# Homework Help: Parametric equations for circle of curvature at given point.

1. Mar 4, 2014

### magimag

Hey guys, I'm new here. I got a problem from my professor that is different from any other problems we have done. I'm stuck and need a little help.

1. The problem statement, all variables and given/known data

r(t) = <cos(t), t, 2sin(t)>

Find parametric equations for the circle of curvature at (0, pi/2, 2)

3. The attempt at a solution
I know the circle of curvature is the intersection of a sphere with a plane.
Hint the professor gave me:
To get the parametric equations, look at the projections of the sphere in each of coordinate planes.

I don't know how I should approach this problem.
Any help to get me going would be much appreciated.

Thank you :)

2. Mar 4, 2014

### HallsofIvy

Have you determined what the normal to the curve and radius of curvature are at $(0, \pi/2, 2)$?

3. Mar 4, 2014

### micromass

The hint by the professor seems a bit weird to me. Can you please give your definition of "circle of curvature" (which I assume is a synonym for osculating circle) and equivalent properties that you can use?

4. Mar 4, 2014

### magimag

I can find the unit tangent, unit normal normal and binormal vectors.
I can also find the equations for the normal and osculating planes.
I looked it up and found out it is one over the curvature.

I will try to explain what I think the circle of curvature is:
I think a circle of curvature is a osculating circle, that is osculating about a point on a curve.
does that make any sense?

5. Mar 4, 2014

### micromass

Can you give the exact definition of "circle of curvature" as it appears in your notes/syllabus/book?

6. Mar 4, 2014

### magimag

The circle that lies in the osculating plane of C at P, has the same tangent as C at P, lies on the concave side of C (toward which N points), and has radius ρ = 1/κ (the reciprocal of the curvature) is called the osculating circle (or the circle of curvature) of C at P. It is the circle that best describes how C behaves near P; it shares the same tangent, normal, and curvature at P.

7. Mar 4, 2014

### micromass

Do you know (or can you prove) that the center of this circle actually lies on the normal line of $P$? (the normal line is the line through $P$ parallel to the normal vector at $P$).

8. Mar 4, 2014

### magimag

I'll come back in 2 hours I have to run to practice.

9. Mar 4, 2014

### magimag

Ok, I do not know how to prove it to you. Sorry I'm very new to proofing. Isn't the center at the origin? I'm not sure how I can draw the curve.

10. Mar 4, 2014

### micromass

First, can you somehow rotate and translate everything such that
1) The point P becomes the origin
2) The osculating plane becomes the xy-plane

Can you figure out which linear map does that? The idea is to find the osculating circle in this case and then do the general case.

11. Mar 4, 2014

### magimag

Ok I'm getting a bit confused. What if I find the normal vector and radius I should be able to find the center?

12. Mar 5, 2014

### micromass

Yes, you should be.

You know the circle of curvature lies in the osculating plane, right?
So the center of the circle lies in the osculating plane, meaning it has the form $\alpha\mathbf{N}(s) + \beta\mathbf{T}(s)$.

The circle then has the equation:

$$\theta\rightarrow (\alpha + R\cos(\theta))\mathbf{N}(s) + (\beta + R\sin(\theta))\mathbf{T}(s)$$

Right? Your job now is to find what $\alpha$, $\beta$ and $R$ are.