Mathematical Immaturity

14.19 Exercises

• A particle moves in space so that its position at time $t$ has cylindrical coordinates $r = \sin t,$ $\theta = t,$ $z = \log \sec t,$ where $0 \leq t < \frac{1}{2}\pi.$ (a) Show that the curve lies on the cylinder with Cartesian equation $x^2 + (y - \frac{1}{2})^2 = \frac{1}{4}.$ (b) Find a formula (in terms of $t$) for the angle which the velocity vector makes with $\,\mathbf{k}.$
• (a) Translate the polar coordinates of $\mathbf{r}(t) = r\mathbf{u}_r$ into rectangular coordinates $\mathbf{r}(t) = r\left(\cos \theta\,\mathbf{i} + \sin \theta\,\mathbf{j}\right).$ (b) Use the double-angle identities for sine and cosine in the derivation of the position vector. Recall that $D(\log\sec t) = \tan t.$
• (a) The curve with cylindrical coordinates $r = \sin t,$ $\theta = t,$ and $z = \log\sec t$ is described by the position vector $\mathbf{r}(t) = r\mathbf{u}_r + z\mathbf{k},$ whose rectangular coordinates $(x, y, z)$ are given by $$ \begin{align*} \\ (x, y, z) &= \left(r\cos\theta, r\sin\theta, \log\sec t\right) \\ \\ &= \left(\sin t \cos t, \sin^2 t, \log\sec t\right). \end{align*} $$ We wish to show that $x$ and $y$ satisfy the equation $x^2 + (y - \frac{1}{2})^2 = \frac{1}{4}.$ Using the double angle identities of the sine and cosine, $$ \begin{align*} \\ \sin 2t &= 2\sin t \cos t \\ \\ \cos 2t &= 1 - 2\sin^2 t \end{align*} $$ $x$ and $y$ become: $$ \begin{align*} \\ x &= \frac{1}{2}\sin 2t, \quad y = \frac{1}{2}\left(1 - \cos 2t\right). \end{align*} $$ Plugging these values into the formula $x^2 + (y - \frac{1}{2})^2,$ we find that: $$ \begin{align*} \\ x^2 + \left(y - \frac{1}{2}\right)^2 &= \frac{1}{4}\left(\sin^2 2t + \cos^2 2t\right) \\ \\ &= \frac{1}{4} \quad \blacksquare \end{align*} $$ (b) From part (a), we know that we can define $\mathbf{r}$ in rectangular coordinates as $$ \begin{align*} \\ \mathbf{r}(t) &= \frac{1}{2}\sin 2t\,\mathbf{i} + \frac{1}{2}\left(1 - \cos 2t\right)\,\mathbf{j} + \log\sec t\,\mathbf{k}. \end{align*} $$ Its derivative with respect to $t$ is then: $$ \begin{align*} \\ \mathbf{v}(t) &= \cos 2t\,\mathbf{i} + \sin 2t\,\mathbf{j} + \tan t\,\mathbf{k}. \end{align*} $$ To find the angle $\phi$ between $\mathbf{v}$ and $\mathbf{k},$ we first find $\cos \phi$ $$ \begin{align*} \\ \cos\phi &= \frac{\mathbf{v}(t)\cdot\mathbf{k}}{\|\mathbf{v}(t)\|} \\ \\ &= \frac{\tan t}{\sqrt{1 + \tan^2 t}} \\ \\ &= \frac{\tan t}{\sec t} \\ \\ &= \sin t \end{align*} $$ where $0 \leq t < \pi/2.$ Then, we apply the inverse cosine and the identity $\sin t = \cos \left(\frac{\pi}{2} - t\right)$to give us: $$ \begin{align*} \\ \arccos \phi &= \arccos( \sin t) \\ \\ &= \arccos \cos\left(\frac{\pi}{2} - t\right) \\ \\ &= \frac{\pi}{2} - t \quad \blacksquare \end{align*} $$