Mathematical Immaturity

14.19 Exercises

• A particle moves in space so that its position at time $t$ has cylindrical coordinates $r = t,$ $\theta = t,$ $z = t.$ It traces out a curve called a conical helix. (a) Find formulas for the velocity $\mathbf{v},$ the acceleration $\mathbf{a},$ and the curvature $\kappa$ at time $t.$ (b) Find a formula for determining the angle between the velocity vector and the generator of the cone at each point of the curve.
• (a) Recall from Section 14.18 that for space curves in cylindrical coordinates, we use the position vector $$ \begin{align*} \\ \mathbf{r} &= r\mathbf{u}_r + z(t)\mathbf{k} \end{align*} $$ (b) Recall from Section 13.18 that the generator of a cone is the moving line $G$ that intersects a fixed line $A$ (the axis of the cone) at a given point $P$ (the vertex of the cone). If we set our axis line $A = \mathbf{k}$ (the vertical axis in cylindrical coordinates), we can show that the line $G =\{s\mathbf{r}(t)\ |\ s, t\ \text{real}\}$ intersects $A$ at some point $P,$ making a fixed angle $\phi$ for all $t.$
• (a) Let the space curve with cylindrical coordinates $r = t,$ $\theta = t,$ and $z = t$ be described by the positional vector $$ \begin{align*} \mathbf{r} &= r\mathbf{u}_r + z(t)\mathbf{k} \\ &= t\left(\cos t\,\mathbf{i} + \sin t\,\mathbf{j} + \mathbf{k}\right) \end{align*} $$ Its velocity and acceleration can be defined in terms of radial and transverse components, with the addition of the component $z'(t)\mathbf{k}:$ $$ \begin{align*} \\ \mathbf{v} &= \frac{dr}{dt}\mathbf{u}_r + r\frac{d\mathbf{u}_r}{dt} + z'(t)\mathbf{k} \\ \\ &= \mathbf{u}_r + t\left(-\sin t\,\mathbf{i} + \cos t\,\mathbf{j}\right) + \mathbf{k} \\ \\ &= \mathbf{u}_r + t\mathbf{u}_{\theta} + \mathbf{k} \\ \\ \mathbf{a} &= - t\mathbf{u}_{r} + 2\mathbf{u}_{\theta} \end{align*} $$ To find curvature $\kappa,$ we use the identity: $$ \begin{align*} \\ \kappa(t) &= \frac{\|\mathbf{a}\times \mathbf{v}\|}{v^3(t)} \end{align*} $$ where $v = \|\mathbf{v}\| = \sqrt{2 + t^2}.$ To find $\|\mathbf{a}\times \mathbf{v}\|$ we use Lagrange's identity to give us: $$ \begin{align*} \\ \|\mathbf{a}\times \mathbf{v}\| &= \sqrt{\|\mathbf{a}\|^2\|\mathbf{v}\|^2 - \left(\mathbf{a} \cdot \mathbf{v}\right)^2} \\ &= \sqrt{(t^2 + 4)(t^2 + 2) - t^2} \\ &= \sqrt{t^4 + 5t^2 + 8} \end{align*} $$ Combining these two values, we see that curvature is $$ \begin{align*} \\ \kappa(t) &= \frac{\sqrt{t^4 + 5t^2 + 8}}{\left(t^2 + 2\right)^{3/2}} \quad \blacksquare \end{align*} $$ (b) We first wish to show that $G = \left\{c\mathbf{r}(t)\ |\ c\ \text{real}\right\}$ is the generator line of the conical helix for all $t.$ It will suffice to show that for some fixed line $A,$ $G$ intersects $A$ at point $P$ making a constant angle with $A$ for all $t.$ Let $A = \left\{s\mathbf{k}\ |\ s\ \text{real}\right\}$ and let $P = O.$ We know that when $t = 0$ that $\mathbf{r}(t) = O$ which means that $G$ intersects $A$ at the origin. To show that $G$ makes a constant angle with $A$ for all $t,$ we find the cosine of the angle $\phi$ between the two lines: $$ \begin{align*} \\ \cos\phi &= \frac{\mathbf{r}(t)\cdot\mathbf{k}}{\|\mathbf{r}(t)\|\|\mathbf{k}\|} \\ \\ &= \frac{z(t)}{t\sqrt{\cos^2 t + \sin^2 t + 1}} \\ \\ &= \frac{t}{t\sqrt{2}} \\ \\ &= \frac{\sqrt{2}}{2} \end{align*} $$ for all $t.$ And since the cosine of $\phi$ is constant for all $t,$ with $0 < \phi < \pi/2,$ it implies that the angle $\phi$ must also be constant. Thus, we have shown that the line $G$ through the origin parallel to $\mathbf{r}(t)$ is the generator of the right circular cone containing the conical helix described by $\mathbf{r}(t).$ Now, to find a formula for the angle $\alpha$ between the generator line $G$ and the velocity vector $\mathbf{v}$ at a given $t,$ we first note that since $G$ is a scalar multiple of $\mathbf{r}(t),$ the cosine of the angle between $G$ and $\mathbf{v}(t)$ is the same as that between $\mathbf{r}(t)$ and $\mathbf{v}(t).$ Then, we take the dot product of the position and velocity vectors at $t:$ $$ \begin{align*} \\ \cos\alpha &= \frac{\mathbf{r}(t)\cdot\mathbf{v}(t)}{\|\mathbf{r}(t)\|\|\mathbf{v}(t)\|} \\ \\ &= \frac{r(t) + z(t)}{\sqrt{2}\,t\left(\sqrt{t^2 + 2}\right)} \\ \\ &= \frac{2t}{\sqrt{2}\,t\left(\sqrt{t^2 + 2}\right)} \\ \\ &= \frac{\sqrt{2}}{\sqrt{t^2 + 2}} \end{align*} $$ where $0 < \alpha < \pi/2.$ Taking the inverse cosine, we get: $$ \begin{align*} \\ \alpha &= \arccos\frac{\sqrt{2}}{\sqrt{t^2 + 2}} \quad \blacksquare \end{align*} $$