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 #2: $$ \begin{align*} \\ \textbf{r}(t) &= \cos t\,\mathbf{i} + \sin t\,\mathbf{j} + e^t\,\mathbf{k} \\ \\ \textbf{v}(t) &= -\text{sin}\,t\,\textbf{i} + \text{cos}\,t\,\textbf{j} + e^t\,\textbf{k} \\ \\ \textbf{a}(t) &= -\text{cos}\,t\,\textbf{i} - \text{sin}\,t\,\textbf{j} + e^t\,\textbf{k} \\ \\ v(t) &= \left(1 + e^{2t}\right)^{1/2} \end{align*} $$
• (a) The unit tangent $T(t)$ is: $$ \begin{align*} \\ T(t) &= \frac{\mathbf{v}(t)}{v(t)} \\ \\ &= \frac{-\sin t\,\mathbf{i} + \cos t\,\mathbf{j} + e^t\,\mathbf{k}}{\sqrt{1 + e^{2t}}} \end{align*} $$ Differentiating the unit tangent with respect to $t$ gives us: $$ \begin{align*} \\ T'(t) &= \frac{\left(1 + e^{2t}\right)\left[-\cos t\,\mathbf{i} -\sin t\,\mathbf{j} + e^t\,\mathbf{k}\right]}{\left(1 + e^{2t}\right)^{3/2}} \\ &- \frac{e^{2t}\left[-\sin t\,\mathbf{i} + \cos t \,\mathbf{j} + e^t\,\mathbf{k}\right]}{\left(1 + e^{2t}\right)^{3/2}} \end{align*} $$ Evaluating at $t = \pi$ $$ \begin{align*} \\ T'(\pi) &= \frac{\left(1 + e^{2\pi}\right)\left[\mathbf{i} + e^{\pi}\,\mathbf{k}\right] - e^{2\pi}\left[-\mathbf{j} + e^{\pi}\,\mathbf{k}\right]}{\left(1 + e^{2\pi}\right)^{3/2}} \\ \\ &= \frac{\left(1 + e^{2\pi}\right)\mathbf{i} + e^{2\pi}\,\mathbf{j} + e^{\pi}\,\mathbf{k}}{\left(1 + e^{2\pi}\right)^{3/2}} \end{align*} $$ Its norm at $t = \pi$ is then: $$ \begin{align*} \\ \left\|T'(\pi)\right\| &= \frac{\left[\left(1 + e^{2\pi}\right)^2 + \left(e^{2\pi}\right)^2 + e^{2\pi}\right]^{1/2}}{\left(1 + e^{2\pi}\right)^{3/2}} \\ \\ &= \frac{\left[\left(1 + e^{2\pi}\right)^2 + e^{2\pi}\left(1 + e^{2\pi}\right)\right]^{1/2}}{\left(1 + e^{2\pi}\right)^{3/2}} \\ &= \frac{\left[\left(1 + e^{2\pi}\right)\left(1 + 2e^{2\pi}\right)\right]^{1/2}}{\left(1 + e^{2\pi}\right)^{3/2}} \end{align*} $$ The principal normal at $t = \pi$ is then: $$ \begin{align*} \\ N(\pi) &= \frac{T'(\pi)}{\|T'(\pi)\|} \\ \\ &= \frac{\left(1 + e^{2\pi}\right)\mathbf{i} + e^{2\pi}\,\mathbf{j} + e^{\pi}\,\mathbf{k}}{\left[\left(1 + e^{2\pi}\right)\left(1 + 2e^{2\pi}\right)\right]^{1/2}} \\ \\ \end{align*} $$ Thus, at $t = \pi$ we have: $$ \begin{align*} \\ T(\pi) = \frac{-\mathbf{j} + e^{\pi}\,\mathbf{k}}{\sqrt{1 + e^{2\pi}}}; \quad N(\pi) = \frac{\left(1 + e^{2\pi}\right)\mathbf{i} + e^{2\pi}\,\mathbf{j} + e^{\pi}\,\mathbf{k}}{\left[\left(1 + e^{2\pi}\right)\left(1 + 2e^{2\pi}\right)\right]^{1/2}} \quad \blacksquare \\ \\ \end{align*} $$ (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: $$ \begin{align*} \\ v'(t) = e^{2t}\left(1 + e^{2t}\right)^{-1/2} \end{align*} $$ Evaluating $v(t)$ and $v'(t)$ at $t = \pi$ $$ \begin{align*} \\ v(\pi) = \sqrt{1 + e^{2\pi}}; \quad v'(\pi) = e^{2\pi}\left(1 + e^{2\pi}\right)^{-1/2} \end{align*} $$ Combining this with the results of (a), we see that acceleration is: $$ \begin{align*} \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\,T'(t) \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\|T'(t)\|\,N(t) \\ \\ &= \left(1 + e^{2\pi}\right)^{-1/2}\left[e^{2\pi}\,T + \left(1 + 2e^{2\pi}\right)^{1/2}\,N\right]\quad \blacksquare \\ \\ \end{align*} $$