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Curvature in terms of the tangent vector

  1. Mar 4, 2009 #1
    My teacher wrote an alternative equation on the board for curvature, and I am wondering how it is true:

    k = | dT/dt / |dR/dt| |

    where T is the unit tangent vector.

    I know k = |R' x R''| / |R'|^3 = |dT/ds|
    but I am not sure about the formula in question. How is it true/derived?
  2. jcsd
  3. Mar 4, 2009 #2
    i think these are equivalent, i think the basic definition of curvature is |dT|/ds,meaning the angle turned over curve length traveled, indeed you can find |dT| is angle turned if T is unit tangent vector,and ds=|dR| where R is the position vector, and all the mentioned formulas can be derived.
  4. Mar 4, 2009 #3
    Check out this web site for clarification.

    http://www.usd.edu/~jflores/MultiCalc02/WebBook/Chapter_14/Graphics/Chapter14_3/Html14_3/14.3%20Arc%20Length%20and%20Curvature.htm [Broken]
    Last edited by a moderator: May 4, 2017
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