The Curvature

(1) Proposition (Arc Length Function) Let be a piecewise smooth curve that is the graph of the vector function described parametrically by ,   and let be a particular point on (base point). Then the length of from the base point to the variable is given by the arc length function

(2) Example (Arc Length Function) Find the arc length of the curve



from to

    Solution.  We have, units because







and so the arc length is

(3) Proposition (Speed as the Derivative of Arc Length) Suppose an object moves along a smooth curve that is the graph of the position function , where is continuous on the interval Then the object has speed for where  

(4) Example (Speed as the Derivative of Arc Length) If a moving object has a position vector function of



then find the speed of the object at time and the distance traveled by the object between times and

    Solution. The speed of the object at time is   because



  

and the distance traveled by the object between times and is because

   .
  

(5) Example (Using Arc Length to Parametrize) Express the vector function in terms of arc length measured from the point corresponding to , in the direction of increasing :

    Solution. We have,
    


Solving for we have Thus

(6) Proposition (Unit Tangent and Unit Normal Vectors) If has a piecewise smooth graph and is represented as in terms of the arc length parameter , then the unit tangent vector and the principal unit normal vector satisfies



where is a scalar function of

(7) Definition (Curvature) Suppose the smooth curve is the graph of the vector function , parametrized in terms of the arc length Then the curvature of is the function where is the unit tangent vector.

(8) Example (Curvature) Find the curvature of a circle.

    Solution. A circle can be parametrized by where is the radius. We have



Solving for we find the component functions to be Thus,



and



where


Thus the curvature of a circle is

(9) Example (Unit Tangent and Curvature) Let be the curve given by



Find the unit tangent vector to and the curvature

    Solution. We have,
    


and









thus,



Then,



because we have



therefore,

(10) Proposition (Cross Product Formula for Curvature) Suppose the smooth curve is the graph of the vector function Then the curvature is given by

(11) Example (Cross Product Formula for Curvature) Given the curve defined by find a unit tangent vector at the point on the curve where , the curvature at , and find the length of the curve from to .

    Solution. We have,
    
,

,

and




Therefore,   
  
  
  
  and  
  
   
   

(12) Proposition (Curvature of Planar Curve) The graph of the function has curvature



where and all exist.

    Proof. Given the vector function and defined by we have Using the formula , we have

(13) Example (Curvature of Planar Curve) Find the maximum curvature on the curve

    Solution. We have and



So,



Applying the first derivative test we have the maximum curvature at

(14) Example (Curvature of One Variable Functions) Find the maximum curvature on the curve

    Solution. We have and



So,



Applying the first derivative test we have the maximum curvature at with curvature of   

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Cite this as:
The Curvature
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
http://www.libraryofmath.com/the-curvature.html