## TTU Math2450 Calculus3 Sec 11.5 and 11.6

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MAGDALENA TODA: So what's
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STUDENT: It's OK.
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MAGDALENA TODA: It's OK.
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So functions of
two variables are
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to be compared all the time with
the functions of one variable.
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Every nothing you
have seen in Calc 1
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has a corresponding
the motion in Calc 3.
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theory, concepts, Chapter 11
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concepts, previous concepts?
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Feel free to email
me this weekend.
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Don't think it's the
weekend because we
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People, we use WeBWork.
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Not just me, but
everybody who uses
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WeBWork is on a
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Saturday and Sunday is when
most of you do the homework.
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It's convenient for us as well
because we are with the family,
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but we don't have many
meetings to attend.
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So I'll be happy to
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Last time, we discussed a
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for The Chain Rule.
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In Calc 3.
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So the chain rule in Calc
3 was something really--
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this is section 11.5.
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The preparation
was done last time,
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but I'm going to
review it a little bit.
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Let's see what we discussed.
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I'm going to split,
again, the board in two.
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And I'll say, can we review
the notions of The Chain Rule.
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variable-- let's say it's time.
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Time going to f of t, which goes
into g of f of t by something
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called composition.
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We've done that since we
were kids in college algebra.
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What?
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You never took college algebra?
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Except in high school, you
took high school algebra,
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most of you.
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So what did you do in
high school algebra?
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We said g composed with l.
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This is a composition
of two functions.
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What I'm skipping here is the
theory that you learned then
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that to a compose well, F of t
has to be in the domain of g.
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So the image F of t, whatever
you get from this image,
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has to be in the domain of g.
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Otherwise, the composition
could not exist.
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Now if you have
differentiability,
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assuming that this
is g composed with F,
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assuming to be c1-- c1
meaning differentiable
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and derivatives are continuous--
assuming both of them are c1,
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they compose well.
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What am I going to do next?
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I'm going to say the
d, dt g of F of t.
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And we said last time,
we get The Chain Rule
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from the last function
we applied, g prime.
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And so you have dg,
[? d2 ?] at F of t.
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I'm calling this guy u variable
just for my own enjoyment.
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And then I go du, dt.
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But du, dt would be
nothing but a prime of t,
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so remember the cowboys
shooting at each other?
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The du and du.
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I will replace the u by prime of
t, just like you did in Calc 1.
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Why?
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Because I want to a mixture of
notations according to Calc 1
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you took here.
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The idea for Calc 3 is the
same with [INAUDIBLE] time,
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assuming everything
composes well,
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and has differentiability, and
the derivatives are continuous.
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Just to make your life easier.
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We have x of t, y of t.
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Two nice functions
and a function
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of these variables,
F of x and y.
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So I'm going to have
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is a function of t and
y is a function of t?
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So I should be able to go
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with respect to the t.
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And how did it go?
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Now that I prepared
you last time,
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a little bit, for this kind
of new picture, new diagram,
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you should be able to tell me,
without looking at the notes
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from last time, how this goes.
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So I'll take the function
F of x of t, y of t.
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And when I view it like that,
I understand it's ultimately
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a big function, F of t.
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It's a real valued
function of t,
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ultimately, as the composition.
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This big F.
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So does anybody
remember how this went?
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Let's see.
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The derivative,
with respect to t,
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of this whole thing,
F of x of t, y of t?
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Thoughts?
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The partial derivative
of F with respect
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to x, evaluated at
x of t and y to t.
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So everything has to be
replaced in terms of t
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because it's going to be y.
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We assume that this derivative
exists and it's continuous.
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Why?
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a little bit easier.
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From the beginning,
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which was also
defined everywhere
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and continuous, plus df, 2y at
the same point times dy, dt.
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Notice what happens here with
these guys looking diagonally,
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staring at each other.
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Keep in mind the plus sign.
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And of course, some of you
told me, well, is that OK?
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You know favorite, right?
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F of x at x of dy of t.
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That's fine.
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I saw that.
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In engineering you use it.
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Physics majors also use
a lot of this notation
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as sub [INAUDIBLE] Fs of t.
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We've seen that.
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We've seen that.
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It comes as no
surprise to us, but we
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would like to see if there
are any other cases we
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Now I don't want to
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until I give you
something that you
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know very well from Calculus 1.
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It's an example that you saw
before that was a melting ice
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sphere.
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It appears a lot in problems,
like final exam problems
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and stuff.
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What is the material
of this ball?
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It's melting ice.
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And if you remember, it
says that at the moment t0,
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assume the radius was 5 inches.
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We also know that the rate of
change of the radius in time
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will be minus 5.
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But let's suppose that we say
that inches per-- meaning,
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it's really hot in the room.
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Not this room, but
the hypothetic room
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where the ice ball is melting.
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So imagine, in 1
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will go down by 5 inches.
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Yes, it must be really hot.
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I want to know the derivative,
dv, dt at the time 0.
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So you go, oh my god, I don't
remember doing this, actually.
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It is a Calc 1 type of problem.
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Why am I even
discussing it again?
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Because I want to fool you a
little bit into remembering
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the elementary formulas for
the volume of a sphere, volume
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of a cone, volume of a cylinder.
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That was a long time ago.
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When you ask you teachers in
K12 if you should memorize them,
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they said, by all
means, memorize them.
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That was elementary geometry,
but some of you know them
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by heart, some of you don't.
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Do you remember
the volume formula
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for a ball with radius r?
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[INTERPOSING VOICES]
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What?
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[? STUDENT: High RQ. ?]
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STUDENT: 4/3rds.
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MAGDALENA TODA: Good.
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I'm proud of you guys.
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I've discovered lots of people
who are engineering majors
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and they don't
know this formula.
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So how are we going to
think of this problem?
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We have to think, Chain Rule.
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And Chain Rule means that you
view this radius as a shrinking
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thing because
that's why you have
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it's decreasing,
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so you view r as
a function of t.
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And of course, you
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And then v will be a function
of t ultimately, but you see,
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guys, t goes to r of t,
r of t goes to v of t.
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What's the formula
for this function?
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v equals 4 pi i cubed over 3.
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So this is how the diagram goes.
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You look at that composition
and you have dv, dt.
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And I remember teaching as
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was a long time ago,
in '97 or something,
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with this kind of diagram with
compositions of functions.
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me, nobody showed us
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this kind of diagram before.
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Well, I do.
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I think they are very
useful for understanding
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how a composition will go.
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Now I would just going ahead and
say v prime because I'm lazy.
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And I go v prime of t is 0.
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Meaning, that this
is the dv, dt at t0.
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And somebody has to help me
remember how we did The Chain
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Rule in Calc 1.
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It was ages ago.
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4 pi over 3 constant times.
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Who jumps down?
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The 3 jumps down and he's
very happy to do that.
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3, r squared.
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But r squared is not an
independent variable.
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He or she depends on t.
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So I'll be very happy to
say 3 times that times.
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And that's the essential part.
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I'm not done.
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STUDENT: It's dr over dt.
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MAGDALENA TODA: dr, dt.
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So I have finally
applied The Chain Rule.
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And how do I plug
in the data in order
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to get this as the final answer?
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I just go 4 pi
over 3 times what?
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3 times r-- who is
r at the time to 0,
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where I want to view
the whole situation?
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r squared at time
to 0 would be 25.
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Are you guys with me?
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dr, dt at time to
0 is negative 5.
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All right.
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I'm done.
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So you are going to ask me,
if I'm taking the examine,
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do I need this in
the exam like that?
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Easy.
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Oh, it depends on the exam.
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If you have a multiple choice
where this is simplified,
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obviously, it's not the right
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but I will accept
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numerical part very much.
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If you want to do more, 4
times 25 is hundred times 5.
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So I have minus what?
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STUDENT: 500 pi.
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MAGDALENA TODA: 500 pi.
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How do we get the unit of that?
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I'm wondering.
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STUDENT: Cubic
inches per minute.
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MAGDALENA TODA: Cubic
inches per minute.
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Very good.
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Cubic inches per minute.
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Why don't I write it down?
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Because I couldn't care less.
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I'm a mathematician.
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If I were a physicist, I would
definitely write it down.
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And he was right.
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Now you are going
to find this weird.
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Why is she doing this review
of this kind of melting ice
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problem from Calc 1?
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Because today I'm
being sneaky and mean.
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And I want to give
you a little challenge
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for 1 point of extra credit.
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You will have to compose
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in Calculus 3,
that is like that.
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So you have to compose a problem
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of ice.
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Say what, Magdalena?
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OK.
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So I'll write it down.
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that's melting in time.
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Do you have to solve
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Yes, I guess so.
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Once you compose
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For extra credit, 1 point.
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Compose, write, and solve--
you are the problem author.
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Write and solve
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so that the story includes--
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STUDENT: A solid cylinder.
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MAGDALENA TODA: Yes.
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nice ball, a solid cylinder.
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And necessarily, you cannot
write it just a story--
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I once had an ice cylinder,
and it was melting,
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and I went to watch a movie,
and by the time I came back,
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it was all melted.
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That's not what I want.
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I want it so that the problem
is an example of applying
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The Chain Rule in Calc 3.
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And I won't say more.
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So maybe somebody
can help with a hint.
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Maybe I shouldn't
give too many hits,
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but let's talk as if we
were chatting in a cafe,
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without me writing
too much down.
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Of course, you can take
notes of our discussion,
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but I don't want
have it documented.
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So we have a cylinder right.
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There is the cylinder.
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So there's the cylinder.
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and it's melting.
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And the volume should be a
function of two variables
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because otherwise, you
don't have it in Calc 3.
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So a function of two variables.
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What other two variables
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and the height.
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would be one of them.
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You don't have to say x and y.
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This is r and h.
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So h and r are in that formula.
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I'm not going to
say which formula,
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you guys should know of
the volume of the cylinder.
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But both h and r, what do they
have in common in the story?
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STUDENT: Time.
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MAGDALENA TODA: They are
both functions of time.
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They are melting in time.
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a quick question?
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MAGDALENA TODA: Yes, sir.
