Find the curvature of the curve defined by the parametric equation $r\left(t\right)=\left(\cos\left(-3t\right),\:\sin\left(-3t\right),\:5t\right)$ at $t=\frac{\pi }{6}$.
So what I did was find the unit tangent vector, which was $\left(-\frac{3}{\sqrt{34}},\:0,\:\frac{5}{\sqrt{34}}\right)$. The derivative of this unit tangent vector at the same point is $(0, 9, 0)$ which had a magnitude of $9$. The magnitude for the derivative of the initial parametric equation was $\sqrt{34}$ as the vector was $(-3, 0, 5)$. So to calculate the curvature, I divided the magnitude of the unit tangent vector by the magnitude of the derivative of the initial parametric equation to get $\frac{9}{\sqrt{34}}$, but this is incorrect.
Any help?
Note that $\|r'(t)\|=\sqrt{34}$. Hence $$ r(s) = \left(\cos(\tfrac{3}{\sqrt{34}}s), -\sin(\tfrac{3}{\sqrt{34}}s),\tfrac{5}{\sqrt{34}}s\right) $$ is a unit speed parametrisation of the curve. Now the curvature is given by $\|T'(s)\|$. Note that $$ T(s)={r'(s)}=\left(-\frac{3}{\sqrt{34}} \sin (\tfrac{3}{\sqrt{34}} s),-\frac{3}{\sqrt{34}}\cos (\tfrac{3}{\sqrt{34}} s),\frac{5}{\sqrt{34}}\right). $$ Deriving this once more gives $$ T'(s) = \left(-\frac{9}{34}\cos (\tfrac{3}{\sqrt{34}} s),\frac{9}{34}\sin(\tfrac{3}{\sqrt{34}} s),0\right). $$ Note that the norm $\|T'(s)\|=\tfrac{9}{34}$ everywhere.