Differential of unit vector for vector r=xi+yj+zk is always zero

[BAT_MAN]

Is this is true that

differential of unit vector for vector r=xi+yj+zk is always zero

 

[Dmitrij]

Re: Unit vector

From my point of view, differential may be always equal to zero only when the fuction is constant in the whole time domain. Differential is known to be the main part of the functions increase. It may be writtten the following way:

R = R(x,y,z) - function of 3 independent variables.
dy = R'(x)dx+R'dy+R'(z)dz - differential of the function

Here R'(x),R',R'(z) denote the private derivatives of the functions taken on x,y,z respectively. If all the derivatives are zero, then the differential is zero as well.

If you use the unit vector (in my opinion unit vector means normalizing the initial vector by dividing on its length: ru = r / |r|, where |r|=sqrt(x^2+y^2+z^2)) nothing changes, therefore all the above results remain fair.

With respect,

Dmitrij

 

[MSRA]

Unit vector

As far as my little knowledge says...
differential of any constant is zero....
and the uunit vector has a constant magnitude of unity..
so it must be true according to my theory...

 

[harsha_jois]

Re: Unit vector

Unit vector has unit magnitude on each coordinate over infinite time. right?
Then if you differentiate it in time domain it has to be Zero.

Added after 1 minutes:

Unit vector has unit magnitude on each coordinate over infinite time. right?
Then if you differentiate it in time domain it has to be Zero.

 

[piash]

Re: Unit vector

yeah it must be zero because it's magnitude is 1 ( a constant) and the derivative of any constant is zero