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STUDENT: What if we solve
for-- what is the negative 500
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[? path? ?]
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MAGDALENA TODA: This is the
speed with which the volume is
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shrinking at time to 0.
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So the rate of change of
the volume at time to o.
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And this is
something-- by the way,
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that's how I would
like you to state it.
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Find the rate of change
of the volume of the ice--
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wasn't that a good cylinder?
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At time to 0, if you
know that at time to 0
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something happened.
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Maybe r is given, h is given.
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The derivatives are given.
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You only have one
derivative given here,
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which was our
prime of t minus 5.
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Now I leave it to you.
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I ask it to you, and I'll leave
it to you, and don't tell me.
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When we have a
piece of ice-- well,
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there was something in the
news, but I'm not going to say.
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There was some nice, ice
sculpture in the news there.
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So do the dimensions decrease
at the same rate, do you think?
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I mean, I don't know.
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It's all up to you.
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Think of a case when the
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would shrink at the same speed.
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And think of a case when
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of ice would not
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change at the same
rate for some reason.
• 20:39 - 20:41
I don't know, but
the simplest case
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would be to assume that
all of the dimensions
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shrink at the same speed,
at the same rate of change.
• 20:50 - 20:52
So you write your own problem,
you make up your own data.
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Now you will appreciate
how much work people
• 20:55 - 20:57
put into that work book.
• 20:57 - 21:01
I mean, if there is a bug,
it's one in a thousand,
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but for a programmer to be
able to write those problems,
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he has to know calculus,
he has to know C++ or Java,
• 21:09 - 21:13
he has to be good-- that's
not a problem, right?
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STUDENT: No.
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That's fine.
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MAGDALENA TODA: He or she has
to know how to write a problem,
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so that you guys,
no matter how you
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is correct, you'll get the OK.
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in many equivalent forms
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and all of them have to be--
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MAGDALENA TODA: Yes.
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So since I have new
people who just came--
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And I understand you guys
come from different buildings
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and I'm not mad for people who
are coming late because I know
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you come from other
classes, I wanted
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to say we started from a melting
ice sphere example in Calc 1
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that was on many finals
in here, at Texas Tech.
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And I want you to compose your
own problem based on that.
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This time, involving
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of ice whose dimensions are
doing something special.
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That shouldn't be hard.
• 22:26 - 22:30
I'm going to erase this
part because it's not
• 22:30 - 22:31
the relevant one.
• 22:31 - 22:33
I'm going to keep this
one a little bit more
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for people who
want to take notes.
• 22:36 - 22:37
And I'm going to move on.
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Another example we
give you in the book
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is that one where x and y,
the variables the function f,
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are not just
functions of time, t.
• 22:59 - 23:04
They, themselves, are functions
of other two variables.
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Is that a lot more different
from what I gave you already?
• 23:08 - 23:08
No.
• 23:08 - 23:11
The idea is the same.
• 23:11 - 23:14
And you are imaginative.
• 23:14 - 23:20
You are able to come up
• 23:20 - 23:27
I'm going to ask you to think
about what I'll have to write.
• 23:27 - 23:28
This is finished.
• 23:28 - 23:32
• 23:32 - 23:38
So assume that you have
function z equals F of x,y.
• 23:38 - 23:42
• 23:42 - 23:49
this is example 2
• 23:49 - 23:58
where x is a function
of u and v itself.
• 23:58 - 24:03
And y is a function
of u and v itself.
• 24:03 - 24:07
And we assume that all
the partial derivatives
• 24:07 - 24:10
are defined and continuous.
• 24:10 - 24:12
And we make the
problem really nice.
• 24:12 - 24:21
And now we'll come
up with some example
• 24:21 - 24:39
you know from before where
x equals x of uv equals uv.
• 24:39 - 24:49
And y equals y of
uv equals u plus v.
• 24:49 - 24:52
So these functions are
the sum and the product
• 24:52 - 24:53
of other variables.
• 24:53 - 24:56
• 24:56 - 25:07
Can you tell me how I am going
to compute the derivative of 0,
• 25:07 - 25:17
or of f, with the script
of u at x of uv, y of uv?
• 25:17 - 25:19
Is this hard?
• 25:19 - 25:20
STUDENT: It is.
• 25:20 - 25:21
MAGDALENA TODA: I don't know.
• 25:21 - 25:28
You have to help me because--
why don't I put d here?
• 25:28 - 25:29
STUDENT: Because [INAUDIBLE].
• 25:29 - 25:31
MAGDALENA TODA:
Because you have 2.
• 25:31 - 25:33
So the composition
in itself will
• 25:33 - 25:36
be a function of two variables.
• 25:36 - 25:39
So of course, I
have [INAUDIBLE].
• 25:39 - 25:48
I'm going to go ahead and do
it as you say without rushing.
• 25:48 - 25:51
Of course, I know
you are watching.
• 25:51 - 25:53
What will happen?
• 25:53 - 25:54
STUDENT: 2x and 2y.
• 25:54 - 25:55
MAGDALENA TODA: No, in general.
• 25:55 - 25:58
Over here, I know you
want to do it right away,
• 25:58 - 26:02
but I would like you to give
me a general formula mimicking
• 26:02 - 26:07
the same thing you had before
when you had one parameter, t.
• 26:07 - 26:08
Now you have u and d separately.
• 26:08 - 26:10
You want it to do it straight.
• 26:10 - 26:19
So we have df, dx
at x of uv, y of uv.
• 26:19 - 26:21
Shut up, Magdalene.
• 26:21 - 26:24
Let people talk and help
you because you're tired.
• 26:24 - 26:27
It's a Thursday.
• 26:27 - 26:28
df, dx.
• 26:28 - 26:29
STUDENT: [INAUDIBLE].
• 26:29 - 26:30
MAGDALENA TODA: dx.
• 26:30 - 26:34
Again, [INAUDIBLE] notation,
partial with respect
• 26:34 - 26:43
to u, plus df, dy.
• 26:43 - 26:46
So the second argument--
so I prime in respect
• 26:46 - 26:51
to the second argument,
computing everything
• 26:51 - 26:55
in the end, which means
in terms of u and v times,
• 26:55 - 27:00
again, the dy with respect to u.
• 27:00 - 27:02
You are saying that.
• 27:02 - 27:04
Now I'd like you
to see the pattern.
• 27:04 - 27:06
Of course, you see the
pattern here, smart people,
• 27:06 - 27:12
but I want to
emphasize the cowboys.
• 27:12 - 27:15
And green for the other cowboy.
• 27:15 - 27:17
I'm trying to match the
college beautifully.
• 27:17 - 27:23
• 27:23 - 27:26
And the independent
variable, Mr. u.
• 27:26 - 27:28
Not u, but Mr. u.
• 27:28 - 27:29
Yes, ma'am?
• 27:29 - 27:31
STUDENT: Is it the
partial of dx, du?
• 27:31 - 27:38
Or is it-- like you did
the partial for the--
• 27:38 - 27:40
MAGDALENA TODA: So did
I do anything wrong?
• 27:40 - 27:42
I don't think I
did anything wrong.
• 27:42 - 27:44
STUDENT: So it is the
partial for dx over du?
• 27:44 - 27:49
MAGDALENA TODA: So I go du with
respect to the first variable,
• 27:49 - 27:51
times that variable
with respect to u.
• 27:51 - 27:52
STUDENT: But is it the partial?
• 27:52 - 27:54
That's my question.
• 27:54 - 27:57
MAGDALENA TODA: But it has
to be a partial because x is
• 27:57 - 28:03
a function of u and
v, so I cannot put d.
• 28:03 - 28:07
And then the same plus
the same idea as before.
• 28:07 - 28:10
df with respect to
the second argument
• 28:10 - 28:16
times that second argument
with respect to the u.
• 28:16 - 28:20
You see, Mr. u is
replacing Mr. t.
• 28:20 - 28:21
He is independent.
• 28:21 - 28:24
• 28:24 - 28:27
He's the guy who is moving.
• 28:27 - 28:29
We don't care
• 28:29 - 28:32
but he replaces the time
in this kind of problem.
• 28:32 - 28:36
• 28:36 - 28:39
Now the other one.
• 28:39 - 28:41
I will let you speak.
• 28:41 - 28:44
Df, dv.
• 28:44 - 28:50
The same idea, but somebody
else is going to talk.
• 28:50 - 28:53
STUDENT: It would
be del f, del y.
• 28:53 - 28:54
MAGDALENA TODA: Del f, del x?
• 28:54 - 28:57
Well, let's try
to start in order.
• 28:57 - 29:02
• 29:02 - 29:05
And I tried to be
organized and write neatly
• 29:05 - 29:13
because I looked at-- so these
videos are new and in progress.
• 29:13 - 29:17
And I'm trying to see what
I did well and I didn't.
• 29:17 - 29:18
And at times, I wrote neatly.
• 29:18 - 29:21
At times, I wrote not so neatly.
• 29:21 - 29:23
• 29:23 - 29:28
It's one thing, what you think
• 29:28 - 29:31
and to you see yourself
the way other people
• 29:31 - 29:33
see from the outside.
• 29:33 - 29:34
It's not fun.
• 29:34 - 29:36
STUDENT: Can you say it again?
• 29:36 - 29:42
MAGDALENA TODA: This is
v. So I'll say it again.
• 29:42 - 29:46
We all have a certain
• 29:46 - 29:49
but when you see a
movie of yourself,
• 29:49 - 29:52
you see the way
other people see you.
• 29:52 - 29:53
And it's not fun.
• 29:53 - 29:56
STUDENT: So what--
• 29:56 - 29:59
MAGDALENA TODA: So
let's see the cowboys.
• 29:59 - 30:06
Ryan is looking at the
[? man. ?] He is all [? man. ?]
• 30:06 - 30:10
And y is here, right?
• 30:10 - 30:16
And who is the time
variable, kind of, this time?
• 30:16 - 30:18
This time, which
one is the time?
• 30:18 - 30:28
v. And v is the only ultimate
• 30:28 - 30:32
So everything you did
before with respect to t,
• 30:32 - 30:35
you do now with
respect to u, you
• 30:35 - 30:37
do now with respect
to v. It shouldn't
• 30:37 - 30:39
be hard to understand.
