Mathematical Immaturity

14.15 Exercises

• Two fixed unit vectors $A$ and $B$ make an angle $\theta$ with each other, where $0 < \theta < \pi.$ A particle moves on a space curve in such a way that its position vector $\mathbf{r}(t)$ and velocity $\mathbf{v}(t)$ are related by the equation $\mathbf{v}(t) = A \times \mathbf{r}(t).$ If $\mathbf{r}(0) = B,$ prove that the curve has constant curvature and compute this curvature in terms of $\theta.$
• Recall the solution from Section 14.9 #15.
• We know from a previous solution (Section 14.9 #15) that a particle with motion satisfying the above conditions has constant speed satisfying the relation $$ \begin{align*} \\ v(t) &= \|A\|\|B\|\sin\theta \end{align*} $$ for all $t.$ And since $A$ and $B$ both have unit length, we can write $v(t) = \sin \theta.$ Then, the unit tangent $T(t)$ is $$ \begin{align*} \\ T(t) &= \frac{A \times \mathbf{r}(t)}{\sin \theta} \end{align*} $$ Its derivative with respect to $t$ is $$ \begin{align*} \\ T'(t) &= \frac{A \times \mathbf{r}'(t)}{\sin \theta} \\ \\ &= \frac{A \times \mathbf{v}(t)}{\sin \theta} \end{align*} $$ Noting that $\sin \theta > 0,$ since $0 < \theta < \pi,$ the norm of $T'(t)$ is $$ \begin{align*} \\ \|T'(t)\| &= \frac{\|A \times \mathbf{v}(t)\|}{\sin \theta} \\ \\ &= \frac{\|A\| \|\mathbf{v}(t)\|\sin\phi}{\sin \theta} \end{align*} $$ where $\phi$ is the angle between $A$ and the velocity vector $\mathbf{v}(t)$ satisfying $0 < \phi < \pi,$ whose cosine is $$ \cos \phi = \frac{A \cdot \mathbf{v}(t)}{\|A\|\|\mathbf{v}(t)\|} $$ But we know that $\mathbf{v}(t) = A \times \mathbf{r}(t),$ which means that $A \cdot \mathbf{v}(t) = 0$ and thus $\cos \phi = 0.$ But since $0 < \phi < \pi$ it must be the case that $\phi = \pi/2$ and $\sin \phi = 1.$ Plugging this back into our equation for the norm of $T'(t)$ we see that $$ \begin{align*} \\ \|T'(t)\| &= \frac{\|A\| \|\mathbf{v}(t)\|\sin\phi}{\sin\theta} \\ \\ &= \frac{\sin\theta}{\sin\theta} \\ \\ &= 1 \end{align*} $$ Then, curvature $\kappa(t)$ is $$ \begin{align*} \kappa(t) &= \frac{\|T'(t)\|}{v(t)} \\ \\ &= \frac{1}{\sin\theta} \quad \blacksquare \end{align*} $$ $($Note: The back-of-book equation says that $\kappa(t) = \frac{1}{\|B\|\sin\theta}$ but we know from the given information that B is of unit length, thus $\|B\|\sin\theta = \sin \theta.)$