Results 1 to 2 of 2

Math Help - Family of curves on a surface (Differential geometry)

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    1

    Family of curves on a surface (Differential geometry)

    Here is a question i am not sure how to tackle, I am not familiar with how to deal with family of curves and don't really have much time to look around for the definition as i am sitting the exam in two days.




    The question is divided into three parts: Here is my attempt any help appreciated.
    1) I am have no idea, i think it is a case of knowing the definition and i don't.

    2) It is simply constraining the local parametrization to the given function so:
     xz-hy=> v*sin(u)=h(1-cos(u))=> h= v*sin(u)/(1-cos(u)) which is a constant.

    3) \psi(u,v)=const is like phi therefore the tangent vectors to the family defined by the psi are of the multiples of \psi_{v}x_{u}-\psi_{u}x_{v}
    So for the families to be orthogonal their tangent must be orthogonal and so (\psi_{v}x_{u}-\psi_{u}x_{v}).(\phi_{v}x_{u}-\phi_{u}x_{v})=0
    Using the fundamental forms E=1=G and F=0 we get \psi_{v}\phi_{v}+\psi_{u}\phi_{u}=0 which after differentiating gives \psi_{v}\sin(u)-\psi_{u}v=0


    And after that i am stuck ... any help would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    543
    Thanks
    12
    Ok, lets see...


    1) Consider a curve in the domain. The gradient of \phi is normal to the tangent (du,dv) of the curve. Since all directions of such surface curves are given by x_{u}du+x_vdv, the functional dependence of x and \phi means that the determinant of their Jacobian matrix is zero, whence the required result.


    2) We have <br />
\psi_{v}\sin(u)-\psi_{u}v=0,<br />
so \sin(u)du=vdv, from which \psi(u,v)=\frac{v^2}{2}-\cos(u)=const.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differential equation of the family of curves y=c(x-c)^2
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: November 21st 2011, 06:49 AM
  2. [SOLVED] Differential Calculus Problem on a family of cubic functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 22nd 2011, 07:01 AM
  3. Family of functions, differential equations
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: January 25th 2009, 02:57 PM
  4. Family of Curves
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 12th 2007, 11:56 AM
  5. Help for family of curves.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 7th 2006, 12:44 PM

Search Tags


/mathhelpforum @mathhelpforum