Mathematical Immaturity

14.15 Exercises

• For the curve whose vector equation is $\mathbf{r}(t) = e^t\,\mathbf{i} + e^{-t}\,\mathbf{j} + \sqrt{2}t\,\mathbf{k},$ show that the curvature is $\kappa(t) = \sqrt{2}/(e^t + e^{-t})^2.$
• From Theorem 14.14 we can express curvature in terms of acceleration, velocity, and speed: $$ \begin{align*} \kappa(t) &= \frac{\|\mathbf{a}(t) \times \mathbf{v}(t)\|}{v^3(t)} \end{align*} $$ And we can use Lagrange's identity to express the norm of the cross product as $$ \begin{align*} \\ \|\mathbf{a}(t) \times \mathbf{v}(t)\|^2 &= \|\mathbf{a}(t)\|^2\|\mathbf{v}(t)\|^2 - \left(\mathbf{a}(t)\cdot\mathbf{v}(t)\right)^2 \end{align*} $$
• To find the curvature of the curve described by $\mathbf{r}(t)$ we must find its velocity, acceleration, and speed. $$ \begin{align*} \\ \mathbf{v}(t) &= e^t\,\mathbf{i} - e^{-t}\,\mathbf{j} + \sqrt{2}\,\mathbf{k} \\ \\ \mathbf{a}(t) &= e^t\,\mathbf{i} + e^{-t}\,\mathbf{j} \\ \\ v(t) &= \sqrt{e^{2t} + 2 + e^{-2t}} \\ &= \sqrt{(e^{t} + e^{-t})^2} \\ &= e^t + e^{-t} \end{align*} $$ Then, we can apply Lagrange's identity to find the norm $\|\mathbf{a}(t) \times \mathbf{v}(t)\|$ $$ \begin{align*} \\ \|\mathbf{a}(t) \times \mathbf{v}(t)\|^2 &= \|\mathbf{a}(t)\|^2\|\mathbf{v}(t)\|^2 - \left(\mathbf{a}(t)\cdot\mathbf{v}(t)\right)^2 \\ \\ &= \left(e^{2t} + e^{-2t}\right)\left(e^{2t} + e^{-2t} + 2\right) - \left(e^{2t} - e^{-2t}\right)^2 \\ &= \left(e^{4t} + 2 + e^{-4t} + 2e^{2t} + 2e^{-2t}\right) \\ &- \left(e^{4t} - 2 + e^{-4t}\right) \\ &= 2e^{2t} + 4 + 2e^{-2t} \\ &= 2\left(e^t + e^{-t}\right)^2 \end{align*} $$ Thus, $\|\mathbf{a}(t) \times \mathbf{v}(t)\| = \sqrt{2}\left(e^t + e^{-t}\right).$ Combining this result with our previously calculated speed function, we get: $$ \begin{align*} \kappa(t) &= \frac{\sqrt{2}\left(e^t + e^{-t}\right)}{\left(e^t + e^{-t}\right)^3} \\ \\ &= \frac{\sqrt{2}}{\left(e^t + e^{-t}\right)^2} \quad \blacksquare \end{align*} $$