TTU Math2450 Calculus3 Sec 10.2 and 10.4 part 1

Derivatives of Vector Value Functions

PROFESSOR: Any
questions about theory

that gave you headaches
regarding homework

you'd like to talk about?

Anything related
to what we covered

from chapter nine and today?

STUDENT: Can we
do some problems?

PROFESSOR: I can
fix from problems

like the ones in the
homework, but also I

can have you tell me what
bothers you in the homework.

STUDENT: Oh, I have [INAUDIBLE].

PROFESSOR: What bothered
me about my own homework

was that I realized that I
did not remind you something

I assume you should
know, which is

the equation of a sphere of
given center and given radius.

And since I trust you so much,
I said, OK they know about it.

And then somebody asked
me by email what that was,

and I said, oh, yeah.

I did not review that in class.

So review the equation
in r3 form that's x, y, z

of the sphere of radius r and
center p of coordinates x0, y0,

z0.

One of you asked me by email,
does-- of course you do,

and then if you know it,
can you help me-- can you

help remind what that was?

STUDENT: x minus x0--

PROFESSOR: x minus x0 squared
plus y minus y0 squared

plus z minus z0 squared
equals R squared.

OK?

When you ask, for
example, what is

the equation of a units sphere,
what do I mean by unit sphere?

STUDENT: Radius--

PROFESSOR: Radius 1, and
center 0, standard unit sphere,

will be.

There is a notation for that
in mathematics called s2.

I'll tell you why its called s2.

x squared plus y squared
plus z squared equals 1.

s2 stands for the dimension.

That means the number
of the-- the number

of degrees of freedom.

you have on a certain manifold.

What is a manifold?

It's a geometric structure.

I'm not going to
go into details.

It's a geometric structure
with some special properties.

I'm not talking about
other fields of algebra,

anthropology.

I'm just talking about geometry
and calculus math 3, which

is multivariable calculus.

Now, how do I think
of degrees of freedom?

Look at the table.

What freedom do I have to move
along one of these sticks?

I have one degree of
freedom in the sense

that it's given by a
parameter like time.

Right?

It's a 1-parameter
manifold in the sense

that maybe I have
a line, maybe I

have the trajectory of the
parking space in terms of time.

The freedom that the bag has
is to move according to time,

and that's considered only
one degree of freedom.

Now if you were on a
plane or another surface,

why would you have more
than one degrees of freedom?

Well, I can move towards
you, or I can move this way.

I can draw a grid the way
the x and y coordinate.

And those are my
degrees of freedom.

Practically, the basis
IJ gives me that kind

of two degrees of freedom.

Right?

If I'm in three coordinates, I
have without other constraints,

because I could be
in three coordinates

and constrained to be on
a cylinder, in which case

I still have two
degrees of freedom.

But if I am a bug
who is free to fly,

I have the freedom to go with
three degrees of freedom,

right?

I have three degrees of
freedom, but if the bug

is moving-- not flying,
moving on a surface,

then he has two
degrees of freedom.

So to again review, lines
and curves in general

are one dimensional
things, because you

have one degree of freedom.

Two dimensional
things are surfaces,

three dimensional things are
spaces, like the Euclidean

space, and we are not
going to go beyond,

at least for the time
being, we are not

going to go beyond that.

However, where anybody is
interested in relativity,

say or let's say four
dimensional spaces, or things

of x, y, z spatial coordinates
and t as a fourth coordinate,

then we can go into higher
dimensions, as well.

OK.

I want to ask you a question.

If somebody gives you on
WeBWorK or outside of WeBWorK,

on the first quiz or
on the final exam,

let's say you have
this equation,

x squared plus y squared plus
z squared plus 2x plus 2y

equals 9.

What is this identified as?

It's a quadric.

Why would this be a quadric?

Well, there is no x, y, y, z.

Those terms are missing.

