Curvature of a Helix

  • Thread starter LRP0790
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Homework Statement



Find the curvature of a helix given by the parametric equation r(t)=<acost, asint, bt> where a and b are real numbers

Homework Equations



I know k=|T'(t)/r'(t)|

The Attempt at a Solution



and I believe the answer to be k=b/(a2+b2)1/2, I just don't know how to get there
 

Answers and Replies

  • #2
HallsofIvy
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First step, write the formula correctly! You can't divide vectors!
Did you mean k= |T'(t)|/|r'(t)|?

If so then if r= <a cos t, a sin t, bt>, r'= <-a sin t, a cos t, b> and it's length is [itex]|r'|= \sqrt{a^2 sin^2 t+ a^2 sin^2 t+ b^2}= \sqrt{a^2+ b^2}[/itex], a constant. That means that T, the unit tangent vector is
[tex]T= \frac{1}{\sqrt{a^2+ b^2}}<-a sin t, a cos t, b>[/tex]

That's easy to differentiate with respect to t (since that whole first fraction is a constant). Do that and take the length of |T'|. Divide by the length of r' which I've already given you.
 

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