yoyosuper7
New member
- Joined
- Feb 26, 2009
- Messages
- 3
I'm taking vector calculus and I have this problem that I can't seem to figure out and I need someone to help me understand or point out what I'm missing, please!!!!!!!!!
Here is the problem:
r(3) = <1, -3, 2> T(3) = <-1/5, 0, 2/5> N(3) = <0, -1, 0> k(3) = 1/3
For t=3 find: the center of the circle of curvature (or osculating circle); the equation of the
oculating plane (the plane determined by T and N); and (most difficult) the equation of the
circle of curvature in parametric form.
When I read the chapter of the book I followed an example they had in 2d. I was able to figure out that the radius of a circle is found by using this formula:
r = 1/k
k is the curvature at a given point, in this case k(3) = 1/3
so the radius of the circle is:
r = 1/(1/3) = 3
However, I need the coordinates for the center of the osculating circle. I have this idea that the radius of the circle is perpendicular to the unit vector at a given value for t, am I correct???
Can someone please help me with this problem??? I'm having some brain malfunction.
Thank you for your time
Andy
Here is the problem:
r(3) = <1, -3, 2> T(3) = <-1/5, 0, 2/5> N(3) = <0, -1, 0> k(3) = 1/3
For t=3 find: the center of the circle of curvature (or osculating circle); the equation of the
oculating plane (the plane determined by T and N); and (most difficult) the equation of the
circle of curvature in parametric form.
When I read the chapter of the book I followed an example they had in 2d. I was able to figure out that the radius of a circle is found by using this formula:
r = 1/k
k is the curvature at a given point, in this case k(3) = 1/3
so the radius of the circle is:
r = 1/(1/3) = 3
However, I need the coordinates for the center of the osculating circle. I have this idea that the radius of the circle is perpendicular to the unit vector at a given value for t, am I correct???
Can someone please help me with this problem??? I'm having some brain malfunction.
Thank you for your time
Andy