## TTU Math2450 Calculus3 Sec 10.2 and 10.4 part 1

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PROFESSOR: Any
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that gave you headaches
regarding homework
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you'd like to talk about?
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Anything related
to what we covered
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from chapter nine and today?
• 0:14 - 0:16
STUDENT: Can we
do some problems?
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PROFESSOR: I can
fix from problems
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like the ones in the
homework, but also I
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can have you tell me what
bothers you in the homework.
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STUDENT: Oh, I have [INAUDIBLE].
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PROFESSOR: What bothered
me about my own homework
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was that I realized that I
did not remind you something
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I assume you should
know, which is
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the equation of a sphere of
given center and given radius.
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And since I trust you so much,
I said, OK they know about it.
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And then somebody asked
me by email what that was,
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and I said, oh, yeah.
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I did not review that in class.
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So review the equation
in r3 form that's x, y, z
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of the sphere of radius r and
center p of coordinates x0, y0,
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z0.
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One of you asked me by email,
does-- of course you do,
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and then if you know it,
can you help me-- can you
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help remind what that was?
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STUDENT: x minus x0--
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PROFESSOR: x minus x0 squared
plus y minus y0 squared
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plus z minus z0 squared
equals R squared.
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OK?
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When you ask, for
example, what is
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the equation of a units sphere,
what do I mean by unit sphere?
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PROFESSOR: Radius 1, and
center 0, standard unit sphere,
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will be.
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There is a notation for that
in mathematics called s2.
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I'll tell you why its called s2.
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x squared plus y squared
plus z squared equals 1.
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s2 stands for the dimension.
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That means the number
of the-- the number
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of degrees of freedom.
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you have on a certain manifold.
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What is a manifold?
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It's a geometric structure.
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I'm not going to
go into details.
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It's a geometric structure
with some special properties.
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I'm not talking about
other fields of algebra,
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anthropology.
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I'm just talking about geometry
and calculus math 3, which
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is multivariable calculus.
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Now, how do I think
of degrees of freedom?
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Look at the table.
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What freedom do I have to move
along one of these sticks?
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I have one degree of
freedom in the sense
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that it's given by a
parameter like time.
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Right?
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It's a 1-parameter
manifold in the sense
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that maybe I have
a line, maybe I
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have the trajectory of the
parking space in terms of time.
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The freedom that the bag has
is to move according to time,
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and that's considered only
one degree of freedom.
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Now if you were on a
plane or another surface,
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why would you have more
than one degrees of freedom?
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Well, I can move towards
you, or I can move this way.
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I can draw a grid the way
the x and y coordinate.
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And those are my
degrees of freedom.
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Practically, the basis
IJ gives me that kind
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of two degrees of freedom.
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Right?
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If I'm in three coordinates, I
have without other constraints,
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because I could be
in three coordinates
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and constrained to be on
a cylinder, in which case
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I still have two
degrees of freedom.
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But if I am a bug
who is free to fly,
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I have the freedom to go with
three degrees of freedom,
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right?
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I have three degrees of
freedom, but if the bug
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is moving-- not flying,
moving on a surface,
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then he has two
degrees of freedom.
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So to again review, lines
and curves in general
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are one dimensional
things, because you
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have one degree of freedom.
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Two dimensional
things are surfaces,
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three dimensional things are
spaces, like the Euclidean
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space, and we are not
going to go beyond,
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at least for the time
being, we are not
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going to go beyond that.
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However, where anybody is
interested in relativity,
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say or let's say four
dimensional spaces, or things
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of x, y, z spatial coordinates
and t as a fourth coordinate,
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then we can go into higher
dimensions, as well.
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OK.
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I want to ask you a question.
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If somebody gives you on
WeBWorK or outside of WeBWorK,
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on the first quiz or
on the final exam,
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let's say you have
this equation,
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x squared plus y squared plus
z squared plus 2x plus 2y
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equals 9.
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What is this identified as?
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Why would this be a quadric?
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Well, there is no x, y, y, z.
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Those terms are missing.
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But I have something of the
type of quadric x squared
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plus By squared plus
c squared plus dxy
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plus exz plus fyz plus, those
are, oh my God, so many.
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Degree two.
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Degree one I would
have ax plus by plus cz
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plus a little d constant, and
whew, that was a long one.
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Right?
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Now, is this of the
type of a project?
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Yes, it is.
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Of course there are some terms
that are missing, good for us.
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How are you going to try to
identify the type of quadric
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by looking at this?
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As you said very well,
I think it's-- you say,
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I think of a sphere, maybe I can
complete the squares, you said.
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How do we complete the squares?
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x squared plus 2x plus
some missing number,
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a magic number-- yes sir?
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STUDENT: So, basically I'll
have to take x plus 2 times 4
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will go outside.
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It's like x min-- x plus 2--
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PROFESSOR: Why x plus 2?
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STUDENT: Because it's 2x--
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STUDENT: It's 2x.
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PROFESSOR: But if
I take x plus 2,
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then that's going to give
me x squared plus 4x plus 4,
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so it's not a good idea.
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STUDENT: On the x plus 1
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PROFESSOR: x plus 1.
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So I'm going to complete
x plus 1 squared.
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What did I invent
that wasn't there?
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STUDENT: 1.
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PROFESSOR: I invented
the 1, and I have
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to compensate for my invention.
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I added the 1, created
the 1 out of nothing,
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so I have to compensate
by subtracting it.
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How much is from here to here?
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Is it exactly the
thing that I underlined
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with a wiggly line, a
light wiggly line thing,
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plus what is the
blue wiggly line,
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the blue wiggly line
that doesn't show--
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I have y plus 1 squared, and
again, I have to compensate
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for what I invented.
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I created a 1 out of nothing,
so this is y squared plus 2y.
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The z squared is all by himself,
and he's crying, I'm so lonely,
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I don't know, there is
nobody like me over there.
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So in the end, I can rewrite
the whole thing as x plus 1
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squared plus y plus 1
squared plus z squared, if I
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want to work them out in
this format, equals what?
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STUDENT: 10.
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PROFESSOR: 11.
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11 is the square
root of 11 squared.
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Like my son said the other day.
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So that the radius
would be square foot
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11 of a sphere of what circle?
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What is the-- or the
sphere of what center?
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STUDENT: Minus 1--
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PROFESSOR: Minus
1, minus 1, and 0.