But I have something of the
type of quadric x squared

plus By squared plus
c squared plus dxy

plus exz plus fyz plus, those
are, oh my God, so many.

Degree two.

Degree one I would
have ax plus by plus cz

plus a little d constant, and
whew, that was a long one.

Right?

Now, is this of the
type of a project?

Yes, it is.

Of course there are some terms
that are missing, good for us.

How are you going to try to
identify the type of quadric

by looking at this?

As you said very well,
I think it's-- you say,

I think of a sphere, maybe I can
complete the squares, you said.

How do we complete the squares?

x squared plus 2x plus
some missing number,

a magic number-- yes sir?

STUDENT: So, basically I'll
have to take x plus 2 times 4

will go outside.

It's like x min-- x plus 2--

PROFESSOR: Why x plus 2?

STUDENT: Because it's 2x--

STUDENT: It's 2x.

PROFESSOR: But if
I take x plus 2,

then that's going to give
me x squared plus 4x plus 4,

so it's not a good idea.

STUDENT: On the x plus 1

PROFESSOR: x plus 1.

So I'm going to complete
x plus 1 squared.

What did I invent
that wasn't there?

STUDENT: 1.

PROFESSOR: I invented
the 1, and I have

to compensate for my invention.

I added the 1, created
the 1 out of nothing,

so I have to compensate
by subtracting it.

How much is from here to here?

Is it exactly the
thing that I underlined

with a wiggly line, a
light wiggly line thing,

plus what is the
blue wiggly line,

the blue wiggly line
that doesn't show--

I have y plus 1 squared, and
again, I have to compensate

for what I invented.

I created a 1 out of nothing,
so this is y squared plus 2y.

The z squared is all by himself,
and he's crying, I'm so lonely,

I don't know, there is
nobody like me over there.

So in the end, I can rewrite
the whole thing as x plus 1

squared plus y plus 1
squared plus z squared, if I

want to work them out in
this format, equals what?

STUDENT: 10.

PROFESSOR: 11.

11 is the square
root of 11 squared.

Like my son said the other day.

So that the radius
would be square foot

11 of a sphere of what circle?

What is the-- or the
sphere of what center?

STUDENT: Minus 1--

PROFESSOR: Minus
1, minus 1, and 0.

So I don't want to insult you.

Of course you know how
to complete squares.

However, I have discovered in an
upper level class at some point

that my students didn't know
how to complete squares, which

was very, very heartbreaking.

All right, now.

Any questions regarding--
while I have a few of yours,

I'm going to wait
a little bit longer

until I give
everybody the chance

to complete the extra credit.

I have the question
by email saying,

you mentioned that
genius guy in your class.

This is a 1-sheeted hyperboloid.

x squared plus y squared minus
z squared minus 1 equals 0.

The question was, by
email, how in the world,

did he figure out what the two
families of generatrices are?

So you have one family
and another family,

and both together generate
the 1-sheeted hyperboloid.

Let me give you a little
bit more of a hint,

but I'm still going to stop.

So last time I said, he
noticed you can root together

the y squared minus 1 and the
x squared minus z squared,

and you can separate them.

So you're going to have x
squared minus z squared equals

1 minus y squared.

You can't hide the
difference of two squares

as product of sum
and difference.

x plus z times x minus z equals
1 plus y times 1 minus y.

So how can you
eventually arrange stuff

to be giving due
to the lines that

are sitting on the surface?

The lines that are
sitting on the surface

are infinitely many,
and I would like

at least a 1-parameter
family of such lines.

You can have choices.

One of the choices
would be-- this

is a product, of
two numbers, right?

So you can write it as an
equality of two fractions.

So you would have something
like x plus z on top, x minus

z below.

Observe that you are
creating singularities here.

So you have to take x minus
z case equals 0 separately,

and then you have, let's
say you have 1 minus y here,

and 1 plus y here.

What else do you have to impose
when you impose x minus z

equals 0.

You cannot have 7 over 0.

