Unit tangent vector vs principal normal vector

[chetzread]

Homework Statement

http://mathwiki.ucdavis.edu/Core/Ca.../The_Unit_Tangent_and_the_Unit_Normal_Vectors
In the link, I can't understand that why the Principal Unit Normal Vector is defined by N(t) = T'(t) / | T'(t) | ,can someone explain...

Homework Equations

The Attempt at a Solution

Since tangent vector is normal to principal normal vector, why the N(t) shouldn't be -1 / ( T'(t) / | T'(t) | ) , since we know that if two vectors ( A and B) are prependicular to each other, then A multiply by B will get -1,
so, N(t) shouldn't be -1 / ( T'(t) / | T'(t) | ),
am i right? why we must differentiate it?

 

[BvU]

How do you divide the number -1 by a vector ? Write it out as vectors and perhaps it will become clearer then...

Oh, and two vectors are perpendicular if their (inner) product is zero.

Two lines in the x,y plane may be perpendicular if the product of their slopes is -1, but that's something else ...

 

[chetzread]

How do you divide the number -1 by a vector ? Write it out as vectors and perhaps it will become clearer then...

Oh, and two vectors are perpendicular if their (inner) product is zero.

Two lines in the x,y plane may be perpendicular if the product of their slopes is -1, but that's something else ...

here's my idea
i mean when we find dy/dx = gradient of tangent , gradient of tangent x gradient of normal = -1 , am i right ?
https://revisionmaths.com/advanced-l...ts-and-normals

 

[BvU]

For a line $y_1 = a+bx$ and a line $y_2= c + dx$ then $bd = -1 \Rightarrow$ lines are perpendicular.
$\displaystyle {dy\over dx}$ is the tangent of the slope. Not the gradient of the tangent.

Numerical example: $y_1 = x$ and $y_2 = -x$ bd = -1 so perpendicular.

Otherwise: In the x,y plane doing analytical geometry, yes.But a vector is something else -- even in the x,y plane !

Vector $\vec v_1 = (a,b)$ and $\vec v_2 = (c,d)$ then $$\vec v_1 \perp \vec v_2 \ \ \Leftrightarrow \ \ \vec v_1 \cdot \vec v_2 = ac + bd = 0 $$.

Numerical example: $(1,1)$ and $(1,-1)$

 

[chetzread]

For a line $y_1 = a+bx$ and a line $y_2= c + dx$ then $bd = -1 \Rightarrow$ lines are perpendicular.
$\displaystyle {dy\over dx}$ is the tangent of the slope. Not the gradient of the tangent.Numerical example: $y_1 = x$ and $y_2 = -x$ bd = -1 so perpendicular.
Otherwise: In the x,y plane doing analytical geometry, yes.

But a vector is something else -- even in the
x,y plane !

Vector $\vec v_1 = (a,b)$ and $\vec v_2 = (c,d)$ then $$\vec v_1 \perp \vec v_2 \ \ \Leftrightarrow \ \ \vec v_1 \cdot \vec v_2 = ac + bd = 0 $$.

Numerical example: $(1,1)$ and $(1,-1)$

So, N(t) is gradient of tangent which us different from gradient of slope in the previous case? So, -1 / (dy/dx) can't be used here?

 

[BvU]

More mission work required. All caps is shouting in PF, not allowed.

Loudly:

not gradient. Don't confuse! Gradient has a different meaning (namely: spatial derivative of a scalar quantity. A gradient is a vector, because you can have different derivatives for any of the n dimensions).​

$\vec N(t)$ is the normalized time derivative of $\vec T$.

Indeed a different beast altogeher. We can make a link (of course -- why try to keep things simple -- but let's do that later).

 

[BvU]

Hehe, when I read my own post I must concede that gradient = tangent of slope in x,y plane in the 1D case y = f(x) . (So still no gradient of tangent ! ) Why ? Well, in the 1D case there's no genuine difference between a vector and a scalar, I think (euphemism for feeling on thin ice...).

Back to the main thread ! Since T is normalized, it doesn't change in size very much. All that can change is its direction -- and that happens in a direction perpendicular to the thing itself.

(I'm giving it away a little bit; sheer excitement. ) FIrst things first: clear so far ?

 

[BvU]

Now, once converted and convinced: welcome to the wonderful world of vector calculus .

As we should agree by now, the $\vec N$ definition is sound if we can underpin it with two properties: $\|\vec N\| = 1$ and $\vec N\cdot\vec T = 0$ .

First one is easy, right ? Can you write it out ? And the second one ?

 

[chetzread]

Now, once converted and convinced: welcome to the wonderful world of vector calculus .

As we should agree by now, the $\vec N$ definition is sound if we can underpin it with
two properties: $\|\vec N\| = 1$ and $\vec N\cdot\vec T = 0$ .

First one is easy, right ? Can you write it out ?
And the second one ?

What do you want me to write? I am confused...

 

[BvU]

Well, I simply want you to write out $\|\vec N\|$ in such a way that it shows that the value is 1

 

[chetzread]

Well, I simply want you to write out $\|\vec N\|$ in such a way that it shows that the value is 1

sorry , why it will equal to 1 ? I really have no idea...

