TTU Math2450 Calculus3 Sec 10.2 and 10.4 part 3

Integration of a Vector Fuction, Position Velocity and Acceleration, Frenet Formula

PROFESSOR TODA: And Calc II.

And I will go ahead and
solve some problems today out

of chapter 10 as a review.

Meaning what?

Meaning, that you have
section 10.1 followed by 10.2

followed by 10.4.

These ones are
required sections,

but I'm putting the material
all together as a compact set.

So, if we cannot officially
cut between, as I told you,

cut between the sections.

One thing that I did
not work examples on,

trusting that you'd
remember it was integration.

In particular, I didn't
cover integration

of vector valued functions
and examples that

are very very important.

Now, do you need to learn
something special for that?

No.

But just like you cannot learn
organic chemistry without

knowing inorganic chemistry,
then you could not know how

to integrate a vector value
function r prime of d to get r

of d unless you know calculus
one and caluculus two, right?

So let's say first
a bunch of formulas

that you use going back
to last week's knowledge

what have we learned?

We work with regular
curves in r3.

And in particular if
they are part of R2,

they are plain curves.

I want to encourage
you to ask questions

about the example
[INAUDIBLE] now.

In the review session we
have applications [INAUDIBLE]

from 2 2 3.

What was a regular curve?

Is anybody willing to tell
me what a regular curve was?

Was it vector value function?

Do you like big r or little r?

STUDENT: Doesn't matter.

PROFESSOR TODA: Big r of t.

Vector value function.

x of t [INAUDIBLE] You know,
I told you that sometimes we

use brackets here.

Sometimes we use round
parentheses depending

how you represent a vector in r3
in our book they use brackets,

but in other calculus books,
they use round parentheses

around it.

So these are the coordinates
of the moving particle in time.

Doesn't have to be a specific
object, could be a fly,

could be just a
particle, anything

in physical motion between this
point a of b equals a and b

of t equals b.

So at time a and
time b you are there.

What have we learned?

We've learned that a regular
curve means its differentiable

and the derivative is
continuous, it's a c1 function.

And what else?

The derivative of
the position vector

called velocity never vanishes.

So it's different from 0
for every t in the interval

that you take, like ab.

That's a regular curve.

Regular curve was something we
talked about at least 5 times.

The point is how do we
see the backwards process?

That means if somebody gives you
the velocity of a vector curve,

they ask you for
the position vector.

So let's see an example.

Integration example
1 says I gave you

the veclocity vector or
a certain law of motion

that I don't know.

I just know the velocity
vector is being 1 over 1

plus t squared.

Should I put the brace here?

An angular bracket?

One over one plus t squared.

And I'm gonna put a cosign
on 2t, and t squared

plus equal to minus t.

And somebody says,
that's all I know for P

on an arbitrary real integral.

And we know via the
0 as being even.

Let's say it's even
as 0 0 and that

takes a little bit of thinking.

I don't know.

How about a 1, which
would be just k.

Using this velocity vector
find me being normal,

which means find
the position vector

corresponding to this velocity.

What is this?

It's actually initial value

STUDENT: [INAUDIBLE]
1, 1, and 1?

PROFESSOR TODA: 0, what is it?

When place 0 in?

STUDENT: Yeah.

[INTERPOSING VOICES]

STUDENT: Are these
the initial conditions

for the location, or--

PROFESSOR TODA: I'm sorry.

I wrote r the intial
condition for the location.

Thank you so much, OK?

I probably would've realized
it as soon as possible.

Not the initial velocity
I wanted to give you,

but the initial position.

All right, so how do
I get to the r of d?

I would say integrate,
and when I integrate,

I have to keep in mind that
I have to add the constants.

Right?

OK.

So from v, v is our priority.

It follows that r will
be-- who tells me?

Do you guys remember the
integral of 1 plus t squared?

STUDENT: [INAUDIBLE]

PROFESSOR TODA: So
that's the inverse.

Or, I'll write it [? arc tan, ?]
and I'm very happy that you

remember that, but there
are many students who don't.

If you feel you don't, that
means that you have to open

the -- where? -- Between
chapters 5 and chapter 7.

You have all these
integration chapters--

the main ones over there.

It's a function definted
on the whole real interval,

so I don't care
to worry about it.

This what we call an IVP,
initial value problem.

So what kind of problem is that?

It's a problem
like somebody would

give you knowing that f
prime of t is the little f,

and knowing that big f
of 0 is the initial value

for your function of find f.

So you have actually an initial
value problem of the calc

that you've seen
in previous class.

arctangent of t plus c1 and then
if you miss the c1 in general,

this can mess up the whole thing
because-- see, in your case,

you're really lucky.

If you plug in the 0 here,
what are you gonna have?

You're gonna have arctangent
of 0, and that is 0.

So in that case c1 is just 0.

And [? three ?] [? not ?] and
if you forgot it would not be

the end of the world, but
if you forgot it in general,

it would be a big problem.

So don't forget
about the constant.

When you integrate-- the
familiar of antiderivatives

is cosine 2t.

I know you know it.

1/2 sine of t.

Am I done?

No, I should say plus C2.

And finally the familiar
of antiderivatives of t

squared plus e to minus t.

STUDENT: 2t minus e
to the negative t.

PROFESSOR TODA: No, integral of.

So what's the integral of--

STUDENT: t 2 squared.

PROFESSOR TODA: t cubed
over 3-- minus, excellent.

Now, do you want one
of you guys almost

kill me during the weekend.

But that's OK.

I mean, this problem
had something

to do with integral minus.

He put that integral of e to the
minus t was equal to minus t.

So pay attention to the sign.

Remember that integral
of e to the at,

the t is to the at over a plus.

Right?

OK, so this is what you
have, a minus plus C3.

Pay attention also to the exam.

Because in the
exams, when you rush,

you make lots of
mistakes like that.

R of 0 is even.

So the initial position
is given as C1.

I'm replacing in my formula.

It's going to be
C1, C2, and what?

When I replace the 0 here,
what am I going to get?

STUDENT: You're going
to get negative 1.

PROFESSOR TODA: Minus 1 plus C3.

Note that I fabricated this
example, so that C3 is not

going to be 0.

I wanted some customs to
be zero and some customs

to not be 0, just for
you to realize it's

important to pay attention.

OK, minus 1 plus C3.

And then I have 0, 0, 1 as
given as initial position.

So what do you get by solving
this linear system that's

very simple?

In general, you can get
more complicated stuff.

C1 is 0, C2 is 0, C3 is a--

STUDENT: 2.

PROFESSOR TODA: 2.

And so it was a piece of cake.

What is my formula?

If you leave it like
that, generally you're

going to get full credit.

What would you need to
do to get full credit?

STUDENT: Rt is equal to R10
plus 1/2 sine of 2t plus tq--

PROFESSOR TODA: Precisely,
and thank you so much

for your help.

So you have R10 of
t, 1/2 sine of 2t

and t cubed over 3 minus
e to the minus e plus 2.

And close, and that's it.

And box your answer.

So I got the long motion back.

Similarly, you could find,
if somebody gives you

the acceleration of a
long motion and asks you

this is the acceleration.

And I give you some
initial values.

And you have to find
first the velocity,

going backwards one step.

And from the velocity,
backwards a second step,

get the position vector.

And that sounds a little
bit more elaborate.

But it doesn't have to
be a long computation.

In general, we do not
focus on giving you

an awfully long computation.

We just want to test your
understanding of the concepts.

And having this in mind,
I picked another example.

I would like to
see what that is.

And the initial velocity
will be given in this case.

This is what I was thinking
a little bit ahead of that.

So somebody gives you the
acceleration in the velocity

vector at 0 and is asking you
to find the velocity vector So

let me give it to you
for t between 0 and 2 pi.

I give you the
acceleration vector,

it will be nice and sassy.

Let's see, that's going to be
cosine of t, sine of t and 0.

And you'll say, oh, I
know how to do those.

Of course you know.

But I want you to pay
attention to the constraints

of integration.

This is why I do this
kind of exercise again.

So what do we have for V of t.

V of 0 is-- somebody will say,
let's give something nice,

and let's say this would be--
I have no idea what I want.

Let's say i, j, and that's it.

How do you do that?

V of t.

Let's integrate together.

You don't like this?

I hope that by now,
you've got used to it.

A bracket, I'm doing a
bracket, like in the book.

So sine t plus a constant.

What's the integral
of sine, class?

V equals sine t plus a constant.

And C3 is a constant.

And there I go.

You say, oh my god,
what am I having?

V of 0-- is as a
vector, I presented it

in the canonical standard
basis as 1, 1, and 0.

So from that one, you
can jump to this one

and say, yes, I'm going to
plug in 0, see what I get.

In the general formula,
when you plug in 0,

you get C1-- what
is cosine of 0?

Minus 1, I have here, plus C2.

And C3, that is always there.

And then V of 0 is
what I got here.

V of 0 has to be compared to
what your initial data was.

So C1 is 1, C2 is 2, and C3 is--

So let me replace it.

I say the answer will be--
cosine t plus 1, sine t plus 2,

and the constants.

But then somebody, who is
really an experimental guy,

says well--

STUDENT: You have it backwards.

It's sine of t plus
1, and then you

have the cosine of t plus 2.

PROFESSOR TODA: Oh, yeah.

Wait a minute.

This is-- I
miscopied looking up.

So I have sine t, I was
supposed to-- minus cosine t

and I'm done.

So thank you for telling me.

So sum t plus 1 minus
cosine t plus 2 and 0

are the functions that I put
here by replacing C1, C2, C3.

And then, somebody
says, wait a minute,

now let me give you V of 0.

Let me give you R of 0.

We have zeroes already there.

And you were supposed
to get R from here.

So what is R of t, the
position vector, find it.

V of t is given.

Actually, it's given by
you, because you found it

at the previous step.

And R of 0 is given as well.

And let's say that would
be-- let's say 1, 1, and 1.

So what do you need to do next?

You have R prime given.

That leaves you to
integrate to get R t.

And R of t is going to be what?

Who is going to tell me
what I have to write down?

Minus cosine t plus t plus--
let's use the constant K1

integration.

And then what?

STUDENT: Sine of t.

PROFESSOR TODA: I think
it's minus sine, right?

Minus sine of t plus 2t
plus K2 and K3, right?

So R of 0 is going to be what?

First of all, we use this
piece of information.

Second of all, we identify
from the formula we got.

So from the formula I
got, just plugging in 0,

it should come out straight
as minus 1 plus K1.

0 for this guy, 0 for the
second term, K2 and K3.

So who is helping me solve
the system really quickly?

K1 is 2.

