Re: Finding Normal Unit Vectors

From: Sylvain Croussette (NO_SPAM_sylvaincroussette_at_yahoo.ca)
Date: 09/11/04 

Date: Fri, 10 Sep 2004 20:46:09 -0400

Will Oram <spamguy@ihatespam.com> dixit:

>I understand how to find a tangential unit vector T of a parametric 3D
>function. What I "don't" understand, however, is how to find a normal
>unit vector N from T.
>
>I am told (from books and MathWorld) that
>
> dT/dt
>N = _________
> |dT/dt|
>
>...does that mean that
>
> a
>N = _____
> |a|
>
>since T is based off velocity? If so, that doesn't seem to correspond
>with N being perpendicular to T.

If I remember correctly, the derivative of a unit vector is always
perpendicular to that unit vector. So if T is unit, then T' is
perpendicular to T. But v(t) is not necessarily unit, its magnitude
is the speed, v'(t) = a(t) means ||a(t)|| not necessarily 1 and a(t)
is not necessarily perpendicular to v(t).
So I would say that
      dT/dt
N = _________
     |dT/dt|
is correct, but is not equal to
      a
N = _____
     |a|
in general.

>
>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).

v(t) is not unit so a(t) will not be perpendicular to v(t).
A unit v(t) would be:
v(t)=<1, 2t, 3t^2>/sqrt(1+t^2+9t^4). Then v'(t)=something but
probably not <0,2,6t>

Ok, Maple says v'(t)=:
x component: -1/2/(1+4*t^2+9*t^4)^(3/2)*(8*t+36*t^3)
y: 2/(1+4*t^2+9*t^4)^(1/2)-t/(1+4*t^2+9*t^4)^(3/2)*(8*t+36*t^3)
z:6*t/(1+4*t^2+9*t^4)^(1/2)-3/2*t^2/(1+4*t^2+9*t^4)^(3/2)*(8*t+36*t^3)

Ouch. Ok. I tell Maple to evaluate these at t=-1 I get:
x=11*sqrt(14)/98
y=-4*sqrt(14)/49
z=-9*sqrt(14)/98
If you do the dot product with <1,-2,3>/sqrt(14) you get 0, which
means this v'(t) is perpendicular to v(t).

>
>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.
>
>But what?
>
>Best,
>
>Will

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