Mathematical Immaturity

14.15 Exercises

• A helix is described by the position function $\mathbf{r}(t) = a\cos \omega t\,\mathbf{i} + a\sin \omega t\,\mathbf{j} + b\omega t\,\mathbf{k}.$ Prove that it has constant curvature $\kappa = a/(a^2 + b^2).$
• Recall from Section 14.14, (equation 14.20) that the length of the curvature vector, or the curvature $\kappa$ of a curve is defined as: $$ \begin{align} \\ \kappa(t) = \frac{\|T'(t)\|}{v(t)} \tag{14.20} \end{align} $$
• If the position function is defined by $$ \begin{align*} \\ \mathbf{r}(t) = a\cos \omega t\,\mathbf{i} + a\sin \omega t\,\mathbf{j} + b\omega t\,\mathbf{k} \end{align*} $$ Then its velocity $\mathbf{v}(t)$ and speed $v(t)$ are defined by $$ \begin{align*} \\ \mathbf{v}(t) &= -a\omega\sin \omega t\,\mathbf{i} + a\omega\cos \omega t\,\mathbf{j} + b\omega\,\mathbf{k} \\ \\ v(t) &= |\omega| \sqrt{a^2 + b^2} \end{align*} $$ Then, the unit tangent $T(t)$ is: $$ \begin{align*} \\ T(t) &= \frac{\mathbf{v}(t)}{v(t)} \\ &= \frac{-a\omega\sin \omega t\,\mathbf{i} + a\omega\cos \omega t\,\mathbf{j} + b\omega\,\mathbf{k}}{|\omega| \sqrt{a^2 + b^2}} \end{align*} $$ Taking its derivative with respect to $t,$ noting that $v(t)$ is constant, we obtain $$ \begin{align*} \\ T'(t) &= \frac{\mathbf{a}(t)}{v(t)} \\ &= \frac{-a\omega^2\cos \omega t\,\mathbf{i} - a\omega^2\sin \omega t\,\mathbf{j}}{|\omega| \sqrt{a^2 + b^2}} \end{align*} $$ Whose norm is $$ \begin{align*} \\ \|T'(t)\| &= \frac{|a| \omega^2}{|\omega|\sqrt{a^2 + b^2}} \end{align*} $$ Then, curvature $\kappa(t)$ is $$ \begin{align*} \\ \kappa(t) &= \frac{\|T'(t)\|}{v(t)} \\ &= \frac{|a| \omega^2}{\omega^2(a^2 + b^2)} \end{align*} $$ Given that $a > 0$ for helical motion (as defined in section 14.6), we get $\kappa = a/(a^2 + b^2). \quad \blacksquare$