1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Curvature/surface area of a tractrix

  1. Feb 18, 2017 #1
    1. The problem statement, all variables and given/known data
    Given ## ds^2 = a^2cot^2 \alpha d\alpha^2 + a^2sin^2 \alpha d\phi^2## where a is some constant. Find:
    a) The gaussian curvature
    b) the surface area of the upper half of the tractrix

    2. Relevant equations
    Stoke theorem: ## \int_S dF = \int_{\partial S} F ##
    and for curvature, we know the ## \Omega^1_2 = K\omega ## where K is the gaussian curvature, and omega is the basis.

    3. The attempt at a solution
    I'll start with the curvature: So to get ## \Omega^1_2 ## I need to first find ##\omega^1_2##
    So, given my line element, I see that my basis is ##\{ a\cot \alpha d alpha, a \sin \alpha d\phi \}##

    So, my torsion conditions are: 1) ## 0 = d\sigma^1+\omega^1_1 \wedge \sigma^1+\omega^1_2 \wedge \sigma^2 ##
    2) ## 0 = d\sigma^2+ \omega^2_1 \wedge sigma^1 + \omega^2_2 \wedge sigma^2 ##
    Plugging in our basis and simplifying, we see that: 1) ## 0 = 0 + 0 + \omega^1_2 \wedge a\sin\alpha d\phi## which forces ##\omega^1_2 = b d\phi## for some b.
    Thus, 2 becomes : ## 0 = a\cos\alpha d\alpha \wedge d\phi + \omega^2_1 \wedge a\cot\alpha d\alpha + 0## which becomes ## 0 = a\cos\alpha d\alpha \wedge d\phi + ab\cot\alpha d\alpha \wedge d\phi ## which forces ##b= -sin\alpha##. Thus, ## \omega^1_2 = -sin\alpha d\phi ## this makes ##\Omega^1_2 = d\omega^1_2 = d(-sin\alpha d\phi) = -cos \alpha d\alpha \wedge d\phi## since ##\omega = \sigma^1 \wedge \sigma^2 ## we see that ## \Omega^2_1 = \frac{-cos\alpha}{a^2cot\alpha sin\alpha} d\alpha \wedge d\phi## which would make ## K = \frac{-1}{a^2}## giving us a constant negative curvature.

    b) This one, in my opinion, is more difficult. My boundaries will be ## \alpha \in \left[0, \frac{\pi}{2} \right]## and ##\phi \in \left[0, 2\pi \right ) ## So, for my F, I know that ## F= \vec{F} \cdot d\vec {r} ## where I can either convert to Cartesian, or try to just say that ## \vec{F} = F_{\alpha} \hat {\alpha} + F_{\phi} \hat {\phi} ## and ## d\vec {r} = a \cot \alpha d\alpha \hat {\alpha} + a \sin \alpha d\phi \hat {\phi} ## making ## F = a F_{\alpha} \cot \alpha d\alpha + a F_{\phi} \sin \alpha d\phi ## which would make ## dF = \left( a \frac{\partial F_\phi}{\partial \alpha}+a F_\phi \cos \phi-a \cot \alpha \frac{\partial F_\alpha}{\partial \phi} \right) d\alpha \wedge d\phi ## using stokes tells us that our surface integral is ## \int_S \left( -a \cot \alpha \frac{\partial F_\alpha}{\partial \phi} + a \frac{\partial F_\phi}{\partial \alpha}+a F_\phi \cos \phi \right) d\alpha \wedge d\phi = \int^{\frac{\pi}{2}}_{0} F_{\alpha} a \cot \alpha d\alpha + \int^{\frac{\pi}{2}}_{0} F_{\phi} a \sin \alpha d\phi ##
    which if I do it, I get

    however, there is a problem with this, I don't know my functions true identities? So must I really go through converting to Cartesian? Or is there something I'm missing? I know in Cartersian that ## x = a\sin \alpha \cos \phi ## ## y = a \sin \alpha \sin \alpha ## and ## z = -a \cos \alpha - ln {\tan{\frac{\alpha}{2}}}## so I could work it out if i really had to, but I think I might be missing something obvious!
     
  2. jcsd
  3. Feb 23, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Curvature/surface area of a tractrix
  1. Surface Area (Replies: 5)

  2. Surface area (Replies: 2)

  3. Surface area (Replies: 2)

  4. Area of surface (Replies: 3)

  5. Surface area (Replies: 2)

Loading...