Infinitesimal arc-length square

[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?

 

[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

 

[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 <br /> (ds)^2=(dx)^2+(1+x^2)(dy)^2 -2x(dy) +(dz)^2<br />?

 

[LCKurtz]

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 <br /> (ds)^2=(dx)^2+(1+x^2)(dy)^2 -2x(dy) +(dz)^2<br />?

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.