# Homework Help: Unit tangent vector vs principal normal vector

1. Jul 25, 2016

1. The problem statement, all variables and given/known data
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....

2. Relevant equations

3. 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?

2. Jul 25, 2016

### 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 ...

3. Jul 25, 2016

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 ?

4. Jul 25, 2016

### 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 $(1,-1)$

5. Jul 25, 2016

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?

6. Jul 25, 2016

### 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).

7. Jul 25, 2016

### 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 ?

8. Jul 25, 2016

### 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 ?

9. Jul 25, 2016

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

10. Jul 25, 2016

### BvU

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

11. Jul 25, 2016

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

12. Jul 25, 2016

### BvU

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

13. Jul 25, 2016

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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14. Jul 25, 2016

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 ?

15. Jul 25, 2016

### 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...

16. Jul 25, 2016

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 ?

17. Jul 25, 2016

### BvU

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).

18. Jul 25, 2016

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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19. Jul 26, 2016

### BvU

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 ?

20. Jul 26, 2016

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) ?

21. Jul 26, 2016

### BvU

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 ).

22. Jul 26, 2016

ok, can you show it please?

23. Jul 26, 2016

### BvU

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

24. Jul 26, 2016

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

25. Jul 26, 2016

### 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)...

Last edited: Jul 26, 2016