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So I don't want to insult you.
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Of course you know how
to complete squares.
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However, I have discovered in an
upper level class at some point
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that my students didn't know
how to complete squares, which
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was very, very heartbreaking.
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All right, now.
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Any questions regarding--
while I have a few of yours,
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I'm going to wait
a little bit longer
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until I give
everybody the chance
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to complete the extra credit.
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I have the question
by email saying,
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you mentioned that
genius guy in your class.
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This is a 1-sheeted hyperboloid.
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x squared plus y squared minus
z squared minus 1 equals 0.
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The question was, by
email, how in the world,
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did he figure out what the two
families of generatrices are?
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So you have one family
and another family,
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and both together generate
the 1-sheeted hyperboloid.
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Let me give you a little
bit more of a hint,
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but I'm still going to stop.
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So last time I said, he
noticed you can root together
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the y squared minus 1 and the
x squared minus z squared,
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and you can separate them.
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So you're going to have x
squared minus z squared equals
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1 minus y squared.
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You can't hide the
difference of two squares
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as product of sum
and difference.
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x plus z times x minus z equals
1 plus y times 1 minus y.
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So how can you
eventually arrange stuff
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to be giving due
to the lines that
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are sitting on the surface?
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The lines that are
sitting on the surface
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are infinitely many,
and I would like
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at least a 1-parameter
family of such lines.
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You can have choices.
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One of the choices
would be-- this
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is a product, of
two numbers, right?
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So you can write it as an
equality of two fractions.
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So you would have something
like x plus z on top, x minus
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z below.
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Observe that you are
creating singularities here.
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So you have to take x minus
z case equals 0 separately,
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and then you have, let's
say you have 1 minus y here,
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and 1 plus y here.
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What else do you have to impose
when you impose x minus z
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equals 0.
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You cannot have 7 over 0.
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That is undefined.
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but if you have 0 over
0, that's still possible.
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So whenever you take x
minus z equals 0 separately,
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that will imply that the
numerator corresponding to it
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will also have to be 0.
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And together these
guys are friends.
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What are they?
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2--
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STUDENT: A system of equations.
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PROFESSOR: It's a
system of equations.
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They both represent planes, and
the intersection of two planes
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is a line.
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It's a particular line, which
is part of the family-- which
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is part of a family.
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OK.
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Now, on the other hand, in case
you have 1 plus y equals 0--
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so if it happens that you
have this extreme case
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that the denominator
will be 0, you absolutely
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have to impose x plus z to be 0,
and then you have another life.
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It's not easy for
me to draw those,
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but I could if you
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to draw those and show you
what the lines look like.
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OK?
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All right.
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So you have two special lines
that are part of that picture.
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They are embedded
in the surface.
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How do you find a
family of planes?
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Oh my god, I only
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but I could have
yet another choice
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of how to pick the parameters.
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Let's take lambda to be
a real number parameter.
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And lambda could be
anything-- if lambda is 0,
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what have I got to have, guys?
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STUDENT: The top.
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PROFESSOR: The top
guys will be 0,
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and I still have 1 minus y
equals 0, a plane, intersected
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with x plus z equals 0,
another plane, so still a line.
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So lambda equals 0 will give
me yet another line, which
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is not written big.
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Are you guys with me?
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Could lambda ever
go to infinity?
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Lambda wants to go to
infinity, and when does lambda
• 15:49 - 15:50
go to infinity?
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STUDENT: When the
bottoms would equal 0--
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PROFESSOR: When both
the bottoms would be 0.
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So this is-- I can call it L
infinity, the line of infinity.
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You see?
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But still those
would be two planes.
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There's an intersection,
it's a line.
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OK.
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Can we write this family--
just one family of lines?
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A line is always an intersection
of two planes, right?
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So which are the planes
that I'm talking about?
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x plus z equals
lambda times 1 plus y.
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This is not in the book,
because, oh my God, this is
• 16:37 - 16:39
too hard for the book, right?
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But it's a nice example to
look at in an honors class.
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1 minus y equals
lambda times x minus z.
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It's not in the book.
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It's not in any book that I know
of at the level of calculus.
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All right, OK.
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What are these animals?
• 16:59 - 17:00
The first animal is a plane.
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The second animal is a plane.
• 17:03 - 17:05
How many planes
are in the picture?
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For each lambda, you have a--
for each lambda value in R,
• 17:12 - 17:16
you have a couple of planes
that intersect along your line.
• 17:16 - 17:18
This is the line L lambda.
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And shut up, Magdalena,
you told people too much.
• 17:22 - 17:25
If you still want them to do
this for 2 extra credit points,
• 17:25 - 17:28
give them the chance
to finish the exercise.
• 17:28 - 17:32
So I zip my lips, but
only after I ask you,
• 17:32 - 17:34
how do you think
you are going to get
• 17:34 - 17:38
the other family of rulers?
• 17:38 - 17:42
The ruling guys are
two families, you see?
• 17:42 - 17:47
So this family is
going in one direction.
• 17:47 - 17:49
How am I going to
get two families?
• 17:49 - 17:52
• 17:52 - 17:55
I have another choice
that-- how did I take this?
• 17:55 - 17:58
More or less, I made my choice.
• 17:58 - 18:02
Just like having two
people that would
• 18:02 - 18:05
be prospective job candidates.
• 18:05 - 18:08
You pick one of them.
• 18:08 - 18:10
STUDENT: Now, we can put 1
minus y in the denominator.
• 18:10 - 18:13
The denominator in
place of 1 plus y.
• 18:13 - 18:17
PROFESSOR: So I could have
done-- I could have taken this,
• 18:17 - 18:20
and put 1 plus y here,
and 1 minus y here.
• 18:20 - 18:21
I'm going to let
you do the rest,
• 18:21 - 18:25
and get the second
family of generators
• 18:25 - 18:28
for the whole surface.
• 18:28 - 18:29
That's enough.
• 18:29 - 18:31
You're not missing your credit.
• 18:31 - 18:36
Just, you wanted help,
and I helped you.
• 18:36 - 18:40
And I'm not mad whatsoever
when you ask me things.
• 18:40 - 18:44
The email I got sounded like--
says, this is not in the book,
• 18:44 - 18:47
or in any book, or
on the internet.
• 18:47 - 18:49
How shall I approach this?
• 18:49 - 18:51
How shall I start thinking
• 18:51 - 18:54
This is a completely
legitimate question.
