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 $N.$ $$ \begin{align*} \\ t = \frac{\pi}{4} \end{align*} $$
• Recall from Section 14.7 #6: $$ \begin{align*} \\ \textbf{r}(t) &= t\mathbf{i} + \sin t\mathbf{j} + (1 - \cos t)\,\mathbf{k} \\ \\ \mathbf{v}(t) &= \mathbf{i} + \cos t\,\mathbf{j} + \sin t \,\mathbf{k} \\ \\ \mathbf{a}(t) &= -\sin t\,\mathbf{j} + \cos t\,\mathbf{k} \\ \\ v(t) &= \sqrt{2} \end{align*} $$
• (a) The unit tangent $T(t)$ is: $$ \begin{align*} \\ T(t) &= \frac{\mathbf{v}(t)}{v(t)} \\ \\ &= \frac{\mathbf{i} + \cos t\,\mathbf{j} + \sin t \,\mathbf{k}} {\sqrt{2}} \end{align*} $$ Evaluating at $t = \pi/4$ $$ \begin{align*} \\ T\left(\frac{\pi}{4}\right) &= \frac{\sqrt{2}}{2}\,\mathbf{i} + \frac{1}{2}\,\mathbf{j} + \frac{1}{2}\,\mathbf{k} \end{align*} $$ Differentiating the unit tangent with respect to $t$ and evaluating at $t = \pi/4,$ noting that $v'(t) = 0$ for all $t$ $$ \begin{align*} \\ T'(t) &= \frac{\mathbf{a}(t)}{v(t)} - \frac{v'(t)}{v^2(t)}\,\mathbf{v}(t) \\ \\ &= \frac{\sqrt{2}}{2} \left(-\sin t\,\mathbf{j} + \cos t\,\mathbf{k}\right) \\ \\ &= -\frac{1}{2}\,\mathbf{j} + \frac{1}{2}\,\mathbf{k} \end{align*} $$ Its norm at $t = \pi/4$ is then: $$ \begin{align*} \\ \left\|T'\left(\frac{\pi}{4}\right)\right\| &= \frac{\sqrt{2}}{2} \end{align*} $$ The principal normal at $t = \pi/4$ is then: $$ \begin{align*} \\ N\left(\frac{\pi}{4}\right) &= \frac{T'\left(\frac{\pi}{4}\right)}{\|T'\left(\frac{\pi}{4}\right)\|} \\ \\ &= \frac{\sqrt{2}}{2}\left(-\mathbf{j} + \,\mathbf{k}\right) \end{align*} $$ Thus, at $t = \pi/4$ we have: $$ \begin{align*} \\ T\left(\frac{\pi}{4}\right) = \frac{1}{2}\sqrt{2}\,\mathbf{i} + \frac{1}{2}\,\mathbf{j} + \frac{1}{2}\,\mathbf{k}; \quad N\left(\frac{\pi}{4}\right) = -\frac{1}{2}\sqrt{2}\,\mathbf{j} + \frac{1}{2}\sqrt{2}\,\mathbf{k} \quad \blacksquare \\ \\ \end{align*} $$ (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 $v(t)$ $v'(t)$ and $\|T(t)\|$ at $t = \pi/4$ $$ \begin{align*} \\ v\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad v'\left(\frac{\pi}{4}\right) = 0, \quad \left\|T\left(\frac{\pi}{4}\right)\right\| = \frac{1}{2}\sqrt{2} \end{align*} $$ Combining this with the results of (a), we see that acceleration is: $$ \begin{align*} \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\,T'(t) \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\|T'(t)\|\,N(t) \\ \\ &= N \quad \blacksquare \\ \\ \end{align*} $$