TTU Math2450 Calculus3 Sec 11.4 part 1

Tangent plains, Aproximations, and Differentiability

PROFESSOR: So let's
forget about this example

and review what we learned
in 11.3, chapter 11.

Chapter 11, again, was
functions of several variables.

In our case, I'll say
functions of two variables.

11.3 taught you, what?

Taught you some
beautiful things.

Practically, if you
understand this picture,

you will remember everything.

This picture is going to
try and [INAUDIBLE] a graph

that's sitting
above here somewhere

in Euclidean free space,
dimensional space.

You have the origin.

And you say I want markers.

No, you don't say
I want markers.

I say I want markers.

We want to fix a point
x0, y0 on the surface,

assuming the surface is smooth.

That x0 of mine
should be projected.

I'm going to try to draw
better than I did last time.

X0, y0 corresponds
to a certain altitude

z0 that is projected like that.

And this is my
[INAUDIBLE] 0 here.

But I don't care much
about that right now.

I care about the
fact that locally, I

represent the function
as a graph-- z of f-- f

of x and y defined
over a domain.

I have a domain
that is an open set.

And you connect to-- that's
more than you need to know.

Could be anything.

Could be a square, could be
a-- this could be something,

a nice patch of them like.

So the projection of my
point here is x0, y0.

I'm going to draw these
parallels as well as I can.

But I cannot draw very well.

But I'm trying.

X0 and y0-- and
remember from last time.

What did we say?

I'm going to draw a plane
of equation x equals x0.

All right, I'll try.

I'll try and do a good job--
x equals x0 is this plane.

STUDENT: Don't you
have the x amount

and the y amounts backward?

Or [INAUDIBLE]

PROFESSOR: No.

STUDENT: [INAUDIBLE]

PROFESSOR: x is this 1
coming towards you like that.

And I also think about
that always, Ryan.

Do I have them backward?

This time, I was lucky.

I didn't have them backward.

So y goes this way.

For y0, let me pick another
color, a more beautiful color.

For y0, my video is not
going to see the y0.

But hopefully, it's going to
see it, this beautiful line.

Spring is coming.

So this is going
to be the plane.

Label it [INAUDIBLE]
y equals 0y.

Now, the green plane cuts the
surface into a plane curve,

of course, because
teasing the plane

that I drew with the line.

And in the plane that I drew
with red-- was it red or pink?

It's Valentine's Day.

It's pink.

OK, so I have it like that.

So what is the pink curve?

The pink curve is the
intersection between z

equals-- x equals x0
plane with my surface.

My surface is black.

I'm going to say s on surface.

And then I have a pink curve.

Let's call it c1.

Because you cannot see
pink on your notes.

You only can imagine that
it's not the same thing.

C2 is y equals y0 plane
intersected with s.

And what have we
learned last time?

Last time, we learned that
we introduce some derivatives

at the point at 0, y is
0, so that they represent

those partial derivatives of
the function z with respect

to x and y.

So we have the partial z sub x
at x0, y0 and the partial and z

sub y at x0, y0.

Do we have a more
elegant definition?

That's elegant enough for
me, thank you very much.

But if I wanted to give the
original definition, what

was that?

That is d of bx at x0, y0, which
is a limit of the difference

quotient.

And this time, we're going to--
not going to do the x of y.

I I'm different today.

So I do h goes to 0.

H is my smallest
displacement of [INAUDIBLE].

Here, I have f of-- now,
who is the variable?

X.

So who is going to say fixed?

Y. So I'm going to say I'm
displacing mister x0 with an h.

And y0 will be fixed minus
f of x0, y0, all over h.

So again, instead of-- instead
of a delta x, I call the h.

And the derivative
with respect to y

will assume that
x0 is a constant.

I saw how well
[INAUDIBLE] explained that

and I'm ambitious.

I want to do an even better
job than [INAUDIBLE].

Hopefully, I might manage.

D of ty equals [INAUDIBLE] h
going to 0 of that CF of-- now,

who's telling me what we have?

Of course, mister
x, y and y, yy.

F of x0, y0 is their constant
waiting for his turn.

H is your parameter.

And then you'll have, what?

H0 is fixed, right?

STUDENT: So h0 is--

PROFESSOR: --fixed.

Y is the variable.

So I go into the direction
of y starting from y0.

And I displace that with
a small quantity, right?

So these are my
partial velocity--

my partial derivatives, I'm
sorry, not partial velocities.

Forget what I said.