• 18:54 - 18:58
How do I start on this problem?
• 18:58 - 18:59
OK.
• 18:59 - 19:02
On the homework-- maybe it's
too easy-- you have two or three
• 19:02 - 19:05
examples involving spheres.
• 19:05 - 19:07
Those will be too easy for you.
• 19:07 - 19:11
I only gave you a very thin
among of homework this time.
• 19:11 - 19:15
You Have plenty of time until
Monday at 1:30 or something PM.
• 19:15 - 19:19
• 19:19 - 19:21
I would like to draw
a little bit more,
• 19:21 - 19:25
because in this homework
and the next homework,
• 19:25 - 19:33
I'm building something special
called the Frenet Trihedron.
• 19:33 - 19:37
And I told you a little bit
• 19:37 - 19:40
but I didn't tell you much.
• 19:40 - 19:43
• 19:43 - 19:46
Many textbooks in
multivariable calculus
• 19:46 - 19:49
don't say much about it,
which I think is a shame.
• 19:49 - 19:53
• 19:53 - 19:57
You have a position
vector that gives you
• 19:57 - 19:59
the equation of a regular curve.
• 19:59 - 20:06
• 20:06 - 20:08
x of t, y of t, z of t.
• 20:08 - 20:10
Again, what was a regular curve?
• 20:10 - 20:14
I'm just doing review of
what we did last time.
• 20:14 - 20:17
A very nice curve
that is differentiable
• 20:17 - 20:22
and whose derivative is
continuous everywhere
• 20:22 - 20:23
on the interval.
• 20:23 - 20:31
But moreover, the r prime
of t never becomes 0.
• 20:31 - 20:36
So continuously differentiable,
and r prime of t
• 20:36 - 20:41
never becomes 0 for any--
do you know this name,
• 20:41 - 20:43
any for every or for any?
• 20:43 - 20:44
OK.
• 20:44 - 20:47
This is the symbolistics
of mathematics.
• 20:47 - 20:50
You know because you
are as nerdy as me.
• 20:50 - 20:52
But everybody else doesn't.
• 20:52 - 20:54
You guys will learn.
• 20:54 - 20:56
This is what
mathematicians like.
• 20:56 - 21:00
You see, mathematicians hate
writing lots of words down.
• 21:00 - 21:05
If we liked writing essays
and lots of blah, blah, blah,
• 21:05 - 21:07
we would do something else.
• 21:07 - 21:09
We wouldn't do mathematics.
• 21:09 - 21:11
We would do debates,
we would do politics,
• 21:11 - 21:14
we would do other things.
• 21:14 - 21:17
Mathematicians like
ideas, but when
• 21:17 - 21:19
it comes to writing
them down, they
• 21:19 - 21:23
want to right them down in
the most compact way possible.
• 21:23 - 21:26
That's why they created
sort of their own language,
• 21:26 - 21:31
and they have all sorts
of logical quantifiers.
• 21:31 - 21:34
And it's like your
secret language
• 21:34 - 21:37
when it comes to your
less nerdy friends.
• 21:37 - 21:45
So you go for every--
for any or for every--
• 21:45 - 21:46
do you know this sign?
• 21:46 - 21:49
• 21:49 - 21:50
There exists.
• 21:50 - 21:55
• 21:55 - 21:57
And do you know this thing?
• 21:57 - 22:01
Because one of the-- huh?
• 22:01 - 22:03
STUDENT: Is that factorial?
• 22:03 - 22:05
PROFESSOR: Factorial,
but in logic,
• 22:05 - 22:09
that means there exists
a unique-- a unique.
• 22:09 - 22:11
So there exists a unique.
• 22:11 - 22:15
There exists a unique number.
• 22:15 - 22:19
There is a unique number.
• 22:19 - 22:21
So we have our own language.
• 22:21 - 22:24
Of course, empty set,
everybody knows that.
• 22:24 - 22:28
And it's used in
mathematical logic a lot.
• 22:28 - 22:33
You know most of the symbols
from unit intersection,
• 22:33 - 22:35
or, and.
• 22:35 - 22:39
I'm going to use some
of those as well.
• 22:39 - 22:40
Coming back to the
Frenet Trihedron,
• 22:40 - 22:44
we have that velocity
vector at every point.
• 22:44 - 22:45
We are happy with it.
• 22:45 - 22:49
We have our prime of t
that is referred from 0.
• 22:49 - 22:51
I said I want to
make it uniform,
• 22:51 - 22:54
and then I divided
by the magnitude,
• 22:54 - 22:58
and I have this wonderful t
vector we just talked about.
• 22:58 - 23:04
Mr. t is r prime over the
magnitude of r prime, which
• 23:04 - 23:07
is called it's peak right?
• 23:07 - 23:10
We divide by its peak.
• 23:10 - 23:13
What's the name of t, again?
• 23:13 - 23:14
STUDENT: Tangent unit--
• 23:14 - 23:16
PROFESSOR: Tangent
unit vector, very good.
• 23:16 - 23:20
How did you remember
that so quickly?
• 23:20 - 23:22
Tangent unit vector.
• 23:22 - 23:28
There is also another
guy who is famous.
• 23:28 - 23:33
I wanted to make him
green, but let's see
• 23:33 - 23:35
if I can make him blue.
• 23:35 - 23:42
t is defined-- should I
write the f on top of here?
• 23:42 - 23:44
Do you know what that is?
• 23:44 - 23:46
STUDENT: I thought n
was the normal vector.
• 23:46 - 23:48
PROFESSOR: t prime
divided by the length of--
• 23:48 - 23:49
STUDENT: Wait.
• 23:49 - 23:53
I thought the vector
n was the normal.
• 23:53 - 23:56
PROFESSOR: n-- there
are many normals.
• 23:56 - 24:01
It's a very good thing, because
we don't say that in the book.
• 24:01 - 24:05
OK, this is the t along my r.
• 24:05 - 24:09
Now when I go through a point,
this is the normal plane,
• 24:09 - 24:10
right?
• 24:10 - 24:15
There are many normals to
the surface-- to the curve.
• 24:15 - 24:16
Which one am I taking?
• 24:16 - 24:20
All of them are perpendicular
to the direction, right?
• 24:20 - 24:20
STUDENT: tf.