K2 is--

STUDENT: 1.

PROFESSOR TODA: K3 is 1.

And I'm going back
to R and replace it.

And that's my final answer
for this two-step problem.

So I have a two-step integration
from the acceleration

to the velocity,
from the velocity

to the position vector.

Minus cosine t plus t plus 2.

Remind me, because I have
a tendency to miscopy,

an I looking in the right place?

Yes.

So I have minus sine t plus
2t plus 1 and K3 is one.

So this is the process you
are supposed to remember

for the rest of the semester.

It's not a hard one.

It's something that
everybody should master.

Is it hard?

How many of you understood this?

Please raise hands.

Oh, no problem, good.

Now would you tell me--
I'm not going to ask you

what kind of motion this is.

It's a little bit close to
a circular motion but not

a circular motion.

However, can you tell
me anything interesting

about the type of trajectory
that I have, in terms

of the acceleration vector?

The acceleration
vector is beautiful,

just like in the
case of the washer.

That was a vector
that-- like this

would be the circular motion.

The acceleration would
be this unique vector

that comes inside.

Is this going outside
or coming inside?

Is it a unit vector?

Yes, it is a unit vector.

So suppose that I'm
looking at the trajectory,

if it were more or
less a motion that has

to do with mixing into a bowl.

Would this go inside or outside?

Towards the outside
or towards the inside?

I plugged j-- depends on
what I'm looking at, in terms

of surface that I'm on, right?

Do you remember
from last time we

had that helix that
was on a cylinder.

And we asked ourselves, how
is that [INAUDIBLE] pointing?

And it was pointing
outside of the cylinder,

in the direction
towards the outside.

Coming back to the
review, there are

several things I'd like to
review but not all of them.

Because some of the
examples we have there,

you understood them really well.

I was very proud
of you, and I saw

that you finished--
almost all of you

finished the
homework number one.

So I was looking outside
at homework number

two that is over
these three sections.

So I was hoping you would ask
me today, between two and three,

if you have any difficulties
with homework two.

That's due February 11.

And then the latest homework
that I posted yesterday, I

don't know how many
of you logged in.

But last night I
posted a homework

that is getting a huge
extended deadline, which

is the 28th of February.

Because somebody's
birthday is February 29.

I was just thinking why would
somebody need be a whole month?

You would need the whole
month to have a good view

of the whole chapter 11.

I sent you the videos
for chapter 11.

And for chapter 11, you
have this huge homework

which is 49 problems.

So please do not,
do not leave it

to the last five
days or six days,

because it's going to kill you.

There are people who
say, I can finish

this in the next five days.

I know you can.

I know you can,
I don't doubt it.

That's why I left
you so much freedom.

But you have-- today is
the second or the third?

So practically you have
25 days to work on this.

On the 28th at 11 PM
it's going to close.

I would work a few
problems every other day.

Because I need a break,
so I would alternate.

But don't leave it--
even if you have help,

especially if you have help,
like a tutor or tutoring

services here that are
free in the department.

Do not leave it
to the last days.

Because you're putting pressure
on yourself, on your brain,

on your tutor, on everybody.

Yes sir.

STUDENT: So that's
homework three?

PROFESSOR TODA:
That's homework three,

and it's a huge homework
over chapter 11.

STUDENT: You said
there are 49 problems?

PROFESSOR TODA: I don't
remember exactly but 47, 49.

I don't remember how many.

STUDENT: Between 45 and 50.

PROFESSOR TODA:
Between 45 and 50, yes.

If you encounter any bug--
although there shouldn't

be bugs, maybe 1 in 1,000.

If you encounter any
bug that the programmer

of those problems may
have accidentally put in,

you let me know.

So I can contact them.

If there is a problem that I
consider shouldn't be there,

I will eliminate that later on.

But hopefully, everything
will be doable,

everything will be fair and
you will be able to solve it.

Any questions?

Particular questions
from the homework?

STUDENT: [INAUDIBLE] is it to
parametrize a circle of a set,

like of a certain
radius on the xy-plane?

PROFESSOR TODA:
Shall we do that?

Do you want me to do that
in general, in xy-plane, OK.

STUDENT: [INAUDIBLE]
in the xy-plane.

PROFESSOR TODA: xy-plane and
then what was the equation?

Was it like a equals sine
of t or a equals sine of bt?

Because it's a
little bit different,

depending on how the
parametrization was given.

What's your name
again, I forgot.

I don't know what to refer you.

STUDENT: Ryder.

PROFESSOR TODA: Was that part
of what's due on the 11th?

STUDENT: It doesn't-- yes, it
doesn't give a revision set.

It says--

PROFESSOR TODA: Let me quickly
read-- find parametrization

of the circle of radius 7 in
the xy-plane, centered at 3, 1,

oriented counterclockwise.

The point 10, 1
should be connected--

STUDENT: Just one more second.

PROFESSOR TODA: Do
you mind if I put it.

I'll take good care of it.

I won't drop it.

So the point-- parametrization
of the circle of radius

7 in the xy-plane,
centered at 3, 1.

So circle centered at-- and
I'll say it x0, 1 0, being 3, 1.

No, because then I'm
solving your problem.

But I'm solving
your problem anyway,

even if I change
change the numbers.

Why don't I change
the numbers, and then

you do it for the given numbers.

Let's say 1, 0.

And it's the same type
of problem, right?

Oriented counterclockwise.

That's important.

So you have circle radius 7.

I think people could
have any other,

because problems are-- sometimes
you get a random assignment.

So you have R
equals 2, let's say.

And you have the point,
how to make up something.

The point corresponding
to t equals

0 will be given as you have
[INAUDIBLE], 1, 0, whatever.

OK?

Use the t as the parameter
for all your answers.

So use t as a parameter
for all your answers,

and the answers are written in
the interactive field as x of t

equals what and y
of t equals what,

and it's waiting for
you to fill them in.

You know.

OK, now I was talking
to [INAUDIBLE].

I'm going to give
this back to you.

Thank you, Ryan.

So when you said it's a
little bit frustrating,

and I agree wit you, that
in this variant of webwork

problems you have to enter
both of them correctly

in order to say yes, correct.

I was used to another library--
the library was outdated

[INAUDIBLE]-- where if I
enter this correctly I get 50%

credit, and if I enter this
incorrectly it's not going

to penalize me.

So I a little bit
complained about it,

and I was shown the
old library where

I can go ahead and go
back and assign problems

where you get the answer
correct for this one

and incorrect for this one,
and you get partial credit.

So I'm probably going
to switch to that.

Let's do that.

This is a very good problem.

I'm glad you brought it up.

What have you learned about
conics in high school?

You've learned about--
well, it depends.

You've learned about ellipse.

You've learned about hyperbola.

You've learned about parabola.

Some of you put them down
for me for extra credit.

I was very happy you did that.

It's a good exercise.

If you have-- Alex, yes?

STUDENT: I was just
thinking, does that say 1, 0?

The point corresponding
to t0 [INAUDIBLE]?

PROFESSOR TODA: I think
that's what I meant.

I don't know, I just
came up with it.

I made it.

1, 0.

I make up all my problems.

STUDENT: But the center
of the circle isn't 1, 0.

PROFESSOR TODA: Oh, oops.

Yes.

Sorry.

So 2, 0.

No--

[INTERPOSING VOICES]

PROFESSOR TODA:
--because the radius.

This is the problem when you
don't think very [INAUDIBLE].

I always like to make
up my own problems.

When an author, when we came up
with the problems in the book,

of course we had to think, draw,
and make sure they made sense.

But when you just come up with
a problem out of the middle

of nowhere-- thank you so much.

Of course, we
would have realized

that was nonsense
in just a minute.

But it's good that you told me.

So x of t, y of t.

Let's find it.

Based on what?

What is the general
equation of a circle?

x minus x0 squared plus y minus
y0 squared equals R squared.

And you have learned
that in high school.

Am I right or not?

You have.

OK.

Good.

Now, in our case what
is x0 and what is y0?

x0 is 1 and y0 is 0.

Because that's
why-- I don't know.

I just made it up.

And I said that's the center.

I'll draw.

I should have drawn
it in the beginning,

and that would have
helped me not come up

with some nonsensical data.

c is 1, 0.

Radius is 2.

So I'm going this way.

What point is this way, guys?

Just by the way.

Because [INAUDIBLE]
is 1, 0, right?

And this way the other
extreme, the antipode is 3, 0.

So that's exactly what
Alexander was saying.

And now it makes sense.

Well, I cannot draw today.

STUDENT: [INAUDIBLE]

PROFESSOR TODA:
It looks horrible.

It looks like an egg that
is shaped-- disabled egg.

OK.

All right.

So the motion of-- the
motion will come like that.

From t equals 0, when I'm
here, counterclockwise,

I have to draw-- any kind of
circle you have in the homework

should be drawn on the board.

If you have a general, you
don't know what the data is.

I want to help you solve
the general problem.

For the original problem,
which is a circle

of center x, 0, y, 0 and
radius R, generic one,

what is the parametrization
without data?

Without specific data.

What is the parametrization?

And I want you to pay
attention very well.

You are paying attention.

You are very careful today.

[INAUDIBLE]

So what do you have?

STUDENT: Cosine.

PROFESSOR TODA:
Before that cosine

there is an R, excellent.

So [INAUDIBLE]
there R cosine of t.

I'm not done.

What do I put here?

STUDENT: Over d.

PROFESSOR TODA: No, no.

I'm continuing.

STUDENT: Plus x0.

PROFESSOR TODA: Plus x0.

And R sine t plus y0.

Who taught me that?

First of all, this
is not unique.

It's not unique.

I could put sine t
here and cosine t here

and it would be the same
type of parametrization.

But we usually put
the cosine first

because we look at the
x-axis corresponding

to the cosine and the y-axis
corresponding to the sine.

If I don't know that,
because I happen to know that

from when I was 16 in high
school, if I don't know that,

what do I know?

I cook up my own
parametrization.

And that's a very good thing.

And I'm glad Ryan
asked about that.

How does one come up with this?

Do we have to memorize?

In mathematics, thank god,
we don't memorize much.

The way we cook up things
is just from, in this case,

from the Pythagorean
theorem of-- no.

Pythagorean theorem
of trigonometry?

The fundamental identity
of trigonometry,

which is the same thing as
the Pythagorean theorem.

What's the fundamental
identity of trigonometry?

Cosine squared plus
sin squared equals 1.

If I have a problem
like that, I must

have that this is R cosine
t and this is R sine t.

Because when I take
the red guys and I

square them and I
add them together,

I'm going to have R squared.

All righty, good.

