Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 3 / 3C (Vector Calculus)

Differentiate the parametrized curve to obtain ${\mathbf{x}}^{\prime }(t)=(1-t^2,2t,1+t^2)$. Its speed obeysSince $1+t^2>0$, the arc length is

Divide the velocity by the speed calculated above. The unit tangent vector isIts derivative isBecause $ds/dt=\sqrt{2}(1+t^2)>0$, the principal normal vector has the direction of $T^{\prime }(t)$, givingThis is a unit vector, and $N\cdot e_z=0$ for every real $t$. The curvature of a space curve is $|T^{\prime }|/(ds/dt)=1/(1+t^2{)}^2>0$, so the principal normal vector is defined everywhere on this curve. The same perpendicularity follows from constant tangent component forces perpendicular principal normals: $T\cdot e_z=1/\sqrt{2}$ is constant, and differentiating this identity forces the normal component along $e_z$ to vanish.