That is undefined.

but if you have 0 over
0, that's still possible.

So whenever you take x
minus z equals 0 separately,

that will imply that the
numerator corresponding to it

will also have to be 0.

And together these
guys are friends.

What are they?

2--

STUDENT: A system of equations.

PROFESSOR: It's a
system of equations.

They both represent planes, and
the intersection of two planes

is a line.

It's a particular line, which
is part of the family-- which

is part of a family.

OK.

Now, on the other hand, in case
you have 1 plus y equals 0--

so if it happens that you
have this extreme case

that the denominator
will be 0, you absolutely

have to impose x plus z to be 0,
and then you have another life.

It's not easy for
me to draw those,

but I could if you
asked me privately

to draw those and show you
what the lines look like.

OK?

All right.

So you have two special lines
that are part of that picture.

They are embedded
in the surface.

How do you find a
family of planes?

Oh my god, I only
had one choice,

but I could have
yet another choice

of how to pick the parameters.

Let's take lambda to be
a real number parameter.

And lambda could be
anything-- if lambda is 0,

what have I got to have, guys?

STUDENT: The top.

PROFESSOR: The top
guys will be 0,

and I still have 1 minus y
equals 0, a plane, intersected

with x plus z equals 0,
another plane, so still a line.

So lambda equals 0 will give
me yet another line, which

is not written big.

Are you guys with me?

Could lambda ever
go to infinity?

Lambda wants to go to
infinity, and when does lambda

go to infinity?

STUDENT: When the
bottoms would equal 0--

PROFESSOR: When both
the bottoms would be 0.

So this is-- I can call it L
infinity, the line of infinity.

You see?

But still those
would be two planes.

There's an intersection,
it's a line.

OK.

Can we write this family--
just one family of lines?

A line is always an intersection
of two planes, right?

So which are the planes
that I'm talking about?

x plus z equals
lambda times 1 plus y.

This is not in the book,
because, oh my God, this is

too hard for the book, right?

But it's a nice example to
look at in an honors class.

1 minus y equals
lambda times x minus z.

It's not in the book.

It's not in any book that I know
of at the level of calculus.

All right, OK.

What are these animals?

The first animal is a plane.

The second animal is a plane.

How many planes
are in the picture?

For each lambda, you have a--
for each lambda value in R,

you have a couple of planes
that intersect along your line.

This is the line L lambda.

And shut up, Magdalena,
you told people too much.

If you still want them to do
this for 2 extra credit points,

give them the chance
to finish the exercise.

So I zip my lips, but
only after I ask you,

how do you think
you are going to get

the other family of rulers?

The ruling guys are
two families, you see?

So this family is
going in one direction.

How am I going to
get two families?

I have another choice
that-- how did I take this?

More or less, I made my choice.

Just like having two
people that would

be prospective job candidates.

You pick one of them.

STUDENT: Now, we can put 1
minus y in the denominator.

The denominator in
place of 1 plus y.

PROFESSOR: So I could have
done-- I could have taken this,

and put 1 plus y here,
and 1 minus y here.

I'm going to let
you do the rest,

and get the second
family of generators

for the whole surface.

That's enough.

You're not missing your credit.

Just, you wanted help,
and I helped you.

And I'm not mad whatsoever
when you ask me things.

The email I got sounded like--
says, this is not in the book,

or in any book, or
on the internet.

How shall I approach this?

How shall I start thinking
about this problem?

This is a completely
legitimate question.

How do I start on this problem?

OK.

On the homework-- maybe it's
too easy-- you have two or three

examples involving spheres.

Those will be too easy for you.

I only gave you a very thin
among of homework this time.

You Have plenty of time until
Monday at 1:30 or something PM.

I would like to draw
a little bit more,

because in this homework
and the next homework,

I'm building something special
called the Frenet Trihedron.

And I told you a little bit
about this Frenet Trihedron,

but I didn't tell you much.

