Unit tangent vector vs principal normal vector

[chetzread]

Ah, again postings cross. We go faster now ! Good. This is still in reply to #33:Yes.

Do you remember $\vec N \equiv \displaystyle { \vec T'\over \|\vec T'\|}$ ?

And do you remember $\|\vec T\| = 1$ , so that $\vec T\cdot\vec T = 1 \$ ?

and
$$
{d\vec T^2\over dt\ }= 2\; \vec T\cdot {d\vec T \over dt}\quad\quad\quad? \quad $$

so that - using $\displaystyle{d\over dt }\; 1 = 0 \$ we come to the unavoidable conclusion that

[and now we smoothly go into responding to #34] :
$$
0 = {d\over dt }\; 1 = {d\over dt }\; \vec T^2 = 2 \vec T \cdot {d\over dt }\vec T = 2 \;\vec T\cdot\vec N \; \|\vec T'\| \quad \Rightarrow \vec N \cdot\vec T = 0 $$
(because | T| = 1 ≠ 0 ).

[edit] Oops, sorry, $\ \|\vec T'\|\$, not $\ \|\vec T\|\$ without the quote. But if $\ \|\vec T'\|\ = 0$ then $\ \vec N = 0 \$ too.

it should be 2T (dott) N = 0 , am i right ? why there's an extra || T || ?

 

[BvU]

$\vec N \equiv \displaystyle { \vec T'\over \|\vec T'\|} \Rightarrow \vec T' = \vec N\; \|\vec T' \| \$.

Before I forget: you can read all about this in Paul1 , Paul2 , Paul3 (I do find the pdf downloads easier on the eyes). And you see that it's at the end of very thick books; no wonder it takes us a while

 

[chetzread]

$\vec N \equiv \displaystyle { \vec T'\over \|\vec T'\|} \Rightarrow \vec T' = \vec N\; \|\vec T' \| \$.

Before I forget: you can read all about this in Paul1 , Paul2 , Paul3 (I do find the pdf downloads easier on the eyes). And you see that it's at the end of very thick books; no wonder it takes us a while

in post #35 , do you mean $\ \|\vec T'\|\ = 0$ , so that
$$
{d\vec T^2\over dt\ } $$ = 0 ?

 

[BvU]

No, it's the other way around: $\displaystyle {d\vec T^2\over dt\ } = 0$ always, because $\vec T^2 = 1$.

We had $$\vec N \equiv \displaystyle { \vec T'\over \|\vec T'\|} \Rightarrow \vec T' = \vec N\; \|\vec T' \| \ $$ with the complication that, if $\ \|\vec T'\|\ =0$ then $\ \vec N\$ does not exist. (actually, it's the same complication as with $\ \vec T\$ when $\ \vec v' = \vec 0\$).

From $\|\vec T\| = 1 \$ we deduce $\ \vec T\cdot \vec T' = 0 \$. Therefore $\ \vec T \cdot \vec N \; \|T'\| \;= 0\$ but that's not the same as showing that $\ \vec T \cdot \vec N = 0\$ which we needed to prove that $\ \vec T \perp \vec N\$.

For that last step, from $\ \vec T \cdot \vec N \; \|T'\| \;= 0\$ to $\ \vec T \cdot \vec N = 0\$we need $\ \|T'\| \;\ne 0\$, so I looked at that separately.

 

[BvU]

I'd like to summarize this with the simplest possible example: uniform circular motion in 2D.

So: $(x,y) = (\cos\omega t , \sin\omega t)$ in cartesian coordinates.

What are $\ \vec v$, $\ \vec T$ and $\ \vec N$ ?​

 

[chetzread]

No, it's the other way around: $\displaystyle {d\vec T^2\over dt\ } = 0$ always, because $\vec T^2 = 1$.

We had $$\vec N \equiv \displaystyle { \vec T'\over \|\vec T'\|} \Rightarrow \vec T' = \vec N\; \|\vec T' \| \ $$ with the complication that, if $\ \|\vec T'\|\ =0$ then $\ \vec N\$ does not exist. (actually, it's the same complication as with $\ \vec T\$ when $\ \vec v' = \vec 0\$).

