Results 1 to 3 of 3

Math Help - Parametric Equation of a Tangent Line off a Helix

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    2

    Parametric Equation of a Tangent Line off a Helix

    Suppose <br />
\vec r = <3cos(t), 3sin(t), 4t><br />
    represents the position of a particle on a helix after t seconds, where z is the height of the particle above the ground in meters.

    When the particle is 12 meters above the ground, the particle leaves the helix and moves along the tangent line to the helix. Find a parametric equation for this tangent line.

    Sorry if I messed up the LaTex, I'm new. So far I have the t will be 3 when it is 12 meters off of the ground and I have this so far for my parametric equation by taking the derivative of the vector:
    y=-3sin(t)
    x=3cos(t)
    z=4

    I'm not sure that this is right at all, and I assume that the tangent line has to cross through the point (3cos(3), 3sin(3),12) and that parametric equation doesn't. Any help would be great, thanks!
    Last edited by blastulation; October 28th 2010 at 09:31 PM. Reason: Yeah, I completely messed up the LaTex. It's pretty easy to figure out what I was trying to say though (hopefully)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    I'm not sure about this, but it would seem to me that the equation would be in the form:

    \vec{l} = \left( \begin{array}{c} x \\ y \\ z\end{array} \right) + \lambda \left( \begin{array}{c} i \\ j \\ k\end{array} \right)

    Where <i, j, k> is your direction vector and <x, y, z> your point from where the line starts and lambda a parameter.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,419
    Thanks
    1855
    Quote Originally Posted by blastulation View Post
    Suppose <br />
\vec r = <3cos(t), 3sin(t), 4t><br />
    represents the position of a particle on a helix after t seconds, where z is the height of the particle above the ground in meters.

    When the particle is 12 meters above the ground, the particle leaves the helix and moves along the tangent line to the helix. Find a parametric equation for this tangent line.

    Sorry if I messed up the LaTex, I'm new. So far I have the t will be 3 when it is 12 meters off of the ground and I have this so far for my parametric equation by taking the derivative of the vector:
    y=-3sin(t)
    x=3cos(t)
    z=4
    No. First, you have swapped x and y (probably at typo) x= -sin(t), y= 3 cos(t). The vector <-3sin(t), 3cos(t), 4> is the tangent vector to any point on the helix. For your point, where z= 12, t= 3 as you said so the tangent vector is <-3sin(3), 3cos(3), 4>. A line with "tangent vector" (direction vector) <A, B, C>, through point (x_0, y_0, z_0) is given by x= At+ x_0, y= Bt+ y_0, z= Ct+ z_0.

    Since the line goes through the point (cos(3), sin(3), 12), your line is given by x= -3sin(3)t+ cos(3), y= 3cos(3)t+ sin(3), z= 4t+ 12.

    I'm not sure that this is right at all, and I assume that the tangent line has to cross through the point (3cos(3), 3sin(3),12) and that parametric equation doesn't. Any help would be great, thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding Tangent Line and Normal of a parametric equation
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 14th 2012, 08:02 AM
  2. Replies: 3
    Last Post: November 4th 2011, 07:42 PM
  3. Replies: 7
    Last Post: December 9th 2010, 10:53 AM
  4. parametric equation for tangent line
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 20th 2009, 06:47 PM
  5. parametric equation and tangent line
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 11th 2008, 11:15 PM

Search Tags


/mathhelpforum @mathhelpforum