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Find the unit normal vector for the vector-valued function r ( t ) = ( t 2 3 t ) i + ( 4 t + 1 ) j and evaluate it at t = 2.

N ( 2 ) = 2 2 ( i j )

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For any smooth curve in three dimensions that is defined by a vector-valued function, we now have formulas for the unit tangent vector T , the unit normal vector N , and the binormal vector B . The unit normal vector and the binormal vector form a plane that is perpendicular to the curve at any point on the curve, called the normal plane    . In addition, these three vectors form a frame of reference in three-dimensional space called the Frenet frame of reference    (also called the TNB frame) ( [link] ). Lat, the plane determined by the vectors T and N forms the osculating plane of C at any point P on the curve.

This figure is the graph of a curve increasing and decreasing. Along the curve at 4 different points are 3 vectors at each point. The first vector is labeled “T” and is tangent to the curve at the point. The second vector is labeled “N” and is normal to the curve at the point. The third vector is labeled “B” and is orthogonal to T and N.
This figure depicts a Frenet frame of reference. At every point P on a three-dimensional curve, the unit tangent, unit normal, and binormal vectors form a three-dimensional frame of reference.

Suppose we form a circle in the osculating plane of C at point P on the curve. Assume that the circle has the same curvature as the curve does at point P and let the circle have radius r. Then, the curvature of the circle is given by 1 / r . We call r the radius of curvature    of the curve, and it is equal to the reciprocal of the curvature. If this circle lies on the concave side of the curve and is tangent to the curve at point P, then this circle is called the osculating circle    of C at P , as shown in the following figure.

This figure is the graph of a curve with a circle in the middle. The bottom of the circle is the same as part of the curve. Inside of the circle is a vector labeled “r”. It starts at point “P” on the circle and points towards the radius. There is also a line segment perpendicular to the radius and tangent to point P.
In this osculating circle, the circle is tangent to curve C at point P and shares the same curvature.

For more information on osculating circles, see this demonstration on curvature and torsion, this article on osculating circles, and this discussion of Serret formulas.

To find the equation of an osculating circle in two dimensions, we need find only the center and radius of the circle.

Finding the equation of an osculating circle

Find the equation of the osculating circle of the helix defined by the function y = x 3 3 x + 1 at t = 1.

[link] shows the graph of y = x 3 3 x + 1.

This figure is the graph of a cubic function y = x^3-3x+1. The curve increases, reaches a maximum at x=-1, decreases passing through the y-axis at 1, then reaching a minimum at x =1 before increasing again.
We want to find the osculating circle of this graph at the point where t = 1.

First, let’s calculate the curvature at x = 1 :

κ = | f ( x ) | ( 1 + [ f ( x ) ] 2 ) 3 / 2 = | 6 x | ( 1 + [ 3 x 2 3 ] 2 ) 3 / 2 .

This gives κ = 6. Therefore, the radius of the osculating circle is given by R = 1 κ = 1 6 . Next, we then calculate the coordinates of the center of the circle. When x = 1 , the slope of the tangent line is zero. Therefore, the center of the osculating circle is directly above the point on the graph with coordinates ( 1 , −1 ) . The center is located at ( 1 , 5 6 ) . The formula for a circle with radius r and center ( h , k ) is given by ( x h ) 2 + ( y k ) 2 = r 2 . Therefore, the equation of the osculating circle is ( x 1 ) 2 + ( y + 5 6 ) 2 = 1 36 . The graph and its osculating circle appears in the following graph.

This figure is the graph of a cubic function y = x^3-3x+1. The curve increases, reaches a maximum at x=-1, decreases passing through the y-axis at 1, then reaching a minimum at x =1 before increasing again. There is a small circle inside of the bend of the cure at x = 1.
The osculating circle has radius R = 1 / 6.
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Find the equation of the osculating circle of the curve defined by the vector-valued function y = 2 x 2 4 x + 5 at x = 1.

κ = 4 [ 1 + ( 4 x 4 ) 2 ] 3 / 2

At the point x = 1 , the curvature is equal to 4. Therefore, the radius of the osculating circle is 1 4 .

A graph of this function appears next:
This figure is the graph of the function y = 2x^2-4x+5. The curve is a parabola opening up with vertex at (1, 3).

The vertex of this parabola is located at the point ( 1 , 3 ) . Furthermore, the center of the osculating circle is directly above the vertex. Therefore, the coordinates of the center are ( 1 , 13 4 ) . The equation of the osculating circle is

( x 1 ) 2 + ( y 13 4 ) 2 = 1 16 .

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Questions & Answers

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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