# Parametric Equation of a Tangent Line off a Helix

• Oct 28th 2010, 08:27 PM
blastulation
Parametric Equation of a Tangent Line off a Helix
Suppose $
\vec r = <3cos(t), 3sin(t), 4t>
$

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!
• Oct 28th 2010, 09:24 PM
Unknown008
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.
• Oct 29th 2010, 06:56 AM
HallsofIvy
Quote:

Originally Posted by blastulation
Suppose $
\vec r = <3cos(t), 3sin(t), 4t>
$

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.

Quote:

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!