Find the torsion of the space curve.
-r(t) = (cosh t)i + tj + (sinh t)k
A) $\tau = - \frac { \tanh ^ { 2 } t } { 2 }$
B) $\tau = - \frac { \operatorname { sech } ^ { 2 } t } { 2 }$
C) $\tau = \frac { \tanh ^ { 2 } t } { 2 }$
D) $\tau = 0$

Question

Find the torsion of the space curve.
-r(t) = (cosh t)i + tj + (sinh t)k 
A) $\tau = - \frac { \tanh ^ { 2 } t } { 2 }$
B) $\tau = - \frac { \operatorname { sech } ^ { 2 } t } { 2 }$
C) $\tau = \frac { \tanh ^ { 2 } t } { 2 }$
D) $\tau = 0$

Quizplus · Accepted Answer

The torsion $\tau$ of a space curve can be found using the formula $\tau = -\frac{\mathbf{r}' \times \mathbf{r}'' \cdot \mathbf{r}'''}{|\mathbf{r}' \times \mathbf{r}''|^2}$, where $\mathbf{r}(t)$ is the position vector of the curve, and primes denote derivatives with respect to $t$. For the given curve $\mathbf{r}(t) = (\cosh t)\mathbf{i} + t\mathbf{j} + (\sinh t)\mathbf{k}$, the derivatives are $\mathbf{r}'(t) = (\sinh t)\mathbf{i} + \mathbf{j} + (\cosh t)\mathbf{k}$, $\mathbf{r}''(t) = (\cosh t)\mathbf{i} + (\sinh t)\mathbf{k}$, and $\mathbf{r}'''(t) = (\sinh t)\mathbf{i} + (\cosh t)\mathbf{k}$. Calculating the cross product $\mathbf{r}' \times \mathbf{r}''$ and then the dot product with $\mathbf{r}'''$, and finally the magnitude squared of $\mathbf{r}' \times \mathbf{r}''$, and substituting into the torsion formula, we find that the torsion simplifies to $\tau = -\frac{\operatorname{sech}^2 t}{2}$, which corresponds to choice B. This calculation involves using the identities for hyperbolic functions and their derivatives.

Find the Torsion of the Space Curve τ=−tanh⁡2t2\tau = - \frac { \tanh ^ { 2 } t } { 2 }τ=−2tanh2t​

Find the Torsion of the Space Curve $\tau = - \frac { \tanh ^ { 2 } t } { 2 }$