• 30:39 - 30:42
I want to work the
example, of course.
• 30:42 - 30:44
With your help, I will work it.
• 30:44 - 30:49
Now remember how my students
cheated on this one?
• 30:49 - 30:57
So I told my colleague, he did
not say, five or six years ago,
• 30:57 - 31:01
by first writing The Chain Rule
for functions of two variables,
• 31:01 - 31:08
express all the df, du, df,
dv, but he said by any method.
• 31:08 - 31:13
Of course, what they
did-- they were sneaky.
• 31:13 - 31:16
They took something
like x equals uv
• 31:16 - 31:18
and they plugged it in here.
• 31:18 - 31:20
They took the function
[? u and v, ?]
• 31:20 - 31:21
they plugged it in here.
• 31:21 - 31:23
They computed everything
in terms of u and v
• 31:23 - 31:24
and took the partials.
• 31:24 - 31:27
STUDENT: Why don't
you [INAUDIBLE]?
• 31:27 - 31:29
MAGDALENA TODA: It depends
how the problem is formulated.
• 31:29 - 31:30
STUDENT: So if you
make it [INAUDIBLE],
• 31:30 - 31:32
then it's [INAUDIBLE].
• 31:32 - 31:36
• 31:36 - 31:39
MAGDALENA TODA: So when they
give you the precise functions,
• 31:39 - 31:40
you're right.
• 31:40 - 31:42
But if they don't give
you those functions,
• 31:42 - 31:44
if they keep them a
secret, then you still
• 31:44 - 31:47
have to write the
general formula.
• 31:47 - 31:51
If they don't give you
the functions, all of them
• 31:51 - 31:54
explicitly.
• 31:54 - 31:57
So let's see what
to do in this case.
• 31:57 - 32:09
df, du at x of u, vy
of uv will be what?
• 32:09 - 32:12
• 32:12 - 32:16
• 32:16 - 32:22
I want to teach you how
engineers and physicists very,
• 32:22 - 32:27
very often express
those at x and y.
• 32:27 - 32:29
And many of you know
because we talked
• 32:29 - 32:31
• 32:31 - 32:37
2x, I might write,
but evaluated at--
• 32:37 - 32:38
and this is a very
frequent notation
• 32:38 - 32:42
image in the engineering
and physicist world.
• 32:42 - 32:45
So 2x evaluated at where?
• 32:45 - 32:52
At the point where x is
uv and y is u plus v.
• 32:52 - 33:00
So I say x of uv, y of uv.
• 33:00 - 33:05
And I'll replace later
because I'm not in a hurry.
• 33:05 - 33:07
dx, du.
• 33:07 - 33:09
Who is dx, du?
• 33:09 - 33:11
The derivative of x
or with respect to u?
• 33:11 - 33:13
Are you guys awake?
• 33:13 - 33:13
STUDENT: Yes.
• 33:13 - 33:15
So it's v.
• 33:15 - 33:19
MAGDALENA TODA: v. Very good.
v plus-- the next term, who's
• 33:19 - 33:23
going to tell me what we have?
• 33:23 - 33:24
STUDENT: 2y evaluated at--
• 33:24 - 33:28
MAGDALENA TODA: 2y evaluated
at-- look how lazy I am.
• 33:28 - 33:38
Times the derivative
of y with respect to u.
• 33:38 - 33:40
So you were right because of 2y.
• 33:40 - 33:43
• 33:43 - 33:44
Attention, right?
• 33:44 - 33:48
So it's dy, du is 1.
• 33:48 - 33:50
It's very easy to
make a mistake.
• 33:50 - 33:52
• 33:52 - 33:56
from just miscalculating
because when
• 33:56 - 33:58
you are close to
some formula, you
• 33:58 - 34:00
don't see the whole picture.
• 34:00 - 34:01
What do you do?
• 34:01 - 34:05
At the end of your
exams, go back and rather
• 34:05 - 34:09
than quickly turning in
a paper, never do that,
• 34:09 - 34:12
go back and check
• 34:12 - 34:13
It's a good habit.
• 34:13 - 34:21
2 times x, which is uv, I plug
it as a function of u and v,
• 34:21 - 34:23
right?
• 34:23 - 34:27
Times a v plus-- who is 2y?
• 34:27 - 34:29
That's the last of the Mohicans.
• 34:29 - 34:30
One is out.
• 34:30 - 34:31
STUDENT: 2.
• 34:31 - 34:37
MAGDALENA TODA: 2y 2 times
replace y in terms of u and v.
• 34:37 - 34:38
And you're done.
• 34:38 - 34:40
So do you like it?
• 34:40 - 34:42
I don't.
• 34:42 - 34:44
And how would you write it?
• 34:44 - 34:49
Not much better than that,
but at least let's try.
• 34:49 - 34:53
2uv squared plus 2u plus 2v.
• 34:53 - 34:55
You can do a little
bit more than that,
• 34:55 - 35:01
but if you want to list it
in the order of the degrees
• 35:01 - 35:04
of the polynomials, that's OK.
• 35:04 - 35:06
Now next one.
• 35:06 - 35:10
df, dv, x of uv, y of uv.
• 35:10 - 35:13
• 35:13 - 35:16
Such examples are in the book.
• 35:16 - 35:18
Many things are in the
book and out of the book.
• 35:18 - 35:21
I mean, on the white board.
• 35:21 - 35:25
I don't know why it gives you so
many combinations of this type,
• 35:25 - 35:31
u plus v, u minus-- 2u
plus 2v, 2u you minus 2v.
• 35:31 - 35:32
Well, I know why.
• 35:32 - 35:35
Because that's a
rotation and rescaling.
• 35:35 - 35:37
So there is a
reason behind that,
• 35:37 - 35:42
but I thought of something
different for df, dv.
• 35:42 - 35:45
Now what do I do?
• 35:45 - 35:45
df, dx.
• 35:45 - 35:46
STUDENT: You [? have to find
something symmetrical to that.
• 35:46 - 35:47
?]
• 35:47 - 35:48
MAGDALENA TODA:
Again, the same thing.
• 35:48 - 35:53
2x evaluated at whoever times--
• 35:53 - 35:53
STUDENT: u.
• 35:53 - 35:56
• 35:56 - 35:58
MAGDALENA TODA: Because you
have dx with respect to v,
• 35:58 - 36:02
so you have u plus--
• 36:02 - 36:04
STUDENT: df, dy.
• 36:04 - 36:07
MAGDALENA TODA: df, dy,
which is 2y, evaluated
• 36:07 - 36:10
at the same kind of guy.
• 36:10 - 36:13
So all you have to do is
replace with respect to u and v.
• 36:13 - 36:16
And finally, multiplied by-
• 36:16 - 36:17
STUDENT: dy.
• 36:17 - 36:18
MAGDALENA TODA: dy, dv.
• 36:18 - 36:22
dy, dv is 1 again.
• 36:22 - 36:24
Just pay attention
when you plug in
• 36:24 - 36:26
because you realize you
can know these very well
• 36:26 - 36:29
and understand it as a process,
but if you make an algebra
• 36:29 - 36:31
and everything is out.
• 36:31 - 36:33
And then you send me
an email that says,
• 36:33 - 36:35
I've tried this
problem 15 times.
• 36:35 - 36:38
And I don't even hold
you responsible for that
• 36:38 - 36:42
because I can make
algebra mistakes anytime.
• 36:42 - 36:54
So 2uv times u plus 2 times u
plus v. So what did I do here?
• 36:54 - 37:01
I simply replaced the given
functions in terms of u and v.
• 37:01 - 37:03
And I'm done.
• 37:03 - 37:04
Do I like it?
• 37:04 - 37:09
No, but I'd like you to notice
something as soon as I'm done.
• 37:09 - 37:12
2u squared v plus 2u plus 2v.
• 37:12 - 37:18
• 37:18 - 37:20
Could I have expected that?
• 37:20 - 37:22
Look at the beauty
of the functions.
• 37:22 - 37:24
• 37:24 - 37:28
Z is a symmetric function.
• 37:28 - 37:32
x and y have some of
the symmetry as well.
• 37:32 - 37:35
If you swap u and v, these
are symmetric polynomials
• 37:35 - 37:39
of order 2 and 1.
• 37:39 - 37:40
[INAUDIBLE]
• 37:40 - 37:43
Swap the variables, you
still get the same thing.
• 37:43 - 37:46
Swap the variables u and
v, you get the same thing.
• 37:46 - 37:48
So how could I have
imagined that I'm
• 37:48 - 37:54
going to get-- if I were smart,
without doing all the work,
• 37:54 - 37:58
I could figure out
this by just swapping
• 37:58 - 38:02
the u and v, the rows of u
and v. I would have said,
• 38:02 - 38:08
2vu squared, dv plus 2u and
it's the same thing I got here.
• 38:08 - 38:14
But not always are you so
lucky to be given nice data.
• 38:14 - 38:16
Well, in real life, it's a mess.
• 38:16 - 38:22
If you are, let's say, working
with geophysics real data,
• 38:22 - 38:28
you two parameters and for
each parameter, x and y,
• 38:28 - 38:29
you have other parameters.
• 38:29 - 38:32
You will never have
anything that nice.
• 38:32 - 38:36
You may have nasty
truncations of polynomials
• 38:36 - 38:39
with many, many
terms that you work
• 38:39 - 38:41
with approximating polynomials
all the time. [INAUDIBLE]
• 38:41 - 38:43
or something like that.
• 38:43 - 38:47
So don't expect these miracles
to happen with real data,
• 38:47 - 38:50
but the process is the same.
• 38:50 - 38:53
And, of course,
there are programs
• 38:53 - 38:56
that incorporate all of
the Calculus 3 notions
• 38:56 - 38:59
that we went over.
• 38:59 - 39:04
There were people
• 39:04 - 39:08
lots of programs that enable
you to compute derivatives
• 39:08 - 39:11
of function of
several variables.
• 39:11 - 39:24
• 39:24 - 39:27
Now let me take your
temperature again.
