1. Vector calculus

If T=ai+bj

since N is orthogonal to T and both has length 1,

so I set

N=bj-ai or N= -bi+aj (which is correct)

How can I make sure that N(s) is in the direction as dT/ds

(dT/ds points in the direction in which T turns as the curve bends)

2. Hello,

The derivative of the unit vector $\displaystyle \vec{i}$ is related to $\displaystyle \vec{j}$ and vice-versa :

$\displaystyle \frac{d\vec{i}}{dt}=\vec{j}$

$\displaystyle \frac{d \vec{j}}{dt}=-\vec{i}$

3. Originally Posted by Moo
Hello,

The derivative of the unit vector $\displaystyle \vec{i}$ is related to $\displaystyle \vec{j}$ and vice-versa :

$\displaystyle \frac{d\vec{i}}{dt}=\vec{j}$

$\displaystyle \frac{d \vec{j}}{dt}=-\vec{i}$

I still don't understand.

And which N is correct.

4. Oh i thought you were saying that the two were correct.

Well, for me it's N= -bi+aj which is correct.

How can I make sure that N(s) is in the direction as dT/ds
Derivate T.

$\displaystyle \vec{T}=a \vec{i}+b \vec{j}$

As a and b are constant, we have :

$\displaystyle \frac{d \vec{T}}{ds}=a \frac{d \vec{i}}{ds} + b \frac{d \vec{j}}{ds}$

According to what i wrote above :

$\displaystyle \frac{d \vec{T}}{ds}=a \vec{j} - b \vec{i}$, which is the same direction as N.

5. Originally Posted by Moo
Oh i thought you were saying that the two were correct.

Well, for me it's N= -bi+aj which is correct.

Derivate T.

$\displaystyle \vec{T}=a \vec{i}+b \vec{j}$

As a and b are constant, we have :

$\displaystyle \frac{d \vec{T}}{ds}=a \frac{d \vec{i}}{ds} + b \frac{d \vec{j}}{ds}$

According to what i wrote above :

$\displaystyle \frac{d \vec{T}}{ds}=a \vec{j} - b \vec{i}$, which is the same direction as N.

a and b are not constants
a and b are functions of variable t

6. Not of s though... So can it be considered as constant if the variable is s ?

What's the exact text ?

7. Originally Posted by Moo
Not of s though... So can it be considered as constant if the variable is s ?

What's the exact text ?
consider this question

r(t)=3tcos(t)j-3tsin(t)k

determine N,T,B

you can get the answer from this website

default

Section 12-2
Problem 10

and you'll see that if
T=aj+bk then N=bj-ak

8. Outch, my eyes >.<

Well, if i got the pitch, T is the unit vector of V, which you can get by dividing V by its norm.
N is the normal vector of T, that is to say the derivative of T in t.

But you can also remember that if you consider a vector of (a,b) coordinates, one of its normal vector will have (b,-a) coordinates (scalar product). Another of its normal vectors will have (-a,b) coordinates. Never mind, the direction remains the same, it's just the sens that will change.

This explains the formula N=bj-ai or N=-bj+ai

Then for this :

How can I make sure that N(s) is in the direction as dT/ds
I need more time... but it's not the thing i prefer to do :s

9. Originally Posted by Moo
Outch, my eyes >.<

Well, if i got the pitch, T is the unit vector of V, which you can get by dividing V by its norm.
N is the normal vector of T, that is to say the derivative of T in t.

But you can also remember that if you consider a vector of (a,b) coordinates, one of its normal vector will have (b,-a) coordinates (scalar product). Another of its normal vectors will have (-a,b) coordinates. Never mind, the direction remains the same, it's just the sens that will change.

This explains the formula N=bj-ai or N=-bj+ai

Then for this :

I need more time... but it's not the thing i prefer to do :s
The question you do not prefer is exactly the question I asked at the first time.

There can be only one N ,which is in the same direction as T'(s).

and I wonder which N is correct

10. Well, at the beginning, it was not clear about what was a, what was b. I don't understand either the variable is s or t...

I find strange things for T'(t)...

11. Ok, i was not looking at the right function... I took V on the website you gave.

r(t)=3tcos(t)j-3tsin(t)k
What do you name N, T and B ??? (just to be sure...)