Mathematical Immaturity

14.9 Exercises

• Exercises 1 through 6 below refer to the motions described in Exercises 1 through 6, respectively, of Section 14.7. For the value of $t$ specified: (a) express the unit tangent $T$ and the principal normal $N$ in terms of $\mathbf{i}, \mathbf{j}, \mathbf{k}$ (b) express the acceleration $\mathbf{a}$ as a linear combination of $T$ and $N.$ $$t = \pi$$
• Recall from Section 14.7 #4: $$ \begin{align*} \\ \textbf{r}(t) &= (t - \sin t)\mathbf{i} + (1 - \cos t)\mathbf{j} + 4\sin \frac{t}{2}\mathbf{k} \\ \\ \textbf{v}(t) &= (1 - \cos t)\,\mathbf{i} + \sin t\,\mathbf{j} + 2\cos \frac{t}{2}\,\mathbf{k} \\ \\ \textbf{a}(t) &= \sin t\,\mathbf{i} + \cos t\,\mathbf{j} - \sin \frac{t}{2}\,\mathbf{k} \\ \\ v(t) &= \sqrt{1 - 2 \cos t + \cos^2 t + \sin^2 t + 4 \cos^2 \frac{t}{2}} \\ \\ &= 2 \end{align*} $$
• (a) The unit tangent $T(t)$ is: $$ \begin{align*} \\ T(t) &= \frac{\mathbf{v}(t)}{v(t)} \\ \\ &= \frac{(1 - \cos t)\,\mathbf{i} + \sin t\,\mathbf{j} + 2\cos \frac{t}{2}\,\mathbf{k}} {2} \end{align*} $$ Evaluating at $t = \pi$ $$ \begin{align*} \\ T(\pi) &= \mathbf{i} \end{align*} $$ Differentiating the unit tangent with respect to $t,$ noting that $v'(t) = 0$ for all $t$ $$ \begin{align*} \\ T'(t) &= \frac{\mathbf{a}(t)}{v(t)} - \frac{v'(t)}{v^2(t)}\,\mathbf{v}(t) \\ \\ &= \frac{\sin t\,\mathbf{i} + \cos t\,\mathbf{j} - \sin \frac{t}{2}\,\mathbf{k}} {2} \end{align*} $$ Evaluating at $t = \pi$ $$ \begin{align*} \\ T'(\pi) &= \frac{-\mathbf{j} - \mathbf{k}} {2} \end{align*} $$ Its norm at $t = \pi$ is then: $$ \begin{align*} \\ \left\|T'(\pi)\right\| &= \frac{\sqrt{2}}{2} \end{align*} $$ The principal normal at $t = \pi$ is then: $$ \begin{align*} \\ N(\pi) &= \frac{T'(\pi)}{\|T'(\pi)\|} \\ \\ &= -\frac{1}{2}\sqrt{2}\left(\,\mathbf{j} + \mathbf{k}\right) \end{align*} $$ Thus, at $t = \pi$ we have: $$ \begin{align*} \\ T(\pi) = \mathbf{i}; \quad N(\pi) = -\frac{1}{2}\sqrt{2}\left(\,\mathbf{j} + \mathbf{k}\right) \quad \blacksquare \\ \\ \end{align*} $$ (b) Acceleration can be written component-wise as: $$ \begin{align*} \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\,T'(t) \end{align*} $$ Differentiating speed $v(t)$ gives us: $$ \begin{align*} \\ v'(t) = 0 \end{align*} $$ Evaluating $v(t)$ and $v'(t)$ at $t = \pi$ $$ \begin{align*} \\ v(\pi) = 2, \quad v'(\pi) = 0 \end{align*} $$ Combining this with the results of (a), we see that acceleration is: $$ \begin{align*} \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\,T'(t) \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\|T'(t)\|\,N(t) \\ \\ &= \sqrt{2}\,N \quad \blacksquare \\ \\ \end{align*} $$