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 $N.$ $$t = 1$$
• Recall from Section 14.7 #5 $$ \begin{align*} \\ \textbf{r}(t) &= 3t^2\mathbf{i} + 2t^3\mathbf{j} + 3t\mathbf{k} \\ \\ \mathbf{v}(t) &= 6t\,\mathbf{i} + 6t^2\,\mathbf{j} +3\,\mathbf{k} \\ \\ \mathbf{a}(t) &= 6\,\mathbf{i} + 12t\,\mathbf{j} \\ \\ v(t) &= 3 + 6t^2 \end{align*} $$
• (a) The unit tangent $T(t)$ is: $$ \begin{align*} \\ T(t) &= \frac{\mathbf{v}(t)}{v(t)} \\ \\ &= \frac{6t\,\mathbf{i} + 6t^2\,\mathbf{j} +3\,\mathbf{k}} {6t^2 + 3} \end{align*} $$ Evaluating at $t = 1$ $$ \begin{align*} \\ T(1) &= \frac{1}{3}\left(2\,\mathbf{i} + 2\,\mathbf{j} + \,\mathbf{k}\right) \end{align*} $$ Differentiating the unit tangent with respect to $t$ and evaluating at $t = 1$ $$ \begin{align*} \\ T'(t) &= \frac{\mathbf{a}(t)}{v(t)} - \frac{v'(t)}{v^2(t)}\,\mathbf{v}(t) \\ \\ &= \frac{6}{9}\,\mathbf{i} + \frac{12}{9}\,\mathbf{j} - \frac{4}{9}\left(2\,\mathbf{i} + 2\,\mathbf{j} + \,\mathbf{k}\right) \\ \\ &= \frac{2}{9}\,\mathbf{i} + \frac{4}{9}\,\mathbf{j} - \frac{4}{9}\,\mathbf{k} \end{align*} $$ Its norm at $t = 1$ is then: $$ \begin{align*} \\ \left\|T'(1)\right\| &= \frac{2}{3} \end{align*} $$ The principal normal at $t = 1$ is then: $$ \begin{align*} \\ N(1) &= \frac{T'(1)}{\|T'(1)\|} \\ \\ &= \frac{1}{3}\left(-\mathbf{i} + 2\,\mathbf{j} - 2\,\mathbf{k}\right) \end{align*} $$ Thus, at $t = 1$ we have: $$ \begin{align*} \\ T(1) = \frac{1}{3}\left(2\,\mathbf{i} + 2\,\mathbf{j} + \,\mathbf{k}\right); \quad N(1) = \frac{1}{3}\left(-\mathbf{i} + 2\,\mathbf{j} - 2\,\mathbf{k}\right) \quad \blacksquare \\ \\ \end{align*} $$ Note: The back-of-book solution contains a typo for $N,$ missing the minus-sign in front of $\mathbf{i}.$ (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: $$ \begin{align*} \\ v'(t) = 12t \end{align*} $$ Evaluating $v(t)$ $v'(t)$ and $\|T(t)\|$ at $t = 1$ $$ \begin{align*} \\ v(1) = 9, \quad v'(1) = 12, \quad \|T(1)\| = \frac{2}{3} \end{align*} $$ Combining this with the results of (a), we see that acceleration is: $$ \begin{align*} \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\,T'(t) \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\|T'(t)\|\,N(t) \\ \\ &= 12\,T + 6\,N \quad \blacksquare \\ \\ \end{align*} $$