I said something that
you will learn later.

What are those?

Those are the slopes at x0, y0
of the tangents at the point

here, OK?

The tangents to the two curves,
the pink one-- the pink one

and the green one, all right?

For the pink one, for the pink
curve, what is the variable?

The variable is the y, right?

So this is c1 is a
curve that depends on y.

And c2 is a curve
that depends on x.

So this comes with x0 fixed.

I better write it like that.

F of x is 0.

Y, instead of c2 of x, I'll
say f of y-- f of x and yz.

So, which slope is which?

The d of dy at this point
is the slope to this one.

Are you guys with me?

The slope of that tangent.

Considered in the
plane where it is.

How about the other one?

S of x will be the slope of this
line in the green plane, OK?

That is considered as a
plane of axis of coordinates.

Good, good-- so we
know what they are.

A quick example to review--
I've given you some really ugly,

nasty functions today.

The last time, you
did a good job.

So today, I'm not
challenging you anymore.

I'm just going to give
you one simple example.

And I'm asked you, what
does this guy look like

and what will the meanings
of z sub x and z sub y be?

What will they be at?

Let's say I think I know what I
want to take at the point 0, 0.

And maybe you're going to
tell me what else it will be.

And eventually at
another point like z

sub a, so coordinates, 1
over square root of 2 and 1

over square root of 2.

And v sub y is same-- 1
over square root of 2,

1 over square root of 2.

Can one draw them and have
a geometric explanation

of what's going on?

Well, I don't want you to
forget the definitions,

but since you absorbed them
with your mind hopefully

and with your eyes, you're not
going to need them anymore.

We should be able to draw
this quadric that you love.

I'm sure you love it.

When it's-- what
does it look like?

STUDENT: [INAUDIBLE]

PROFESSOR: Wait a minute, you're
not awake or I'm not awake.

So if you do x squared
plus y squared,

don't write it down please.

It would be that.

And what is this?

STUDENT: That's a [INAUDIBLE].

PROFESSOR: A circular
paraboloid-- you are correct.

We've done that before.

I'd say it looks
like an egg shell,

but it's actually--
this is a parabola

if it's going to infinity.

And you said a bunch of circles.

Yes, sir.

STUDENT: So is it an
upside down graph?

PROFESSOR: It's an
upside down paraboloid.

STUDENT: [INAUDIBLE]

PROFESSOR: So, very
good-- how do we do that?

We make this guy
look in the mirror.

This is the lake.

The lake is xy plane.

So this guy is
looking in the mirror.

Take his image and
shift it just like he

said-- shift it one unit up.

This is one.

You're going to have
another paraboloid.

So from this construction,
I'm going to draw.

And he's going to look
like you took a cup

and you put it upside down.

But it's more like
an eggshell, right?

It's not a cup because
a cup is supposed

to have a flat bottom, right?

But this is like an eggshell.

And I'll draw.

And for this fellow, we
have a beautiful picture

that looks like this hopefully
But I'm going to try and draw.

STUDENT: Are you looking
from a top to bottom?

PROFESSOR: We can look it
whatever you want to look.

That's a very good thing.

You're getting too close
to what I wanted to go.

We'll discuss in one minute.

So you can imagine this
is a hill full of snow.

Although in two days,
we have Valentine's Day

and there is no snow.

But assume that we
go to New Mexico

and we find a hill
that more or less looks

like a perfect hill like that.

And we start thinking
of skiing down the hill.

Where am I at 0, 0?

I am on top of the hill.

I'm on top of the
hill and I decide

to analyze the slope
of the tangents

to the surface in the
direction of-- who is this?

Like now, and you
make me nervous.

So in the direction of
y, I have one slope.

In the direction of x, I have
another slope in general.

Only in this case, they
are the same slope.

And what is that same slope if
I'm here on top of the hill?

This is me-- well, I don't
know, one of you guys.

That looks horrible.

What's going to happen?

We don't want to think about it.

But it definitely is too steep.

So this will be the slope
of the line in the direction

with respect to y.

So I'm going to think
f sub y and f sub

x if I change my skis go
this direction and I go down.

So I could go down this
way and break my neck.

Or I could go down this way
and break my neck as well.

OK, it has to go like-- right?

Can you tell me what
these guys will be?

I'm going to put them in pink
because they're beautiful.

STUDENT: 0 [INAUDIBLE].

PROFESSOR: Thank God,
they are beautiful.

Larry, what does it mean?