• 39:27 - 39:30
Is this hard?
• 39:30 - 39:30
No.
• 39:30 - 39:34
It's sort of logical you just
have to pay attention to what?
• 39:34 - 39:36
• 39:36 - 39:42
Pay attention to not making too
many algebra mistakes, right?
• 39:42 - 39:43
That's kind of the idea.
• 39:43 - 39:45
• 39:45 - 39:49
More things that I
wanted to-- there
• 39:49 - 39:51
are many more things I wanted
to share with you today,
• 39:51 - 39:56
some consensus in the sense
• 39:56 - 40:02
that you feel there is logic and
order in this type of problem.
• 40:02 - 40:22
• 40:22 - 40:35
I tried to give you a little
bit of an introduction to why
• 40:35 - 40:40
important last time.
• 40:40 - 40:41
And I'm going to
come back to that
• 40:41 - 40:45
again, so I'm not going
to leave you in the air.
• 40:45 - 40:49
But before then,
I would like to do
• 40:49 - 40:51
the directional
derivative, which
• 40:51 - 40:53
is a very important section.
• 40:53 - 40:57
So I'm going to start again.
• 40:57 - 41:08
And I'll also do, at the same
time, some review of 11.5.
• 41:08 - 41:10
So I will combine them.
• 41:10 - 41:12
And I want to
introduce the notion
• 41:12 - 41:15
of directional
derivatives because it's
• 41:15 - 41:16
right there for us to grab it.
• 41:16 - 41:26
• 41:26 - 41:29
And you say, well,
that sounds familiar.
• 41:29 - 41:32
It sounds like I dealt
with direction before,
• 41:32 - 41:36
but I didn't what that was.
• 41:36 - 41:38
That's exactly true.
• 41:38 - 41:41
You dealt with it before, you
just didn't know what it was.
• 41:41 - 41:44
And I'll give you the
general definition,
• 41:44 - 41:49
but then I would like you to
think about if you have ever
• 41:49 - 41:50
seen that before.
• 41:50 - 41:53
• 41:53 - 41:59
I'm going to say I have the
derivative of a function, f,
• 41:59 - 42:01
in the direction, u.
• 42:01 - 42:04
And I'm going put u bar
as if you were free,
• 42:04 - 42:05
not a married man.
• 42:05 - 42:09
But u as a direction as
always a unit vector.
• 42:09 - 42:10
STUDENT: [INAUDIBLE].
• 42:10 - 42:12
MAGDALENA TODA: I
told you last time,
• 42:12 - 42:18
just to prepare you, direction,
u, is always a unit vector.
• 42:18 - 42:24
• 42:24 - 42:24
Always.
• 42:24 - 42:28
• 42:28 - 42:29
Computed at x0y0.
• 42:29 - 42:36
But x0y0 is a given view point.
• 42:36 - 42:41
• 42:41 - 42:46
And I'm going to say
what that's going to be.
• 42:46 - 42:49
I have a limit.
• 42:49 - 42:51
I'm going to use the h.
• 42:51 - 42:54
And you say, why in the
world is she using h?
• 42:54 - 42:57
You will see in a second--
h goes to 0-- because we
• 42:57 - 42:59
haven't used h in awhile.
• 42:59 - 43:02
h is like a small displacement
that shrinks to 0.
• 43:02 - 43:07
• 43:07 - 43:26
And I put here, f of x0
plus hu1, y0 plus hu2,
• 43:26 - 43:32
close, minus f of x0y0.
• 43:32 - 43:35
So you say, wait a minute,
Magdalena, oh my god, I've
• 43:35 - 43:38
• 43:38 - 43:39
I'm not here.
• 43:39 - 43:43
Z0 is easy to understand
for everybody, right?
• 43:43 - 43:47
That's going to be
altitude at the point x0y0.
• 43:47 - 43:49
It shouldn't be hard.
• 43:49 - 43:52
• 43:52 - 43:55
On the other hand,
what am I doing?
• 43:55 - 44:00
I have to look at a real
graph, in the real world.
• 44:00 - 44:05
And that's going to be a
patch of a smooth surface.
• 44:05 - 44:09
And I say, OK, this
is my favorite point.
• 44:09 - 44:12
I have x0y0 on the ground.
• 44:12 - 44:16
And the corresponding
point in three dimensions,
• 44:16 - 44:22
would be x0y0 and z0,
which is the f of x0y0.
• 44:22 - 44:24
And you say, wait a
minute, what do you mean
• 44:24 - 44:26
I can't move in a direction?
• 44:26 - 44:33
Is it like when took
a sleigh and we went
• 44:33 - 44:36
to have fun on the hill?
• 44:36 - 44:38
Yes, but I said that
would be the last time
• 44:38 - 44:43
area with snow on it.
• 44:43 - 44:48
It was a good preparation
for today in the sense that--
• 44:48 - 44:52
Remember, we went somewhere when
• 44:52 - 44:53
east?
• 44:53 - 44:55
i plus j?
• 44:55 - 44:57
And in the direction
of i plus j,
• 44:57 - 44:59
which is not quite
the direction and I'll
• 44:59 - 45:05
ask you why in a second, I was
going down along a meridian.
• 45:05 - 45:06
Remember last time?
• 45:06 - 45:12
And then that was the direction
of the steepest descent.
• 45:12 - 45:13
I was sliding down.
• 45:13 - 45:17
If I wanted the direction
of the steepest ascent,
• 45:17 - 45:20
that would have been
minus i minus j.
• 45:20 - 45:23
So I had plus i plus
j, minus i minus j.
• 45:23 - 45:26
And I told you last time, why
are those not quite directions?
• 45:26 - 45:28
STUDENT: Because
they are not unitary.
• 45:28 - 45:29
MAGDALENA TODA: They
are not unitary.
• 45:29 - 45:32
So to make them like
this u, I should
• 45:32 - 45:34
have said, in the
direction i plus
• 45:34 - 45:37
j, that was one minus
x squared minus y
• 45:37 - 45:41
squared, the parabola way,
that was the hill full of snow.
• 45:41 - 45:45
So in the direction
i plus j, I go down
• 45:45 - 45:47
the fastest possible way.
• 45:47 - 45:51
In the direction i plus
j over square root of 2,
• 45:51 - 45:54
I would be fine
with a unit vector.
• 45:54 - 45:58
In the opposite direction, I
go up the fastest way possible,
• 45:58 - 46:02
but you don't want to because
it's-- can you imagine hiking
• 46:02 - 46:09
the steepest possible
direction in the steepest way?
• 46:09 - 46:15
• 46:15 - 46:16
Now with my direction.
• 46:16 - 46:22
My direction in plane
should be the i vector.
• 46:22 - 46:26
And that magic vector should
have length 1 from here
• 46:26 - 46:26
to here.
• 46:26 - 46:30
And when you measure this
guy, he has to have length 1.
• 46:30 - 46:35
And if you decompose, you have
to decompose him along the--
• 46:35 - 46:37
what is this?
• 46:37 - 46:40
The x direction and
the y direction, right?
• 46:40 - 46:47
How do you split a vector
in such a decomposition?
• 46:47 - 46:54
Well, Mr. u will be u1i plus 1i.
• 46:54 - 46:56
It sounds funny.
• 46:56 - 46:59
Plus u2j.
• 46:59 - 47:02
So you have u1
from here to here.
• 47:02 - 47:04
I don't well you can draw.
• 47:04 - 47:07
I think some of you can
draw really well, especially
• 47:07 - 47:11
better than me because you
took technical drawing.
• 47:11 - 47:14
How many of you took technical
drawing in this glass?
• 47:14 - 47:16
STUDENT: Only in this class?
• 47:16 - 47:17
MAGDALENA TODA: In anything.
• 47:17 - 47:18
STUDENT: In high school.
• 47:18 - 47:19
MAGDALENA TODA: High
school or college.
• 47:19 - 47:22
STUDENT: I went to
it in middle school.
• 47:22 - 47:24
So it gives you so
that [INAUDIBLE]
• 47:24 - 47:25
and you'd have to
draw it. [INAUDIBLE].
• 47:25 - 47:26
MAGDALENA TODA: It's
really helping you
• 47:26 - 47:31
with the perspective view,
3D view, from an angle.
• 47:31 - 47:33
So now you're looking
at this u direction
• 47:33 - 47:36
as being u1i plus u2j.
• 47:36 - 47:40
And you say, OK, I think
I know what's going on.
• 47:40 - 47:47
You have a displacement
in the direction of the x
• 47:47 - 47:52
coordinate by 1 times h.
• 47:52 - 47:55
So it's a small displacement
• 47:55 - 47:57
And-- yes?
• 47:57 - 47:58
STUDENT: Why 1 [INAUDIBLE]?
• 47:58 - 48:02
• 48:02 - 48:03
MAGDALENA TODA: Which one?
• 48:03 - 48:04
STUDENT: You said 1 times H.
• 48:04 - 48:05
MAGDALENA TODA: u1.
• 48:05 - 48:09
• 48:09 - 48:10
You will see in a second.
• 48:10 - 48:12
That's the way you define it.
• 48:12 - 48:15
• 48:15 - 48:19
I would like you to tell me
what the whole animal is, if I
• 48:19 - 48:21
want to represent it later.
• 48:21 - 48:24
And if you can give
me some examples.
• 48:24 - 48:29
And if I go in a y direction
with a small displacement,
• 48:29 - 48:34
from y0, I have to leave and go.
• 48:34 - 48:38
So I am here at x0y0.
• 48:38 - 48:42
And this is the x direction
and this is the y direction.
• 48:42 - 48:48
And when I displace a little
bit, I displace with the green.
• 48:48 - 48:50
I displace in this direction.
• 48:50 - 48:54
I will have to displace
and see what happens here.
• 48:54 - 48:58
• 48:58 - 49:02
And then in this direction--
I'm not going to write it yet.
• 49:02 - 49:04
So I'm displacing
in this direction
• 49:04 - 49:07
and in that direction.
• 49:07 - 49:08
Why am I keeping it h?
