# Unit tangent vector vs principal normal vector

## Homework Statement

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

## 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
Homework Helper
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 ...

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 ?

BvU
Homework Helper
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)##

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
Homework Helper
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
Homework Helper
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
Homework Helper
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 ?

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

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

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

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

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 dont 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
Homework Helper
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).

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

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BvU
Homework Helper
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 ?

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 dont 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
Homework Helper
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 ).

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
Homework Helper
The time derivative of ##\ \vec T^2## is ?