Results 1 to 10 of 10

Math Help - Equation for Tangent Plane

  1. #1
    Member Ranger SVO's Avatar
    Joined
    Apr 2006
    From
    Abilene Tx
    Posts
    90

    Equation for Tangent Plane

    2. Find an equation for the plane tangent to the torus (donut)

    X(s,t) = (5+2*cost)*coss, (5+2*cost)*sins, 2*sint

    at the point (5-sqrt(3))/sqrt(2), (5-sqrt(3))/sqrt(2), 1

    I'm trying to get a head start on the lecture Monday. I'm still trying to figure out how to get ya'll some of the tuition that I've paid, your doing a great job

    As always I thank you for taking your time and helping
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Ranger SVO's Avatar
    Joined
    Apr 2006
    From
    Abilene Tx
    Posts
    90
    Am I going in the right direction?

    Because Z = 2*sint and Z = 1 then t = pi/6 or 5pi/6

    now X and Y both equal (5-sqrt(3))/sqrt(2) so going back to X(s,t) I can see that sins and coss need to equal each other. This happens at pi/4 or 5pi/4

    So s and t are pi/4 and 5pi/6

    Am I headed in the right direction???????????????????
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    I was thinking about this question and I have never seen how one can draw tangent plane to a differenciable surface defined parametrically (like here).

    My suggestion is to try to "eliminate the parameter" and express f(x,y,z)=C. Meaning remove s,t so you end with some function of x,y,z. Henceforth, you can use the property of the gradient.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    Yes, you're going in the right direction. Your values for s and t are correct.
    The equation of the plane tangent to X at the point P(a,b,c) is given by:

    \vec S = \vec p + \lambda \left. {\frac{{\partial \vec X}}{{\partial s}}} \right|_{\vec p}  + \mu \left. {\frac{{\partial \vec X}}{{\partial t}}} \right|_{\vec p}

    Here, lambda and mu are scalars, the partial derivatives which have to be evaluated in p are the directional vectors.
    This works as long as both directional vectors are non-zero and not parallel, because then they span the tangent plane.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2006
    From
    Texas
    Posts
    5
    I still dont understand the steps we need to take to find the equation for the tangent plane...?!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member Ranger SVO's Avatar
    Joined
    Apr 2006
    From
    Abilene Tx
    Posts
    90
    Now that we now s, t here is what we need to do.

    We need the derivative for X(s,t) with respect to s and with respect to t

    Once we do that, evauate at the points s and t

    Ts and Tt (below) are already evauated at s and t (please excuse any bad notation)
    Attached Thumbnails Attached Thumbnails Equation for Tangent Plane-ts.jpg  
    Last edited by Ranger SVO; November 28th 2006 at 05:15 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2006
    From
    Texas
    Posts
    5
    Okay, so apparently I'm loosing my mind & cant remember what to do after taking the cross product of T(s) and T(t) with (x,y,z).

    I came out with:
    -sqrt(6)*(5-sqrt(3))/2 X + sqrt(6)*(sqrt(3)-5)/2 Y + (5-sqrt(3)) Z

    If you can remember , what did you do next? (I know I cant end up with a scalar because the problem asks for an equation.)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    @ Ranger SVO: your directional vectors are correct, now fill in the formula for the tangent plane I gave.

    @ AbbyNormal: where are you stuck, what's the problem? Do you understand the equation of the tangen plane?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Nov 2006
    From
    Texas
    Posts
    5
    No, I dont understand the equation of the tangent plane.
    Ranger told me to get it in terms of (x,y,z) through the cross product, but I dont understand what I need to do with that / why it needs to be in cartesian coordinates.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    You don't have to switch to cartesian coordinates, if you use the equation I gave you can use the parametric equation.
    The tangent plane is spanned by two (linearly independent) tangent vectors, the directional vectors.
    These can be found as the partial derivatives of the surface with respect to the parameters s and t.
    Evaluating these partial derivatives at the given (s,t) yields the directional vectors, now just plug it in the equation.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. equation of plane tangent
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 11th 2011, 04:57 AM
  2. Equation of a tangent plane
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 5th 2010, 05:04 PM
  3. Equation of the tangent plane
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 11th 2010, 10:38 AM
  4. Equation of tangent plane
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 30th 2010, 05:30 PM
  5. Equation of tangent plane
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 3rd 2009, 04:46 PM

Search Tags


/mathhelpforum @mathhelpforum