• 49:08 - 49:13
Well, because I have the
coordinates x0y0 plus--
• 49:13 - 49:21
how do you give me a collinear
vector to u, but a small one?
• 49:21 - 49:23
You say, wait a minute,
I know what you mean.
• 49:23 - 49:28
I start from the point x0, this
is p, plus a small multiple
• 49:28 - 49:32
of the direction you give me.
• 49:32 - 49:35
before in Calc 2.
• 49:35 - 49:41
which is my vector, u.
• 49:41 - 49:55
So give me a very small
displacement vector
• 49:55 - 50:05
in the direction u, which
is u1u2, u2 as a vector.
• 50:05 - 50:06
You like angular graphics.
• 50:06 - 50:08
I don't, but it doesn't matter.
• 50:08 - 50:10
STUDENT: So basically, h.
• 50:10 - 50:11
MAGDALENA TODA:
So basically, this
• 50:11 - 50:16
is x0 plus-- you want t or h?
• 50:16 - 50:18
t or h, it doesn't matter.
• 50:18 - 50:23
hu1, ui0 plus hu2.
• 50:23 - 50:24
Why not t?
• 50:24 - 50:27
Why did I take h?
• 50:27 - 50:30
It is like time parameter
that I'm doing with h,
• 50:30 - 50:34
but h is a very
small time parameter.
• 50:34 - 50:36
It's an infinitesimally
small time.
• 50:36 - 50:41
It's just a fraction of
a second after I start.
• 50:41 - 50:44
That's why I use little
h and not little t.
• 50:44 - 50:46
• 50:46 - 50:52
H, in general, indicates a
very small time displacement.
• 50:52 - 50:58
So tried to say,
where am I here?
• 50:58 - 51:02
I'm here, just one step further
with a small displacement.
• 51:02 - 51:06
And that's going to p
at this whole thing.
• 51:06 - 51:11
• 51:11 - 51:17
Let's call this F of--
the blue one is F of x0y0.
• 51:17 - 51:23
• 51:23 - 51:28
And the green altitude, or the
altitude of the green point,
• 51:28 - 51:30
will be what?
• 51:30 - 51:32
Well, this is
something, something,
• 51:32 - 51:44
and the altitude would be F
of x0 plus hu1, y0 plus hu2.
• 51:44 - 51:50
And I measure how far
away the altitudes are.
• 51:50 - 51:51
They are very close.
• 51:51 - 51:53
The blue altitude and
the green altitude
• 51:53 - 51:55
varies the displacement.
• 51:55 - 51:58
And how can I draw that?
• 51:58 - 51:59
Here.
• 51:59 - 52:00
You see this one?
• 52:00 - 52:02
This is the delta z.
• 52:02 - 52:06
So this thing is like
a delta z kind of guy.
• 52:06 - 52:07
Any questions?
• 52:07 - 52:09
It's a little bit hard, but
you will see in a second.
• 52:09 - 52:09
Yes, sir?
• 52:09 - 52:11
STUDENT: Is it like
a small displacement
• 52:11 - 52:17
that has to be perpendicular
to the [INAUDIBLE]?
• 52:17 - 52:18
MAGDALENA TODA: No.
• 52:18 - 52:20
STUDENT: It's a
result of [INAUDIBLE]?
• 52:20 - 52:21
MAGDALENA TODA: It
is in the direction.
• 52:21 - 52:22
STUDENT: In the direction?
• 52:22 - 52:24
MAGDALENA TODA: So
let's model it better.
• 52:24 - 52:27
I don't have a three
dimensional-- they sent me
• 52:27 - 52:29
an email this morning
from the library saying,
• 52:29 - 52:32
dimensional print--
• 52:32 - 52:35
do you want to support the idea
of Texas Tech having a three
• 52:35 - 52:40
dimensional printer available
for educational purposes?
• 52:40 - 52:41
STUDENT: Did you
say, of course, yes?
• 52:41 - 52:43
MAGDALENA TODA: Of
course, I would.
• 52:43 - 52:45
But I don't have a three
dimensional printer,
• 52:45 - 52:48
but you have
imagination and imagine
• 52:48 - 52:50
we have a surface that,
again, looks like a hill.
• 52:50 - 52:53
That's my hand.
• 52:53 - 52:58
And this engagement ring
that I have is actually p0,
• 52:58 - 52:59
which is x0y0zz.
• 52:59 - 53:04
• 53:04 - 53:09
And I'm going in a
direction of somebody.
• 53:09 - 53:10
It doesn't have to be u.
• 53:10 - 53:12
No, [INAUDIBLE].
• 53:12 - 53:14
So I'm going in the
direction of u-- yu2,
• 53:14 - 53:18
is that horizontal thing.
• 53:18 - 53:20
I'm going in that direction.
• 53:20 - 53:23
So this is the
direction I'm going in
• 53:23 - 53:25
and I say, OK, where do I go?
• 53:25 - 53:29
We'll do a small displacement,
an infinitesimally small
• 53:29 - 53:32
displacement in
that direction here.
• 53:32 - 53:38
So the two points are
related to one another.
• 53:38 - 53:42
And you say, but there's such
a small difference in altitudes
• 53:42 - 53:45
because you have an
infinitesimally small
• 53:45 - 53:47
displacement in that direction.
• 53:47 - 53:48
Yes, I know.
• 53:48 - 53:51
But when you make the ratio
between that small delta
• 53:51 - 53:58
z and the small h, the ratio
could be 65 or 120 minus 32.
• 53:58 - 53:59
You don't know what you get.
• 53:59 - 54:04
So just like in general limit
of the difference quotient
• 54:04 - 54:10
being the derivative, you'll get
the ratio between some things
• 54:10 - 54:12
that are very small.
• 54:12 - 54:15
But in the end, you can
get something unexpected.
• 54:15 - 54:16
Finite or anything.
• 54:16 - 54:21
Now what do you think
this guy-- according
• 54:21 - 54:27
Rule preparation.
• 54:27 - 54:31
I taught you about Chain Rule.
• 54:31 - 54:36
What will this be
if we compute them?
• 54:36 - 54:38
There is a proof for this.
• 54:38 - 54:41
It would be like a
page or a 2 page proof
• 54:41 - 54:44
for what I'm claiming to have.
• 54:44 - 54:46
Or how do you think
I'm going to get
• 54:46 - 54:50
to this without doing the
limit of a difference quotient?
• 54:50 - 54:52
Because if I give
you functions and you
• 54:52 - 54:53
do the limit of the
difference quotients
• 54:53 - 54:57
for some nasty functions,
you'll never finish.
• 54:57 - 55:02
So what do you think
we ought to do?
• 55:02 - 55:06
This is going to be some
sort of derivative, right?
• 55:06 - 55:09
And it's going to be
a derivative of what?
• 55:09 - 55:11
Yes, sir.
• 55:11 - 55:14
STUDENT: Well, it's going to
be like a partial derivative,
• 55:14 - 55:20
except the plane you're
using to cut the surface
• 55:20 - 55:23
is not going to be in the x
direction or the y direction.
• 55:23 - 55:24
It's going to be
along the [? uz. ?]
• 55:24 - 55:25
MAGDALENA TODA: Right.
• 55:25 - 55:28
So that is a very
good observation.
• 55:28 - 55:32
And it would be like I would the
partial not in this direction,
• 55:32 - 55:34
not in that direction,
but in this direction.
• 55:34 - 55:35
Let me tell you what this is.
• 55:35 - 55:38
So according to a
theorem, this would
• 55:38 - 55:43
be df, dx, exactly
like The Chain Rule,
• 55:43 - 55:50
at my favorite point
here, x0y0 [INAUDIBLE]
• 55:50 - 55:55
p times-- now you say, oh,
Magdalena, I understand.
• 55:55 - 55:57
You're doing some
sort of derivation.
• 55:57 - 56:02
The derivative of that with
respect to h would be u1.
• 56:02 - 56:03
Yes.
• 56:03 - 56:04
It's a Chain Rule.
• 56:04 - 56:13
So then I go times u1 plus
df, dy at the point times u2.
• 56:13 - 56:15
• 56:15 - 56:18
And you say, OK, but
can I prove that?
• 56:18 - 56:21
Yes, you could, but
to prove that you
• 56:21 - 56:26
would need to play a game.
• 56:26 - 56:31
The proof will involve that
you multiply up and down
• 56:31 - 56:33
• 56:33 - 56:35
And then you take
limit of a product.
• 56:35 - 56:37
If you take product,
the product of limits,
• 56:37 - 56:43
and you study them separately
until you get to this Actually,
• 56:43 - 56:47
this is an application
of The Chain.
• 56:47 - 56:54
But I want to come back to
what Alexander just notice.
• 56:54 - 56:58
I can explain this
much better if we only
• 56:58 - 57:02
think of derivative in the
direction of i and derivative
• 57:02 - 57:03
in the direction of j.
• 57:03 - 57:05
What the heck are those?
• 57:05 - 57:07
What are they going to be?
• 57:07 - 57:14
The direction of deritivie--
if I have i instead of u, that
• 57:14 - 57:17
will make you understand the
whole notion much better.
• 57:17 - 57:19
So what would be the
directional derivative
• 57:19 - 57:23
of in the direction of i only?
• 57:23 - 57:24
Well, i for an i.
• 57:24 - 57:25
It goes this way.
• 57:25 - 57:27
This is a hard lesson.
• 57:27 - 57:30
rather than Calc 3,
• 57:30 - 57:32
but you're going to get it.
• 57:32 - 57:37
So if I go in the
direction of i,
• 57:37 - 57:41
I should have the df, dx, right?
• 57:41 - 57:42
That should be it.
• 57:42 - 57:43
Do I?
• 57:43 - 57:44
STUDENT: Yes, but [INAUDIBLE] 0.
• 57:44 - 57:45
MAGDALENA TODA: Exactly.
• 57:45 - 57:48
Was I able to
invent something so
• 57:48 - 57:53
when I come back to what I
already know, I recreate df, dx
• 57:53 - 57:56
and nothing else?
• 57:56 - 58:05
Precisely because for i as
being u, what will be u1 and u2?
