# Infinitesimal arc-length square

1. Nov 30, 2009

### andlook

Hello

I am trying to understand the "infinitesimal arc-length square." So (ds)^2=(dx)^2+(dy)^2+(dz)^2. What does this means?

And then what does (ds)^2=(dx)^2+(1+x^2)(dy)^2 -2x(dy) +(dz)^2 mean? And how does this apply to a space?

2. Nov 30, 2009

### LCKurtz

If you have a parameterized curve

$$\vec r(t) = \langle x(t), y(t), z(t) \rangle,\, t \in [a,b]$$

you calculate the arc length by taking the limit of polygonal approcimations. If

$$a = t_0 < t_1 < ...< t_n = b$$

the polygonal approximation to the arc length is

$$P_n = \sum_{i=1}^n |\vec r(t_i) - \vec r(t_{i-1})|$$

Here the absolute value signs represent Euclidean distance so

$$|\vec r(t_i) - \vec r(t_{i-1})| = \sqrt{(x(t_i)-x(t_{i-1}))^2+(y(t_i)-y(t_{i-1}))^2+(z(t_i)-z(t_{i-1}))^2}$$

This is sometimes written the the delta notation:

$$\Delta s_i \approx \Delta P_i =\sqrt{(\Delta x_i)^2 + (\Delta y_i)^2 + (\Delta z_i})^2$$

Passing to the limit suggests the notation $ds^2 = dx^2+dy^2+dz^2$

3. Nov 30, 2009

### andlook

Thanks, that's great.

I'm not really sure if this question would be well defined, but how would I then apply this method to the second metric $$(ds)^2=(dx)^2+(1+x^2)(dy)^2 -2x(dy) +(dz)^2$$?

4. Nov 30, 2009

### LCKurtz

I can't say I have seen that kind of expression. The type of line integral you usually see in calculus is one of these two:

$$\int_C \vec F \cdot\, d\vec R\ or\ \int_C \delta(x,y,z)\ ds$$

The first might represent work in a force field and the second the mass of a wire. Perhaps you are thinking of some other setting or application.