• 24:20 - 24:22
PROFESSOR: So I take
this one, or this one,
• 24:22 - 24:25
or this one, or this one, or
this one, or this one, there.
• 24:25 - 24:27
I have to make up my mind.
• 24:27 - 24:31
And that's how people came up
with the so-called principal
• 24:31 - 24:33
unit normal.
• 24:33 - 24:36
And this is the one
• 24:36 - 24:39
And you are right, it is normal.
• 24:39 - 24:42
Principal unit normal.
• 24:42 - 24:45
Remember this very
well for your exam,
• 24:45 - 24:48
because it's a very
important notion.
• 24:48 - 24:50
How do I get to that?
• 24:50 - 24:54
I take t, I differentiate
it, and I divide
• 24:54 - 24:59
by the lengths of t prime.
• 24:59 - 25:07
Now, can you prove to me
that indeed this fellow
• 25:07 - 25:10
is perpendicular to t?
• 25:10 - 25:12
Can you do that?
• 25:12 - 25:14
STUDENT: That n is
perpendicular to t?
• 25:14 - 25:16
PROFESSOR: Mm-hmm.
• 25:16 - 25:18
So a little exercise.
• 25:18 - 25:23
• 25:23 - 25:31
Prove that-- Prove that I don't
have a good marker anymore.
• 25:31 - 25:38
Prove that n, the unit
principal vector field,
• 25:38 - 25:45
is perpendicular-- you
see, I'm a mathematician.
• 25:45 - 25:49
I swear, I hate to write down
the whole word perpendicular.
• 25:49 - 25:52
I would love to
say, perpendicular.
• 25:52 - 25:58
That's how I write perpendicular
really fast-- to t fore
• 25:58 - 26:01
every value of t.
• 26:01 - 26:03
For every value of t.
• 26:03 - 26:04
OK.
• 26:04 - 26:06
How in the world can I do that?
• 26:06 - 26:09
I have to think about it.
• 26:09 - 26:12
This is hard.
• 26:12 - 26:13
Wish me luck.
• 26:13 - 26:16
So do I know
anything about Mr. t?
• 26:16 - 26:18
What do I know about Mr. t?
• 26:18 - 26:20
I'll take it and I'll
differentiate it later.
• 26:20 - 26:25
It Mr. t is magic in the
sense that he's a unit vector.
• 26:25 - 26:28
I'm going to write that down.
• 26:28 - 26:32
t in absolute value equals 1.
• 26:32 - 26:33
It's beautiful.
• 26:33 - 26:37
If I squared that-- and
you're going to say,
• 26:37 - 26:39
why would you want
to square that?
• 26:39 - 26:40
You're going to see in a minute.
• 26:40 - 26:43
If I squared that,
then I'm going
• 26:43 - 26:51
to have the dot product
between t and itself equals 1.
• 26:51 - 26:53
• 26:53 - 26:57
Can somebody tell me why the
dot product between t and itself
• 26:57 - 27:01
is the square of a length of t?
• 27:01 - 27:05
What's the definition
of the dot product?
• 27:05 - 27:08
Magnitude of the first
vector, times the magnitude
• 27:08 - 27:11
of the second vector--
there i am already--
• 27:11 - 27:15
times the cosine of the
angle between the two vectors
• 27:15 - 27:17
Duh, that's 0.
• 27:17 - 27:20
So cosine of 0 is 1, I'm done.
• 27:20 - 27:21
Right?
• 27:21 - 27:27
Now, I have a vector function
times a vector function--
• 27:27 - 27:31
this is crazy, right-- equals 1.
• 27:31 - 27:34
I'm going to go ahead
and differentiate.
• 27:34 - 27:38
Keep in mind that
this is a product.
• 27:38 - 27:40
What's the product?
• 27:40 - 27:42
One of my professors,
colleagues,
• 27:42 - 27:45
was telling me, now,
let's be serious.
• 27:45 - 27:49
In five years, how many
of your engineering majors
• 27:49 - 27:51
will remember the product?
• 27:51 - 27:53
I really was
• 27:53 - 27:57
I hope everybody, if
they were my students,
• 27:57 - 27:59
because we are going to
have enough practice.
• 27:59 - 28:02
So the prime rule in
Calc 1 said that if you
• 28:02 - 28:05
have f of t times g of
t, you have a product.
• 28:05 - 28:08
You prime that product,
and never write
• 28:08 - 28:13
f prime times g prime unless you
want me to call you around 2 AM
• 28:13 - 28:15
to say you should never do that.
• 28:15 - 28:20
• 28:20 - 28:24
So how does the
product rule work?
• 28:24 - 28:28
The first one prime
times the second unprime
• 28:28 - 28:32
plus the first one unprime
times the second prime.
• 28:32 - 28:35
My students know
the product rule.
• 28:35 - 28:37
I don't care if the rest
of the world doesn't.
• 28:37 - 28:40
I don't care about any
community college who
• 28:40 - 28:43
would say, I don't want the
product rule to be known,
• 28:43 - 28:45
you can differentiate
with a calculator.
• 28:45 - 28:46
That's a no, no, no.
• 28:46 - 28:50
You don't know calculus if you
don't know the product rule.
• 28:50 - 28:53
So the product rule is
a blessing from God.
• 28:53 - 28:58
It helps everywhere in physics,
in mechanics, in engineering.
• 28:58 - 29:01
It really helps in
differential geometry
• 29:01 - 29:04
with the directional
derivative, the Lie derivative.
• 29:04 - 29:08
It helps you understand all
the upper level mathematics.
• 29:08 - 29:12
Now here you have t prime,
the first prime times
• 29:12 - 29:16
the second unprime, plus the
first unprime times the second
• 29:16 - 29:17
prime.
• 29:17 - 29:21
It's the same as for
regular scalar functions.
• 29:21 - 29:24
What's the derivative of 1?
• 29:24 - 29:24
STUDENT: 0.
• 29:24 - 29:26
PROFESSOR: 0.
• 29:26 - 29:27
Look at this guy!
• 29:27 - 29:29
Doesn't he look funny?
• 29:29 - 29:33
It is the dot product community.
• 29:33 - 29:34
Yes it is, by definition.
• 29:34 - 29:40
So you have twice T
times T prime equals 0.
• 29:40 - 29:44
This 2 is-- stinking
guy, let's divide by 2.
• 29:44 - 29:45
• 29:45 - 29:47
What does this say?