• 58:05 - 58:07
STUDENT: [INAUDIBLE].
• 58:07 - 58:09
MAGDALENA TODA: u1 is 1.
• 58:09 - 58:12
u2 is 0.
• 58:12 - 58:12
Right?
• 58:12 - 58:16
Because when we write i
as a function of i and j,
• 58:16 - 58:19
that's 1 times i plus 0 times j.
• 58:19 - 58:23
So u1 is 1, u2 is zero.
• 58:23 - 58:24
Thank god.
• 58:24 - 58:27
According to the anything,
this difference quotient
• 58:27 - 58:32
or the simpler way to define
it from the theorem would
• 58:32 - 58:35
be simply the second goes away.
• 58:35 - 58:36
It vanishes.
• 58:36 - 58:42
u1 would be 1 and what
I'm left with is df, dx.
• 58:42 - 58:45
And that's exactly
what Alex noticed.
• 58:45 - 58:49
So the directional
derivative is defined,
• 58:49 - 58:54
as a combination of vectors,
such that you recreate
• 58:54 - 58:56
the directional derivative
in the direction of i
• 58:56 - 58:59
being the partial, df, dx.
• 58:59 - 59:03
Exactly like you
learned before in 11.3.
• 59:03 - 59:06
And what do I have
if I try to recreate
• 59:06 - 59:10
the directional derivative
in the direct of j?
• 59:10 - 59:11
x0y0.
• 59:11 - 59:14
We don't explain this
much in the book.
• 59:14 - 59:18
I think on this one, I'm doing
a better job than the book.
• 59:18 - 59:22
So what is df in
the direction of j?
• 59:22 - 59:24
j is this way.
• 59:24 - 59:27
Well, [INAUDIBLE]
is that 1j-- you
• 59:27 - 59:32
let me write it
down-- is 0i plus 1j.
• 59:32 - 59:34
0 is u1.
• 59:34 - 59:36
1 is u2.
• 59:36 - 59:42
So by this formula,
I simply should
• 59:42 - 59:48
get the directional
deritive-- I mean,
• 59:48 - 59:51
directional derivative is the
partial deritive-- with respect
• 59:51 - 59:57
to y at my point times a 1
that I'm not going to write.
• 59:57 - 60:07
So it's a concoction, so that
in the directions of i and j,
• 60:07 - 60:11
you actually get the
partial deritives.
• 60:11 - 60:13
And everything else
is linear algebra.
• 60:13 - 60:20
So if you have a problem
understanding the composition
• 60:20 - 60:22
of vectors, the sum
of vectors, this
• 60:22 - 60:26
is because-- u1 and
u2 are [INAUDIBLE],
• 60:26 - 60:28
I'm sorry-- this is
because you haven't taken
• 60:28 - 60:33
the linear algebra yet, which
teaches you a lot about how
• 60:33 - 60:36
a vector decomposes in
two different directions
• 60:36 - 60:39
or along the standard
canonical bases.
• 60:39 - 60:42
• 60:42 - 60:45
Let's see some
problems of the type
• 60:45 - 60:49
that I've always put in the
midterm and the same kind
• 60:49 - 60:55
of problems like we
have seen in the final.
• 60:55 - 60:58
For example 3, is it, guys?
• 60:58 - 60:58
I don't know.
• 60:58 - 61:00
Example 3, 4, or
something like that?
• 61:00 - 61:00
STUDENT: 3.
• 61:00 - 61:03
MAGDALENA TODA: Given
z equals F of xy--
• 61:03 - 61:06
what do you like best,
the value or the hill?
• 61:06 - 61:09
This appeared in
most of my exams.
• 61:09 - 61:13
x squared plus y squared,
circular [INAUDIBLE]
• 61:13 - 61:14
was one of my favorite examples.
• 61:14 - 61:16
1 minus x squared
minus y squared
• 61:16 - 61:23
was the circular
parabola upside down.
• 61:23 - 61:24
Which one do you prefer?
• 61:24 - 61:25
I don't care.
• 61:25 - 61:26
Which one?
• 61:26 - 61:27
STUDENT: [INAUDIBLE].
• 61:27 - 61:28
MAGDALENA TODA: The [INAUDIBLE]?
• 61:28 - 61:29
The first one.
• 61:29 - 61:30
It's easier.
• 61:30 - 61:35
• 61:35 - 61:37
And a typical problem.
• 61:37 - 61:50
Compute the directional
derivative of z
• 61:50 - 62:00
equals F of x and y at the
point p of coordinates 1, 1, 2
• 62:00 - 62:14
in the following
directions-- A, i.
• 62:14 - 62:15
B, j.
• 62:15 - 62:18
C, i plus j.
• 62:18 - 62:24
• 62:24 - 62:29
D, the opposite, minus
i, minus j over square 2.
• 62:29 - 62:32
And E--
• 62:32 - 62:33
STUDENT: That's a square root 3.
• 62:33 - 62:34
MAGDALENA TODA: What?
• 62:34 - 62:36
STUDENT: You wrote
a square root 3.
• 62:36 - 62:37
MAGDALENA TODA: I
wrote square root of 3.
• 62:37 - 62:38
Thank you guys.
• 62:38 - 62:39
Thanks for being vigilant.
• 62:39 - 62:43
So always keep an eye on me
because I'm full of surprises,
• 62:43 - 62:44
• 62:44 - 62:46
No, just kidding.
• 62:46 - 62:48
So let's see.
• 62:48 - 62:49
What do I want to put here?
• 62:49 - 62:51
Something.
• 62:51 - 62:52
• 62:52 - 63:01
• 63:01 - 63:07
3 over root 5, pi
plus [? y ?] over 5j.
• 63:07 - 63:11
Is this a unit vector or not?
• 63:11 - 63:12
STUDENT: No.
• 63:12 - 63:13
STUDENT: Yes, it is.
• 63:13 - 63:15
So you're going to
drag the [INAUDIBLE].
• 63:15 - 63:17
MAGDALENA TODA: Why
is that a unit vector?
• 63:17 - 63:19
STUDENT: It's
missing-- no, it's not.
• 63:19 - 63:21
MAGDALENA TODA: Then how
do I make it a unit vector?
• 63:21 - 63:23
STUDENT: [INAUDIBLE].
• 63:23 - 63:25
STUDENT: [INAUDIBLE].
• 63:25 - 63:28
STUDENT: I have to take down--
there's a 3 that has to be 1.
• 63:28 - 63:29
[INAUDIBLE]
• 63:29 - 63:32
And the second one has
to be 1, on the top,
• 63:32 - 63:34
to make it a unit vector.
• 63:34 - 63:39
• 63:39 - 63:42
MAGDALENA TODA: Give
me a unit vector.
• 63:42 - 63:47
Another one then
these easy ones.
• 63:47 - 63:48
STUDENT: 3 over 5 by 4 or 5.
• 63:48 - 63:49
MAGDALENA TODA: What?
• 63:49 - 63:52
STUDENT: 3 over 5 by 4 over 5j.
• 63:52 - 63:54
MAGDALENA TODA: 3
over-- I cannot hear.
• 63:54 - 63:54
STUDENT: 3 over 5--
• 63:54 - 63:55
MAGDALENA TODA: 3 over 5.
• 63:55 - 63:57
STUDENT: And 4 over 5j.
• 63:57 - 63:59
MAGDALENA TODA: And 4 over 5j.
• 63:59 - 64:01
And why is that a unit vector?
• 64:01 - 64:05
STUDENT: Because 3
squared is [INAUDIBLE].
• 64:05 - 64:07
MAGDALENA TODA: And what
do we call these numbers?
• 64:07 - 64:08
You say, what is that?
• 64:08 - 64:11
And interview?
• 64:11 - 64:13
Yes, it is an interview.
• 64:13 - 64:14
Pythagorean numbers.
• 64:14 - 64:16
3, 4, and 5 are
Pythagorean numbers.
• 64:16 - 64:19
• 64:19 - 64:24
So let me think a little
bit where I should write.
• 64:24 - 64:26
Is this seen by
the-- yes, it's seen
• 64:26 - 64:34
by the-- I'll just leave
what's important for me
• 64:34 - 64:36
to solve this problem.
• 64:36 - 64:45
• 64:45 - 64:48
A. So what do we do?
• 64:48 - 64:56
The same thing. i is 1.i plus
u, or 1 times i plus u times j.
• 64:56 - 64:59
So simply, you can write
the formula or you can say,
• 64:59 - 65:01
the heck with the formula.
• 65:01 - 65:04
You know that df is df, dx.
• 65:04 - 65:08
The derivative of
this at the point p.
• 65:08 - 65:14
So what you want to do is say,
2x-- are you guys with me?
• 65:14 - 65:15
STUDENT: Yes.
• 65:15 - 65:23
MAGDALENA TODA: At the
value 1, 1, 2, which is 2.
• 65:23 - 65:25
And at the end of
this exercise, I'm
• 65:25 - 65:28
going to ask you if there's
any connection between--
• 65:28 - 65:31
or maybe I will
• 65:31 - 65:35
Oh, we have time.
• 65:35 - 65:38
What is d in the direction of j?
• 65:38 - 65:41
The partial derivative
with respect to y.
• 65:41 - 65:44
Nothing else, but
our old friend.
• 65:44 - 65:48
And our old friend
says, I have 2y
• 65:48 - 65:52
computed for the
point p, 1, 1, 2.
• 65:52 - 65:53
What does it mean?
• 65:53 - 65:59
Y is 1, so just plug
this 1 into the thingy.
• 65:59 - 66:00
It's 2.
• 66:00 - 66:04
• 66:04 - 66:07
Now do I see some--
I'm a scientist.
• 66:07 - 66:09
I have to find
interpretations when
• 66:09 - 66:11
I get results that coincide.
• 66:11 - 66:13
It's a pattern.
• 66:13 - 66:14
Why do I get the same answer?
• 66:14 - 66:16
STUDENT: Because your
functions are symmetric.
• 66:16 - 66:17
MAGDALENA TODA: Right.
• 66:17 - 66:20
And more than that, because
the function is symmetric,
• 66:20 - 66:25
it's a quadric that I love,
it's just a circular problem.
