# Unit vector tangential to curve

1. Aug 28, 2007

### hypn0tika

Find the expression for the unit vector tangential to the curve given in spherical coordinates by r = 1, $$\phi$$ = 2$$\theta$$, 0 $$\leq$$ $$\theta$$ $$\leq$$ $$\pi$$

the equation for the differential length vector for spherical coordinates:
dl = drar + rd$$\theta$$a$$\theta$$ + rsin($$\theta$$)d$$\phi$$a$$\phi$$
where ar, a$$\theta$$, and a$$\phi$$ are unit vectors

i used the equations for the curve to determine that:
dr = 0, d$$\phi$$ = 2d$$\theta$$

by substituting into the differential length vector, i got:
rd$$\theta$$[a$$\theta$$ + 2sin($$\theta$$)a$$\phi$$]
i know this is not a unit vector, but i am lost. i think there would need to be either a cosine or a way to get rid of that sine in the differential length vector equation for the expression to be a unit vector.

Last edited: Aug 28, 2007
2. Aug 28, 2007

### Dick

I'm lost, too. Mostly because I don't get your notation. d(phi) isn't a unit vector, it's a 1-form. And I don't have any idea what a 'differential length vector' is. Maybe somebody does, but in the meantime, have you just considered treating this as a parametric equation in cartesian coordinates and working from there?

Last edited: Aug 28, 2007
3. Aug 28, 2007

### hypn0tika

my fault. when i said d(theta) and d(phi) were unit vectors i mean a(theta), a(phi).

4. Aug 29, 2007

### Dick

Ok, I'm starting to get the notation. aphi e.g. is the vector $$\frac{\partial}{\partial \phi}$$ etc. I would not call them unit vectors (they don't have unit length). You haven't written an expression for the tangent vector itself yet. What is it? The curve's (r,theta,phi) coordinates are (1,theta,theta*2).