Circle of Curvature: Find Equation for Space Curve

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Homework Help Overview

The discussion revolves around finding the equation for the circle of curvature for a given space curve defined by r(t) = t i + sin(t) j at the point (π/2, 1). Participants are exploring concepts related to curvature, tangent and normal vectors, and the geometric properties of circles in relation to curves.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need for defining a circle in the plane, including the radius and center. There are questions about the calculation of curvature (κ) and its implications for determining the circle's center. The relationship between the unit normal vector and the circle's center is also explored.

Discussion Status

Participants are actively engaging with the problem, raising questions about the definitions and calculations involved. Some guidance has been provided regarding the relationship between the normal vector and the center of the circle, as well as considerations about whether the curve is unit speed. There is no explicit consensus, but various interpretations and approaches are being discussed.

Contextual Notes

There is uncertainty regarding whether the curve is unit speed and how this affects calculations. Participants are also considering the implications of signed versus absolute curvature in their discussions.

nns91
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Homework Statement


I am given a space curve r(t)= t i + sin(t) j and point (pi/2,1). They ask me to find an equation for the circle of curvature.


Homework Equations



Kappa, T, N, not sure


The Attempt at a Solution



So I have found the radius of curvature which is row= 1/kappa= 1 at the given point. I have also found out T which is the unit tangent vector. Do I have to find the unit normal vector also ?

How do I write the equation ? will it just be (x-pi/2)^2 + (y-1)^2 = 1 ??
 
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Hi nns91

what do you need to define a circle in the plane?
- a radius
- a circle centre
The equation of a cirlce of centre (a,b) and radius r is
(x-a)^2 + (y-b)^2 = r^2

You have your radius from
r = \frac{1}{\kappa}
Also are you sure \kappa is 1? how did you get this?

how do you find the centre of the circle? It is not (pi/2,1). This is a point on your curve. Think about the unit normal direction.
 
you should also consider whether you curve is unit speed or whether you need to normalise...
 
I calculate Kappa and got like sqrt(1+cos^2(t)) / sin(t). Then I substitute pi/2 for t and get 1. Am I right ??

I still don't get the relationship between center of circle and the unit normal direction.

My curve is not a unite speed curve I think since it is r(t) instead of v(t)
 
try drawing your curve and imagine where the circle sits, the tangent of the curve will match the tangent of the circle

the normal direction will point toward (or away) from the centre of the circle

you curve is unit speed instanaeously at that point (calulate |dr/dt|, but not in general so you may need to be careful with your calcs and check what this affects... I'm not totally sure without looking back at the equations which I don't have handy

Also I'm not too sure how you got it, but I think your curvature value is correct, if you are speaking about absolute curvature. If we are talking about signed curvature need to be careful as it could be + or - depending on how things are defined.
 
Last edited:
WIll the normal direction point toward the center of the circle ?
 
either directly toward or directly away depending on your parameterisation & curvature definition
 
Then how do I find the center of the circle base on the normal vector ?
 
you have a point on the circle (ie the point on your curve) a direction to the centre (normal vector) and a radius (from you curvature)... should have everything you need
 
  • #10
so the radius is the distance from the center to the point, so do I use the distance formula then ?
 

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