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 $N.$ $$t = 2$$
• Recall from Section 14.7 #1: \begin{align*} \textbf{r}(t) &= (3t - t^3)\,\mathbf{i} + 3t^3\,\mathbf{j} + (3t + t^3)\,\mathbf{k} \\ \textbf{v}(t) &= (3 - 3t^2)\,\textbf{i} + 6t\,\textbf{j} + (3 + 3t^2)\textbf{k} \\ \textbf{a}(t) &= -6t\,\textbf{i} + 6\,\textbf{j} + 6t\,\textbf{k} \\ v(t) &= 3\sqrt{2}(1 + t^2) \end{align*}
• (a) The unit tangent $T(t)$ is: \begin{align*} T(t) &= \frac{\mathbf{v}(t)}{v(t)} \\ &= \frac{(3-3t^2)\,\mathbf{i} + 6t\,\mathbf{j} + (3+3t^2)\,\mathbf{k}}{3\sqrt{2}(1+t^2)} \\ &= \frac{1}{\sqrt{2}(1+t^2)}\Big[(1-t^2)\,\mathbf{i} + 2t\,\mathbf{j} + (1+t^2)\,\mathbf{k}\Big]. \end{align*} Differentiating the unit tangent with respect to $t$ gives us: $$ \begin{align*} T'(t) &= \frac{d}{dt}\left\{\frac{1}{\sqrt{2}(1+t^2)}\Big[(1-t^2)\,\mathbf{i} + 2t\,\mathbf{j} + (1+t^2)\,\mathbf{k}\Big]\right\} \\ &= \left(-\frac{4t}{\sqrt{2}(1+t^2)^2}\right)\mathbf{i} + \left(\frac{2(1-t^2)}{\sqrt{2}(1+t^2)^2}\right)\mathbf{j} + 0\,\mathbf{k}. \end{align*} $$ Whose norm is: $$ \begin{align*} \|T'(t)\| &= \frac{\sqrt{2}}{1+t^2}. \end{align*} $$ The principal normal $N(t)$ is then: $$ \begin{align*} N(t) &= \frac{T'(t)}{\|T'(t)\|} \\ &= \left(-\frac{2t}{1+t^2}\right)\mathbf{i} + \left(\frac{1-t^2}{1+t^2}\right)\mathbf{j} + 0\,\mathbf{k}. \end{align*} $$ Evaluating these at $t = 2$ we get: $$ \begin{align*} T(2) &= \frac{1}{\sqrt{2}(1+2^2)}\Big[(1-2^2)\,\mathbf{i} + 2(2)\,\mathbf{j} + (1+2^2)\,\mathbf{k}\Big] \\ &= \frac{1}{5\sqrt{2}}\Big[(-3)\,\mathbf{i} + 4\,\mathbf{j} + 5\,\mathbf{k}\Big], \\ \\ N(2) &= \left(-\frac{2(2)}{1+2^2}\right)\mathbf{i} + \left(\frac{1-2^2}{1+2^2}\right)\mathbf{j} + 0\,\mathbf{k} \\ &= -\frac{4}{5}\,\mathbf{i} -\frac{3}{5}\,\mathbf{j} + 0\,\mathbf{k} \quad \blacksquare. \end{align*} $$ (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: $$ v'(t) = \frac{d}{dt}\Big[3\sqrt{2}(1+t^2)\Big] = 6\sqrt{2}\,t. $$ Combining this with the results of (a), we see that acceleration is: $$ \begin{align*} \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\,T'(t) \\ &=6\sqrt{2}\,t\,\left[\frac{1}{\sqrt{2}(1+t^2)}\Big((1-t^2)\,\mathbf{i} + 2t\,\mathbf{j} + (1+t^2)\,\mathbf{k}\Big)\right] \\ &\quad + 3\sqrt{2}(1+t^2)\left[\left(-\frac{4t}{\sqrt{2}(1+t^2)^2}\right)\mathbf{i} + \left(\frac{2(1-t^2)}{\sqrt{2}(1+t^2)^2}\right)\mathbf{j}\right] \\ &= \frac{6t}{(1+t^2)}\Big[(1-t^2)\,\mathbf{i} + 2t\,\mathbf{j} + (1+t^2)\,\mathbf{k}\Big] \\ &\quad + \frac{3}{(1+t^2)}\Big[(-4t)\,\mathbf{i} + 2(1-t^2)\,\mathbf{j}\Big] \\ &= 6\sqrt{2}\,t\,T(t) + 6\,N(t) \\ &= 12\sqrt{2}\,T + 6\,N \quad \blacksquare \end{align*} $$