Re: Finding Normal Unit Vectors

From: Lynn Kurtz (kurtzDELETE-THIS_at_asu.edu)
Date: 09/13/04 

Date: Mon, 13 Sep 2004 03:34:14 GMT

On Fri, 10 Sep 2004 17:49:04 -0400, Will Oram <spamguy@ihatespam.com>
wrote:

>
>Let r(t) equal, say, <t, t^2, t^3>, and I want to find the tangential
>and normal at t = -1 (I vow this does not correspond to homework
>problems or the like, etc etc.; this problem is one I generated for this
>purpose).
>
>Then v(t) = <1, 2t, 3t^2>, and a(t) = <0, 2, 6t>. I find T by dividing v
>by its scalar, getting <1, -2, 3>/sqrt(14).
>
>If I find N by the method above, I get <0, 2, -6>/sqrt(40). It doesn't
>take much effort to see N and T are not perpendicular as they should be
>(dot product != 0), therefore I'm not doing something right.

So you have v(-1) = < -1, 2, 3 > and a(-1) = < 0, 2, -6>.

To get the principal normal you need to compute the vector projection
of a(-1) on on v(-1), which is the tangential component of the
acceleration vector. Call this a_T. If you subtract this from the
acceleration vector you will get the normal component of the
acceleration vector a_N, which is what you want.

You have the unit tangent vector at your point already:

T = <1, -2, 3>/sqrt(14).

To get a_T use:

a_T = (a dot T)T
= {(< 0, 2, -6> dot <1, -2, 3>/sqrt(14)}<1, -2, 3>(1/sqrt(14))
= (1/14)(0 - 4 - 18)<1, -2, 3> = (-11/7)<1, -2, 3>
= <-11/7, 22/7, -33/7 >.

Now a_N = a - a_T
= < 0, 2, -6> - < -11/7, 22/7, -33/7 >
= < 11/7, -8/7, -9/7 > .

You can check that this is perpendicular to T. Notice that on a
problem like this, where you need the answers only at the value of t =
-1, it is much simpler to just compute v and a at -1 and work with the
resulting constant vectors than to work out the formula generally for
variable t and plug t = -1 in at the end.

--Lynn

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