So no matter what
kind of data you have,

you should be able to come
up with this on your own.

And what else is
going to be happening?

When I solve for x of-- the
point corresponding to t

equals 0.

x of 0 and y of 0 will
therefore be what?

It will be R plus x0.

This is going to be what?

Just y0.

Does anybody give them to me?

STUDENT: 3, 0.

PROFESSOR TODA: Alexander
gave me the correct ones.

They will be 3 and 0.

Are you guys with me?

They could be anything,
anything that makes sense.

All right, for example somebody
would say, I'm starting here.

I give you other points.

Then you put them in, you
plug in that initial point,

meaning that you're
starting your motion here.

And you do go around
the circle one

because, you take [INAUDIBLE]
only between 0 and 2 pi.

Alexander.

STUDENT: I have [INAUDIBLE].

PROFESSOR TODA: OK.

STUDENT: [INAUDIBLE]

PROFESSOR TODA: No, I thought
that I misprinted something

again.

STUDENT: No, I was about to
say something really dumb.

PROFESSOR TODA: OK.

So how do we make sense
of what we have here?

Well, y0 corresponds
to what I said.

So this is a
superfluous equation.

I don't need that.

What do I know from that?

R will be 2.

x1 is 1.

I have a superfluous equation.

I have to get identities
in that case, right?

OK, now.

What is going to be my--
my bunch of equations

will be x of t equals 2
cosine t plus 1 and y of t

equals-- I don't
like this marker.

I hate it.

Where did I get it?

In the math department.

And it's a new one.

I got it as a new one.

It's not working.

OK, y of t.

The blue contrast is invisible.

I have 2 sine t.

Okey dokey.

When you finish a
problem, always quickly

verify if what you
got makes sense.

And obviously if I
look at everything,

it's matching the whole point.

Right?

OK.

Now going back to-- this is
reminding me of something in 3d

that I wanted to talk
to you today about.

This is [INAUDIBLE].

I'm going to give
you, in a similar way

with this simple
problem, I'm going

to give you something
more complicated

and say find the
parametrization of a helix.

And you say, well,
I'm happy that this

isn't a made-up problem again.

I have to be a little
bit more careful

with these made-up problems
so that they make sense.

Of a helix R of t such that
it is contained or it lies,

it lies on the circular
cylinder x squared

plus y squared equals 4.

Why is that a cylinder?

The z's missing, so it's
going to be a cylinder whose

main axis is the z axis.

Right?

Are you guys with me?

I think we are on the same page.

And you cannot solve the
problem just with this data.

Do you agree with me?

And knowing that, the
curvature of the helix is k

equals 2/5 at every point.

And of course it's an oxymoron.

Because what I
proved last time is

that the curvature of
a helix is a constant.

So remember, we got the
curvature of a helix

as being a constant.

STUDENT: What's that last
word of the sentence?

It's "the curvature
is at every" what?

PROFESSOR TODA: At every point.

I'm sorry I said, it very--
I abbreviated [INAUDIBLE].

So at every point you
have the same curvature.

When you draw a
helix you say, wait,

the helix is bent uniformly.

If you were to play with a
spring taken from am old bed,

you would go with your
hands along the spring.

And then you say, oh,
it bends about the same.

Yes, it does.

And that means the
curvature is the same.

How would you
solve this problem?

This problem is hard,
because you cannot integrate

the curvature.

Well, what is the curvature?

The curvature would be--

STUDENT: Absolute value.

PROFESSOR TODA: Just
absolute value of R

double prime if it were in s.

And you cannot integrate.

If somebody gave you
the vector equation

of double prime of
this, them you say,

yes, I can integrate
one step going back.

I get R prime of s.

Then I go back to R of s.

But this is a little
bit complicated.

I'm giving you a scalar.

You have to be a little bit
aware of what you did last time

and try to remember
what we did last time.

What did we do last time?

I would not give you
a problem like that

on the final, because it would
assume that you have solved

the problem we did last
time in terms of R of t

equals A equals sine t.

A sine t and [? vt. ?]

And we said, this is the
standard parametrized helix

that sits on a cylinder of
radius A and has the phb.

So the distance between
consecutive spirals

really matters.

That really makes
the difference.

STUDENT: I have a question.

PROFESSOR TODA: You wanted
to ask me something.

STUDENT: Is s always
the reciprocal of t?

Are they always--

PROFESSOR TODA:
No, not reciprocal.

You mean s of t is a function
is from t0 to t of the speed.

R prime and t-- d tau, right?

Tau not t. [INAUDIBLE].

t and s are
different parameters.

Different times.

Different parameter times.

And you say--

STUDENT: Isn't s
the parameter time

when [INAUDIBLE] parametrized?

PROFESSOR TODA: Very good.

So what is the magic s?

I'm proud of you.

This is the important
thing to remember.

t could be any time.

I start measuring
wherever I want.

I can set my watch to start now.

It could be crazy.

Doesn't have to be uniform.

Motion, I don't care.

s is a friend of
yours that says,

I am that special time
so that according to me

the speed will become one.

So for a physicist to measure
the speed with respect to this,

parameter s time, the speed
will always become one.

That is the arclength
time and position.

How you get from one
another, I told you last time

that for both of them
you have-- this is R of t

and this is little r of s.

And there is a composition.

s can be viewed as
a function of t,

and t can be viewed
as a function of s.

As functions they are
inverse to one another.

So going back to who they
are, a very good question,

because this is a review
anyway, [? who wants ?]

s as a function of t for
this particular problem?

I hope you remember, we were
like-- have you seen this movie

with Mickey Mouse going
on a mountain that

was more like a cylinder.

And this is the train
going at a constant slope.

And one of my colleagues,
actually, he's at Stanford,

was telling me that he
gave his students in Calc 1

to prove, formally prove,
that the angle formed

by the law of motion
by the velocity vector,

with the horizontal plane
passing through the particle,

is always a constant.

I didn't think about doing
in now, but of course we can.

We could do that.

So maybe the next
thing would be, like,

if you [INAUDIBLE]
an extra problem, can

we show that angle between the
velocity vector on the helix

and the horizontal plane through
that point is a constant.

STUDENT: Wouldn't it
just be, because B of t

is just a constant times t?

PROFESSOR TODA: Yeah.

We'll get to that.

We'll get to that in a second.

So he reminded me of an old
movie from like 70 years ago,

with Mickey Mouse and the train.

And the train going
up at the same speed.

You have to maintain
the same speed.

Because if you risk it
not, then you sort of

are getting trouble.

So you never stop.

If you stop you go back.

So it's a regular curve.

What I have here is
that such a curve.

Regular curve, never stop.

Get up with a constant speed.

Do you guys remember the
speed from last time?

We'll square root the a
squared plus b squared.

When we did the
velocity thingie.

And I get square root a
squared plus b squared times t.

Now, today I would like
to ask you one question.

What if-- Ryan brought this up.

It's very good.

b is a constant.

What if b would
not be a constant,

or maybe could be worse?

For example, instead of having
another linear function with t,

but something that contains
higher powers of t.

Then you don't go at the
constant speed anymore.

You can say goodbye
to the cartoon.

Yes, sir?

STUDENT: And then
it's [INAUDIBLE].

One that goes [INAUDIBLE].

PROFESSOR TODA: I
mean, it's still--

STUDENT: s is not
multiplied by a constant.

The function between t and
s is not a constant one.

PROFESSOR TODA: It's going to
be a different parameterization,

different speed.

Sometimes-- OK, you
have to understand.

Let's say I have a cone.

And I'm going slow
at first, and I

go faster and faster
and faster and faster

to the end of the cone.

But then I have the
same physical curve,

and I parameterized
[INAUDIBLE] the length.

And I say, no, I'm a mechanic.

Or I'm the engineer
of the strain.

I can make the motion
have a constant speed.

So even if the helix
is no longer circular,

and it's sort of a crazy helix
going on top of the mountain,

as an engineer I
can just say, oh no,

I want cruise control
for my little train.

And I will go at the same speed.

See, the problem is
the slope a constant.

And thinking of
what they did that

stand for, because
it didn't stand

for [INAUDIBLE] in honors.

We can do it in honors as well.

We'll do it in a second.

Now, k obviously is what?

Some of you have
very good memory,

and like the memory of a
medical doctor, which is great.

Some of you don't.

But if you don't you just go
back and look at the notes.

What I'm trying to
do, but I don't know,

it's also a matter
of money-- I don't

want to use the math
department copier-- I'd

like to make a stack of notes.

So that's why I'm collecting
these notes, to bring them back

to you.

For free!

I'm not going to
sell them to you.

I'm [INAUDIBLE].

So that you can have those
with you whenever you want,

or put them in a spiral,
punch holes in them,

and have them for
review at any time.

Reminds me of what that
was-- that was in the notes.

a over a squared plus b squared.

So who can tell me, a
and b really quickly,

so we don't waste too
much time, Mr. a is--?

STUDENT: So this is another way

STUDENT: 2.

PROFESSOR TODA: 2.

STUDENT: So is this another
way of defining k in k of s?

PROFESSOR TODA: Actually--

STUDENT: That's the general
curvature for [INAUDIBLE].

PROFESSOR TODA: You know
what is the magic thing?

Even if-- the curvature
is an invariant.

It doesn't depend the
reparametrization.

There is a way maybe I'm going
to teach you, although this

is not in the book.

What are the formulas
corresponding

to the [INAUDIBLE] t and v that
depend on curvature and torsion

and the speed along the curve.

And if you analyze the notion
of curvature, [INAUDIBLE],

no matter what your
parameter will be, t, s, tau,

God knows what, k will
still be the same number.

So k is viewed as an
invariant with respect

to the parametrization.

STUDENT: So then that a over
a squared plus b squared,

that's another way of finding k?

PROFESSOR TODA: Say it again?

STUDENT: So using a over
a squared plus b squared

is another way of finding k?

PROFESSOR TODA: No.

Somebody gave you k.

And then you say, if it's
a standard parametrization,

and then I get 2/5,
can I be sure a is 2?

I'm sure a is 2 from nothing.

This is what makes me aware
that a is 2 the first place.

Because its the radius
of the cylinder.

This is x squared, x and y.

You see, x squared plus
y squared is a squared.

This is where I get a from.

a is 2.

I replace it in here
and I say, all righty,

so I only have one
choice. a is 2 and b is?

STUDENT: [INAUDIBLE]

PROFESSOR TODA: But can b
plus-- So what I'm saying,

a is 2, right?

We know that from this.

If I block in here I have 4
and somebody says plus minus 1.

No.

b is always positive.

So you remember the
last time we discussed

about the standard
parametrization.

