What is the Curvature of a Helix Given by a Parametric Equation?

[LRP0790]

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

 

[HallsofIvy]

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 |r&#039;|= \sqrt{a^2 sin^2 t+ a^2 sin^2 t+ b^2}= \sqrt{a^2+ b^2}, a constant. That means that T, the unit tangent vector is
T= \frac{1}{\sqrt{a^2+ b^2}}&lt;-a sin t, a cos t, b&gt;

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.