How Does the Definition of Arc Length Lead to Its Integral Formula?

[seanc12]

Homework Statement

Confirm that th definition of th arc length ds$$^{2}$$=dr.dr leads to the formula
L=$$\int$$$$_{C}$$$$\frac{dr}{dt}$$dt

Homework Equations

$$\frac{dr}{dt}$$=$$\frac{dx}{dt}$$+$$\frac{dy}{dt}$$+$$\frac{dz}{dt}$$dt

The Attempt at a Solution

I am really unsure of what to do here. I have tried starting at each of the given equations and trying to meet somewhere in the middle. I won't post all my working becauses most if it is gibberish and I am still trying to get this LaTeX thing to work properly for me. Can someone give me some hints in what I am trying to get here.

the best I've gotten is expanding the dot product to get

ds^2=dx^2+dy^2+dz^2

 

[HallsofIvy]

Since nothing at all is said about xyz-coordinates in the original problem, I don't see why you are changing to dx, dy, and dz.

(And certainly dr/dt= dx/dt+ dy/dt+ dz/dt is NOT true. What is true is that
$$\frac{dr}{dt}= \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2+ \left(\frac{dz}{dt}\right)^2}$$)

From $ds^2= dr\cdot\dr$, take the square root of both sides.

 

[seanc12]

Hi HallsofIvy, thanks for the reply.

(And certainly dr/dt= dx/dt+ dy/dt+ dz/dt is NOT true

Yes ofcourse, how could i have been so silly :P wasn't paying attention properly thanks for pointing that out.

$$\frac{dr}{dt}= \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2+ \left(\frac{dz}{dt}\right)^2}$$

I understand the RHS, but howcome it is = to dr/dt instead of ds/dt?

Also as a note, the C on the integral should be a subscript but it doesn't seem to want to go there

 

[jhae2.718]

Also as a note, the C on the integral should be a subscript but it doesn't seem to want to go there

You need to put the entire expression in one set of tex tags, e.g. $$\int_a^b dx$$ gives:
$$\int_a^b dx$$

I understand the RHS, but howcome it is = to dr/dt instead of ds/dt?

That's the magnitude of $dr/dt$ in terms of dx,dy,and dz.

 

[seanc12]

$$\int_C \frac{d <u>r</u>}{dt}dt$$

Thanks,

So from there can i go ahead and substitute the $$\frac{d <u>r</u>}{dt}$$ back into the integral? or am i missing a few steps? It seems a little easy. Also, where does the C come into it?

 

[seanc12]

Also, howcome the underlines don't show up in my LaTeX?