91Ó°ÊÓ

Find the derivative \( \mathbf{r}'(t) \) of the curve \( \mathbf{r}(t) = (\cosh t) \mathbf{i} - (\sinh t) \mathbf{j} + t \mathbf{k} \). The derivatives are \( \frac{d}{dt}(\cosh t) = \sinh t \), \( \frac{d}{dt}(-\sinh t) = -\cosh t \), and \( \frac{d}{dt}(t) = 1 \). Thus, \( \mathbf{r}'(t) = (\sinh t) \mathbf{i} - (\cosh t) \mathbf{j} + \mathbf{k} \).

The unit tangent vector \( \mathbf{T}(t) \) is given by \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \). First, find \( \|\mathbf{r}'(t)\| \) as follows: \( \|\mathbf{r}'(t)\| = \sqrt{(\sinh t)^2 + (\cosh t)^2 + 1^2} \). Use the identity \( \cosh^2 t - \sinh^2 t = 1 \) to simplify this as \( \|\mathbf{r}'(t)\| = \sqrt{1 + 1} = \sqrt{2} \). Therefore, \( \mathbf{T}(t) = \frac{1}{\sqrt{2}}((\sinh t) \mathbf{i} - (\cosh t) \mathbf{j} + \mathbf{k}) \).

Find the second derivative \( \mathbf{r}''(t) \) of the function to proceed. We have \( \frac{d}{dt}(\sinh t) = \cosh t \) and \( \frac{d}{dt}(-\cosh t) = -\sinh t \). Thus, \( \mathbf{r}''(t) = (\cosh t) \mathbf{i} + (\sinh t) \mathbf{j} \).

Calculate \( \mathbf{T}'(t) = \frac{d}{dt}\left(\frac{1}{\sqrt{2}}(\sinh t \mathbf{i} - \cosh t \mathbf{j} + \mathbf{k})\right) \). After differentiating, we have \( \mathbf{T}'(t) = \frac{1}{\sqrt{2}}(\cosh t \mathbf{i} + \sinh t \mathbf{j}) \).

The unit normal vector \( \mathbf{N}(t) \) is given by \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} \). Note that \( \|\mathbf{T}'(t)\| = \frac{1}{\sqrt{2}}\sqrt{(\cosh t)^2 + (\sinh t)^2} = 1 \). Hence, \( \mathbf{N}(t) = \mathbf{T}'(t) = \frac{1}{\sqrt{2}}(\cosh t \mathbf{i} + \sinh t \mathbf{j}) \).

The curvature \( \kappa(t) \) is given by \( \kappa(t) = \frac{\|\mathbf{r}''(t) \times \mathbf{r}'(t)\|}{\|\mathbf{r}'(t)\|^3} \). Compute the cross product \( \mathbf{r}''(t) \times \mathbf{r}'(t) \). Using \( \mathbf{r}''(t) = \cosh t \mathbf{i} + \sinh t \mathbf{j} \) and \( \mathbf{r}'(t) = \sinh t \mathbf{i} - \cosh t \mathbf{j} + \mathbf{k} \):\[\mathbf{r}''(t) \times \mathbf{r}'(t) = \left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\cosh t & \sinh t & 0 \\sinh t & -\cosh t & 1\end{array}\right| = \left\|\begin{array}{cc}\sinh t & 0 \\-\cosh t & 1\end{array}\right\| \mathbf{i} + \left\|\begin{array}{cc}\cosh t & 0 \\sinh t & 1\end{array}\right\| \mathbf{j} + \left\|\begin{array}{cc}\cosh t & \sinh t \\sinh t & -\cosh t\end{array}\right\| \mathbf{k} \]The resulting calculation leads to \( \mathbf{k} \), thus \( \|\mathbf{r}''(t) \times \mathbf{r}'(t)\| = 1 \). Finally, compute \( \kappa(t) = \frac{1}{(\sqrt{2})^3} = \frac{1}{2\sqrt{2}} \).