From $\|\vec T\| = 1 \$ we deduce $\ \vec T\cdot \vec T' = 0 \$. Therefore $\ \vec T \cdot \vec N \; \|T'\| \;= 0\$ but that's not the same as showing that $\ \vec T \cdot \vec N = 0\$ which we needed to prove that $\ \vec T \perp \vec N\$.

For that last step, from $\ \vec T \cdot \vec N \; \|T'\| \;= 0\$ to $\ \vec T \cdot \vec N = 0\$we need $\ \|T'\| \;\ne 0\$, so I looked at that separately.

if we need to show that $\ \vec T \cdot \vec N = 0\$ , then , we need ||T'|| = 1 , am i right ?
So that we can ignore the ||T'|| in $\ \vec T \cdot \vec N \; \|T'\| \;= 0\$ , to get $\ \vec T \cdot \vec N \= 0\$

 

[BvU]

No, we only need $\ \|T'\| \;\ne 0\$ (fortunately).

 

[chetzread]

No, we only need $\ \|T'\| \;\ne 0\$ (fortunately).

so , we are ignoring the value of $\ \|T'\| \$ to get T (dot)N only , rather than $\ \vec T \cdot \vec N \; \|T'\| \;= 0\$ ?

 

[BvU]

We are not ignoring it: we divide left and right by it. That can only be done if it's not zero -- hence the caution.

 

[chetzread]

We are not ignoring it: we divide left and right by it. That can only be done if it's not zero -- hence the caution.

can you explain further ? i still couldn't gt it

 

[BvU]

$$ \vec T \cdot\vec T' = 0 \quad \Rightarrow \quad A \; \vec T \cdot\vec T' = 0 $$ for any $A$.

Multiply left and right with $A = 1/\|\vec T'\|$ to get $$\quad\quad \Rightarrow \vec T \cdot\vec N = 0 $$

 

[chetzread]

$$ \vec T \cdot\vec T' = 0 \quad \Rightarrow \quad A \; \vec T \cdot\vec T' = 0 $$ for any $A$.

Multiply left and right with $A = 1/\|\vec T'\|$ to get $$\quad\quad \Rightarrow \vec T \cdot\vec N = 0 $$

ok , since anything including $A = 1/\|\vec T'\|$ multiply by 0 = 0 ?
we multiply RHS of ]$$ \vec T \cdot\vec T' = 0$$ by 0 will get 0 ...

 

[chetzread]

​

my working is in 317.jpg[/QUOTE]

Can someone explain how to turn the formula of curvature T'(t) / r'(t) into | r'(t) x r"(t) | / | (r't)^3 | ?my working is in 317.jpg

Well, can someone help to explain which part gone wrong now?

 

[BvU]

my working is in 317.jpg
Well, can someone help to explain which part gone wrong now?

We'll come to that. The example from post #40 can even help us there:

I'd like to summarize this with the simplest possible example: uniform circular motion in 2D.

So: $(x,y) = (\cos\omega t , \sin\omega t)$ in cartesian coordinates.

What are $\ \vec v$, $\ \vec T$ and $\ \vec N$ ?​

 

[chetzread]

I'd like to summarize this with the simplest possible example: uniform circular motion in 2D.

So: $(x,y) = (\cos\omega t , \sin\omega t)$ in cartesian coordinates.

What are $\ \vec v$, $\ \vec T$ and $\ \vec N$ ?​

i am not sure what does $\ \vec v$ mean ? do you mean $\ \vec v = r\cos\omega t , r\sin\omega t$

 

[BvU]

No. Given is: $\vec r(t) = (x(t),y(t)) = (\cos \omega t, \sin \omega t)$.

What is $\vec v(t) \equiv \;\displaystyle {d\vec r(t)\over dt}$ ?

Have you never done such things before ?