But somebody will say,
but what if I put a minus?

What if I'm going
to put a minus?

That's an excellent question.

What's going to happen
if you put minus t?

[INTERPOSING VOICES]

PROFESSOR TODA: Exactly.

In the opposite direction.

Instead of going
up, you go down.

All right.

Now, I'm gonna-- what else?

Ah, I said, let's do this.

Let's prove that the
angle is a constant,

the angle that's
made by the velocity

vector of the train with the
horizontal plane is a constant.

Is this hard?

Nah.

Yes, sir?

STUDENT: Are we still going
to find R of t given only k?

PROFESSOR TODA: But didn't we?

We did.

R of t was 2 cosine
t, 2 sine t, and t.

All right?

OK, so we are done.

What did I say?

I said that let's
prove-- it's a proof.

Let's prove that the angle made
by the velocity to the train--

to the train?-- to the direction
of motion, which is the helix.

And the horizontal
plane is a constant.

Is this hard?

How are we going to do that?

Now I start waking up,
because I was very tired.

STUDENT: [INAUDIBLE]

PROFESSOR TODA: Excuse me.

STUDENT: [INAUDIBLE]

PROFESSOR TODA: So you see,
the helix contains this point.

And I'm looking at
the velocity vector

that is standard to the helix.

And I'll call that R prime.

And then you say,
yea, but how am I

going to compute that angle?

What is that angle?

STUDENT: It's a function of b.

PROFESSOR TODA: It will be.

But we have to do it rigorously.

So what's going to happen
for me to draw that angle?

First of all, I should
take-- from the tip

of the vector I should
draw perpendicular

to the horizontal plane
passing through the point.

And I'll get P prime.

God knows why.

I don't know why, I don't know
why. [? Q. ?] And this is PR,

and P-- not PR.

PR is too much
[INAUDIBLE] radius, M.

OK, so then you would
take PQ and then

you would measure this angle.

Well, you have to be a
little bit smarter than that,

because you can
prove something else.

This is the complement of
another angle that you love.

And using chapter 9 you can
do that angle in no time.

So this is the
complement of the angle

formed by the velocity vector
of prime with the normal.

But not the normal principle
normal to the curve,

but the normal to the plane.

And what is the
normal to the plane?

Let's call the principal normal
n to the curve big N bar.

So in order to avoid confusion,
I'll write this little n.

How about that?

Do you guys know-- like
they do in mechanics.

If you have two normals,
they call that 1n.

1 is little n, and
stuff like that.

So this is the complement.

If I were able to prove
that that complement

is a constant-- this is the
Stanford [? property-- ?] then

I will be happy.

Is it hard?

No, for god's sake.

Who is little n?

Little n would be-- is
that the normal to a plane

that you love?

What is your plane?

STUDENT: xy plane.

PROFESSOR TODA: Your
plane is horizontal plane.

STUDENT: xy.

PROFESSOR TODA: Yes, xy plane.

Or xy plane shifted,
shifted, shifted, shifted.

That's the normal change?

No.

Who is the normal?

STUDENT: [INAUDIBLE]

PROFESSOR TODA: [INAUDIBLE].

STUDENT: 0, 0, 1.

PROFESSOR TODA: 0, 0, 1.

OK.

When I put 0 I was [INAUDIBLE].

So this is k.

All right.

And what is our prime?

I was-- that was
a piece of cake.

We did it last time minus a
sine t, a equals sine t and b.

Let's find that angle.

Well, I don't know.

You have to teach me, because
you have chapter 9 fresher

in your memory than I have it.

Are you taking attendance also?

Are you writing your name down?

Oh, no problem whatsoever.

STUDENT: We didn't get it.

PROFESSOR TODA:
You didn't get it.

Circulate it.

All right, so who is going
to help me with the angle?

What is the angle between
two vectors, guys?

That should be review from what
we just covered in chapter 9.

Let me call them
u and v. And who's

going to tell me how
I get that angle?

STUDENT: [INAUDIBLE] is equal
to the inverse cosine of the dot

product of [? the magnitude. ?]

PROFESSOR TODA: Do you
like me to write arc

cosine or cosine [INAUDIBLE].

Doesn't matter.

Arc cosine of--

STUDENT: The dot products.

PROFESSOR TODA: The dot
product between u and v.

STUDENT: Over magnitude.

PROFESSOR TODA: Divided by the
product of their magnitudes.

Look, I will change the
order, because you're not

going to like it.

Doesn't matter.

OK?

So the angle phi between
my favorite vectors

here is going to be
simply the dot product.

That's a blessing.

It's a constant.

STUDENT: So you're
doing the dot product

between the normal [INAUDIBLE]?

PROFESSOR TODA:
Between this and that.

So this is u and this
is v. So the dot product

would be 0 plus v.
So the dot product

is arc cosine of v, which,
thank god, is a constant.

I don't have to do
anything anymore.

I'm done with the proof
bit, because arc cosine

of a constant will
be a constant.

OK?

All right.

So I have v over what?

What is the length
of this vector?

1. [INAUDIBLE].

What's the length
of that vector?

Square root of a
squared plus b squared.

All right?

STUDENT: How did
you [INAUDIBLE].

PROFESSOR TODA: So now
let me ask you one thing.

What kind of function
is arc cosine?

Of course I said arc cosine
of a constant is a constant.

What kind of a
function is arc cosine?

I'm doing review with you
because I think it's useful.

Arc cosine is defined on
what with values in what?

STUDENT: Repeat the question?

PROFESSOR TODA: Arc cosine.

Or cosine inverse,
like Ryan prefers.

Cosine inverse is
the same thing.

It's a function defined
by where to where?

Cosine is defined
from where to where?

From R to minus 1.

It's a cosine of t.

t could be any real number.

The range is minus 1, 1.

Close the interval.

STUDENT: So it's-- so
I just wonder why--

PROFESSOR TODA: Minus
1 to 1, close interval.

But pay attention, please.

Because it cannot go back to R.
It has to be a 1 to 1 function.

You cannot have an inverse
function if you don't take

a restriction of a
function to be 1 to 1.

And we took that
restriction of a function.

And do you remember what it was?

[INTERPOSING VOICES]

PROFESSOR TODA: 0 to pi.

Now, on this one
I'm really happy.

Because I asked
several people-- people

come to my office to get
all sorts of transcripts,

[INAUDIBLE].

And in trigonometry
I asked one student,

so you took trigonometry.

So do you remember that?

He didn't remember that.

So I'm glad you do.

How about when I had
the sine inverse?

How was my restriction so that
would be a 1 to 1 function?

It's got to go
from minus 1 to 1.

What is the range?

[INTERPOSING VOICES]

PROFESSOR TODA: Minus pi over 2.

You guys know your trig.

Good.

That's a very good thing.

You were in high school
when you learned that?

Here at Lubbock High?

STUDENT: Yes.

PROFESSOR TODA: Great.

Good job, Lubbock High.

But many students, I caught
them, who wanted credit

for trig who didn't know that.

Good.

So since arc cosine is a
function that is of 0, pi,

for example, what if my--
let me give you an example.

What was last time, guys?

a was 1. b was 1.

For one example.

In that case, 1 with 5b.

STUDENT: [INAUDIBLE] ask you
for the example you just did?

PROFESSOR TODA: No last time.

STUDENT: A was 3 and b was--

PROFESSOR TODA: So what would
that be, in this case 5?

STUDENT: That would be
b over the square root--

STUDENT: 3 over pi.

PROFESSOR TODA: a is 1 and b
is 1, like we did last time.

STUDENT: [INAUDIBLE]
2, which is--

PROFESSOR TODA: Plug
in 1 is a, b is 1.

What is this?

STUDENT: It's just pi over 4.

PROFESSOR TODA: Pi over 4.

So pi will be our cosine, of
1 over square root 2, which

is 45 degree angle, which is--
you said pi over 4, right?

[INAUDIBLE].

So exactly, you would
have that over here.

This is where the
cosine [INAUDIBLE].

Now you see, guys, the way we
have, the way I assume a and b,

the way anybody-- the
book also introduces

a and b to be positive numbers.

Can you tell me what kind
of angle phi will be,

not only restricted to 0 pi?

Well, a is positive.

b is positive.

a doesn't matter.

The whole thing
will be positive.

Arc cosine of a
positive number--

STUDENT: Between
0 and pi over 2.

PROFESSOR TODA: That is.

Yeah, so it has to be
between 0 and pi over 2.

So it's going to be
only this quadrant.

Does that make sense?

Yes, think with the
imagination of your eyes,

or the eyes of your imagination.

OK.

You have a cylinder.

And you are moving
along that cylinder.

And this is how you turn.

You turn with that little train.

Du-du-du-du-du, you go up.

When you turn the
velocity vector and you

look at the-- mm.

STUDENT: The normal.

PROFESSOR TODA: The normal!

Thank you.

The z axis, you always have an
angle between 0 and pi over 2.

So it makes sense.

I'm going to go ahead and
erase the whole thing.

So we reviewed, more or less, s
of t, integration, derivation,

moving from position vector
to velocity to acceleration

and back, acceleration to
velocity to position vector,

the meaning of arclength.

There are some things I
would like to tell you,

because Ryan asked me a few more
questions about the curvature.

The curvature
formula depends very

much on the type of formula
you used for the curve.

So you say, wait,
wait, wait, Magdelena,

you told us-- you
are confusing us.

You told us that the
curvature is uniquely

defined as the magnitude
of the acceleration vector

when the law of motion
is an arclength.

And that is correct.

So suppose my original law of
motion was R of t [INAUDIBLE]

time, any time, t,
any time parameter.

I'm making a face.

But then from that we switch
to something beautiful,

which is called the
arclength parametrization.

Why am I so happy?

Because in this parametrization
the magnitude of the speed

is 1.

And I define k to
be the magnitude

of R double prime of s, right?

The acceleration only in
the arclength [? time ?]

parameterization.

And then this was
the definition.

A. Can you prove-- what?

Can you prove the
following formula?

T prime of s equals
k times N of s.

This is famous for people
who do-- not for everybody.

But imagine you have
an engineer who does

research of the laws of motion.

Maybe he works for
the railways and he's

looking at skew
curves, or he is one

of those people who
project the ski slopes,

or all sorts of winter sports
slope or something, that

involve a lot of
curvatures and torsions.

That guy has to know
the Frenet formula.

So this is the famous
first Frenet formula.

Frenet was a mathematician
who gave the name to the TNB

vectors, the trihedron.

You have the T was what?

The T was the tangent
[INAUDIBLE] vector.

The N was the
principal unit normal.