Many textbooks in
multivariable calculus

don't say much about it,
which I think is a shame.

You have a position
vector that gives you

the equation of a regular curve.

x of t, y of t, z of t.

Again, what was a regular curve?

I'm just doing review of
what we did last time.

A very nice curve
that is differentiable

and whose derivative is
continuous everywhere

on the interval.

But moreover, the r prime
of t never becomes 0.

So continuously differentiable,
and r prime of t

never becomes 0 for any--
do you know this name,

any for every or for any?

OK.

This is the symbolistics
of mathematics.

You know because you
are as nerdy as me.

But everybody else doesn't.

You guys will learn.

This is what
mathematicians like.

You see, mathematicians hate
writing lots of words down.

If we liked writing essays
and lots of blah, blah, blah,

we would do something else.

We wouldn't do mathematics.

We would do debates,
we would do politics,

we would do other things.

Mathematicians like
ideas, but when

it comes to writing
them down, they

want to right them down in
the most compact way possible.

That's why they created
sort of their own language,

and they have all sorts
of logical quantifiers.

And it's like your
secret language

when it comes to your
less nerdy friends.

So you go for every--
for any or for every--

do you know this sign?

There exists.

And do you know this thing?

Because one of the-- huh?

STUDENT: Is that factorial?

PROFESSOR: Factorial,
but in logic,

that means there exists
a unique-- a unique.

So there exists a unique.

There exists a unique number.

There is a unique number.

So we have our own language.

Of course, empty set,
everybody knows that.

And it's used in
mathematical logic a lot.

You know most of the symbols
from unit intersection,

or, and.

I'm going to use some
of those as well.

Coming back to the
Frenet Trihedron,

we have that velocity
vector at every point.

We are happy with it.

We have our prime of t
that is referred from 0.

I said I want to
make it uniform,

and then I divided
by the magnitude,

and I have this wonderful t
vector we just talked about.

Mr. t is r prime over the
magnitude of r prime, which

is called it's peak right?

We divide by its peak.

What's the name of t, again?

STUDENT: Tangent unit--

PROFESSOR: Tangent
unit vector, very good.

How did you remember
that so quickly?

Tangent unit vector.

There is also another
guy who is famous.

I wanted to make him
green, but let's see

if I can make him blue.

t is defined-- should I
write the f on top of here?

Do you know what that is?

STUDENT: I thought n
was the normal vector.

PROFESSOR: t prime
divided by the length of--

STUDENT: Wait.

I thought the vector
n was the normal.

PROFESSOR: n-- there
are many normals.

It's a very good thing, because
we don't say that in the book.

OK, this is the t along my r.

Now when I go through a point,
this is the normal plane,

right?

There are many normals to
the surface-- to the curve.

Which one am I taking?

All of them are perpendicular
to the direction, right?

STUDENT: tf.

PROFESSOR: So I take
this one, or this one,

or this one, or this one, or
this one, or this one, there.

I have to make up my mind.

And that's how people came up
with the so-called principal

unit normal.

And this is the one
I'm talking about.

And you are right, it is normal.

Principal unit normal.

Remember this very
well for your exam,

because it's a very
important notion.

How do I get to that?

I take t, I differentiate
it, and I divide

by the lengths of t prime.

Now, can you prove to me
that indeed this fellow

is perpendicular to t?

Can you do that?

STUDENT: That n is
perpendicular to t?

PROFESSOR: Mm-hmm.

So a little exercise.

Prove that-- Prove that I don't
have a good marker anymore.

Prove that n, the unit
principal vector field,

is perpendicular-- you
see, I'm a mathematician.

I swear, I hate to write down
the whole word perpendicular.

I would love to
say, perpendicular.

That's how I write perpendicular
really fast-- to t fore

every value of t.

For every value of t.

OK.

How in the world can I do that?

I have to think about it.

This is hard.

Wish me luck.

So do I know
anything about Mr. t?

