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 $N.$ $$ \begin{align*} \\ t = 0 \end{align*} $$
• Recall from Section 14.7 #3: $$ \begin{align*} \\ \textbf{r}(t) &= 3t\cos t\mathbf{i} + 3t\sin t\mathbf{j} + 4t\,\mathbf{k} \\ \\ \mathbf{v}(t) &= (3\,\text{cos}\,t - 3t\,\text{sin}\,t)\, \mathbf{i} + (3\,\text{sin}\,t + 3t\,\text{cos}\,t)\,\mathbf{j} + 4\,\textbf{k} \\ \\ \mathbf{a}(t) &= (-6\,\text{sin}\,t - 3t\,\text{cos}\,t)\,\mathbf{i} + (6\,\text{cos}\,t - 3t\,\text{sin}\,t)\,\mathbf{j} \\ \\ v(t) &= \sqrt{9t^2 + 25} \end{align*} $$
• (a) The unit tangent $T(t)$ is: $$ \begin{align*} \\ T(t) &= \frac{\mathbf{v}(t)}{v(t)} \\ \\ &= \frac{3(\cos t - t\sin t)\,\mathbf{i} + 3(\sin t + t\cos t)\,\mathbf{j} + 4\,\mathbf{k}}{\sqrt{9t^2 + 25}} \end{align*} $$ Evaluating at $t = 0$ $$ \begin{align*} \\ T(0) &= \frac{3\,\mathbf{i} + 4\,\mathbf{k}}{5} \end{align*} $$ Differentiating the unit tangent with respect to $t$ gives us: $$ \begin{align*} \\ T'(t) &= \frac{3(-2\sin t - t\cos t)\,\mathbf{i} + 3(2\cos t - t\sin t)\,\mathbf{j}}{\sqrt{9t^2 + 25}} \\ &-9t(9t^2 + 25)^{-3/2}\,\mathbf{v}(t) \end{align*} $$ Evaluating at $t = 0$ $$ \begin{align*} \\ T'(0) &= \frac{6}{5}\,\mathbf{j} \end{align*} $$ Its norm at $t = 0$ is then: $$ \begin{align*} \\ \left\|T'(0)\right\| &= \frac{6}{5} \end{align*} $$ The principal normal at $t = 0$ is then: $$ \begin{align*} \\ N(0) &= \frac{T'(0)}{\|T'(0)\|} \\ &= \mathbf{j} \end{align*} $$ Thus, at $t = 0$ we have: $$ \begin{align*} \\ T(0) = \frac{3}{5}\,\mathbf{i} + \frac{4}{5}\,\mathbf{k}; \quad N(0) = \mathbf{j} \quad \blacksquare \\ \\ \end{align*} $$ (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: $$ \begin{align*} \\ v'(t) = 9t(9t^2 + 25)^{-1/2} \end{align*} $$ Evaluating $v(t)$ and $v'(t)$ at $t = 0$ $$ \begin{align*} \\ v(0) = 5, \quad v'(0) = 0 \end{align*} $$ Combining this with the results of (a), we see that acceleration is: $$ \begin{align*} \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\,T'(t) \\ \mathbf{a}(t) &= v'(t)\,T(t) + v(t)\|T'(t)\|\,N(t) \\ \\ &= 6\,N \\ \\ \end{align*} $$