In those videos that I'm
watching that I also sent you--

I like most of them.

I like the Khan Academy
more than everything.

Also I like the one that
was made by Dr. [? Gock ?]

But Dr. [? Gock ?] made a
little bit of a mistake.

A conceptual mistake.

We all make mistakes by
misprinting or misreading

or goofy mistake.

But he said this is
the normal vector.

This is not-- it's the
principle normal vectors.

There are many normals.

There is only one
tangent direction,

but in terms of normals
there are many that

are-- all of these are normals.

All the perpendicular in
the plane-- [INAUDIBLE]

so this is my law of motion,
T. All this plane is normal.

So any of these
vectors is a normal.

The one we choose and
defined as T prime

over T prime [INAUDIBLE]
absolute values

called the principal normal.

It's like the principal
of a high school.

He is important.

So T and B-- B goes
down, or goes-- down.

Well, yeah, because B is T cross
N. So when you find the Frenet

Trihedron, TNB, it's like that.

T, N, and B. What's special,
why do we call it the frame,

is that every
[? payer ?] of vectors

are mutually orthogonal.

And they are all unit vectors.

This is the famous Frenet frame.

Now, Mr. Frenet was a smart guy.

He found-- I don't know whether
he was adopting mathematics

or not.

Doesn't matter.

He found a bunch of formulas,
of which this is the first one.

And it's called a
first Frenet formula.

That's one thing
I want to ask you.

And then I'm going to give you
more formulas for curvatures,

depending on how you
define your curve.

So for example, base B
based on the definition one

can prove that for a curve
that is not parametrizing

arclength-- you say, ugh,
forget about parametrization

in arclength.

This time you're
assuming, I want to know!

I'm coming to this
because Ryan asked.

I want to know, what is
the formula directly?

Is there a direct
formula that comes

from here for the curvature?

Yeah, but it's a lot
more complicated.

When I was a freshman, maybe
a freshman or a sophomore,

I don't remember, when
I was asked to memorize

that, I did not memorize it.

Then when I started working
as a faculty member,

I saw that I am supposed
to ask it from my students.

So this is going to be
R prime plus product

R double prime in magnitude
over R prime cubed.

So how am I supposed
to remember that?

It's not so easy.

Are you cold there?

It's cold there.

I don't know how
these roofs are made.

Velocity times acceleration.

This is what I try
to teach myself.

I was old already, 26 or 27.

Velocity times
acceleration, cross product,

take the magnitude,
divide by the speed, cube.

Oh my god.

So I was supposed to know
that when I was 18 or 19.

Now, I was teaching majors
of mechanical engineering.

They knew that by heart.

I didn't, so I had to learn it.

So if one is too
lazy or it's simply

inconvenient to try to
reparametrize from R of T

being arclength parametrization
R of s and do that thing here,

one can just plug in and
find the curvature like that.

For example, guys,
as Ryan asked,

if I have A cosine, [INAUDIBLE],
and I do this with respect

to T, can I get k
without-- k will not

depend on T or s or tau.

It will always be the same.

I will still get A
over A squared plus B

squared, no matter what.

So even if I use this
formula for my helix,

I'm going to get the same thing.

I'll get A over A
squared plus B squared,

because curvature
is an invariant.

There is another invariant
that's-- the other invariant,

of course, in space
is called torsion.

We want to talk a little
bit about that later.

So is this hard?

No.

It shouldn't be hard.

And you guys should be able
to help me on that, hopefully.

How do we prove that?

STUDENT: N is G
prime [INAUDIBLE].

PROFESSOR TODA:
That's right, proof.

And that's a very good
start, wouldn't you say?

So what were the definitions?

Let me start from
the definition of T.

That's going to be-- I
am in hard planes, right?

So you say, wait, why do
you write it as a quotient?

You're being silly.

You are in arclength, Magdalena.

I am.

I am.

I just pretend that
I cannot see that.

So if I'm in
arclength, that means

that the denominator is 1.

So I'm being silly.

So R prime of s is
T. Say it again.

R prime of s is T. OK.

Now, did we know that
T and N are orthogonal?

How did we know that T
and N were orthogonal?

We proved that last
time, actually.

T and N are orthogonal.

How do I write
that? [INAUDIBLE].

Meaning that T is
perpendicular to N, right?

From the definition.

You said it right, Sandra.

But why is it from
the definition

that I can jump to
conclusions and say, oh,

since I have T prime here, then
this is perpendicular to T?

Well, we did that last time.

STUDENT: Two parallel vectors.

PROFESSOR TODA: We did
it-- how did we do it?

We did this last.

We said T dot T equals 1.

Prime the whole thing.

T prime times T plus T times T
prime, T dot T prime will be 0.

So T and T prime are
perpendicular always.

Right?

OK, so the whole thing is a
colinear vector to T prime.

It's just T prime
times the scalar.

So he must be
perpendicular to T.

So T and N are perpendicular.

So I do have the
direction of motion.

I know that I must
have some scalar here.

How do I prove that this
scalar is the curvature?

So if I have-- if they
are colinear-- why are

they colinear?

T perpendicular to T
prime implies that T prime

is colinear to N. Say it again.

If T and T prime are
perpendicular to one another,

that means T prime is
calling it to the normal.

So here I may have
alph-- no alpha.

I don't know!

Alpha over [INAUDIBLE]
sounds like a curve.

Give me some function.

STUDENT: u of s?

PROFESSOR TODA: Gamma of s.

u of s, I don't know.

So how did I conclude that?

From T perpendicular to T prime.

Now from here on, you
have to tell me why

gamma must be exactly kappa.

Well, let's take
T prime from here.

T prime from here
will give me what?

T prime is our prime prime.

Say what?

Our prime prime.

What is our prime prime?

Our [? problem ?] prime of s.

STUDENT: You have one
too many primes inside.

PROFESSOR TODA: Oh my god.

Yeah.

So R prime prime.

So T prime in
absolute value will

be exactly R double prime of s.

Oh, OK.

Note that from here also T
prime of s in absolute value,

in magnitude, I'm sorry,
has to be gamma of s.

Why is that?

Because the magnitude of N is 1.

N is unique vector
by definition.

So these two guys
have to coincide.

So R double prime,
the best thing

that I need to do,
it must coincide

with the scalar gamma of s.

So who is the
mysterious gamma of s?

He has no chance
but being this guy.

But this guy has a name.

This guy, he's the curvature
[? cap ?] of s by definition.

Remember, Ryan, this
is the definition.

So by definition the
curvature was the magnitude

of the acceleration
in arclength.

OK.

Both of these guys are
T prime in magnitude.

So they must be equal
from here and here.

It implies that my
gamma must be kappa.

And I prove the formula.

OK.

How do you say
something is proved?

Because this is what we wanted.

We wanted to replace this
generic scalar function

to prove that this is
just the curvature.

QED.

That's exactly what
we wanted to prove.

Now, whatever scalar
function you have here,

that must be the curvature.

Very smart guy, this Mr. Frenet.

I'm now going to take a break.

If you want to go use the
bathroom really quickly,

feel free to do it.

I'm just going to
clean the board,

and I'll keep going
in a few minutes.

STUDENT: [INAUDIBLE]

PROFESSOR TODA: I
will do it-- well,

actually I want to do a
different example, simple one,

which is a plain curve, and show
that the curvature has a very

pretty formula that you
could [INAUDIBLE] memorize,

that in essence is the same.

But it depends on
y equals f of x.

[INAUDIBLE] So if
somebody gives you

a plane called y
equals f of x, can you

write that curvature
[INAUDIBLE] function of f?

And you can.

And again, I was deep in
that when I was 18 or 19

as a freshman.

But unfortunately for me I
didn't learn it at that time.

And several years later when
I started teaching engineers,

well, they are
mostly mechanical.

And mechanical
engineering [INAUDIBLE].

They knew those, and they needed
those in every research paper.

So I had to learn it
together with them.

I'll worry about [INAUDIBLE].

STUDENT: Can you do a really
ugly one, like [INAUDIBLE]?

PROFESSOR TODA: I can
do some ugly ones.

And once you know the
general parametrization,

it will give you a curvature.

Now I'm testing your memory.

Let's see what you remember.

Um-- don't look at the notes.

A positive function,
absolute-- actually,

magnitude of what vector?

STUDENT: R prime.

PROFESSOR TODA: R prime velocity
plus acceleration speed cubed.

Right?

OK.

Now, can we take advantage
of what we just learned

and find-- you find
with me, of course, not

as professor and student,
but like a group of students

together.

Let's find a simple
formula corresponding

to the curvature
of a plane curve.

And the plane curve
could be [INAUDIBLE]

in two different ways,
just because I want

you to practice more on that.

Either given as a general
parametrization-- guys,

what is the general
parametrization

I'm talking about
for a plane curve?

x of t, y of t, right?

x equals x of t.

y equals y of t.

So one should not have
to do that all the time,

not have to do that for a
simplification like a playing

card.

We have to find another
formula that's pretty, right?

Well, maybe it's not as pretty.

But when is it really pretty?

I bet it's going to be really
pretty if you have a plane

curve even as you're used
to in an explicit form--

I keep going.

No stop. [INAUDIBLE].

I think it's better.

We make better use
of time this way.

Or y equals f of x.

This is an explicit way to
write the equation of a curve.

OK, so what do we need to do?

That should be really easy.

R of t being the first case of
our general parametrization,

x equals x of t, y equals y of
t will be-- who tells me, guys,

that-- this is in your hands.

Now you convinced me
that, for whatever reason,

you [INAUDIBLE].

You became friends
with these curves.

I don't know when.

I guess in the process
of doing homework.

Am I right?

I think you did not quite like
them before or the last week.

But I think you're
friends with them now.

x of t, y of t.

Let people talk.

STUDENT: 0.

PROFESSOR TODA: So.

Great.

And then R prime of t will be
x prime of t, y prime of t,

and 0.

I assume this to
be always non-zero.

I have a regular curve.

R double prime will be--
x double prime where

double prime-- we
did the review today

of the lasting acceleration.

Now, your friends over
here, are they nice or mean?

I hope they are not so mean.

The cross product is
a friendly fellow.

You have i, j, k, and
then the second row

would be x prime, y prime, 0.

The last row would be x double
prime, y double prime, 0.

And it's a piece of cake.

OK, piece of cake,
piece of cake.

But I want to know
what the answer is.

So you have exactly 15 seconds
to answer this question.

Who is R prime plus R double
prime as a [? coordinate. ?]

[INTERPOSING VOICES]

PROFESSOR TODA: Good.

x prime, y double prime minus x
double prime, y prime times k.

