The curvature of a line defined in terms of s

51 Views Asked by At

I have a curve defined in terms of arclength $s$ :

$$ x=\arctan(s) $$ $$ y=\frac{\sqrt{2}}{2}(s^2+1) $$ $$ z=s-\arctan(s) $$

So to compute its curvature, I started by writing its first and second derivatives (which are supposed to correspond to $\hat{T}$ and $\kappa \hat{N}$ as coordinates are directly expressed in terms of s) : $$ \langle \frac{1}{(1+s^2)};\sqrt{2}s; \frac{s^2}{(1+s^2)} \rangle $$ $$ \langle \frac{-2s}{(1+s^2)^2}; \sqrt{2}; \frac{-2s}{(1+s^2)^2} \rangle $$ Now, $\kappa$ is supposed to be $\Vert \frac{d \hat{T}}{ds} \Vert$

And computing it I find : $$ \sqrt{ \frac{8s^2+2(1+s^2)^4}{(1+s^2)^4}} $$

BUT the textbook I use (Schaum's outline of vector analysis, problem 3.48) gives the answer : $$ \kappa= \frac{\sqrt{2}}{s^2+1} $$

They don't give the details of the computation... What am I missing?

Thanks!