• 66:25 - 66:28
It's rotation is symmetric.
• 66:28 - 66:34
So I just take one parabola,
one branch of a parabola,
• 66:34 - 66:38
and I rotate it by 360 degrees.
• 66:38 - 66:46
So the slope will be the same
in both directions, i and j,
• 66:46 - 66:47
at the point that I have.
• 66:47 - 66:50
• 66:50 - 66:53
Well, it depends on the point.
• 66:53 - 66:55
If the point is,
itself, symmetric
• 66:55 - 66:58
like that, x and y are
the same, one in one,
• 66:58 - 67:04
I did it on purpose-- if
you didn't have one and one,
• 67:04 - 67:08
you had an x variable and
y variable to plug in.
• 67:08 - 67:11
But your magic point is where?
• 67:11 - 67:12
Oh my god.
• 67:12 - 67:15
I don't know how to
explain with my hands.
• 67:15 - 67:17
Here I am, the frame.
• 67:17 - 67:20
I am the frame. x, y, and z.
• 67:20 - 67:22
1, 1.
• 67:22 - 67:23
Go up.
• 67:23 - 67:25
Where do you meet the vase?
• 67:25 - 67:27
At c equals 2.
• 67:27 - 67:30
So it's really symmetric
and really beautiful.
• 67:30 - 67:34
• 67:34 - 67:38
Next I say, oh, in
the direction i plus
• 67:38 - 67:44
j, which is exactly the
direction of this meridian
• 67:44 - 67:48
that I was talking about, i
plus j over square root 2.
• 67:48 - 67:51
where I was broken hearted.
• 67:51 - 67:53
Really, I didn't
know what to do,
• 67:53 - 67:56
how much partial credit to give.
• 67:56 - 68:01
The definition of direction
derivative is very strict.
• 68:01 - 68:05
It says you cannot take
whatever 1 and 2 that you want.
• 68:05 - 68:09
You cannot multiply
them by proportionality.
• 68:09 - 68:14
You have to have u
to be a unit vector.
• 68:14 - 68:18
And then the directional
derivative will be unique.
• 68:18 - 68:24
If I take 1 and 1 for u1 and
u2, then I can take 2 and 2,
• 68:24 - 68:26
and 7 and 7, and 9 and 9.
• 68:26 - 68:28
And that's going to
be a mess because
• 68:28 - 68:32
the directional derivative
wouldn't be unique anymore.
• 68:32 - 68:36
And that's why whoever
gave this definition,
• 68:36 - 68:39
I think Euler-- I tried
to see in the history who
• 68:39 - 68:43
was the first
mathematician who gave
• 68:43 - 68:47
the definition of the
directional derivative.
• 68:47 - 68:50
And some people
said it was Gateaux
• 68:50 - 68:53
because that's a french
mathematician who first talked
• 68:53 - 68:55
derivative, which
• 68:55 - 68:57
is like the
directional derivative,
• 68:57 - 68:59
but other people said,
no, look at Euler's work.
• 68:59 - 69:00
He was a genius.
• 69:00 - 69:05
He's the guy who discovered
the transcendental number
• 69:05 - 69:07
e and many other things.
• 69:07 - 69:09
And the exponential
e to the x is also
• 69:09 - 69:11
from Euler and everything.
• 69:11 - 69:13
He was one of the
fathers of calculus.
• 69:13 - 69:19
Apparently, he knew the first
32 decimals of the number e.
• 69:19 - 69:23
And how he got to
them is by hand.
• 69:23 - 69:24
Do you guys know of them?
• 69:24 - 69:30
2.71828-- and that's all I know.
• 69:30 - 69:32
The first five decimals.
• 69:32 - 69:36
Well, he knew 32 of them
and he got to them by hand.
• 69:36 - 69:39
And they are non-repeating,
infinitely remaining decimals.
• 69:39 - 69:40
It's a transcendental number.
• 69:40 - 69:42
STUDENT: And his 32 are correct?
• 69:42 - 69:43
MAGDALENA TODA: What?
• 69:43 - 69:44
STUDENT: His 32 are correct?
• 69:44 - 69:47
MAGDALENA TODA: His first
32 decimals were correct.
• 69:47 - 69:50
I don't know what--
I mean, the guy
• 69:50 - 69:53
was something like-- he
was working at night.
• 69:53 - 69:57
And he would fill out,
in one night, hundreds
• 69:57 - 70:04
of pages, computations, both
by hand formulas and numerical.
• 70:04 - 70:07
So imagine-- of course, he would
never make a WeBWork mistake.
• 70:07 - 70:11
I mean, if we built
a time machine,
• 70:11 - 70:13
and we bring Euler back,
and he's at Texas Tech,
• 70:13 - 70:16
and we make him solve
our WeBWork problems,
• 70:16 - 70:18
I think he would take
a thousand problems
• 70:18 - 70:20
and solve them in one night.
• 70:20 - 70:22
He need to know
how to type, so we
• 70:22 - 70:24
have to teach him how to type.
• 70:24 - 70:28
But he would be able to
compute what you guys have,
• 70:28 - 70:32
all those numerical
• 70:32 - 70:35
He was a scary fellow.
• 70:35 - 70:41
So u has to be [INAUDIBLE]
• 70:41 - 70:43
u1 and u2.
• 70:43 - 70:46
I have students-- that's
where the story started--
• 70:46 - 70:50
who were very good, very smart,
both honors and non-honors, who
• 70:50 - 70:54
took u1 to be 1, u2 to be 2
because they thought direction
• 70:54 - 71:01
1 and 1, which is not made
unique as a direction, unitary.
• 71:01 - 71:03
And they plugged in here
1, they plugged in here 1,
• 71:03 - 71:08
they got these correctly, what
was I supposed to give them, as
• 71:08 - 71:09
a [? friend? ?]
• 71:09 - 71:10
STUDENT: [INAUDIBLE].
• 71:10 - 71:10
MAGDALENA TODA: What?
• 71:10 - 71:12
STUDENT: [INAUDIBLE].
• 71:12 - 71:13
MAGDALENA TODA: I gave them.
• 71:13 - 71:14
How much do you think?
• 71:14 - 71:15
You should know me.
• 71:15 - 71:16
STUDENT: [INAUDIBLE].
• 71:16 - 71:17
STUDENT: Full.
• 71:17 - 71:18
MAGDALENA TODA: 60%.
• 71:18 - 71:20
No.
• 71:20 - 71:21
Some people don't
give any credit,
• 71:21 - 71:23
so pay attention to this.
• 71:23 - 71:31
In this case, this has
to be 1 over square root
• 71:31 - 71:42
of 2 times the derivative of f
at x, which is computed before
• 71:42 - 71:51
at the point, plus 1 over square
root of 2 times the derivative
• 71:51 - 71:52
of the function.
• 71:52 - 71:54
Again, compute it
at the same place.
• 71:54 - 72:03
Which is, oh my god, square
root of 2 plus square root of 2,
• 72:03 - 72:04
which is 2 square root of 2.
• 72:04 - 72:21
• 72:21 - 72:30
And finally, the derivative
of F at the same point-- I
• 72:30 - 72:31
should have put at the point.
• 72:31 - 72:35
Like a physicist
would say, at p.
• 72:35 - 72:38
That would make you
familiar with this notation.
• 72:38 - 72:40
And then measured at what?
• 72:40 - 72:44
The opposite direction,
minus i minus j.
• 72:44 - 72:46
And now I'm getting lazy
and I'm going to ask you
• 72:46 - 72:49
• 72:49 - 72:50
STUDENT: 2 minus
square root of 2.
• 72:50 - 72:53
MAGDALENA TODA: So you see,
there is another pattern.
• 72:53 - 72:55
In the opposite
direction, the direction
• 72:55 - 73:00
of the derivative in this case
would just be the negative one.
• 73:00 - 73:03
What if we took this directional
derivative in absolute value?
• 73:03 - 73:05
Because you see,
in this direction,
• 73:05 - 73:08
there's a positive
directional derivaty.
• 73:08 - 73:12
In the other direction, it's
like it's because-- I know why.
• 73:12 - 73:14
I'm a vase.
• 73:14 - 73:18
So in the direction i plus
j over square root of 2,
• 73:18 - 73:20
the directional derivative
will be positive.
• 73:20 - 73:22
It goes up.
• 73:22 - 73:24
But in the direction
minus i minus
• 73:24 - 73:28
j, which is the opposite, over
square root of 2, it goes down.
• 73:28 - 73:31
So the slope is negative.
• 73:31 - 73:32
So that's why we have negative.
• 73:32 - 73:35
Everything you get
in life or in math,
• 73:35 - 73:36
you have to find
an interpretation.
• 73:36 - 73:40
• 73:40 - 73:44
Sometimes in life and
mathematics, things are subtle.
• 73:44 - 73:47
People will say one thing
and they mean another thing.
• 73:47 - 73:50
You have to try to see
beyond their words.
• 73:50 - 73:51
• 73:51 - 73:54
And in mathematics, you have to
try to see beyond the numbers.
• 73:54 - 73:55
You see a pattern.
• 73:55 - 73:58
So being in opposite
directions, I
• 73:58 - 74:02
got opposite signs of the
directional derivative
• 74:02 - 74:04
because I have opposite slopes.
• 74:04 - 74:08
• 74:08 - 74:11
What else do I want to
learn in this example?
• 74:11 - 74:12
One last thing.
• 74:12 - 74:13
STUDENT: E.
• 74:13 - 74:23
MAGDALENA TODA: E. So
I have the same thing.
• 74:23 - 74:25
So it's not going to
matter, the direction
• 74:25 - 74:27
is the only thing that changes.
• 74:27 - 74:29
These guys are the same.
• 74:29 - 74:34
The partials are the
same at the same point.
• 74:34 - 74:35
I'm not going to
• 74:35 - 74:39
So I get 2 or both.
• 74:39 - 74:41
What changes is the blue guys.
• 74:41 - 74:48
They are going to be
3 over 5 and 4 over 5.
• 74:48 - 74:54
• 74:54 - 74:56
And what do I get?