 

[chetzread]

No. Given is: $\vec r(t) = (x(t),y(t)) = (\cos \omega t, \sin \omega t)$.

What is $\vec v(t) \equiv \;\displaystyle {d\vec r(t)\over dt}$ ?

Have you never done such things before ?

here it is .

 

[BvU]

Good thing I asked. In your $\vec v$ I can still distinguish an $\hat\imath$ and a $\hat\jmath$, so they can pass as vectors. An alternative notation would be $\vec v(t) = (-\omega\sin\omega t, \omega\cos\omega t)\$.

I'm not sure what you mean with $V$, but if it's $\vec r'(t)\over \|\vec r'(t)\| \$, then so far we have used the name $\vec T$. Why not keep that name ? And: it is a vector, not a number !

 

[chetzread]

Good thing I asked. In your $\vec v$ I can still distinguish an $\hat\imath$ and a $\hat\jmath$, so they can pass as vectors. An alternative notation would be $\vec v(t) = (-\omega\sin\omega t, \omega\cos\omega t)\$.

I'm not sure what you mean with $V$, but if it's $\vec r'(t)\over \|\vec r'(t)\| \$, then so far we have used the name $\vec T$. Why not keep that name ? And: it is a vector, not a number !

V is not r'(t) ?what is V now? I'm confused...

 

[BvU]

How can you be confused about something you introduced yourself ? line 2 in your post has a $v$ (small v) but you mean a vector. Learn yourself to clearly designate vectors as vectors and scalars as scalars.

Line 3 has a $V$ big V (right ?) what do you mean with that ?

 

[chetzread]

How can you be confused about something you introduced yourself ? line 2 in your post has a $v$ (small v) but you mean a vector. Learn yourself to clearly designate vectors as vectors and scalars as scalars.

Line 3 has a $V$ big V (right ?) what do you mean with that ?

they should be the same v , by the way, what is $v$ ? i am still blurred.

 

[chetzread]

No. Given is: $\vec r(t) = (x(t),y(t)) = (\cos \omega t, \sin \omega t)$.

What is $\vec v(t) \equiv \;\displaystyle {d\vec r(t)\over dt}$ ?

Have you never done such things before ?

v is r'(t ) ?
then v is same as T ?

 

[chetzread]

No. Given is: $\vec r(t) = (x(t),y(t)) = (\cos \omega t, \sin \omega t)$.

What is $\vec v(t) \equiv \;\displaystyle {d\vec r(t)\over dt}$ ?

Have you never done such things before ?

v is r'(t ) ?
then v is same as T ?

 

[chetzread]

No. Given is: $\vec r(t) = (x(t),y(t)) = (\cos \omega t, \sin \omega t)$.

What is $\vec v(t) \equiv \;\displaystyle {d\vec r(t)\over dt}$ ?

Have you never done such things before ?

v is r'(t ) ?
then v is same as T ?

 

[chetzread]

No. Given is: $\vec r(t) = (x(t),y(t)) = (\cos \omega t, \sin \omega t)$.

What is $\vec v(t) \equiv \;\displaystyle {d\vec r(t)\over dt}$ ?

Have you never done such things before ?

v is r'(t ) ?
then v is same as T ?

 

[BvU]

by the way, what is $v$

As I said, learn yourself to clearly designate vectors as vectors and scalars as scalars.

There are a few possibilities for the notation.

The one in your textbook is

• vectors are bold
• vector norms are written with double vertical lines ($\ \|\$)

and is not very suitable for handwriting. If your class doesn't have it's own standards, I would advise:

• vectors are written with an arrow above ($\ \vec r\$)
• a single | is enough for a norm ($\ |\vec r| \$ )

Once you are more experienced you can make it easier by dropping the | : $\vec r$ is the vector, $r$ is the norm. Not now, later.
Stay with your standard. Use it consistently.

You are going too fast without a drivers license. Re-post #57 with a decent notation.

In answer to your #56: what is $v$ :

It is common to designate a position vector as $\vec r$.

