91Ó°ÊÓ

Consider

$r\left(t\right)=\mathrm{Î\pm }{e}^{\mathrm{γ}t}\left(\cos t\right)i+\mathrm{Î\pm }{e}^{\mathrm{γ}t}\left(\sin t\right)j+\mathrm{β}{e}^{\mathrm{γ}t}k,tâ‰\yen 0$, known as a concho-spiral.

The tangent vector to $r\left(t\right)$is $r^{\text{'}}\left(t\right)$.

Vector $k=⟨0,0,1âŸ\copyright$

Let the angle between the vectors $r^{\text{'}}\left(t\right)$and vector $k$be $\mathrm{θ}$. Then

$\cos \mathrm{θ}=\frac{r^{\text{'}}\left(t\right)·k}{||r^{\text{'}}\left(t\right)||‖k‖}$

$\cos \mathrm{θ}=\frac{\mathrm{β}\mathrm{γ}{e}^{\mathrm{γ}t}}{\sqrt{{\mathrm{Î\pm }}^2{e}^{2\mathrm{γ}t}(-\sin t+\mathrm{γ}\cos t{)}^2+{\mathrm{Î\pm }}^2{e}^{2\mathrm{γ}t}(\cos t+\mathrm{γ}\sin t{)}^2+{\mathrm{β}}^2{\mathrm{γ}}^2{e}^{\mathrm{γ}t}}}$

$\cos \mathrm{θ}=\frac{\mathrm{β}\mathrm{γ}{e}^{yt}}{\sqrt{{\mathrm{Î\pm }}^2{e}^{2\mathrm{γ}t}\left({\sin }^{2}t+{\mathrm{γ}}^2{\cos }^{2}t-2\mathrm{γ}\sin t\cos t+{\cos }^{2}t+{\mathrm{γ}}^2{\sin }^{2}t+2\mathrm{γ}\sin t\cos t\right)+{\mathrm{β}}^2{\mathrm{γ}}^2{e}^{2rt}}}$

$\cos \mathrm{θ}=\frac{\mathrm{β}\mathrm{γ}{e}^{\mathrm{γ}t}}{\sqrt{{\mathrm{Î\pm }}^2{e}^{2\mathrm{γ}t}\left(1+{\mathrm{γ}}^2\right)+{\mathrm{β}}^2{\mathrm{γ}}^2{e}^{2\mathrm{γ}t}}}$

$\cos \mathrm{θ}=\frac{\mathrm{β}\mathrm{γ}{e}^{\mathrm{γ}t}}{\sqrt{e^{2\mathrm{γ}t}\left({\mathrm{Î\pm }}^2+{\mathrm{Î\pm }}^2{\mathrm{γ}}^2+{\mathrm{β}}^2{\mathrm{γ}}^2\right)}}$

$=\frac{\mathrm{β}\mathrm{γ}}{\sqrt{{\mathrm{Î\pm }}^2+{\mathrm{Î\pm }}^2{\mathrm{γ}}^2+{\mathrm{β}}^2{\mathrm{γ}}^2}}$

$\mathrm{θ}={\cos }^{-1}\left(\frac{\mathrm{β}\mathrm{γ}}{\sqrt{{\mathrm{Î\pm }}^2+{\mathrm{Î\pm }}^2{\mathrm{γ}}^2+{\mathrm{β}}^2{\mathrm{γ}}^2}}\right)$which is a constant.

Thus the angle between the tangent vector to a concho-spiral and the vector $\mathrm{k}$is a constant.