That means that the two
tangents, the tangents

to the curves, are horizontal.

And if I were to draw the plane
between those two tangents--

one tangent is in pink
pen, our is in green.

Today, I'm all about colors.

I'm in a good mood.

And that's going to be the
so-called tangent plane--

tangent plane to the surface
at x0, y0, which is the origin.

That was a nice point.

That is a nice point.

Not all the points
will be [INAUDIBLE]

and nice but beautiful.

[INAUDIBLE] I take the
nice-- well, not so nice,

I don't know.

You'll have to figure it out.

How do I get-- well, first
of all, where is this point?

If I take x to be 1 over square
2 and y to be 1 over square 2

and I plug them in,
what's z going to be?

STUDENT: 0.

PROFESSOR: 0, and I
did that on purpose.

Because in that case, I'm
going to be on flat line again.

This look like [INAUDIBLE].

Except [INAUDIBLE]
is not z equal 0.

What is [INAUDIBLE] like?

z equals--

STUDENT: [INAUDIBLE]

PROFESSOR: Huh?

STUDENT: I don't know.

PROFESSOR: Do you want to
go in meters or in feet?

STUDENT: [INAUDIBLE].

It's about a mile.

PROFESSOR: Yes, I don't know.

I thought it's about one
kilometer, 1,000 something

meters.

But somebody said it's more so.

It's flat land, and I'd say
about a mile above the sea

level.

All right, now, I am going to
be in flat land right here,

1 over a root 2, 1
over a root 2, and 0.

What happened here?

Here, I just already
broke my neck, you know.

Well, if I came
in this direction,

I would need to draw a
prospective trajectory that

was hopefully not mine.

And the tangent would--
the tangent, the slope

of the tangent, would be funny.

Let's see what you need to do.

You need to say, OK, prime
with respect to x, minus 2x.

And then at the point x
equals 1 over [INAUDIBLE] 2 y

equals 1 over 2,
you just plug in.

And what do you have?

STUDENT: Square
root of [INAUDIBLE].

PROFESSOR: Negative
square root of 2-- my god

that is really bad as a slope.

It's a steep slope.

And this one-- how
about this one?

Same idea, symmetric function.

And it's going to be exactly
the same-- very steep slope.

Why are they negative numbers?

Because the slope is
going down, right?

That's the kind of slope I
have in both directions--

one and-- all right.

If I were to draw
this thing continuing,

how would I represent
those slopes?

This circle-- this circle is
just making my life harder.

But I would need to imagine
those slopes as being

like I'm here, all right?

Are you guys with me?

And I will need to draw
x0-- well, what is that?

1 over root 2 and 1 over root
2 And I would draw two planes.

And I would have two curves.

And when you slice
up, imagine this

would be a piece of cheese.

STUDENT: [INAUDIBLE]

PROFESSOR: And you cut--

STUDENT: [INAUDIBLE]

PROFESSOR: Right?

STUDENT: Yeah.

PROFESSOR: And you cut
in this other side.

Well, this is the one
that's facing you.

You cut like that.

And when you cut like
this, it's facing--

STUDENT: [INAUDIBLE]

PROFESSOR: Hm?

But anyway, let's not
draw the other one.

It's hard, right?

STUDENT: [INAUDIBLE] angle
like this-- just the piece

of the corner of the cheese.

PROFESSOR: Right.

STUDENT: The corner
is facing you.

PROFESSOR: So yeah, so it's--
the corner is facing you.

STUDENT: So basically,
you [INAUDIBLE] this.

PROFESSOR: But-- exactly, but--

STUDENT: Like this.

PROFESSOR: Yeah, well--
yeah, it's hard to draw.

So practically,
this is what you're

looking at it is slope that's
negative in both directions.

So you're going to go
this way or this way.

And it's much steeper than
you imagine [INAUDIBLE].

OK, they are equal.

I'm trying to draw them equal.

I don't know how
equal they can be.

One belongs to one plane
just like you said.

This belongs to this plane.

And the green one belongs to
the plane that's facing you.

So the slope goes this way.

But the two slopes are equal.

You have to have a little
bit of imagination.

We would need some cheese
to make a mountain of cheese

and cut them and slice them.

We'll eat everything
after, yeah.

All right, let's move on to
something more challenging

now that we got to
the tangent plane.

So if somebody would
say, wait a minute,

you said this is the tangent
plane to the surface.

You just introduced
a new notion.

You were fooling us.

I'm fooling you guys.