What do I know about Mr. t?

I'll take it and I'll
differentiate it later.

It Mr. t is magic in the
sense that he's a unit vector.

I'm going to write that down.

t in absolute value equals 1.

It's beautiful.

If I squared that-- and
you're going to say,

why would you want
to square that?

You're going to see in a minute.

If I squared that,
then I'm going

to have the dot product
between t and itself equals 1.

Can somebody tell me why the
dot product between t and itself

is the square of a length of t?

What's the definition
of the dot product?

Magnitude of the first
vector, times the magnitude

of the second vector--
there i am already--

times the cosine of the
angle between the two vectors

Duh, that's 0.

So cosine of 0 is 1, I'm done.

Right?

Now, I have a vector function
times a vector function--

this is crazy, right-- equals 1.

I'm going to go ahead
and differentiate.

Keep in mind that
this is a product.

What's the product?

One of my professors,
colleagues,

was telling me, now,
let's be serious.

In five years, how many
of your engineering majors

will remember the product?

I really was
thinking about this.

I hope everybody, if
they were my students,

because we are going to
have enough practice.

So the prime rule in
Calc 1 said that if you

have f of t times g of
t, you have a product.

You prime that product,
and never write

f prime times g prime unless you
want me to call you around 2 AM

to say you should never do that.

So how does the
product rule work?

The first one prime
times the second unprime

plus the first one unprime
times the second prime.

My students know
the product rule.

I don't care if the rest
of the world doesn't.

I don't care about any
community college who

would say, I don't want the
product rule to be known,

you can differentiate
with a calculator.

That's a no, no, no.

You don't know calculus if you
don't know the product rule.

So the product rule is
a blessing from God.

It helps everywhere in physics,
in mechanics, in engineering.

It really helps in
differential geometry

with the directional
derivative, the Lie derivative.

It helps you understand all
the upper level mathematics.

Now here you have t prime,
the first prime times

the second unprime, plus the
first unprime times the second

prime.

It's the same as for
regular scalar functions.

What's the derivative of 1?

STUDENT: 0.

PROFESSOR: 0.

Look at this guy!

Doesn't he look funny?

It is the dot product community.

Yes it is, by definition.

So you have twice T
times T prime equals 0.

This 2 is-- stinking
guy, let's divide by 2.

Forget about that.

What does this say?

The dot product of T times--
I mean by T prime is 0.

When are two vectors
giving you dot product 0?

STUDENT: When they're
perpendicular.

PROFESSOR: So if both
of them are non-zero,

they have to be like that.

They have to be like this,
perpendicular, right?

So it follows that t has to
be perpendicular to T prime.

And now, that's why n
is perpendicular to t.

But, because n is
collinear to t prime.

Hello.

n is collinear to t prime.

So this is t prime.

Is t prime unitary?

I'm going to measure it.

No it's not.

t prime.

So if I want to
make it unitary, I'm

going to chop my-- no,
I'm not going to chop.

I just take it, t prime,
and divide by its magnitude.

Then I'm going to get that
vector n, which is unitary.

So from here it follows that t
and n are indeed perpendicular,

and your colleague over there
said, hey, it has to be normal.

That's perpendicular
to t, but which one?

A special one, because
I have many normals.

Now, this special one is
easy to find like that.

Where shall I put here--
I'll draw him very nicely.

I'll draw him.

Now you guys have to
imagine-- am I drawing

well enough for you?

I don't even know.

t and n should be perpendicular.

Can you imagine them having that
90 degree angle between them?

OK.

Now there is a magic one that
you don't even have to define.

And yes sir?

STUDENT: In this
thing, can [INAUDIBLE]

this T vector [INAUDIBLE]
written by the definition

thing?

PROFESSOR: No.

STUDENT: N vector
times the magnitude

of t vector derivative?

PROFESSOR: So
technically you have

t prime would be the
magnitude of t prime times n.

