# Osculating Circle for vector-valued function?

• doctorwhoo
In summary, the conversation discusses finding the center of a unique circle that passes through a given point on an ellipse and has a radius equal to the curvature of the ellipse at that point. The center of the circle is located on the concave side of the curve, which is the inside of the ellipse. The direction of the tangent and normal vectors at the given point can help determine the location of the center without needing complicated calculations.
doctorwhoo
Hi! This is my first time on Physics Forum. (which shows how desperate I am on figuring out this question).

## Homework Statement

r(t) = <3sin(t),4cos(t)>

There is a unique circle with the following properties:

1. It passes through the point r(∏/2)
2. At the point r(∏/2), the tangent line to r(t) and the tangent line to the circle are the same.
3. The radius of the circle is 1/κ, where κ is the curvature of r(t) at t = ∏/2
4. The center of the circle lies on the concave side of the surve

## Homework Equations

κ = |dT/ds|
T = r'(t)/||r'(t)||
N = T'(t)/||T'(t)||
Not sure if I need more equations for this??

## The Attempt at a Solution

So I know that the graph they gave is an elipse, so I can easily visualize what the hint #2 says. I know how to find curvature, so that gives me the radius of the circle. I just need the center. Based on the info above, the center lines on the concave side of the curve. So that means it's in the same direction as N(t). But not sure where to go from there. Any help would be greatly appreciated.

Did you graph your ellipse? You know r(1) and r'(1) and you know N(1) is perpendicular to r'(1) and points to the concave side. Draw a picture and if you know the radius, it will be obvious where the center is.

Yeah I graphed the ellipse. By 1, do you mean pi/2?

I guess the main doubt I have is that I'm not sure what the concave side means. The whole ellipse is looks convex.

So I have the radius. I tried to solve N(t), but that yielded something hopelessly convoluted, and I think I only need to know the direction of N(t) (which is perpendicular to the tangent vector). So does the center lie on the bottom of the ellipse? This is confusing me.

LCKurtz said:
Did you graph your ellipse? You know r(1) and r'(1) and you know N(1) is perpendicular to r'(1) and points to the concave side. Draw a picture and if you know the radius, it will be obvious where the center is.

doctorwhoo said:
Yeah I graphed the ellipse. By 1, do you mean pi/2?

I guess the main doubt I have is that I'm not sure what the concave side means. The whole ellipse is looks convex.

Yes I meant ##\pi/2##, sorry. An ellipse looks convex if you view it from its exterior, but concave if you view it from its interior. So the concave side of the curve is the inside of it. Your normal points inward.

doctorwhoo said:
So I have the radius. I tried to solve N(t), but that yielded something hopelessly convoluted, and I think I only need to know the direction of N(t) (which is perpendicular to the tangent vector). So does the center lie on the bottom of the ellipse? This is confusing me.

Where is the point when ##t=0## and ##t=\pi/2##? I don't think it is at the "bottom"" of the ellipse. At ##t=\pi/2## what direction is the tangent vector? At ##t=\pi/2##, the normal is perpendicular to the tangent and points inward. What direction is that on this graph? You don't need to do those complicated calculations to figure that out.

## 1. What is an osculating circle for a vector-valued function?

An osculating circle is a circle that is tangent to a curve at a specific point and has the same curvature as the curve at that point. For a vector-valued function, the osculating circle is the circle that best approximates the direction and curvature of the function at a given point.

## 2. How is the osculating circle calculated for a vector-valued function?

The osculating circle is calculated using the derivative of the function at a specific point. The center of the osculating circle is the point on the curve, and the radius of the circle is the reciprocal of the curvature at that point.

## 3. What is the significance of the osculating circle for vector-valued functions?

The osculating circle helps us understand the behavior of a vector-valued function at a particular point. It can also be used to approximate the function and make predictions about its behavior in the future.

## 4. Can the osculating circle change along a vector-valued function?

Yes, the osculating circle can change along a vector-valued function as the curvature of the function changes. The osculating circle will be different at each point along the curve, unless the function has a constant curvature.

## 5. How is the osculating circle related to other geometric concepts?

The osculating circle is closely related to the tangent line and the normal vector at a specific point on a curve. The tangent line is perpendicular to the normal vector, and the osculating circle is tangent to both the curve and the normal vector at that point.

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