Differential in the arc-length formula for curves

[mnb96]

Hello,
given a parametric curve \mathbf{r}(s)=x(s)\mathbf{i} + y(s)\mathbf{j} + z(s)\mathbf{k}, my textbook says that tangent vector having unit-magnitude is given by \mathbf{r}(s)=x'(s)\mathbf{i} + y'(s)\mathbf{j} + z'(s)\mathbf{k}

I don't understand the proof that it has unit magnitude:

\sqrt{(\frac{dx}{ds})^2 + (\frac{dy}{ds})^2 + (\frac{dz}{ds})^2 }

= \sqrt{\frac{ (dx)^2 + (dy)^2 + (dz)^2}{(ds)^2} }

= 1

I can't really follow any of the given steps.
*) What is the reasoning for stepping from line-1 to line-2 ?
**) And finally how did they obtain the identity (ds)^2=(dx)^2 + (dy)^2 + (dz)^2

 

[Matthollyw00d]

s here is your arc length parameter.
s(t)=\int_{a}^{t}||r'(u)||du=\int_{a}^{t}\sqrt{(x'(u))^2+(y'(u))^2+(z'(u))^2}

So by the fundamental theorem of calculus we have
\frac{ds}{dt}=||r'(t)||
Or in differential form ds=||r'(t)||dt. The differential form is written like this as if we omitted the integral sign.

So we have ds=||r'(t)||dt=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2}dt=\sqrt{dx^2+dy^2+dz^2}

Does that help? I think what's confusing you is when you can use the differential form. Differentials by themselves are more like identities, that you implement when integrating or simplifying (and in this case they're being used to simplify).

 

[mnb96]

Ah I see!
I forgot that s represented the arc-length parameter! That was important.
and...

...I think what's confusing you is when you can use the differential form.

Yes. When I see the entire proof like the one you gave here, it totally makes sense to me. But when someone shoots rightaway identities in differential form I get easily confused. Unfortunately what is giving me frustration is that I cannot clearly spot the exact concept that I understood poorly.

Do you have a suggestion on textbooks/exercises/tutorials that could help me fill this gap?

Thanks a lot!

 

[Matthollyw00d]

http://en.wikipedia.org/wiki/Differential_of_a_function"
This might be helpful, some of it might be perhaps too advanced though. Unfortunately, I don't really have any good sources off the top of my head, sorry.