STUDENT: Yes.

PROFESSOR: But keep in mind
that sometimes is tricky,

because this is, in
general, not a constant.

Always keep it in mind,
it's not a constant.

We'll have some examples later.

There is a magic
guy called binormal.

That binormal is the
normal to both t and n.

And he's defined as
t plus n because it's

normal to both of them.

So I'm going to write this
b vector is t cross n.

Now I'm asking you to draw it.

Can anybody come to
the board and draw it

for 0.01 extra credit?

Yes, sir?

STUDENT: [INAUDIBLE]

PROFESSOR: Draw that on the
picture like t and n, t and n,

t is the-- who the heck
is t? t is the red one,

and blue is the n.

So does it go down or up?

We should be perpendicular
to both of them.

Is b unitary or not?

If you have two unit vectors,
will the cross product

be a unit vector?

Only if the two vectors
are perpendicular,

it is going to be, right?

So you have-- well, I
think it goes that--

in which direction does it go?

Because

STUDENT: It should
not be how we have it.

PROFESSOR: No, no, no.

Because this is--

STUDENT: Yeah.

I'm using--

PROFESSOR: So t
goes over n, so I'm

going to try-- it is
like that, sort of.

STUDENT: Into the chord?

PROFESSOR: So again, it's
not very clear because

of my stinking art, here.

It's really not nice art.

t, and this is n.

And if I go t going over n.

T going over n goes up or down?

STUDENT: Down.

PROFESSOR: Goes down.

So it's going to look
more like this, feet.

Now guys, when we--
thank you so much.

So you've like a
0.01 extra credit.

OK.

Tangent, normal, and
binormal form a corner.

Yes, sir?

STUDENT: Is rt-- rt is
the function at the--

for the flag that's flying?

PROFESSOR: The r of t
is the position vector

of the flag that was
flying that he was drunk.

STUDENT: Why wasn't the
derivative of it perpendicular?

Why isn't t perpendicular to rt?

PROFESSOR: If--
well, good question.

We'll talk about it.

If the length of r
would be a constant,

can we prove that r and r
prime are perpendicular?

Let's do that as
another exercise.

All right?

So tnb looks like a corner.

Look at the corner that the
video cannot see over there.

TN and B are mutually octagonal.

I'm going to draw them.

This is an arbitrary
point on a curve,

and this is t, which is
always tangent to the curve,

and this is n.

Let's say that's the
unit principle normal.

And t cross n will
go, again, down.

I don't know.

I have an obsession
with me going down.

This is called the
Frenet Trihedron.

And I have a proposal
for a problem

that maybe I should give
my students in the future.

Show that for a circle,
playing in space, I don't know.

The position vector and the
velocity vector are always how?

Friends.

Let's say friends.

No, come on, I'm kidding.

How are they?

STUDENT: Perpendicular.

PROFESSOR: How do you do that?

Is it hard?

We should be smart
enough to do that, right?

I have a circle.

That circle has what-- what
is the property of a circle?

Euclid defined that-- this is
one of the axioms of Euclid.

Does anybody know which axiom?

That there exists
such a set of points

that are all at the same
distance from a given point

called center.

So that is a circle, right?

That's what Mr. Euclid said.

He was a genius.

So no matter where I put that
circle, I can take r of t

in magnitude measured
from the origin

from the center of the circle.

Keep in mind, always the
center of the circle.

I put it at the origin of the
space-- origin of the universe.

No, origin of the
space, actually.

R of T magnitude
would be a constant.

Give me a constant, guys.

OK?

It doesn't matter.

Let me draw.

I want to draw in plane, OK?

Because I'm getting tired.
x y, and this is r of t,

and the magnitude of this r of
t is the radius of the circle.

Right?

So let's say, this is
the radius of the circle.

How in the world do I
prove the same idea?

Who helps me prove
that r is always

perpendicular to r prime?

Which way do you want to move,
counterclockwise or clockwise?