• 29:47 - 29:54
The dot product of T times--
I mean by T prime is 0.
• 29:54 - 29:57
When are two vectors
giving you dot product 0?
• 29:57 - 29:59
STUDENT: When they're
perpendicular.
• 29:59 - 30:00
• 30:00 - 30:02
PROFESSOR: So if both
of them are non-zero,
• 30:02 - 30:03
they have to be like that.
• 30:03 - 30:06
They have to be like this,
perpendicular, right?
• 30:06 - 30:12
So it follows that t has to
be perpendicular to T prime.
• 30:12 - 30:16
And now, that's why n
is perpendicular to t.
• 30:16 - 30:19
But, because n is
collinear to t prime.
• 30:19 - 30:20
Hello.
• 30:20 - 30:22
n is collinear to t prime.
• 30:22 - 30:26
So this is t prime.
• 30:26 - 30:28
Is t prime unitary?
• 30:28 - 30:29
I'm going to measure it.
• 30:29 - 30:31
No it's not.
• 30:31 - 30:32
t prime.
• 30:32 - 30:34
So if I want to
make it unitary, I'm
• 30:34 - 30:36
going to chop my-- no,
I'm not going to chop.
• 30:36 - 30:40
I just take it, t prime,
and divide by its magnitude.
• 30:40 - 30:43
Then I'm going to get that
vector n, which is unitary.
• 30:43 - 30:48
So from here it follows that t
and n are indeed perpendicular,
• 30:48 - 30:53
and your colleague over there
said, hey, it has to be normal.
• 30:53 - 30:55
That's perpendicular
to t, but which one?
• 30:55 - 30:58
A special one, because
I have many normals.
• 30:58 - 31:02
Now, this special one is
easy to find like that.
• 31:02 - 31:06
Where shall I put here--
I'll draw him very nicely.
• 31:06 - 31:09
• 31:09 - 31:10
I'll draw him.
• 31:10 - 31:13
Now you guys have to
imagine-- am I drawing
• 31:13 - 31:15
well enough for you?
• 31:15 - 31:16
I don't even know.
• 31:16 - 31:18
t and n should be perpendicular.
• 31:18 - 31:22
Can you imagine them having that
90 degree angle between them?
• 31:22 - 31:22
OK.
• 31:22 - 31:27
Now there is a magic one that
you don't even have to define.
• 31:27 - 31:29
And yes sir?
• 31:29 - 31:31
STUDENT: In this
thing, can [INAUDIBLE]
• 31:31 - 31:34
this T vector [INAUDIBLE]
written by the definition
• 31:34 - 31:36
thing?
• 31:36 - 31:38
PROFESSOR: No.
• 31:38 - 31:39
STUDENT: N vector
times the magnitude
• 31:39 - 31:42
of t vector derivative?
• 31:42 - 31:47
PROFESSOR: So
technically you have
• 31:47 - 31:51
t prime would be the
magnitude of t prime times n.
• 31:51 - 31:52
STUDENT: Yes.
• 31:52 - 31:54
PROFESSOR: But keep in mind
that sometimes is tricky,
• 31:54 - 31:57
because this is, in
general, not a constant.
• 31:57 - 31:59
Always keep it in mind,
it's not a constant.
• 31:59 - 32:02
We'll have some examples later.
• 32:02 - 32:05
There is a magic
guy called binormal.
• 32:05 - 32:10
That binormal is the
normal to both t and n.
• 32:10 - 32:12
And he's defined as
t plus n because it's
• 32:12 - 32:14
normal to both of them.
• 32:14 - 32:18
So I'm going to write this
b vector is t cross n.
• 32:18 - 32:22
Now I'm asking you to draw it.
• 32:22 - 32:24
Can anybody come to
the board and draw it
• 32:24 - 32:27
for 0.01 extra credit?
• 32:27 - 32:30
Yes, sir?
• 32:30 - 32:31
STUDENT: [INAUDIBLE]
• 32:31 - 32:35
PROFESSOR: Draw that on the
picture like t and n, t and n,
• 32:35 - 32:38
t is the-- who the heck
is t? t is the red one,
• 32:38 - 32:41
and blue is the n.
• 32:41 - 32:43
So does it go down or up?
• 32:43 - 32:46
We should be perpendicular
to both of them.
• 32:46 - 32:49
Is b unitary or not?
• 32:49 - 32:52
If you have two unit vectors,
will the cross product
• 32:52 - 32:53
be a unit vector?
• 32:53 - 32:56
• 32:56 - 33:00
Only if the two vectors
are perpendicular,
• 33:00 - 33:05
it is going to be, right?
• 33:05 - 33:12
So you have-- well, I
think it goes that--
• 33:12 - 33:14
in which direction does it go?
• 33:14 - 33:15
Because
• 33:15 - 33:16
STUDENT: It should
not be how we have it.
• 33:16 - 33:17
PROFESSOR: No, no, no.
• 33:17 - 33:18
Because this is--
• 33:18 - 33:19
STUDENT: Yeah.
• 33:19 - 33:20
I'm using--
• 33:20 - 33:23
PROFESSOR: So t
goes over n, so I'm
• 33:23 - 33:27
going to try-- it is
like that, sort of.
• 33:27 - 33:29
STUDENT: Into the chord?
• 33:29 - 33:31
PROFESSOR: So again, it's
not very clear because
• 33:31 - 33:34
of my stinking art, here.
• 33:34 - 33:36
It's really not nice art.
• 33:36 - 33:40
t, and this is n.
• 33:40 - 33:44
And if I go t going over n.
• 33:44 - 33:48
T going over n goes up or down?
• 33:48 - 33:48
STUDENT: Down.
• 33:48 - 33:49
PROFESSOR: Goes down.
• 33:49 - 33:52
So it's going to look
more like this, feet.
• 33:52 - 33:55
Now guys, when we--
thank you so much.
• 33:55 - 33:58
So you've like a
0.01 extra credit.
• 33:58 - 34:01
OK.
• 34:01 - 34:03
Tangent, normal, and
binormal form a corner.
• 34:03 - 34:04
Yes, sir?
• 34:04 - 34:07
STUDENT: Is rt-- rt is
the function at the--
• 34:07 - 34:09
for the flag that's flying?
• 34:09 - 34:12
PROFESSOR: The r of t
is the position vector
• 34:12 - 34:15
of the flag that was
flying that he was drunk.