And it doesn't matter
when I take the magnitude,

because magnitude of k is 1.

So I discovered some.

This is how mathematicians like
to discover new formulas based

on the formulas they
[? knew. ?] They

have a lot of satisfaction.

Look what I got.

Of course, they in general have
more complicated things to do,

and they have to
check and recheck.

But every piece of a
computation is a challenge.

And that gives
people satisfaction.

And when they make a mistake, it
brings a lot of tears as well.

So what-- could be written
on the bottom, what's

the speed cubed?

Speed is coming from this guy.

So the speed of the velocity,
the magnitude of the velocity

is the speed.

And that-- going
to give you square.

I'm not going to write
down [INAUDIBLE].

Square root of x squared,
x prime squared times

y prime squared,
and I cube that.

Many people, and I saw
that in engineering, they

don't like to put that
square root anymore.

And they just write x prime
squared plus y prime squared

to the what power?

STUDENT: 3/2.

PROFESSOR TODA: 3/2.

So this is very useful
for engineering styles,

when you have to deal
with plane curves, motions

in plane curves.

But now what do you
have in the case,

in the happy case, when
you have y equals f of x?

I'm going to do
that in a second.

I want to keep this
formula on the board.

What's the simplest
parametrization?

Because that's why we
need it, to look over

parametrizations
again and again.

R of t for this plane
curve will be-- what is t?

x is t, right?

x is t, y is f of t.

Piece of cake.

So you have t and f of t.

And how many of you watched
the videos that I sent you?

Do you prefer Khan
Academy, or do you

prefer the guys, [INAUDIBLE]
guys who are lecturing?

The professors who are lecturing
in front of a board or in front

of a-- what is that?

A projector screen?

I like all of them.

I think they're very good.

I think you can learn
a lot from three

or four different
instructors at the same time.

That's ideal.

I guess that you have
this chance only now

in the past few years.

Because 20 years ago, if you're
didn't like your instructor

or just you couldn't stand
them, you had no other chance.

There was no
YouTube, no internet,

no way to learn from others.

R prime of t would
be 1 f prime of t.

But instead of t I'll
out x, because x is t.

I don't care.

R double prime of t would
be 0, f double prime of x.

So I feel that, hey, I know
what's going to come up.

And I'm ready.

Well, we are ready
to write it down.

This is going to be Mr. x prime.

This is going to be
replacing Mr. y prime.

This is going to replace
Mr. a double prime.

This is going to be replacing
Mr. y double prime of x.

Oh, OK, all right.

So k, our old friend from
here will become what?

And I'd better shut up,
because I'm talking too much.

STUDENT: [INAUDIBLE]
double prime [INAUDIBLE].

PROFESSOR TODA: That is
the absolute value, mm-hmm.

[? n ?] double prime
of x, and nothing else.

Right, guys?

Are you with me?

Divided by--

STUDENT: [INAUDIBLE]

PROFESSOR TODA: Should
I add square root?

I love square roots.

I'm crazy about them.

So you go 1 plus f
prime squared cubed.

So that's going to
be-- any questions?

Are you guys with me?

That's going to be the
formula that I'm going

to use in the next example.

In case somebody
wants to know-- I got

this question from one of you.

Suppose we get a
parametrization of a circle

in the midterm or the final.

Somebody says, I have x
of t, just like we did it

today, a cosine t plus 0.

And y of t equals
a sine t plus y 0.

What is this, guys?

This is a circle, a center
at 0, y, 0, and radius a.

Can use a better formula-- that
anticipated my action today--

to actually prove that k
is going to be [? 1/a? ?]

Precisely.

Can we do that in the exam?

Yes.

So while I told
you a long time ago

that engineers and
mathematicians observed

hundreds of years
ago-- actually,

somebody said, no,
you're not right.

The Egyptians already saw that.

They had the notion of
inverse proportionality

in Egypt, which makes sense
if you look at the pyramids.

So one look at the radius,
it says if the radius is 2,

then the curvature
is not very bent.

So the curvature's inverse
proportion [INAUDIBLE]

the radius.

So if this is 2, we said
the curvature's 1/2.

If you take a big
circle, the bigger

the radius, the
smaller the bending

of the arc of the circle,
the smaller of the curvature.

Apparently the ancient
world knew that already.

They Egyptians knew that.

The Greeks knew that.

But I think they
never formalized it--

not that I know.

So if you are asked to
do this in any exam,

do you think that
would be a problem?

Of course we would do review.

Because people are going to
forget this formula, or even

the definition.

You can compute k
for this formula.

And we are going to
get k to the 1/a.

This is a piece
of cake, actually.

You may not believe me, but
once you plug in the equations

it's very easy.

Or you can do it
from the definition

that gives you k of s.

You'll reparametrize
this in arclength.

You can do that as well.

And you still get 1/a.

The question that
I got by email,

and I get a lot of email.

I told you, that
keeps me busy a lot,

about 200 emails every day.

I really like the emails
I get from students,

because I get emails from
all sorts of sources--

Got some spam also.

Anyway, what I'm trying to say,
I got this question last time

saying, if on the midterm
we get such a question,

can we say simply, curvature's
1/a, a is the radius.

Is that enough?

Depends on how the
problem was formulated.

Most likely I'm going to make
it through that or show that.

Even if you state something,
like, yes, it's 1/a,

with a little argument,
it's inverse proportional

to the radius, I will
still give partial credit.

For any argument that
is valid, especially

if it's based on
empirical observation,

I do give some extra
credit, even if you didn't

use the specific formula.

Let's see one example.

Let's take y equals e to the x.

No, let's take e
to the negative x.

Doesn't matter.

y equals e to the negative x.

And let's make x
between 0 and 1.

I'll say, write the curvature.

Write the equation or the
formula of the curvature.

And I know it's 2 o'clock
and I am answering questions.

This was a question that one of
you had during the short break

we took.

Can we do such a problem?

Like she said.

Yes, I [INAUDIBLE]
to the negative

x because I want
to catch somebody

not knowing the derivative.

I don't know why I'm doing this.

Right?

So if I were to draw that, OK,
try and draw that, but not now.

Now, what formula
are you going to use?

Of course, you could
do this in many ways.

All those formulas are
equivalent for the curvature.

What's the simplest
way to do it?

Do y prime.

Minus it to the minus x.

Note here in this problem that
even if you mess up and forget

the minus sign, you still
get the final answer correct.

But I may subtract a few points
if I see something nonsensical.

y double prime equals--

[INTERPOSING VOICES]

--plus e to the minus x.

And what is the
curvature k of t?

STUDENT: y prime over--

PROFESSOR TODA: Oh, I
didn't say one more thing.

I want the curvature, but
I also want the curvature

in three separate moments,
in the beginning, in the end,

and in the middle.

STUDENT: Don't we
need to parametrize it

so we can [INAUDIBLE]
x prime [INAUDIBLE]?

PROFESSOR TODA: No.

Did I erase it?

STUDENT: Yeah, you did.

PROFESSOR TODA: [INAUDIBLE].

And one of my colleagues
said, Magda, you are smart,

but you are like one
of those people who,

in the anecdotes
about math professors,

gets out of their office
and starts walking

and stops a student.

Was I going this
way or that way?

And that's me.

And I'm sorry about that.

I should not have erased that.

I'm going to go
ahead and rewrite it,

because I'm a goofball.

So the one that I wanted to use
k of x will be f double prime.

STUDENT: And cubed.

PROFESSOR TODA: Cubed!

Thank you.

So that 3/2, remember it,
[INAUDIBLE] 3/2 [INAUDIBLE]

square root cubed.

Now, for this one, is it hard?

No.

That's a piece of cake.

I said I like it in
general, but I also

like it-- find the curvature
of this curve in the beginning.

You travel on me.

From time 0 to 1
o'clock, whatever.

One second.

That's saying this is in seconds
to make it more physical.

I want the k at 0, I want k
at 1/2, and I want k at 1.

And I'd like you to
compare those values.

And I'll give you one
more task after that.

But let me start working.

So you say you help me on that.

[INAUDIBLE]

Minus x over square
root of 1 plus--

STUDENT: [INAUDIBLE]

PROFESSOR TODA: Right.

So can I write this differently,
a little bit differently?

Like what?

I don't want to square
each of the minus 2x.

Can I do that?

And then the whole thing
I can say to the 3/2

or I can use the square root,
whichever is your favorite.

Now, what is k of 0?

STUDENT: 0.

Or 1.

PROFESSOR TODA: Really?

STUDENT: 1/2.

3/2.

PROFESSOR TODA: So
let's take this slowly.

Because we can all make
mistakes, goofy mistakes.

That doesn't mean
we're not smart.

We're very smart, right?

But it's just a matter of
book-keeping and paying

attention, being attentive.

OK.

When I take 0 and replace--
this is drying fast.

I'm trying to draw it.

I have 1 over 1
plus 1 to the 3/2.

I have a student in one exam
who was just-- I don't know.

He was rushing.

He didn't realize that
he had to take it slowly.

He was extremely smart, though.

1 over-- you have
that 1 plus 1 is 2.

2 to the 1/2 would be
square root of 2 cubed.

It would be exactly
2 square root of 2.

And more you can write
this as rationalized.

Now, I have a question for you.

[INAUDIBLE]

I'm When we were kids, if you
remember-- you are too young.

Maybe you don't remember.

But I remember when I was a kid,
my teacher would always ask me,

rationalize your answer.

Rationalize your answer.

Put the rational number
in the denominator.

Why do you think that was?

For hundreds of years
people did that.

STUDENT: [INAUDIBLE]

PROFESSOR TODA: Because they
didn't have a calculator.

So we used to, even I used to be
able to get the square root out

by hand.

Has anybody taught you how to
compute square root by hand?

You know that.

Who taught you?

STUDENT: I don't remember it.

My seventh grade
teacher taught us.

PROFESSOR TODA:
There is a technique

of taking groups of twos
and then fitting the--

and they still teach that.

I was amazed, they
still teach that

in half of the Asian countries.

And it's hard, but kids
in fifth and sixth grade

have that practice, which some
of us learned and forgot about.

So imagine that how people would
have done this, and of course,

square root of 2 is easy.

1.4142, blah blah blah.

Divide by 2.

You can do it by hand.

At least a good approximation.

But imagine having a nasty
square root there to compute,

and then you would divide
by that natural number.

You have to rely on your
own computation to do it.

There were no calculators.

How about k of 1?

How is that?

What is that?

e to the minus 1.

That's a little bit
harder to compute, right?

1 plus [INAUDIBLE].

What is that going to be?

Minus 2.

Replace it by 1 to the 3/2.

I would like you to go
home and do the following.