STUDENT: Counterclockwise.

PROFESSOR: Counterclockwise.

Because if you are
a real scientist,

I'm proud of you guys.

It's clear from the
picture that r prime

would be perpendicular to r.

Why is that?

How am I going to do that?

Now, mimic everything I--
don't look at your notes,

and try to tell me how
I show that quickly.

What am I going to do?

So all I know, all
that gave me was r of t

equals k in magnitude constant.

For every t, this same constant.

What's next?

What do I want to do next?

STUDENT: Square it?

PROFESSOR: Square
it, differentiate it.

I can also go ahead
and differentiate it

without squaring
it, but that's going

to be a little bit of more pain.

So square it, differentiate it.

I'm too lazy.

When I differentiate,
what am I going to get?

From the product rule, twice
r dot r primed of t equals 0.

Well, I'm done.

Because it means that for
every t that radius-- not

the radius, guys, I'm sorry.

The position vector will be
perpendicular to the velocity

vector.

Now, if I draw the
trajectory of my drunken flag

this [INAUDIBLE]
is not true, right?

This is crazy.

Of course this is r,
and this is r prime,

and there is an arbitrary
angle between r and r prime.

The good thing is that
the arbitrary angle always

exists, and is
continuous as a function.

I never have that
angle disappear.

That's way I want that
prime never to become 0.

Because if the bag was
stopping its motion,

goodbye angle, goodbye
analysis, right?

OK.

Very nice.

So don't give me more ideas.

You smart people, if
you give me more ideas,

I'm going to come up with
all sorts of problems.

And this is actually one
of the first problems

you learn in a graduate
level geometry class.

Let me give you another
piece of information

that you're going
to love, which could

be one of those
types of combined

problems on a final
exam or midterm,

A, B, C, D, E. The
curvature of a curve

is a measure of how
the curve will bend.

Say what?

The curvature of a
curve is a measure

of the bending of that curve.

By definition, you have
to take it like that.

If the curve is parameterized
in arc length-- somebody

remind me what that is.

What does it mean?

That is r of s such
that-- what does it mean,

parameterizing arc length--

STUDENT: r prime of s.

PROFESSOR: r primed of
s in magnitude is 1.

The speed 1.

It's a speed 1 curve.

Then, the curvature of this
curve is defined as k of s

equals the magnitude of
the acceleration vector

will respect the S.
Say what, Magdalena?

I can also write
it magnitude of d--

oh my gosh, second derivative
with respect s of r.

I'll do it right now. d2r ds2.

And I know you get
a headache when

I solve, when I write that,
because you are not used to it.

A quick and beautiful example
that can be on the homework,

and would also be on the
exam, maybe on all the exams,

I don't know.

Compute the curvature of a
circle of radius a Say what?

Compute the curvature of a
circle of radius a And you say,

wait a minute.

For a circle of
radius a in plane--

why can I assume it's in plane?

Because if the circle
is a planar curve,

I can always assume
it to be in plane.

And it has radius a I
can find infinitely many

parameterizations.

So what, am I crazy?

Well, yes, I am, but
that's another story.

Now, if I want to
parameterize, I

have to parameterize
in arc length.

If I do anything else,
that means I'm stupid.

So, r of s will be what?

Can somebody tell me
how I parameterize

a curve in arc length
for a-- what is this guy?

A circle of radius a.

Yeah, I cannot do it.

I'm not smart enough.

So I'll say R of T will be
a cosine t, a sine t and 0.

And here I stop, because
I had a headache.

t is from 0 to 2 pi, and
I think this a is making

my life miserable,
because it's telling me,

you don't have
speed 1, Magdalena.

Drive to Amarillo
and back, you're

not going to get speed 1.

Why don't I have speed 1?

Think about it.

Bear with me.

Minus a sine t equals sine t, 0.

Bad.

What is the speed?

a.

If you do the math,
the speed will be a.

