:41
STUDENT: How about e to the i t?
• 128:41 - 128:42
MAGDALENA TODA: Plus b sine t.
• 128:42 - 128:43
I tell you in a second.
• 128:43 - 128:47
All right, we have to
do an e to the i t.
• 128:47 - 128:48
OK.
• 128:48 - 128:51
So you guessed that all the
solutions will be combinations
• 128:51 - 128:56
like-- on the monitor when you
have cosine and sine, if you
• 128:56 - 128:59
• 128:59 - 129:02
you get something like the
monitor thing at the hospital.
• 129:02 - 129:04
So any kind of
oscillation like that
• 129:04 - 129:08
is a combination of this kind.
• 129:08 - 129:13
Maybe with some different
phases and amplitudes.
• 129:13 - 129:17
You have cosine of 70 or
cosine of 5t or something.
• 129:17 - 129:19
But let me show
you what they are
• 129:19 - 129:25
going to show you [INAUDIBLE]
for the harmonic oscillator
• 129:25 - 129:27
equation how the method goes.
• 129:27 - 129:29
You solve for the
characteristic equation.
• 129:29 - 129:35
So you have r squared
plus 1 equals 0.
• 129:35 - 129:39
Now, here's where most of
the students in 3350 fail.
• 129:39 - 129:40
They understand that.
• 129:40 - 129:44
And some of them say, OK,
this has no solutions.
• 129:44 - 129:47
Some of them even say this
has solutions plus minus 1.
• 129:47 - 129:49
I mean, crazy stuff.
• 129:49 - 129:51
Now, what are the
solutions of that?
• 129:51 - 129:53
Because the theory
in this case says
• 129:53 - 129:57
imaginary, then y1
• 129:57 - 130:01
would be e to the ax cosine bx.
• 130:01 - 130:05
And y2 will be e
to the ax sine bx
• 130:05 - 130:09
solutions are a plus minus ib.
• 130:09 - 130:14
It has a lot to do with
Euler's formula in a way.
• 130:14 - 130:21
So if you knew the theory in
3350 and not be just very smart
• 130:21 - 130:24
and get these by yourselves
by guessing them,
• 130:24 - 130:27
how are you supposed
to know that?
• 130:27 - 130:31
Well, r squared
equals minus 1, right?
• 130:31 - 130:34
The square root of minus 1 is i.
• 130:34 - 130:35
STUDENT: Or negative.
• 130:35 - 130:36
MAGDALENA TODA: Or negative i.
• 130:36 - 130:41
So r1 is 0 plus minus i.
• 130:41 - 130:42
So who is a?
• 130:42 - 130:44
a is 0.
• 130:44 - 130:46
Who is b?
• 130:46 - 130:47
b is 1.
• 130:47 - 130:53
So the solutions are e to
the 0x equal cosine 1x and e
• 130:53 - 130:59
to the 0x sine 1x, which
is cosine x, sine x.
• 130:59 - 131:03
Now you know why you can
do everything formalized
• 131:03 - 131:06
and you get all these
solutions from a method.
• 131:06 - 131:10
This method is an
entire chapter.
• 131:10 - 131:12
It's so much easier than in 350.
• 131:12 - 131:15
So much easier than Calculus 3.
• 131:15 - 131:16
You will say this is easy.
• 131:16 - 131:18
It's a pleasure.
• 131:18 - 131:23
fourth of the semester
• 131:23 - 131:25
just on this method.
• 131:25 - 131:26
So now you don't have
to take it anymore.
• 131:26 - 131:29
You can learn it all by
yourself and you're going
• 131:29 - 131:33
to be ready for the next thing.
• 131:33 - 131:35
So I'm just giving you courage.
• 131:35 - 131:40
If you do really, really well in
Calc 3, 3350 will be a breeze.
• 131:40 - 131:42
You can breeze through that.
• 131:42 - 131:46
You only have the probability
in stats for most engineers
• 131:46 - 131:49
to take.
• 131:49 - 131:54
Math is not so complicated.
Title:
TTU Math2450 Calculus3 Sec 13.3
Description:

Conservative vector fields

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Video Language:
English
 jackie.luft edited English subtitles for TTU Math2450 Calculus3 Sec 13.3