[INAUDIBLE]-- Not now, not now.

We stay a little
bit longer together.

k of 0, k of 1/2, and k of 1.

Which one is bigger?

And one last question about
that, how much extra credit

should I give you?

One point?

One point if you turn this in.

Um, yeah.

Four, [? maybe ?] two points.

Compare all these
three values, and find

the maximum and the
minimum of kappa of t,

kappa of x, for
the interval where

x is in the interval 0, 1.

0, closed 1.

Close it.

Now, don't ask me,
because it's extra credit.

One question was, by email,
can I ask my tutor to help me?

As long as your tutor doesn't
write down your solution,

you are in good shape.

Your tutor should help you
understand some constants,

spend time with you.

But they should not write
your assignment themselves.

OK?

So it's not a big deal.

Not I want to tell you one
secret that I normally don't

tell my Calculus 3 students.

But the more I get
to know you, the more

I realize that you are worth
me telling you about that.

STUDENT: [INAUDIBLE]

PROFESSOR TODA: No.

There is a beautiful
theory that engineers

use when they start the motions
of curves and parametrizations

in space.

And that includes
the Frenet formulas.

And you already
know the first one.

And I was debating, I was just
reviewing what I taught you,

and I was happy with
what I taught you.

And I said, they know
about position vector.

They know about
velocity, acceleration.

They know how to get back
and forth from one another.

They know our claim.

They know how to
[? reparameterize our ?]

claims.

They know the [INAUDIBLE]
and B. They know already

the first Frenet formula.

They know the curvature.

What else can I teach them?

I want to show you--
one of you asked me,

is this all that we should know?

This is all that a regular
student should know in Calculus

3, but there is more.

And you are honor students.

And I want to show you some
beautiful equations here.

So do you remember that
if I introduce r of s

as a curving arclength,
that is a regular curve.

I said there is a certain famous
formula that is T prime of s

called-- leave space.

Leave a little bit of space.

You'll see why.

It's a surprise.

k times-- why
don't I say k of s?

Because I want to point
out that k is an invariant.

Even if you have
another parameter,

would be the same function.

But yes, as a function of s,
would be k times N bar, bar.

More bars because
they are free vectors.

They are not bound
to a certain point.

They're not married
to a certain point.

They are free to shift
by parallelism in space.

However, I'm going to review
them as bound at the point

where they are.

So they-- no way they
are married to the point

that they belong to.

Maybe the [? bend ?]
will change.

I don't know how it's
going to change like crazy.

Something like that.

At every point you have a T, an
N, and it's a 90 degree angle.

Then you have the binormal,
which makes a 90 degree

angle-- [INAUDIBLE].

So the way you should
imagine these corners

would be something
like that, right?

90-90-90.

It's just hard to draw them.

Between the vectors you have--
If you draw T and N, am I

right, that is coming out?

No.

I have to switch them.

T and N. Now, am I right?

Now I'm thinking of
the [? faucet. ?]

If I move T-- yeah,
now it's coming out.

So this is not getting
into the formula.

So this is the first formula.

You say, so what?

You've taught that.

We proved it together.

What do you want from us?

I want to teach you
two more formulas.

N prime.

And I'd like you to
leave more space here.

So you have like an empty field
here and an empty field here

[INAUDIBLE].

If you were to compute
T prime, the magic thing

is that T prime is a vector.

N prime is a vector.

B prime is a vector.

They're all vectors.

They are the derivatives
of the vectors T and NB.

And you say, why would I
care about the derivatives

of the vectors T and NB?

I'll tell you in a second.

So if you were to
compute in prime,

you're going to get here.

Minus k of s times T of s.

Leave room.

Leave room, because there
is no component that

depends on N. No such component
that that depends on N.

This is [INAUDIBLE].

There is nothing in
N. And then in the end

you'll say, plus tau of s
times B. There is missing--

something is.

And finally, if
you take B prime,

there is nothing
here, nothing here.

In the middle you have
minus tau of s times N of s.

And now you know that nobody
else but you knows that.

The other regular sections
don't know these formulas.

What do you observe about
this bunch of equations?

Say, oh, wait a minute.

First of all, why did
you put it like that?

Looks like a cross.

It is a cross.

It is like one is shaped in the
name of the Father, of the Son,

and so on.

So does it have anything
to do with religion?

No.

But it's going to help you
memorize better the equations.

These are the famous
Frenet equations.

You only saw the first one.

What do they represent?

If somebody asks you, what is k?

What it is k of s?

What's the curvature?

You go to a party.

There are only nerds.

It's you.

Some people taking advanced
calculus or some people

from Physics, and they say, OK,
have you heard of the Frenet

motion, Frenet
formulas, and you say,

I know everything about it.

What if they ask you, what
is the curvature of k?

You say, curvature measures
how a curve is bent.

And they say, yeah, but the
Frenet formula tells you

more about that.

Not only k shows you
how bent the curve is.

But k is a measure of
how fast T changes.

And he sees why.

Practically, if you take
the [INAUDIBLE] to the bat,

this is the speed of T. So how
fast the teaching will change.

That will be magnitude,
will be just k.

Because magnitude of N is 1.

So note that k of s is
the length of T prime.

This measures the change
in T. So how fast T varies.

What does the torsion represent?

Well, how fast the
binormal varies.

But if you want to
think of a helix,

and it's a little
bit hard to imagine,

the curvature measures how
bent a certain curve is.

And it measures how
bent a plane curve is.

For example, for the circle you
have radius a, 1/a, and so on.

But there must be
also a function that

shows you how a curve twists.

Because you have not
just a plane curve where

you care about curvature only.

But in the space curve you
care how the curves twist.

How fast do they move
away from a certain plane?

Now, if I were to draw-- is
it hard to memorize these?

No.

I memorized them easily
based on the fact

that everything looks
like a decomposition

of a vector in terms of
T, N, and B. So in my mind

it was like, I take any vector
I want, B. And this is T,

this is N, and this is B.
Just the weight was IJK.

Instead if I, I have T. Instead
of J, I have N. Instead of K,

I have B. They are
still unit vectors.

So locally at the
point I have this frame

and I have any vector.

This vector-- I'm a physicist.

So let's say I'm going to
represent that as v1 times

the T plus v2
times-- instead of J,

we'll use that N plus
B3 times-- that's

the last element of the bases.

Instead of k I have v.
So it's the same here.

You try to pick a
vector and decompose

that in terms of T, N, and B.
Will I put that on the final?

No.

But I would like you to
remember it, especially

if you are an engineering
major or physics major,

that there is this
kind of Frenet frame.

For those of you who are taking
a-- for differential equations,

you already do some matrices
and built-in systems

of equations, systems of
differential equations.

I'm not going to get there.

But suppose you don't know
differential equations,

but you know a little
bit of linear algebra.

And I know you know how
to multiply matrices.

You know how I know
you multiply matrices,

no matter how much
mathematics you learn.

And most of you, you are not in
general algebra this semester.

Only two of you are
in general algebra.

When I took a C++ course,
the first homework I got was

to program a matrix
multiplication.

I have to give in matrices.

I have to program that in C++.

And freshmen knew that.

So that means you know how
to write this as a matrix

multiplication.

Can anybody help me?

So T, N, B is the magic triple.

T, N, B's the magic corner.

T, N, and B are the Three
Musketeers who are all

orthogonal to one another.

And then I do derivative
with respect to s.

If I want to be
elegant, I'll put d/ds.

OK.

How am I going to
fill in this matrix?

So somebody who wants to know
about differential equations,

this would be a--

STUDENT: 0, k, 0.

PROFESSOR TODA: Very good.

0, k, 0, minus k 0
tau, 0 minus tau 0.

This is called the
skew symmetric matrix.

Such matrices are very
important in robotics.

If you've ever been
to a robotics team,

like one of those
projects, you should

know that when we study
motions of-- let's say

that my arm performs
two rotations in a row.

All these motions
are described based

on some groups of rotations.

And if I go into details,
it's going to be really hard.

But practically
in such a setting

we have to deal with matrices
that either have determined

one, like all rotations
actually have,

or have some other
properties, like this guy.

What's the determinant
of this guy?

What do you guys think?

Just look at it.

STUDENT: 0?

PROFESSOR TODA: 0.

It has determinant 0.

And moreover, it
looks in the mirror.

So this comes from
a group of motion,

which is little s over 3,
the linear algebra, actually.

So when k is looking
in the mirror,

it becomes minus k tau,
is becoming minus tau.

It is antisymmetric
or skew symmetric.

Skew symmetric or
antisymmetric is the same.

STUDENT: Antisymmetric,
skew symmetric matrix.

PROFESSOR TODA: Skew
symmetric or antisymmetric

is exactly the same thing.

They are synonyms.

So it looks in the mirror
and picks up the minus sign,

has 0 in the bag.

What am I going to put here?

You already got the idea.

So when Ryan gave
me this, he meant

that he knew what I'm going
to put here, as a vector,

as a column vector.

STUDENT: [INAUDIBLE]

PROFESSOR TODA: No, no no.

How do I multiply?

TNB, right?

So guys, how do you
multiply matrices?

You go first row
and first column.

So you go like this.

0 times T plus k times 10
plus 0 times B. Here it is.

So I'm teaching you
a little bit more

than-- if you are going to
stick with linear algebra

and stick with
differential equations,

this is a good introduction
to more of those mathematics.

Yes, sir?

STUDENT: Why don't
you use Cramer's rule?

PROFESSOR TODA: Uh?

STUDENT: Why don't you
use the Cramer's rule?

PROFESSOR TODA:
The Cramer's rule?

STUDENT: Yeah. [INAUDIBLE].

PROFESSOR TODA: No.

First of all, Crarmer's rule is
to solve systems of equations

that don't involve derivatives,
like a linear system

like Ax equals B.
I'm going to have,

for example, 3x1
plus 2x3 equals 1.

5x1 plus x2 plus x3
equals something else.

So for that I can
use Cramer's rule.

But look at that!

This is really complicated.

It's a dynamical system.

At every moment of time
the vectors are changing.

So it's a crazy [INAUDIBLE].

Like A of t times
something, so some vector

that is also depending on
time equals the derivative

of that vector that [INAUDIBLE].

So that's a OD system that
you should learn in 3351.

So I don't know what
your degree plan is,

but most of you in
engineering will

take my class, 2316 in algebra,
OD1 3350 where they teach you

about differential equations.

These are all differential
equations, all three of them.

In 3351 you learn
about this system

which is a system of
differential equation.

And then you
practically say, now I

know everything I need to
know in math, and you say,

goodbye math.

