(adsbygoogle = window.adsbygoogle || []).push({}); Space Curves --> Unit Tangent Vector and Curvature

Here is the original question:

Consider the space curve r(t) = (e^t)*cos(t)i + (e^t)*sin(t)j + k. Find the unit tangent vector T(0) and the curvature of r(t) at the point (0,e^(pi/2),1).

I believe I have found the unit tangent vector, T(0), correctly: (1/sqrt(2))i + (1/sqrt(2))j

Is this correct? Also, how do I find the curvature at that particular point?

Thanks!

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Space Curves -> Unit Tangent Vector and Curvature

Loading...

Similar Threads for Space Curves Unit | Date |
---|---|

Strategies for drawing space curves? | Jun 5, 2014 |

Dirac delta in curved space | Sep 28, 2012 |

Gradient and equation for curve in space | Sep 18, 2012 |

Normal/Tangent to space curve | Jun 18, 2010 |

Torsion of space curves, why dB/ds is perpendicular to tangent | Mar 3, 2010 |

**Physics Forums - The Fusion of Science and Community**