It's not April 1, but this
kind of a not a neat thing.

I just tried to introduce
you into the section 11.4.

So if you have a piece
of a curve that's smooth

and you have a point
x0, y0, can you

find out the equation
of the tangent plane?

Pi, and this is s form surface.

How can I find the equation
of the tangent plane?

That x0, y0-- 12 is
going to be also z0.

But what I mean that x0,
y0 is in on the floor

as a projection.

So I'm always
looking at the graph.

And that's why.

The moment I stop
looking at the graph,

things will be different.

But I'm looking at the graph
of independent variables x, y.

And that's why those guys
are always on the floor.

A and z would be a function
to keep in the variable.

Now, does anybody know?

Because I know you guys
are reading in advance

and you have better
teachers than me.

You have the internet.

You have the links.

You have YouTube.

You have Khan Academy.

I know from a bunch of you
that you have already gone

over half of the chapter 11.

I just hope that now you
can compare what you learned

with what I'm teaching
you, And I'm not

expecting you to go in
advance, but several of you

already know this formula.

We talked about it in
office hours on yesterday.

Because Tuesday, I
didn't have office hours.

I had a coordinator meeting.

So what equation corresponds
to the tangent plate?

STUDENT: [INAUDIBLE]

PROFESSOR: Several
of you know it.

You know what I hated?

It's fine that you know it.

I'm proud of you guys
and I'll write it.

But when I was a freshman--
or what the heck was I?

A sophomore I think-- no, I was
a freshman when they fed me.

They spoon-fed me this equation.

And I didn't understand
anything at the time.

I hated the fact that
the Professor painted it

on the board just like
that out of the blue.

I want to see a proof.

And he was able to-- I think
he could have done a good job.

But he didn't.

He showed us a bunch
of justifications

like if you generally have
a surface in implicit form,

I told you that
the gradient of F

represents the normal
connection, right?

And he prepared us pretty
good for what could

have been the proof of that.

He said, OK, guys.

You know the duration
of the normal

as even as the gradient over
the next of the gradient,

if you want unit normal.

How did he do that?

Well, he had a
bunch of examples.

He had the sphere.

He showed us that
for the sphere,

you have the normal,
which is the continuation

of the position vector.

Then he said, OK, you
can have approximations

of a surface that is smooth and
round with oscillating spheres

just the way you have for a
curve, a resonating circle,

a resonating circle-- that's
called oscillating circle.

Resonating circle-- in that
case, what will the normal be?

Well, the normal
will have to depend

on the radius of the circle.

So you have a principal normal
or a normal if it's a plane

curve.

And it's easy to
understand that's the same

as the gradient.

So we have enough
justification for the direction

of the gradient of such
a function is always

normal-- normal to the
surface, normal to all

the curves on the surface.

If we want to find
that without swallowing

this like I had to when I
was a student, it's not hard.

And let me show
you how we do it.

We start from the graph, right?

Z equals f of x and y.

And we say, well, Magdalena,
but this is a graph.

It's not an implicit equation.

And I'll say, yes it is.

Let me show you how I make
it an implicit equation.

I move z to the other side.

I put 0 equals f of xy minus z.

Now it is an implicit equation.

So you say you cheated.

Yes, I did.

I have cheated.

It's funny that whenever
somebody gives you a graph,

you can rewrite that
graph immediately

as an implicit equation.

So that implicit equation
is of the form big F of xyz

now equals a
constant, which is 0.

F of xy is your old
friend and minus z.

Now, can you tell me what is
the normal to this surface?

Yeah, give me a splash
in a minute like that.

So what is the gradient of f?

Gradient of f will
be the normal.

I don't care if
it's unit or not.

To heck with the unit or normal.

I'm going to say I wanted
prime with respect to x, y,

and z respectively.

And what is the gradient?

Is the vector.

Big F sub x comma big F
sub y comma big F sub z.

We see that last time.

So the gradient of a
function is the vector

whose coordinates are
the partial velocity--

your friends form last time.

Can we represent this again?

I don't know.

You need to help me.

Who is big F prime
with respect to x?

There is no x here.

Thank God that's
like a constant.

I just have to take this
little one, f, and prime it

with respect to x.

And that's exactly what that's
going to be-- little f sub x.

What is big F with respect to y?

STUDENT: [INAUDIBLE]

PROFESSOR: Little f sub
y prime with respect

to y-- differentiated
with respect to y.

And finally, if I differentiated
with respect to z,

there is no z here, right?