If you guys wanted
to learn more,

of course I would be very
happy to learn that, hey, I

like math, I'd like
to be a double major.

I'd like to be not just an
engineering, but also math

major if you really like it.

Many people already
have a minor.

Many of you have a
minor in your plan.

Like for that minor
you only need--

STUDENT: One extra math course.

PROFESSOR TODA: One
extra math course.

For example, with 3350 you
don't need 3351 for a minor.

Why?

Because you are taking the
probability in stats anyway.

You have to.

They force you to do that, 3342.

So if you take 3351 it's on top
of the minor that we give you.

I know because that's what I do.

I look at the degree plans.

And I work closely to the
math adviser, with Patty.

She has all the [INAUDIBLE].

STUDENT: So is [INAUDIBLE]?

PROFESSOR TODA: You mean double?

Double degree?

We have this already in place.

We've had it for many years.

It's an excellent plan.

162 hours it is now.

It used to be 159.

Double major, computer
science and mathematics.

And I could say they were
some of the most successful

in terms of finding jobs.

What would you take
on top of that?

Well, as a math major you
have a few more courses

to take one top of that.

You can link your computer
science with the mathematics,

for example, by taking
numerical analysis.

If you love computers
and you like calculus

and you want to put
together all the information

you have in both, then
numerical analysis

would be your best bet.

And they require that in
both computer science degree

if you are a double major,
and your math degree.

So the good thing is that some
things count for both degrees.

And so with those 160
hours you are very happy.

Oh, I'm done, I got
a few more hours.

Many math majors
already have around 130.

They're not supposed to.

They are supposed
to stop at 120.

So why not go the extra 20 hours
and get two degrees in one?

STUDENT: It's a semester.

PROFESSOR TODA: Yeah.

Of course, it's a lot more work.

But we have people
who like-- really they

are nerdy people who loved
computer science from when

they were three or four.

And they also like math.

And they say, OK,
I want to do both.

OK, a little bit more
and I'll let you go.

Now I want you to ask
me other questions

you may have had about the
homework, anything that

gave you headache, anything that
you feel you need a little bit

more of an explanation about.

Yes?

STUDENT: I just have one.

In WeBWork, what
is the easiest way

to take the square
root of something?

STUDENT: sqrt.

PROFESSOR TODA: sqrt
is what you type.

But of course you can
also go to the caret 1/2.

Something non-technical?

Any question, yes sir,
from the homework?

Or in relation to [INAUDIBLE]?

STUDENT: I don't understand
why is the tangent unit vector,

it's just the slope off
of that line, right?

The drunk bug?

Whatever line the
drunk bug is on?

PROFESSOR TODA: So it
would be the tangent

to the directional
motion, which is a curve.

And normalized to
have length one.

Because otherwise our
prime is-- you may say,

why do you need T to be unitary?

OK, computations become
horrible unless your speed

is 1 or 5 or 9.

If the speed is a constant,
everything else becomes easier.

So that's one reason.

STUDENT: And why
is the derivative

of T then perpendicular?

Why does it always turn into--

PROFESSOR TODA:
Perpendicular to T?

We've done that last time,
but I'm glad to do it again.

And I forgot what we
wrote in the book,

and I also saw in
the book this thing

that if you have R, in
absolute value, constant--

and I've done that
with you guys--

prove that R and R prime had
every point perpendicular.

So if you have-- we've
done that before.

Now, what do you do then?

T [INAUDIBLE] T is 1.

The scalar [INAUDIBLE]
the product.

T prime times T plus
T prime T prime.

So 0.

And T is perpendicular
to T prime,

because that means T
or T prime equals 0.

When you run in a
circle, you say--

OK, let's run in a circle.

I say, this is my T. I can feel
that there is something that's

trying to bend me this way.

That is my acceleration.

And I have to-- but I don't
know-- how familiar are you

with the winter sports?

In many winter sports, the
Frenet Trihedron is crucial.

Imagine that you have
one of those slopes,

and all of the sudden the
torsion becomes too weak.

That means it becomes dangerous.

That means that the
vehicle you're in,

the snow vehicle or any kind
of-- your skis, [INAUDIBLE],

if the torsion of your body
moving can become too big,

that will be a problem.

So you have to redesign
that some more.

And this is what they do.

You know there have
been many accidents.

And many times they say,
even in Formula One,

the people who project
a certain racetrack,

like a track in
Indianapolis or Montecarlo

or whatever, they
have to have in mind

that Frenet frame every second.

So there are
simulators showing how

the Frenet frame is changing.

There are programs that
measure the curvature

in a torsion for those
simulators at every point.

Neither the curvature
nor the torsion

can exceed a certain value.

Otherwise it becomes dangerous.

You say, oh, I thought
only the speed is a danger.

Nope.

It's also the way that the
motion, if it's a skew curve,

it's really complicated.

Because you twist and turn
and bend in many ways.

And it can become
really dangerous.

Speed is not [INAUDIBLE].

STUDENT: So the torsion was
the twists in the track?

PROFESSOR TODA: The
torsion is the twist.

And by the way, keep your idea.

You wanted to ask
something more?

When you twist-- suppose you
have something like a race car.

And the race car is at
the walls of the track.

And here's-- when you have
a very abrupt curvature

and torsion, and you can have
that in Formula One as well,

why do they build one wall
a lot higher than the other?

Because the poor car-- I
don't know how passionate you

are about Formula
One or car races--

the poor car is going
to be close to the wall.

It's going to bend like that,
that wall would be round.

And as a builder, you have to
build the wall really high.

Because that kind of high
speed, high velocity,

high curvature, the poor
car's going szhhhhh-- then

again on a normal track.

Imagine what happens if the
wall is not high enough.

The wheels of the car
will go up and get over.

And it's going to be a disaster.

So that engineer ha to study
all the parametric equations

and the Frenet frame and
deep down make a simulator,

compute how tall the walls
should be in order for the car

not to get over on the other
side or get off the track.

It's really complicated stuff.

It's all mathematics
and physics,

but all the applications are
run by engineers and-- yes, sir?

STUDENT: What's the difference
[INAUDIBLE] centrifugal force?

PROFESSOR TODA: The
centrifugal force

is related to our double prime.

Our double prime is related
to N and T at the same time.

So at some point, let me ask you
one last question and I'm done.

What's the relationship between
acceleration or double prime?

And are they the same thing?

And when are they
not the same thing?

Because you say, OK,
practically the centrifugal--

STUDENT: They're
the same on a curve.

PROFESSOR TODA:
They are the same--

STUDENT: Like on a circle.

PROFESSOR TODA: On a circle!

And you are getting so close.

It's hot, hot, hot.

On a circle and on a helix they
are the same up to a constant.

So what do you think the
magic answer will be?

N was what, guys?

N was-- remind me again.

That was T prime over
absolute value of T prime.

But that doesn't mean,
does not equal, in general,

does not equal to
R double prime.

When is it equal?

In general it's not equal.

When is it equal?

If you are in aclength, you
see the advantage of aclength.

It's wonderful.

In arclength, T is R prime of s.

And in arclength that means T
prime is R double prime of s.

And in arclength
I just told you,

T prime is the first
Frenet formula.

It'll be curvature times the N.

So the acceleration
practically and the N

will be the same in arclength,
up to a scalar multiplication.

But what if your speed
is not even constant?

Then God help you.

Because the acceleration
R double prime and N

are not colinear.

So if I were to draw-- and
that's my last picture--

let me give you a
wild motion here.

You start slow and then you go
crazy and fast and slow down.

Just like most of the
physical models from the bugs

and the flies and so on.

In that kind of crazy motion you
have a T and N at every point.

[INAUDIBLE]

[? v ?] will be down.

And T is here.

So can you draw arc
double prime for me?

It will still be
towards the inside.

But it's still going to
coincide with N. Maybe this one.

What's the magic thing is
that T, N, and R double prime

are in the same plane always.

That's another
secret other students

don't know in Calculus 3.

That same thing is
called osculating plane.

We have a few magic
names for these things.

So T and N, the plane that
is-- how shall I say that?

I don't know.

The plane given by T and N
is called osculating plane.

The acceleration is
always on that plane.

So imagine T and N are
in the same shaded plane.

R double prime is
in the same plane.

OK?

Now, can you guess
the other two names?

So this is T, this
is N. And B is up.

This is my body's direction.

T and N, look at me.

T, N, and B. I'm the
Frenet Trihedron.

Which one is the
osculating plane?

It's the horizontal xy plane.

OK, do you know-- maybe you're
a mechanical engineering major,

and after that I
will let you go.

No extra credit,
though for this task.

Maybe I'm going to start asking
questions and give you $1.

I used to do that a lot
in differential equations,

like ask a hard question,
whoever gets it first,

give her a dollar.

Until a point when they asked
me to teach Honors 3350 when

I started having three or four
people answering the question

at the same time.

And that was a
significant expense,

because I had to give $4
away at the same time.

STUDENT: I feel like
you should've just

split it between--

PROFESSOR TODA: So that's
normal and binormal.

This is me, the binormal,
and this is the normal.

Does anybody know the
name of this plane,

between normal and bionormal?

This would be this plane.

STUDENT: The skew [INAUDIBLE].

PROFESSOR TODA:
Normal and binormal.

They call that normal plane.

So it's tricky if you are not
a mechanical engineering major.

But some of you are maybe
and will learn that later.

Any other questions for me?

Now, in my office I'm
going to do review.

I was wondering
if you have time,

I don't know if you have
time to come to my office,

but should you have any kind
of homework related question,

I'll be very happy
to answer it now.

3:00 to 5:00.

Now, one time I
had a student who

only had seven questions left.

He came to my office and
he left with no homework.

We finished all of them.

And I felt guilty.

But at the same, he
said, well, no, it's

better I came to you instead
of going to my tutor.

It was fine.

So we can try some
problems together today

if you want between 3:00 and
5:00, if you have the time.

Some of you don't have the time.

All right?

If you don't have
the time today,

and you would like to
be helped [INAUDIBLE],

click Email Instructor.

I'm going to get the
questions [INAUDIBLE].

You're welcome to
ask me anything

at any time over there.

[CLASSROOM CHATTER]

PROFESSOR TODA: I have
somebody who's taking notes.

STUDENT: Yeah, I know.

And that's why I was like--

PROFESSOR TODA: He's
going to make a copy

and I'll give you a copy.

STUDENT: Yeah.

My Cal 1 teacher,
Dr. [INAUDIBLE].

STUDENT: Thank you.

PROFESSOR TODA: Yes, yeah.

Have a nice day.

STUDENT: --got really mad
when I don't take notes.

Because he felt like
I was not, I guess--