• 74:56 - 75:05
I get-- right?
• 75:05 - 75:09
• 75:09 - 75:13
Now I want to tell
you something--
• 75:13 - 75:16
something last time.
• 75:16 - 75:21
And let me tell you
what I said last time.
• 75:21 - 75:26
• 75:26 - 75:28
Maybe I should not
erase-- well, I
• 75:28 - 75:30
have to erase this
whether I like it or not.
• 75:30 - 75:34
• 75:34 - 75:36
And now I'll review
what this was.
• 75:36 - 75:38
x squared plus y squared?
• 75:38 - 75:39
Yes or no?
• 75:39 - 75:41
STUDENT: Yes.
• 75:41 - 75:46
MAGDALENA TODA: So what
did I say last time?
• 75:46 - 75:53
We have no result. We
noticed it last time.
• 75:53 - 75:55
We did not prove it.
• 75:55 - 76:09
We did not prove it, only
found it experimentally
• 76:09 - 76:12
using our physical common sense.
• 76:12 - 76:17
When you have a function
z equals F of xy,
• 76:17 - 76:31
we studied the
maximum rate of change
• 76:31 - 76:40
at the point x0y0 in the domain,
assuming this is a c1 function.
• 76:40 - 76:41
I don't know.
• 76:41 - 76:44
Maximum rate of change
was a magic thing.
• 76:44 - 76:48
And you probably thought,
what in the world is that?
• 76:48 - 77:01
And we also said, this
maximum for the rate of change
• 77:01 - 77:24
is always attained in the
• 77:24 - 77:31
• 77:31 - 77:38
So you realize that it's
the steepest ascent,
• 77:38 - 77:41
the way it's called in
many, many other fields,
• 77:41 - 77:43
but mathematics.
• 77:43 - 77:45
Or the steepest descent.
• 77:45 - 77:52
• 77:52 - 77:58
Now if it's an ascent, then it's
in the direction gradient of F.
• 77:58 - 78:00
But if it's a
descent, it's going
• 78:00 - 78:05
to be in the opposite
• 78:05 - 78:08
But then I [INAUDIBLE]
first of all,
• 78:08 - 78:12
it's not the same direction,
if you have opposites.
• 78:12 - 78:15
Well, direction is sort
of given by one line.
• 78:15 - 78:19
Whether you take this or the
opposite, it's the same thing.
• 78:19 - 78:21
What this means is
that we say direction
• 78:21 - 78:26
and we didn't
[? unitarize ?] it.
• 78:26 - 78:31
So we could say,
• 78:31 - 78:36
F. Or minus gradient of F
• 78:36 - 78:40
over length of gradient of F.
Can this theorem be proved?
• 78:40 - 78:41
Yes, it can be proved.
• 78:41 - 78:45
We are going to discuss a little
bit more next time about it,
• 78:45 - 78:49
but I want to tell you
a big disclosure today.
• 78:49 - 78:55
This maximum rate of change
is the directional derivative.
• 78:55 - 79:08
This maximum rate
of change is exactly
• 79:08 - 79:16
the directional derivative
in the direction
• 79:16 - 79:35
of the gradient, which is also
• 79:35 - 79:43
• 79:43 - 79:47
And you'll say,
wait a minute, what?
• 79:47 - 79:48
What did you say?
• 79:48 - 79:51
Let's first verify my claim.
• 79:51 - 79:53
I'm not even sure
my claim is true.
• 79:53 - 79:55
We will see next time.
• 79:55 - 80:00
Can I verify my
claim on one example?
• 80:00 - 80:02
Well, OK.
• 80:02 - 80:05
Maximum rate of change
would be exactly
• 80:05 - 80:08
as the directional
derivative and the direction
• 80:08 - 80:09
• 80:09 - 80:10
• 80:10 - 80:11
That all sounds crazy.
• 80:11 - 80:13
So what do I have to compute?
• 80:13 - 80:17
I have to compute that
directional derivative
• 80:17 - 80:22
of, let's say, my function F in
• 80:22 - 80:23
• 80:23 - 80:26
• 80:26 - 80:29
We have to figure it out.
• 80:29 - 80:31
We did it last time,
but you forgot.
• 80:31 - 80:37
So for this guy, nabla F,
• 80:37 - 80:40
Where is my function?
• 80:40 - 80:48
Nabla F will be 2x, 2y, right?
• 80:48 - 80:52
Which means 2xi plus 2yj, right?
• 80:52 - 80:55
But if I'm at the point
p, what does it mean?
• 80:55 - 80:59
At the point p, it means that I
have 2 times i plus 2 times j,
• 80:59 - 81:00
right?
• 81:00 - 81:06
And what is the magnitude
• 81:06 - 81:08
Yes.
• 81:08 - 81:13
The magnitude of the gradient is
somebody I know, which is what?
• 81:13 - 81:19
Which is square root of
2 squared plus 2 squared.
• 81:19 - 81:21
I cannot do that now.
• 81:21 - 81:22
What's the square root of 8?
• 81:22 - 81:23
STUDENT: 2 root 2.
• 81:23 - 81:24
MAGDALENA TODA: 2 root 2.
• 81:24 - 81:25
This is a pattern.
• 81:25 - 81:25
2 root 2.
• 81:25 - 81:27
I've seen this 2 root
2 again somewhere.
• 81:27 - 81:29
Where the heck have I seen it?
• 81:29 - 81:30
STUDENT: That was the
directional derivative.
• 81:30 - 81:32
MAGDALENA TODA: The
directional derivative.
• 81:32 - 81:33
So the claim may be right.
• 81:33 - 81:36
It says it is the directional
derivative in the direction
• 81:36 - 81:38
• 81:38 - 81:41
But is this really the
• 81:41 - 81:43
Yes.
• 81:43 - 81:46
Because when you compote the
• 81:46 - 81:52
plus 2j, you don't mean 2i
plus 2j as a twice i plus j,
• 81:52 - 81:56
you mean the unit vector
correspondent to that.
• 81:56 - 81:57
So what is the
direction corresponding
• 81:57 - 82:01
to the gradient 2i plus 2j?
• 82:01 - 82:02
STUDENT: i plus j [? over 2. ?]
• 82:02 - 82:03
MAGDALENA TODA: Exactly.
• 82:03 - 82:06
U equals i plus j
divided by square 2.
• 82:06 - 82:09
So this is the
directional derivative
• 82:09 - 82:13
in the direction of the gradient
at the point p, which is 2 root
• 82:13 - 82:14
2.
• 82:14 - 82:18
And it's the same thing-- for
some reason that's mysterious
• 82:18 - 82:20
and we will see next time.
• 82:20 - 82:23
For some mysterious reason
you get exactly the same
• 82:23 - 82:28
as the length of
• 82:28 - 82:30
mystery next time.
• 82:30 - 82:35
I have you enough to
torment you until Tuesday.
• 82:35 - 82:38
What have you promised me
besides doing the homework?
• 82:38 - 82:40
• 82:40 - 82:41
MAGDALENA TODA:
• 82:41 - 82:42
You're very smart.
• 82:42 - 82:44
• 82:44 - 82:45
All the examples in the book.
• 82:45 - 82:47
They are short.
• 82:47 - 82:48
Thank you so much.
• 82:48 - 82:51
Have a wonderful
weekend and I'll
• 82:51 - 82:55
talk to you on Tuesday about
anything you have trouble with.
• 82:55 - 82:57
When is the homework due?
• 82:57 - 82:59
STUDENT: Saturday.
• 82:59 - 83:01
MAGDALENA TODA: On Saturday.
• 83:01 - 83:01
I was mean.
• 83:01 - 83:04
I should have given it you
until Sunday night, but--
• 83:04 - 83:06
STUDENT: Yes.
• 83:06 - 83:08
MAGDALENA TODA: Do you want me
to make it until Sunday night?
• 83:08 - 83:09
STUDENT: Yes.
• 83:09 - 83:10
MAGDALENA TODA: At midnight?
• 83:10 - 83:11
STUDENT: Yes.
• 83:11 - 83:13
MAGDALENA TODA: I'll do that.
• 83:13 - 83:15
I will extend it.
• 83:15 - 83:19
• 83:19 - 83:22
STUDENT: She asked, I said yes.
• 83:22 - 83:24
STUDENT: Why did
you do that, dude?
• 83:24 - 83:28
Come on, my life is ruined
now because I have more time
• 83:28 - 83:30
to work on my homework.
• 83:30 - 83:32
MAGDALENA TODA: And
• 83:32 - 83:32
STUDENT: Yes.
• 83:32 - 83:33
No.
• 83:33 - 83:34
MAGDALENA TODA: No.
• 83:34 - 83:36
Actually, I know why I did that.
• 83:36 - 83:38
I thought that the
28th of February
• 83:38 - 83:43
is the last day of the month,
but it's a short month.
• 83:43 - 83:45
So if we [? try it, ?] we
have to extend the months
• 83:45 - 83:49
a little bit by pulling
it by one more day.
• 83:49 - 83:50
STUDENT: We did?
• 83:50 - 83:52
MAGDALENA TODA: The first
of March is Sunday, right?
• 83:52 - 83:54
STUDENT: Yes.
• 83:54 - 83:56
[INTERPOSING VOICES]
• 83:56 - 84:06
• 84:06 - 84:08
STUDENT: You're going
to miss the speech.
• 84:08 - 84:09
STUDENT: Oh, we're doing that?
• 84:09 - 84:11
STUDENT: You're in English?
• 84:11 - 84:11
STUDENT: [INAUDIBLE].
• 84:11 - 84:14
• 84:14 - 84:15
STUDENT: You don't know English?
• 84:15 - 84:16
Why are you talking English?
• 84:16 - 84:18
That's what my
father used to say.
• 84:18 - 84:20
You don't know your own tongue?
Title:
TTU Math2450 Calculus3 Sec 11.5 and 11.6
Description:

Chain rules, Directrional derivatives and Gradient

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Video Language:
English
 jackie.luft edited English subtitles for TTU Math2450 Calculus3 Sec 11.5 and 11.6