There is no z.

So that's like a constant.

Prime [INAUDIBLE] 0 and minus 1.

So I know the gradient.

I know the normal.

This is the normal.

Now, if somebody gives you
the normal, there you are.

You have the normal to the
surface-- normal to surface.

What does it mean?

Equals normal to the tangent
plane to the surface.

Normal or perpendicular
to the tangent plane-

to the plane-- of the surface.

At that point--
point is the point p.

All right, so if you were to
study a surface that's-- do you

have a [INAUDIBLE]?

STUDENT: Uh, no.

Do you?

PROFESSOR: OK.

OK, I want to study
the tangent plane

at this point to the surface.

Well, that's flat, Magdalena.

You have no imagination.

The tangent plane is this plane,
is the same as the surface.

So, no fun-- no fun.

How about I pick my
favorite plane here

and I take-- what is-- OK.

I have-- this is
Children Internationals.

I have a little girl
abroad that I'm sponsoring.

So you have a point
here and a plane

that passes through that point.

This is the tangent plane.

And my finger is the normal.

And the normal, we call
that normal to the surface

when it's normal to
the tangent plane.

At every point, this
is what the normal is.

All right, can we write
that based on chapter nine?

Now I will see what you remember
from chapter nine if anything

at all.

All right, how do we
write the tangent plane

if we know the normal?

OK, review-- if the normal
vector is ai plus bj plus ck,

that means the plane that
is perpendicular to it

is of what form?

Ax plus by plus cz
plus d equals 0, right?

You've learned that
in chapter nine.

Most of you learned that
last semester in Calculus 2

at the end.

Now, if my normal is f sub
x, f sub y, and minus 1,

those are ABC for God's sake.

Well, good.

Big A, big B, big C
at the given point.

So I'm going to have f sub
x at the given point d times

x plus f sub y at any
given point d times y.

Who is c?

C is minus 1.

Minus 1 times z is--
say you're being silly.

Magdalena, why do
you write minus 1?

Just because I'm having fun.

And plus, d equals 0.

And you say, well,
wait, wait, wait.

This starts looking like that
but it's not the same thing.

All right, what?

How do you get to d?

Now, actually, the
plane perpendicular

to n that passes
through a given point

can be written
much faster, right?

So if a plane is perpendicular
to a certain line,

how do we write if
we know a point?

If we know a point
in the normal ABC--

I have to go backwards to
read it backwards-- then

the plane is going
to be x minus x0

plus b times y times y0 plus
c times z minus c0 equals 0.

So who is the d?

The d is all the constant
that gets out of here.

So the point x0, y0, z0
has to verify the plane.

And that's why when
you plug in x0, y0, z0,

you get 0 plus 0
plus 0 equals 0.

That's what it means for a
point to verify the plane.

When you take the x0, y0, z0 and
you plug it into the equation,

you have to have an
identity 0 equals 0.

So this can be rewritten
zx plus by plus cz

just like we did there plus a d.

And who in the world is the d?

The d will be exactly minus
ax0 minus by0 minus cz0.

If that makes you uncomfortable,
this is in chapter nine.

Look at the equation of a
plane and the normal to it.

Now I know that I can do
better than that if I'm smart.

So again, I collect the ABC.

Now I know my ABC.

I put them in here.

So I have f sub x at
the point in time.

Oh, OK, x minus x0
plus, who is my b?

F sub y computed at the
point p times y minus y0.

And, what?

Minus, right?

Minus-- minus 1.

I'm not going to write minus 1.

You're going to make fun of me.

Minus z minus cz.

And my proof is done.

QED-- what does it mean, QED?

In Latin.

QED means I proved
what I wanted to prove.

Do you know what it stands for?

Did you take Latin, any of you?

You took Latin?

Quod erat demonstrandum.

So this was to be proved.

That's exactly what
it was to be proved.

That, what, that
c minus z0, which

was my fellow over here pretty
in pink, is going to be f sub x

times x minus x0 plus yf
sub y times y minus y0.

So now you know why the equation
of the tangent plane is that.

I proved it more or less,
making some assumptions,

some axioms as assumption.

But you don't know
how to use it.

So let's use it.

So for the same valley--
not valley, hill--

it was full of snow.

Z equals 1 minus x
squared-- what was you

guys have forgotten?

OK, 1 minus x squared
minus y squared.

Find the tangent plane
at the following points.

Ah, x0, y0 to be origin.

And you say, did you
say that that's trivial?