So length of our
prime of t will be a.

Somebody help me.

Get me out of trouble.

Who is this?

I want to do it in arc length.

Otherwise, how can
I do the curvature?

So somebody tell
me how to get to s.

What the heck is that?

s of t is integral from
0 to t of-- who tells me?

The speed, right?

Was it not the displacement,
the arc length traveled along,

and the curve is integral
in time of the speed.

OK?

So I have-- what is that?

Speed is?

STUDENT: Um--

PROFESSOR: a.

So a time t, am I right,
guys? s is a times t.

So what do I have to do?

Take Mr. t, shake his hand,
and replace him with s over a.

OK.

So instead of r of t, I'll say--
what other letters do I have?

Not r.

Rho of s.

I love rho.

Rho is the Greek [INAUDIBLE].

Is this finally an arc length?

Cosine of-- what
is t, guys, again?

s over a.

s over a, a sine
s over a, and 0.

This is the parameterization
in arc length.

This is an arc length
parameterization of the circle.

And then what is this
definition of curvature?

It's here.

Do that rho once, twice.

Prime it twice,
and do the length.

So rho prime.

Oh my God is it hard.

a times minus sine of s over a.

Am I done, though?

Chain rule.

Pay attention, Magdalena.

Don't screwed up with this one.

1 over a.

Good.

Next.

a cosine of s over a.

Chain rule.

Don't forget,
multiply by 1 over a.

OK, that makes my life easier.

We simplify.

Thank God a simplifies
here, a simplifies there,

so that is that derivative.

What's the second derivative?

Rho double prime of s will
be-- somebody help me, OK?

Because this is a
lot of derivation.

STUDENT: --cosine--

PROFESSOR: Thank you, sir.

Minus cosine of s over a.

STUDENT: Times 1 over a.

PROFESSOR: Times 1 over a,
comma, minus sine of s over a.

That's all I have left
in my life, right?

Minus sine of s over a times
1 over a from the chain rule.

I have to pay attention and see.

What's the magnitude of this?

The magnitude of this of this
animal will be the curvature.

Oh, my God.

So what is k?

k of s will be--
could somebody tell me

what magnitude I get after I
square all these individuals,

sum them up, and take
the square root of them?

STUDENT: [INAUDIBLE]

PROFESSOR: Square root
of 1 over 1 squared.

And I get 1 over a.

You are too fast for
me, you teach me that.

No, I'm just kidding.

I knew it was 1 over a.

Now, how did
engineers know that?

Actually, for hundreds of years,
mathematicians, engineers,

and physicists knew that.

And that's the last thing
I want to teach you today.

We have two circles.

This is of, let's say, radius
1/2, and this is radius 2.

The engineer, mathematician,
physicist, whoever they are,

they knew that the curvature
is inverse proportional

to the radius.

That radius is 1/2.

The curvature will
be 2 in this case.

The radius is 2, the
curvature will be 1/2.

Does that make sense, this
inverse proportionality?

The bigger the radius,
the lesser the curvature,

that less bent you are.

The more fat-- well, OK.

I'm not going to say anything
politically incorrect.

So this is really curved because
the radius is really small.

This less curved,
almost-- at infinity,

this curvature
becomes 0, because

at infinity, that radius
explodes to plus infinity bag

theory.

Then you have 1 over
infinity will be 0,

and that will be the curvature
of a circle of infinite radius.

Right?

So we learned something today.

We learned about the
curvature of a circle, which

is something.

But this is the same
way for any curve.

You reparameterize.

Now you understand why you need
to reparameterize in arc length

s.

You take the acceleration
in arc length.

You get the magnitude.

That measures how
bent the curve is.

Next time, you're going to
do how bent the helix is.

OK?

At every point.

Enjoy your WeBWorK homework.

Ask me anytime, and
ask me also Thursday.

Do not have a block about
your homework questions.

You can ask me anytime
by email, or in person.