Results 1 to 2 of 2

Math Help - Help for family of curves.

  1. #1
    Member
    Joined
    Nov 2005
    Posts
    111

    Help for family of curves.

    Hi please can I have some help with this problem:
    It is about first order ODE.

    Give the family of orthogonal trajectories of the family of parabolas with vertical axis and vertex at the point ( 4 , 2 ).

    The solution is
    (x-4)^2 +(y-2)^2=C

    But I dont know how to find it.
    Thank you
    B
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by braddy
    Hi please can I have some help with this problem:
    It is about first order ODE.

    Give the family of orthogonal trajectories of the family of parabolas with vertical axis and vertex at the point ( 4 , 2 ).

    The solution is
    (x-4)^2 +(y-2)^2=C

    But I dont know how to find it.
    Thank you
    B
    This is going to involve a lot of TeX, so will take rather more time than
    I have available to do it in one go, so I will do this incrementally.

    (PH you were wise to leave your post to be checked, you seem to have
    gotten the wrong end of the stick, as we say over here).

    You want to find a family of functions/curves which are orthogonal
    at every point to the parabola from the given family passing through
    the point.

    Lets start by deriving the equations of the family of parabolas with vertex
    at ( 4,2) and axes parallel to the vertical axis. Well they
    are:

    <br />
(y-2)=\lambda (x-4)^2<br />
,

    where \lambda parameterises the family. Now if one of these
    parabolas passess through the point (u,v), we have:

    <br />
v-2= \lambda (u-4)^2<br />
,

    so:

    <br />
\lambda=\frac{v-2}{(u-4)^2}<br />

    and so the parabola through (u,v) is:

    <br />
y-2= \frac{v-2}{(u-4)^2} (x-4)^2<br />

    We are interested in the slope of this parabola at (u,v) as
    we are interested in the slope of the curve in the orthoganal family through
    the point, which as it is orthogonal to the parabola has a slope of minus the
    reciprical of the slope of the parabola at this point.

    For the parabola we have:

    <br />
\frac{dy}{dx}=\frac{v-2}{(u-4)^2}\ 2\ (x-4)<br />

    so:

    <br />
\frac{dy}{dx}\left|_{{x=u}\atop{y=v}}=2\ \frac{v-2}{u-4}<br />
.

    Hence the slope of the member of the orthogonal family through (u,v) is:

    <br />
\frac{dy}{dx}\left|_{{x=u} \atop {y=v}}=-\frac{1}{2}\ \frac{u-4}{v-2}<br />
,

    so the ODE satisfied by the orthogonal trajectory is:

    <br />
\frac{dy}{dx}=-\frac{1}{2}\ \frac{x-4}{y-2}<br />
.

    Which is of variables seperable type, and I think you will find that the
    solution is a circle with centre (4,2).

    RonL
    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, 05:49 AM
  2. Family of curves on a surface (Differential geometry)
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 27th 2008, 12:18 PM
  3. Family of five
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 23rd 2007, 05:50 AM
  4. Family of Curves
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 12th 2007, 10:56 AM
  5. sketching curves-transformation curves f(x)!!!!
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 2nd 2006, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum