# Differential Geometry: the Osculating Circle

• theshareef
In summary, the centre of the osculating circle of a unit speed curve, f, at some u in J is the intersection of the line normal to f'(u) through f(u) and the normal line to g'(u) through g(u).
theshareef

## Homework Statement

let f: J --> R^2 be a unit speed curve curve and define it's tangentially equidistant campanion by g(u) = f(t) + r*f'(t) for a fixed r>0. Show that the centre of the osculating circle of f at some u in J is the intersection of the line normal to f'(u) through f(u) and the normal line to g'(u) through g(u).

## Homework Equations

T(u) = f'(u) as f is unit speed (ie. parametrized by arclength)

N(u) = vector normal to f'(u) = R(T(u)) where R is a rotation in the plane by pi/2 in the counterclockwise direction.

The centre of the osculating circle is f(u) + [1/k]*(N(u)), where k is the signed curvature of f.

k := f''(u) dot product with R(f'(u)) for a unit speed curve.

The Frenet-Serret equations also tell us that: N'(u) = -k*T(u)

## The Attempt at a Solution

My attempts have thusfar consisted of applying the defition to find the normal line to f'(u) and g'(u) and setting them equal to each other then trying to reduce with seriously no luck in proving the claim no matter how I try to manipulate them. I've tried realizing that:

g'(u) = f'(u) + r*(f''(u)) ==> R(g'(u)) = vector pointing in direction normal to g'(u) = R(f'(u)) + r*R(f''(u)) = N(u) + r*(R(f'(u))' = N(u) + r*N'(u) = N(u) + r*(-k*f'(u))

The last few steps can be justified by the fact that R is linear ==> its derivative is itself so that we can manipulate the term R(f''(u)) as shown (I hope) which at least brings the curvature into the fold and intuitevly makes sense in terms of how it is expressed.

No matter what I've tried to do (even going so far as to seemingly uselessly express the desired point in terms of the orthonormal basis (T, N)) has got me no where. Any ideas I'm missing that could get me on the right track?

Last edited:
First of all thanks to everyone who took time to read and consider this problem in order to help me out, I appreciate it.

So I have a method to prove the claim. I'll sketch the ideas that lead to the proof quickly right now for this who are interested and post something more detailed when I have time for those who need a little more help understanding what to do.

You start as I did above by finding vectors in the direction of the lines desired. It is key to discover that r*R(f'') = -r*k*T, which arises when you are finding a vector perpendicular to g', and understand why this is true.

Then you simply express the line through f(s) perpendicular to f'(s) in the usual way: f(s) + c*N(s), where c is an arbitrary, real number parameter. Similarily, the line through g(s) which is perpendicular to g'(s) is g(s) + d*(N(s) - r*k*T).

Then you express a point on each of these lines in term of the orthonormal basis, {T, N}, set the T component of line 1 equal to line 2 and the N component of line 1 equal to line 2. Simply solve the N component first, plug that information into the T component equality, send what you get into the equations you defined for the lines and the claim follows.

Last edited:
Hi, thank you for taking the time to post this. I do have a question though - how did you go about expressing points on those vectors in terms of the orthonormal basis? This is the only part I'm stuck on, although I realize it's probably quite silly.

Thanks.

Hi guys.
This is a nice problem. But I think to prove it it's not so simple.
With some time in my hands I will work on it.

I see in the equations you have list, there is not the formula to compute the curvature by the derivatives of the curve. I guess you are aware of this curve because you cite other advanced formulas likeFrenet-Serret, so I think you already know it.
Why you don't use it ? Maybe you've seen it not relevant.
In any case I will you it to get some result. I already tried but I got nowhere.

See you soon.

## What is Differential Geometry?

Differential Geometry is a branch of mathematics that studies the properties and behavior of curves and surfaces in various dimensions. It combines principles of geometry, calculus, and linear algebra to study the shape and curvature of objects in space.

## What is the Osculating Circle?

The Osculating Circle, also known as the circle of curvature, is a circle that best approximates the curve at a specific point. It is defined as the circle that touches the curve at that point and has the same curvature as the curve at that point.

## What is the significance of the Osculating Circle?

The Osculating Circle is important in Differential Geometry as it provides a way to measure the curvature of a curve at a specific point. It also helps in understanding the behavior of a curve and its relationship with its tangent line at that point.

## How is the Osculating Circle calculated?

The Osculating Circle is calculated using the first and second derivatives of the curve at the specific point. The center of the circle is located at the point of tangency and the radius is equal to the reciprocal of the curvature at that point.

## What are some real-world applications of the Osculating Circle?

The Osculating Circle has many practical applications, such as in engineering, physics, and computer graphics. It is used to design and analyze curved structures, such as roads, bridges, and roller coasters. It also helps in understanding the motion of objects in space and creating realistic 3D animations.

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