• 34:15 - 34:21
STUDENT: Why wasn't the
derivative of it perpendicular?
• 34:21 - 34:24
Why isn't t perpendicular to rt?
• 34:24 - 34:26
PROFESSOR: If--
well, good question.
• 34:26 - 34:29
• 34:29 - 34:31
We'll talk about it.
• 34:31 - 34:35
If the length of r
would be a constant,
• 34:35 - 34:39
can we prove that r and r
prime are perpendicular?
• 34:39 - 34:41
Let's do that as
another exercise.
• 34:41 - 34:43
All right?
• 34:43 - 34:46
So tnb looks like a corner.
• 34:46 - 34:52
Look at the corner that the
video cannot see over there.
• 34:52 - 34:54
TN and B are mutually octagonal.
• 34:54 - 34:56
• 34:56 - 34:58
I'm going to draw them.
• 34:58 - 35:01
This is an arbitrary
point on a curve,
• 35:01 - 35:04
and this is t, which is
always tangent to the curve,
• 35:04 - 35:06
and this is n.
• 35:06 - 35:08
Let's say that's the
unit principle normal.
• 35:08 - 35:11
And t cross n will
go, again, down.
• 35:11 - 35:12
I don't know.
• 35:12 - 35:15
I have an obsession
with me going down.
• 35:15 - 35:16
This is called the
Frenet Trihedron.
• 35:16 - 35:20
• 35:20 - 35:23
And I have a proposal
for a problem
• 35:23 - 35:35
that maybe I should give
my students in the future.
• 35:35 - 35:46
Show that for a circle,
playing in space, I don't know.
• 35:46 - 36:07
The position vector and the
velocity vector are always how?
• 36:07 - 36:08
Friends.
• 36:08 - 36:09
Let's say friends.
• 36:09 - 36:12
No, come on, I'm kidding.
• 36:12 - 36:13
How are they?
• 36:13 - 36:15
STUDENT: Perpendicular.
• 36:15 - 36:17
PROFESSOR: How do you do that?
• 36:17 - 36:18
Is it hard?
• 36:18 - 36:21
We should be smart
enough to do that, right?
• 36:21 - 36:22
I have a circle.
• 36:22 - 36:26
That circle has what-- what
is the property of a circle?
• 36:26 - 36:30
Euclid defined that-- this is
one of the axioms of Euclid.
• 36:30 - 36:32
Does anybody know which axiom?
• 36:32 - 36:36
That there exists
such a set of points
• 36:36 - 36:39
that are all at the same
distance from a given point
• 36:39 - 36:41
called center.
• 36:41 - 36:43
So that is a circle, right?
• 36:43 - 36:44
That's what Mr. Euclid said.
• 36:44 - 36:45
He was a genius.
• 36:45 - 36:52
So no matter where I put that
circle, I can take r of t
• 36:52 - 36:55
in magnitude measured
from the origin
• 36:55 - 36:57
from the center of the circle.
• 36:57 - 37:01
Keep in mind, always the
center of the circle.
• 37:01 - 37:06
I put it at the origin of the
space-- origin of the universe.
• 37:06 - 37:08
No, origin of the
space, actually.
• 37:08 - 37:13
R of T magnitude
would be a constant.
• 37:13 - 37:14
Give me a constant, guys.
• 37:14 - 37:14
OK?
• 37:14 - 37:16
It doesn't matter.
• 37:16 - 37:18
Let me draw.
• 37:18 - 37:20
I want to draw in plane, OK?
• 37:20 - 37:26
Because I'm getting tired.
x y, and this is r of t,
• 37:26 - 37:30
and the magnitude of this r of
t is the radius of the circle.
• 37:30 - 37:33
Right?
• 37:33 - 37:37
So let's say, this is
the radius of the circle.
• 37:37 - 37:41
• 37:41 - 37:44
How in the world do I
prove the same idea?
• 37:44 - 37:48
Who helps me prove
that r is always
• 37:48 - 37:51
perpendicular to r prime?
• 37:51 - 37:54
Which way do you want to move,
counterclockwise or clockwise?
• 37:54 - 37:55
STUDENT: Counterclockwise.
• 37:55 - 37:56
PROFESSOR: Counterclockwise.
• 37:56 - 37:58
Because if you are
a real scientist,
• 37:58 - 38:00
I'm proud of you guys.
• 38:00 - 38:02
It's clear from the
picture that r prime
• 38:02 - 38:05
would be perpendicular to r.
• 38:05 - 38:06
Why is that?
• 38:06 - 38:08
How am I going to do that?
• 38:08 - 38:12
Now, mimic everything I--
don't look at your notes,
• 38:12 - 38:17
and try to tell me how
I show that quickly.
• 38:17 - 38:18
What am I going to do?
• 38:18 - 38:23
So all I know, all
that gave me was r of t
• 38:23 - 38:28
equals k in magnitude constant.
• 38:28 - 38:31
For every t, this same constant.
• 38:31 - 38:32
What's next?
• 38:32 - 38:35
What do I want to do next?
• 38:35 - 38:36
STUDENT: Square it?
• 38:36 - 38:38
PROFESSOR: Square
it, differentiate it.
• 38:38 - 38:40
I can also go ahead
and differentiate it
• 38:40 - 38:42
without squaring
it, but that's going
• 38:42 - 38:47
to be a little bit of more pain.
• 38:47 - 38:52
So square it, differentiate it.
• 38:52 - 38:53
I'm too lazy.
• 38:53 - 38:56
When I differentiate,
what am I going to get?
• 38:56 - 39:05
From the product rule, twice
r dot r primed of t equals 0.
• 39:05 - 39:07
Well, I'm done.
• 39:07 - 39:12
Because it means that for
every t that radius-- not
• 39:12 - 39:13
the radius, guys, I'm sorry.
• 39:13 - 39:17
The position vector will be
perpendicular to the velocity
• 39:17 - 39:18
vector.
• 39:18 - 39:22
Now, if I draw the
trajectory of my drunken flag
• 39:22 - 39:25
this [INAUDIBLE]
is not true, right?
• 39:25 - 39:27
This is crazy.
• 39:27 - 39:30
Of course this is r,
and this is r prime,
• 39:30 - 39:36
and there is an arbitrary
angle between r and r prime.
• 39:36 - 39:38
The good thing is that
the arbitrary angle always
• 39:38 - 39:41
exists, and is
continuous as a function.