In Cartesian coordinates a 2D $\ \ \vec r = (x,y) \$ and a 3D $\ \ \vec r = (x,y,z)$.
x and y may depend on time and you get e.g. $\ \ \vec r(t) = (x(t),y(t)) \$
​

The time derivative is the velocity vector, commonly designated as $\vec v$. So $\vec v = \displaystyle d\vec r\over dt$.

In Cartesian coordinates a 2D $\ \ \vec v = (v_x,v_y) = \left (\displaystyle {dx(t) \over dt}, \displaystyle {dy(t) \over dt} \right) \$,
often abbreviated with a quote for ${d\over dt}$. E.g. a 3D $\ \ \vec v = (x',y',z')$.
​

The time derivative of the velocity vector is the acceleration vector, commonly designated as $\vec a$. So $\vec a = \displaystyle {d\vec v\over dt}$.

In Cartesian coordinates a 2D $\vec a = (a_x,a_y) = (x'', y'')$
​

I could retype all 385 pages of Paul's excellent pdf, but I suggest you read through it on your own -- just to get familiar. And make a lot of exercises to build up experience.

 

[BvU]

Now I start to double-post too.

 

[chetzread]

http://tutorial.math.lamar.edu/Classes/CalcIII/Curvature.aspx
well , can you explain how to get the circled part ? my problem is this.

 

[BvU]

I spent quite some time typing #61. #63 is identical to post #1 in the thread that was merged into this one. It looks as if you don't want to learn the language of simple vectors and vector calculus, so: yes I can but you wouldn't understand. Is there a good reason for that ?

 

[chetzread]

I spent quite some time typing #61. #63 is identical to post #1 in the thread that was merged into this one. It looks as if you don't want to learn the language of simple vectors and vector calculus, so: yes I can but you wouldn't understand. Is there a good reason for that ?

I spent quite some time typing #61. #63 is identical to post #1 in the thread that was merged into this one. It looks as if you don't want to learn the language of simple vectors and vector calculus, so: yes I can but you wouldn't understand. Is there a good reason for that ?

 

[chetzread]

Ok, here is after correction

 

[BvU]

So far
we have $\vec r\qquad$

don't forget the arrow over the $r$ -- or some other way to clearly mark it as a vector
don't forget the brackets
$\qquad \vec r(t) = (cos\omega t,\sin\omega t)\ \$ which was the given data.​

we have $\vec r'\qquad$ which is $\vec v$ (so $\vec v$ is NOT your ${r'\over|r'|}$ !)

don't forget the arrow over the $r'$ -- or some other way to clearly mark it as a vector
don't forget the brackets
$\qquad \vec r'(t) = (- \omega \sin\omega t,\omega\cos\omega t)\ \ = \omega (\sin\omega t,\cos\omega t)\ \$ .​

We see that $|\vec r| = 1$ (right ?) and we see that $|\vec r'| = |\omega|$ (right ?)

But then you go all wrong by thinking that $\vec T$ is not a vector.
And what you do to find $\vec N$ is a mystery to me. Again, you write it as a number, which it is NOT. $\vec N$ is a vector !

 

[BvU]

http://tutorial.math.lamar.edu/Classes/CalcIII/Curvature.aspx
well , can you explain how to get the circled part ? my problem is this.

Do you know how to calculate the norm of a vector product such as $\ \vec r'\times \vec r'' \$?

 

[chetzread]

But then you go all wrong by thinking that ⃗T\vec T is not a vector.
And what you do to find ⃗N\vec N is a mystery to me. Again, you write it as a number, which it is NOT. ⃗N\vec N is a vector !

sorry , for T , it should be (-w sin wt i + w coswt j ) / w
, similarly ,
for N , it should be = (-(w^2)((coswt)^2) i - (w^2)((sinwt)^2) j ) / (w^2)
Again , here's the formula have in my book

 

[chetzread]

Do you know how to calculate the norm of a vector product such as $\ \vec r'\times \vec r'' \$?

do you mean express it in form of unit vector ?