 

[BvU]

$$\|\vec N\|\ \equiv \sqrt {\vec N\cdot\vec N}$$ $$ {\vec N\cdot\vec N} = { \vec T' \over \|\vec T'\|} \cdot { \vec T' \over \|\vec T'\|} = { \vec T' \cdot \vec T'\over \|\vec T' \|^2 } ={ \|\vec T' \|^2 \over \|\vec T' \|^2 } = 1$$

 

[chetzread]

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

 

[chetzread]

$$\|\vec N\|\ \equiv \sqrt {\vec N\cdot\vec N}$$ $$ {\vec N\cdot\vec N} = { \vec T' \over \|\vec T'\|} \cdot { \vec T' \over \|\vec T'\|} = { \vec T' \cdot \vec T'\over \|\vec T' \|^2 } ={ \|\vec T' \|^2 \over \|\vec T' \|^2 } = 1$$

is T' . T' = || T' ^2 || ?
|| T' ^2 || shows the magnitude , right ? and T' . T' only shows the product of dot product , why they are equal ?

 

[BvU]

Later. I would like to know if my last post is clear to you -- and why you didn't have any idea what I meant ...

Then we go on with showing that $\vec N\cdot\vec T=0$.

And then we move on to curvature...

 

[chetzread]

Later. I would like to know if my last post is clear to you -- and why you didn't have any idea what I meant ...

Then we go on with showing that $\vec N\cdot\vec T=0$.

And then we move on to curvature...

do you mean post#12 ?
i just don't understand why
T' . T' = || T' ^2 || ?
|| T' ^2 || shows the magnitude , right ? and T' . T' only shows the product of dot product , why they are equal ?

 

[BvU]

is T' . T' = || T' ^2 || ?
|| T' ^2 || shows the magnitude , right ? and T' . T' only shows the product of dot product , why they are equal ?

No. $\ \ \vec T'\cdot\vec T'$ is usually written as $\vec T^2$. It is a scalar (a number). So $$
\vec T'\cdot\vec T' = \|\vec T'\|^2 $$ the square of te norm.

Remember Pythagoras, remember what a vector dot product is, try a few numerical examples (the norm of $(3,4)$ for example).

 

[chetzread]

No. $\ \ \vec T'\cdot\vec T'$ is usually written as $\vec T^2$. It is a scalar (a number). So $$
\vec T'\cdot\vec T' = \|\vec T'\|^2 $$ the square of te norm.

Remember Pythagoras, remember what a vector dot product is, try a few numerical examples (the norm of $(3,4)$ for example).

following the same trend as | N | , i gt |T | as 1 , here's my working , how could N dot T = 0 ?
i gt 1 dot 1 = 1

 

[BvU]

i gt 1 dot 1 = 1

No, that's for the magnitudes. and an inner product is $|\vec a \cdot\vec b|=|\vec a||\vec b|\cos\alpha$ and $\alpha = \pi/2$ makes $|\vec a \cdot\vec b|=0$ .

A hint escaped me in post #7. Don't mind giving away the trick, if you solemnly promise never to forget .

Taraaaaa:

Of what is $\vec N\cdot \vec T$ the time derivative ?​

 

[chetzread]

No, that's for the magnitudes. and an inner product is $|\vec a \cdot\vec b|=|\vec a||\vec b|\cos\alpha$ and $\alpha = \pi/2$ makes $|\vec a \cdot\vec b|=0$ .

A hint escaped me in post #7. Don't mind giving away the trick, if you solemnly promise never to forget .

Taraaaaa:

Of what is $\vec N\cdot \vec T$ the time derivative ?​

to be honest , i don't understand the question
Of what is $\vec N\cdot \vec T$ the time derivative ?
I am not a native English speaker . do you mean what is the derivative of N.T with respect to t(time) ?

 

[BvU]

I am not a native English speaker .

Neither am I.

do you mean what is the derivative of N.T with respect to t(time) ?

No, the opposite: (*) $\vec N\cdot \vec T$ is the time derivative of something. If $\vec N\cdot \vec T = 0$, then that something ought to be a constant. If we turn that around we are in business.

( (*) In fact we are going to show that the numerator in the expression for $\ \vec N\cdot \vec T \$ is 0).

Cryptic, eh ? Don't mind showing (but you have to promise solemnly first ).

 

[chetzread]

Neither am I.
No, the opposite: (*) $\vec N\cdot \vec T$ is the time derivative of something. If $\vec N\cdot \vec T = 0$, then that something ought to be a constant. If we turn that around we are in business.

( (*) In fact we are going to show that the numerator in the expression for $\ \vec N\cdot \vec T \$ is 0).

Cryptic, eh ? Don't mind showing (but you have to promise solemnly first ).

ok, can you show it please?

 

[BvU]

The time derivative of $\ \vec T^2$ is ?

 

[chetzread]

The time derivative of $\ \vec T^2$ is ?

it's 2$\ \vec T$dt , where t =time

 

[BvU]

Almost : the chain rule is $(f^2)' = 2 f f'$ -- by the way: a time derivative is a ${d\over dt}$, a differential quotient, so you can't end up with a differential (like your dt at the end)...

 

[BvU]

No

yet ?

 

[chetzread]

Almost : the chain rule is $(f^2)' = 2 f f'$ -- by the way: a time derivative is a ${d\over dt}$, a differential quotient, so you can't end up with a differential (like your dt at the end)...

so , it's 2TT' ?

 

[BvU]

Yes ! Recognize the numerator of $\ \vec N\cdot \vec T \$ ?

 

[chetzread]

Yes ! Recognize the numerator of $\ \vec N\cdot \vec T \$ ?

numerator ? what do you mean ? why did you ask me to differntiate $\ vec T^2 \$ with t (time) ?

 

[chetzread]

Yes ! Recognize the numerator of $\ \vec N\cdot \vec T \$ ?

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 ...

 

[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.