• 39:41 - 39:43
I never have that
angle disappear.
• 39:43 - 39:47
That's way I want that
prime never to become 0.
• 39:47 - 39:49
Because if the bag was
stopping its motion,
• 39:49 - 39:54
goodbye angle, goodbye
analysis, right?
• 39:54 - 39:55
OK.
• 39:55 - 39:56
Very nice.
• 39:56 - 39:57
So don't give me more ideas.
• 39:57 - 40:00
You smart people, if
you give me more ideas,
• 40:00 - 40:03
I'm going to come up with
all sorts of problems.
• 40:03 - 40:05
And this is actually one
of the first problems
• 40:05 - 40:09
you learn in a graduate
level geometry class.
• 40:09 - 40:14
• 40:14 - 40:17
Let me give you another
piece of information
• 40:17 - 40:20
that you're going
to love, which could
• 40:20 - 40:22
be one of those
types of combined
• 40:22 - 40:25
problems on a final
exam or midterm,
• 40:25 - 40:30
A, B, C, D, E. The
curvature of a curve
• 40:30 - 40:34
is a measure of how
the curve will bend.
• 40:34 - 40:36
Say what?
• 40:36 - 40:46
The curvature of a
curve is a measure
• 40:46 - 40:49
of the bending of that curve.
• 40:49 - 40:59
• 40:59 - 41:04
By definition, you have
to take it like that.
• 41:04 - 41:21
If the curve is parameterized
in arc length-- somebody
• 41:21 - 41:23
remind me what that is.
• 41:23 - 41:25
What does it mean?
• 41:25 - 41:32
That is r of s such
that-- what does it mean,
• 41:32 - 41:34
parameterizing arc length--
• 41:34 - 41:35
STUDENT: r prime of s.
• 41:35 - 41:37
PROFESSOR: r primed of
s in magnitude is 1.
• 41:37 - 41:38
The speed 1.
• 41:38 - 41:39
It's a speed 1 curve.
• 41:39 - 41:43
• 41:43 - 42:01
Then, the curvature of this
curve is defined as k of s
• 42:01 - 42:06
equals the magnitude of
the acceleration vector
• 42:06 - 42:09
will respect the S.
Say what, Magdalena?
• 42:09 - 42:13
I can also write
it magnitude of d--
• 42:13 - 42:17
oh my gosh, second derivative
with respect s of r.
• 42:17 - 42:20
I'll do it right now. d2r ds2.
• 42:20 - 42:22
And I know you get
• 42:22 - 42:26
I solve, when I write that,
because you are not used to it.
• 42:26 - 42:33
A quick and beautiful example
that can be on the homework,
• 42:33 - 42:39
and would also be on the
exam, maybe on all the exams,
• 42:39 - 42:42
I don't know.
• 42:42 - 42:48
Compute the curvature of a
circle of radius a Say what?
• 42:48 - 43:03
Compute the curvature of a
circle of radius a And you say,
• 43:03 - 43:04
wait a minute.
• 43:04 - 43:07
For a circle of
radius a in plane--
• 43:07 - 43:09
why can I assume it's in plane?
• 43:09 - 43:13
Because if the circle
is a planar curve,
• 43:13 - 43:16
I can always assume
it to be in plane.
• 43:16 - 43:19
And it has radius a I
can find infinitely many
• 43:19 - 43:20
parameterizations.
• 43:20 - 43:23
So what, am I crazy?
• 43:23 - 43:25
Well, yes, I am, but
that's another story.
• 43:25 - 43:28
Now, if I want to
parameterize, I
• 43:28 - 43:31
have to parameterize
in arc length.
• 43:31 - 43:34
If I do anything else,
that means I'm stupid.
• 43:34 - 43:39
So, r of s will be what?
• 43:39 - 43:41
Can somebody tell me
how I parameterize
• 43:41 - 43:46
a curve in arc length
for a-- what is this guy?
• 43:46 - 43:49
A circle of radius a.
• 43:49 - 43:51
Yeah, I cannot do it.
• 43:51 - 43:52
I'm not smart enough.
• 43:52 - 43:59
So I'll say R of T will be
a cosine t, a sine t and 0.
• 43:59 - 44:03
And here I stop, because
• 44:03 - 44:09
t is from 0 to 2 pi, and
I think this a is making
• 44:09 - 44:15
my life miserable,
because it's telling me,
• 44:15 - 44:17
you don't have
speed 1, Magdalena.
• 44:17 - 44:19
Drive to Amarillo
and back, you're
• 44:19 - 44:22
not going to get speed 1.
• 44:22 - 44:23
Why don't I have speed 1?
• 44:23 - 44:24
• 44:24 - 44:25
Bear with me.
• 44:25 - 44:29
Minus a sine t equals sine t, 0.
• 44:29 - 44:29
• 44:29 - 44:31
What is the speed?
• 44:31 - 44:33
a.
• 44:33 - 44:36
If you do the math,
the speed will be a.
• 44:36 - 44:40
So length of our
prime of t will be a.
• 44:40 - 44:41
Somebody help me.
• 44:41 - 44:42
Get me out of trouble.
• 44:42 - 44:43
Who is this?
• 44:43 - 44:45
I want to do it in arc length.
• 44:45 - 44:48
Otherwise, how can
I do the curvature?
• 44:48 - 44:51
So somebody tell
me how to get to s.
• 44:51 - 44:52
What the heck is that?
• 44:52 - 44:59
s of t is integral from
0 to t of-- who tells me?
• 44:59 - 45:00
The speed, right?
• 45:00 - 45:04
Was it not the displacement,
the arc length traveled along,
• 45:04 - 45:08
and the curve is integral
in time of the speed.
• 45:08 - 45:12
• 45:12 - 45:13
OK?
• 45:13 - 45:16
So I have-- what is that?
• 45:16 - 45:18
Speed is?
• 45:18 - 45:18
STUDENT: Um--
• 45:18 - 45:19
PROFESSOR: a.
• 45:19 - 45:23
So a time t, am I right,
guys? s is a times t.
• 45:23 - 45:25
So what do I have to do?
• 45:25 - 45:31
Take Mr. t, shake his hand,
and replace him with s over a.
• 45:31 - 45:32
OK.
• 45:32 - 45:40
So instead of r of t, I'll say--
what other letters do I have?
• 45:40 - 45:40
Not r.