Yes, it is trivial.

But I'm going to do
it one more time.

And what was my
[INAUDIBLE] point before?

STUDENT: [INAUDIBLE]

PROFESSOR: 1 over
2 and 1 over 2.

OK, and what will be the
corresponding point in 3D?

1 over 2, 1 over 2, I plug in.

Ah, yes.

And with this, I hope
to finish the day so we

can go to our other businesses.

Is this hard?

Now, I was not able-- I
have to be honest with you.

I was not able to memorize the
equation of a tangent plane

when I was-- when I was young,
like a freshman and sophomore.

I wasn't ready to
understand that this

is a linear approximation
of a curved something.

This practically like
the Taylor equation

for functions of
two variables when

you neglect the quadratic
third term and so on.

You just take the--
I'll teach you

next time when this is, a first
order linear approximation.

All right, can we do
this really quickly?

It's going to be
a piece of cake.

Let's see.

Again, how do we do that?

This is f of x and y.

We computed that again.

F of 0, 0 was this 0.

Guys, if I say something
silly, will you stop me?

F of f sub x-- f
of y at 0, 0 is 0.

So I have two slopes.

Those are my hands.

The slopes of my hands are 0.

So the tangent plane will
be z minus z0 equals 0.

What is the 0?

STUDENT: 1

PROFESSOR: 1, excellent.

STUDENT: [INAUDIBLE]

PROFESSOR: Why is that 1?

0 and 0 give me 1.

So that was the picture
that I had z equals 1

as the tangent plane at
the point corresponding

to the origin.

That look like the
north pole, 0, 0, 1.

OK, no.

It's the top of a hill.

And finally, one last
thing [INAUDIBLE].

Maybe you can do
this by yourselves,

but I will shut up if I can.

I can't in general,
but I'll shut up.

Let's see-- f sub x at 1
over root 2, 1 over root 2.

Why was that?

What is f sub x?

STUDENT: The square root of--
negative square root of 2.

PROFESSOR: Right,
we've done that before.

And you got exactly what
you said-- [INAUDIBLE]

2 f sub y at the same point.

I am too lazy to write it
down again-- minus root 2.

And how do we actually
express the final answer

so we can go home and
whatever-- to the next class?

Is it hard?

No.

What's the answer?

Z minus-- now, attention.

What is z0?

STUDENT: 0.

PROFESSOR: 0, right.

Why is that?

Because when I plug 1 over
a 2, 1 over a 2, I got 0.

0-- do I have to write it down?

No, not unless I
want to be silly.

But if you do write
down everything

and you don't simplify
the equation of the plane,

we don't penalize you in
any way in the final, OK?

So if you show your work like
that, you're going to be fine.

What is that 1 over 2?

Plus minus root 2 times
y minus 1 over root 2.

Is it elegant?

No, it's not elegant at all.

So as the last row for
today, one final line.

Can we make it
look more elegant?

Do we care to make
it more elegant?

Definitely some of you care.

Z will be minus root 2x.

I want to be consistent and
keep the same style in y.

And yet the constant
goes wherever

it wants to go at the end.

What's that constant?

STUDENT: 2 [INAUDIBLE].

PROFESSOR: So you
see what you have.

You have this times that.

It's a 1, this then that is a 1.

1 plus 1 is 2.

All right, are you
happy with this?

I'm not.

I'm happy.

You-- if this were
a multiple choice,

you would be able to
recognize it right away.

What's the standardized general
equation of a plane, though?

Something x plus something y
plus something z plus something

equals 0.

So if you wanted to
make me very happy,

you would still move everybody
to the left hand side.

Do you want equal to or minus 3?

Yes, it does.

STUDENT: [INAUDIBLE]

PROFESSOR: Huh?

Negative 2-- is that OK?

Is that fine?

Are you guys done?

Is this hard?

Mm-mm.

It's hard?

No.

Who said it's hard?

So-- so I would work more
tangent planes next time.

But I think it's something
that we can practice on.

And do expect one exercise
like that from one

of those, God knows,
15, 16 on the final.

I'm not sure about the midterm.

I like this type of problem.

So you might even see
something with tangent planes

on the midterm-- normal to
a surface tangent plane.

It's a good topic.

It's really pretty.

For people who like to draw,
it's also nice to draw them.

But do you have to?

No.

Some of you don't like to.

OK, so now I say thank
you for the attendance

and I'll see you next time
on Thursday-- on Tuesday.

Happy Valentine's Day.