Parametric equations for circle of curvature at given point.

In summary, the professor gave me a hint that I need to find the normal and radius of curvature at (0, \pi/2, 2). I can find the unit tangent, unit normal normal and binormal vectors, and the equations for the normal and osculating planes. Radius of curvature is something I have never heard about, and 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? I can find the unit tangent, unit normal normal and binormal vectors
  • #1
magimag
11
0
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.

Homework Statement



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

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

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 :)
 
Physics news on Phys.org
  • #2
Have you determined what the normal to the curve and radius of curvature are at [itex](0, \pi/2, 2)[/itex]?
 
  • #3
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
I can find the unit tangent, unit normal normal and binormal vectors.
I can also find the equations for the normal and osculating planes.
Radius of curvature is something I have never heard about.
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
magimag said:
I can find the unit tangent, unit normal normal and binormal vectors.
I can also find the equations for the normal and osculating planes.
Radius of curvature is something I have never heard about.
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?

Can you give the exact definition of "circle of curvature" as it appears in your notes/syllabus/book?
 
  • #6
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
magimag said:
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.

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
I'll come back in 2 hours I have to run to practice.
 
  • #9
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
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
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
magimag said:
Ok I'm getting a bit confused. What if I find the normal vector and radius I should be able to find the center?

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:

[tex]\theta\rightarrow (\alpha + R\cos(\theta))\mathbf{N}(s) + (\beta + R\sin(\theta))\mathbf{T}(s)[/tex]

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

Related to Parametric equations for circle of curvature at given point.

1. What are parametric equations for a circle of curvature at a given point?

The parametric equations for a circle of curvature at a given point are x = a + rcosθ and y = b + rsinθ, where a and b are the coordinates of the given point, r is the radius of curvature, and θ is the angle of curvature.

2. How do these equations relate to the curvature of a curve at a given point?

The parametric equations for a circle of curvature at a given point describe the shape and size of the circle that best approximates the curvature of a curve at that point. The radius of this circle is equal to the radius of curvature of the curve at that point.

3. How do you determine the radius of curvature using these equations?

The radius of curvature can be determined by finding the value of r in the parametric equations. This value represents the distance from the given point to the center of the circle of curvature, and is equal to the radius of the circle.

4. Can these equations be used for any type of curve?

Yes, these equations can be used for any type of curve, as long as it is differentiable at the given point. This includes both parametric and non-parametric curves.

5. Are there any limitations to using these equations?

One limitation to using these equations is that they only provide an approximation of the curvature at a given point, as the curve may not perfectly align with the circle of curvature. Additionally, these equations may be more complex and difficult to use for non-parametric curves.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
658
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
719
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top