Differential Geometry: Coordinate Patches

[SNOOTCHIEBOOCHEE]

Sorry i wasnt able to get help in the homework department. figured id try here.

Homework Statement

For a coordinate patch x: U--->\Re^{3}show thatu^{1}is arc length on the u^{1} curves iff g_{11} \equiv 1

The Attempt at a Solution

So i know arc legth of a curve \alpha (t) = \frac{ds}{dt} = \sum g_{ij} \frac {d\alpha^{i}}{dt} \frac {d\alpha^{j}}{dt} (well that's actually arclength squared but whatever).

But I am not sure how to write this for just a u^{1} curve. A u^{1} curve throught the point P= x(a,b) is \alpha(u^{1})= x(u^{1},b)

But i have no idea how to find this arclength applies to u^1 curves.

Furthermore i know some stuff about our metric g_{ij}(u^{1}, u^{2})= <x_{i}(u^{1}, u^{2}), x_{j}(u^{1}, u^{2})

But i do not know how to use that to show that u^1 must be arclength but here is what i have so far:

g_{11}(u^{1}, b)= <x_{1}(u^{1}, u^{2}), x_{2}(u^{1}, u^{2})> We know that x_{1}= (1,0) and that is as far as i got :/

Any help appreciated.

 

[lavinia]

A curve tangent to a coordinate direction only has one metric tensor component that is not zero - I think.