• 45:40 - 45:41
Rho of s.
• 45:41 - 45:42
I love rho.
• 45:42 - 45:44
Rho is the Greek [INAUDIBLE].
• 45:44 - 45:46
Is this finally an arc length?
• 45:46 - 45:51
Cosine of-- what
is t, guys, again?
• 45:51 - 45:53
s over a.
• 45:53 - 45:58
s over a, a sine
s over a, and 0.
• 45:58 - 46:02
This is the parameterization
in arc length.
• 46:02 - 46:08
This is an arc length
parameterization of the circle.
• 46:08 - 46:11
And then what is this
definition of curvature?
• 46:11 - 46:14
It's here.
• 46:14 - 46:17
Do that rho once, twice.
• 46:17 - 46:20
Prime it twice,
and do the length.
• 46:20 - 46:21
So rho prime.
• 46:21 - 46:25
Oh my God is it hard.
• 46:25 - 46:29
a times minus sine of s over a.
• 46:29 - 46:30
Am I done, though?
• 46:30 - 46:31
Chain rule.
• 46:31 - 46:32
Pay attention, Magdalena.
• 46:32 - 46:34
Don't screwed up with this one.
• 46:34 - 46:36
1 over a.
• 46:36 - 46:38
Good.
• 46:38 - 46:39
Next.
• 46:39 - 46:42
a cosine of s over a.
• 46:42 - 46:43
Chain rule.
• 46:43 - 46:44
Don't forget,
multiply by 1 over a.
• 46:44 - 46:47
OK, that makes my life easier.
• 46:47 - 46:48
We simplify.
• 46:48 - 46:52
Thank God a simplifies
here, a simplifies there,
• 46:52 - 46:54
so that is that derivative.
• 46:54 - 46:56
What's the second derivative?
• 46:56 - 47:01
Rho double prime of s will
be-- somebody help me, OK?
• 47:01 - 47:03
Because this is a
lot of derivation.
• 47:03 - 47:03
STUDENT: --cosine--
• 47:03 - 47:05
PROFESSOR: Thank you, sir.
• 47:05 - 47:07
Minus cosine of s over a.
• 47:07 - 47:08
STUDENT: Times 1 over a.
• 47:08 - 47:13
PROFESSOR: Times 1 over a,
comma, minus sine of s over a.
• 47:13 - 47:16
That's all I have left
in my life, right?
• 47:16 - 47:20
Minus sine of s over a times
1 over a from the chain rule.
• 47:20 - 47:23
I have to pay attention and see.
• 47:23 - 47:24
What's the magnitude of this?
• 47:24 - 47:29
The magnitude of this of this
animal will be the curvature.
• 47:29 - 47:30
Oh, my God.
• 47:30 - 47:32
So what is k?
• 47:32 - 47:35
k of s will be--
could somebody tell me
• 47:35 - 47:40
what magnitude I get after I
square all these individuals,
• 47:40 - 47:43
sum them up, and take
the square root of them?
• 47:43 - 47:44
STUDENT: [INAUDIBLE]
• 47:44 - 47:50
PROFESSOR: Square root
of 1 over 1 squared.
• 47:50 - 47:51
And I get 1 over a.
• 47:51 - 47:53
You are too fast for
me, you teach me that.
• 47:53 - 47:54
No, I'm just kidding.
• 47:54 - 47:56
I knew it was 1 over a.
• 47:56 - 47:59
Now, how did
engineers know that?
• 47:59 - 48:02
Actually, for hundreds of years,
mathematicians, engineers,
• 48:02 - 48:04
and physicists knew that.
• 48:04 - 48:07
And that's the last thing
I want to teach you today.
• 48:07 - 48:10
We have two circles.
• 48:10 - 48:17
This is of, let's say, radius
1/2, and this is radius 2.
• 48:17 - 48:21
The engineer, mathematician,
physicist, whoever they are,
• 48:21 - 48:26
they knew that the curvature
is inverse proportional
• 48:26 - 48:28
• 48:28 - 48:30
That radius is 1/2.
• 48:30 - 48:34
The curvature will
be 2 in this case.
• 48:34 - 48:38
The radius is 2, the
curvature will be 1/2.
• 48:38 - 48:42
Does that make sense, this
inverse proportionality?
• 48:42 - 48:46
The bigger the radius,
the lesser the curvature,
• 48:46 - 48:48
that less bent you are.
• 48:48 - 48:50
The more fat-- well, OK.
• 48:50 - 48:53
I'm not going to say anything
politically incorrect.
• 48:53 - 48:59
So this is really curved because
the radius is really small.
• 48:59 - 49:03
This less curved,
almost-- at infinity,
• 49:03 - 49:06
this curvature
becomes 0, because
• 49:06 - 49:09
at infinity, that radius
explodes to plus infinity bag
• 49:09 - 49:10
theory.
• 49:10 - 49:14
Then you have 1 over
infinity will be 0,
• 49:14 - 49:19
and that will be the curvature
of a circle of infinite radius.
• 49:19 - 49:20
Right?
• 49:20 - 49:23
So we learned something today.
• 49:23 - 49:25
We learned about the
curvature of a circle, which
• 49:25 - 49:26
is something.
• 49:26 - 49:31
But this is the same
way for any curve.
• 49:31 - 49:32
You reparameterize.
• 49:32 - 49:34
Now you understand why you need
to reparameterize in arc length
• 49:34 - 49:36
s.
• 49:36 - 49:38
You take the acceleration
in arc length.
• 49:38 - 49:39
You get the magnitude.
• 49:39 - 49:42
That measures how
bent the curve is.
• 49:42 - 49:47
Next time, you're going to
do how bent the helix is.
• 49:47 - 49:48
OK?
• 49:48 - 49:49
At every point.
• 49:49 - 49:51
Enjoy your WeBWorK homework.
• 49:51 - 49:55
Ask me anytime, and
ask me also Thursday.
• 49:55 - 49:59
Do not have a block about
• 49:59 - 50:05
You can ask me anytime
by email, or in person.
• 50:05 - 50:10
Title:
TTU Math2450 Calculus3 Sec 10.2 and 10.4 part 1
Description:

Derivatives of Vector Value Functions

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Video Language:
English
Duration:
50:11
 jackie.luft edited English subtitles for TTU Math2450 Calculus3 Sec 10.2 and 10.4 part 1