Osculating Circle for a Curve

JoAdams5000 · #1 $Old$ June 25th, 2009, 10:35 PM

I lost trying to find the osculating circle for the curve defined by r(t) = <sint, cost, -pi + t> at the point t=pi.

Ive calculated the radius of the circle to be 1 and I know I need to somehow use the unit normal vector at point t=pi, which is <0, 1/sqrt(2), 0> to solve for the osculating circle's centre, but I do not know how to do that.

Thans for your help!

CaptainBlack · #2 $Old$ June 26th, 2009, 12:35 AM

Quote:

Originally Posted by JoAdams5000 $View Post$

I lost trying to find the osculating circle for the curve defined by r(t) = <sint, cost, -pi + t> at the point t=pi.

Ive calculated the radius of the circle to be 1 and I know I need to somehow use the unit normal vector at point t=pi, which is <0, 1/sqrt(2), 0> to solve for the osculating circle's centre, but I do not know how to do that.

Thans for your help!

That is not a unit vector.

CB

JoAdams5000 · #3 $Old$ June 26th, 2009, 11:07 AM

Right...i calculated that wrong...

The unit normal vector is actually [0, -1, 0] from the point where t=pi.

How can I use this to calculate the center of the osculating circle if possible?

earboth · #4 $Old$ June 27th, 2009, 06:50 AM

Quote:

Originally Posted by JoAdams5000 $View Post$

I lost trying to find the osculating circle for the curve defined by r(t) = <sint, cost, -pi + t> at the point t=pi.

Ive calculated the radius of the circle to be 1 and I know I need to somehow use the unit normal vector at point t=pi, which is <0, 1/sqrt(2), 0> to solve for the osculating circle's centre, but I do not know how to do that.

Thans for your help!

1. The stationary vector of point P when $t = \pi$ is

$\vec p=(0,-1,0)$ .

2. I've got the normal unit vector as $\vec n = (0,1,0)$

3. The sum of these two vectors will yield the stationary vector of the center of the circle:

$\vec p + \vec n = (0,0,0)$

4. The tangent at the curve in P and the normal vector produce a plain in which the osculating circle must be placed. The equation of the tangent is:

$t: \overrightarrow{r(t)}=(0,-1,0)+t \cdot (-1,0,1)$

earboth · #5 $Old$ June 27th, 2009, 01:16 PM

In addition to my previous post I've attached a drawing containing

the curve: $\overrightarrow{r(t)}=(\sin(t), \cos(t), -\pi+t)$

the tangent: $t: \overrightarrow{r(t)}=(0,-1,0)+t \cdot (-1,0,1)$

the circle: $\overrightarrow{r(t)}=(\sin(t), \cos(t), -\sin(t))$