Unit tangent vector vs principal normal vector

[BvU]

[in reply to #29 which was edited afterwards] :

No, I did NOT ask you to differentiate $\ \vec N\cdot \vec T^2 \$, I asked: what gives $\ \vec N\cdot \vec T \$ when it is differentiated wrt time.

Then I asked you what the time derivative of $\vec T^2$ is. After a while we came to $2\;\vec T\cdot\vec T'$

(I know, you used the shorthand 2TT' that is used by very experienced folks, but I like to take small steps and stick to the explicit vector notation -- just checking: you do understand that indeed $$ {d\vec T^2\over dt\ }= 2\; \vec T\cdot {d\vec T \over dt}\quad\quad\quad? \quad ) $$​

In that case we put 1 and 1 together, connect the dots, have our Aha ! moment and see that $$ {d\vec T^2\over dt } \propto \ \vec N\cdot \vec T \ ! $$

 

[BvU]

do you mean $\vec N \cdot \vec T \ = 0\$? but , I am shown that N = 1 , i am still not convinced that N.T = 0 ...

Quick succession of posts - delay on the line causes crossings.

Yes, we are working towards $\vec N\cdot \vec T \ = 0$

No, we have not shown that $\vec N = 1$ (can't be that a general vector is equal to a number). We have shown that $\ \|\vec N\| = 1 \$.

 

[chetzread]

Quick succession of posts - delay on the line causes crossings.

Yes, we are working towards $\vec N\cdot \vec T \ = 0$

No, we have not shown that $\vec N = 1$ (can't be that a general vector is equal to a number). We have shown that $\ \|\vec N\| = 1 \$.

deleted

 

[chetzread]

[in reply to #29 which was edited afterwards] :

No, I did NOT ask you to differentiate $\ \vec N\cdot \vec T^2 \$, I asked: what gives $\ \vec N\cdot \vec T \$ when it is differentiated wrt time.

Then I asked you what the time derivative of $\vec T^2$ is. After a while we came to $2\;\vec T\cdot\vec T'$

(I know, you used the shorthand 2TT' that is used by very experienced folks, but I like to take small steps and stick to the explicit vector notation -- just checking: you do understand that indeed $$ {d\vec T^2\over dt\ }= 2\; \vec T\cdot {d\vec T \over dt}\quad\quad\quad? \quad ) $$​

In that case we put 1 and 1 together, connect the dots, have our Aha ! moment and see that $$ {d\vec T^2\over dt } \propto \ \vec N\cdot \vec T \ ! $$

ok , i can understand that ...but , i still can't understand why N.T = 0 ?

we only reach this
$$ {d\vec T^2\over dt } \propto \ \vec N\cdot \vec T \ ! $$ , but still haven't reach N.T = 0 ? do i left out something ?

 

[BvU]

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

Can you explain further?

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

just checking: you do understand that indeed

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

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.

 

[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 ?