Torsion at a Point on a Curve Examples 1

Recall from the Torsion at a Point on a Curve page that if $\vec{r}(t) = (x(t), y(t), z(t))$ is a vector-valued function, then the torsion can be calculated with either of the following formulas (noting that $(\vec{r'}(t) \times \vec{r''}(t)) \cdot \vec{r'''}(t)$ is the scalar triple product of $\vec{r'}(t)$, $\vec{r''}(t)$, and $\vec{r'''}(t)$):

• Torsion: $\tau (t) = \frac{(\vec{r'}(t) \times \vec{r''}(t)) \cdot \vec{r'''}(t)}{\| \vec{r'}(t) \times \vec{r''}(t) \|^2}$ or $\tau (t) = \frac{\begin{vmatrix} x'(t) & y'(t) & z'(t)\\ x''(t) & y''(t) & z''(t)\\ x'''(t) & y'''(t) & z'''(t) \end{vmatrix}}{\| \vec{r'}(t) \times \vec{r''}(t) \|^2}$

We will now look at some examples of computing torsion.

Example 1

Find the torsion of the vector-valued function $\vec{r}(t) = t^2 \vec{i} + t^3 \vec{j} + t^4 \vec{k}$. Use this formula to find the torsion at $t = 1$.

We need to differentiate $\vec{r'}(t)$ three times. Thus, $\vec{r'}(t) = (2t, 3t^2, 4t^3)$, $\vec{r''}(t) = (2, 6t, 12t^2)$, and $\vec{r'''}(t) = (0, 6, 24t)$.

Now we will compute $\vec{r'}(t) \times \vec{r''}(t)$ as follows:

(1)

\begin{align} \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2t & 3t^2 & 4t^3\\ 2 & 6t & 12t^2 \end{vmatrix} = (36t^4 - 24t^4) \vec{i} - (24t^3 - 8t^3) \vec{j} + (12t^2 - 6t^2)\vec{k} = (12t^4, -16t^3, 6t^2) \end{align}

Now we will compute the dot product $(\vec{r'}(t) \times \vec{r''}(t)) \cdot \vec{r'''}(t)$ as follows:

(2)

\begin{align} \quad \quad (\vec{r'}(t) \times \vec{r''}(t)) \cdot \vec{r'''}(t) = (12t^4, -16t^3, 6t^2) \cdot (0, 6, 24t) = -96t^3 + 144t^3 = 48t^3 \end{align}

Now we need to length/magnitude of $\vec{r'}(t) \times \vec{r''}(t)$ which is $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{(12t^4)^2 + (-16t^3)^2 + (6t^2)^2} = \sqrt{144t^8 +256t^6 + 36t^4}$. Therefore, $\| \vec{r'}(t) \times \vec{r''} \|^2 = 144t^8 + 256t^6 + 36t^4$.

Putting this all together and we have that:

(3)

\begin{align} \tau(t) = \frac{48t^3}{144t^8 + 256t^6 + 36t^4} \end{align}

Plugging in $t = 1$ and we have that $\tau (1) = \frac{48}{\sqrt{436}}$.

Example 2

Find the torsion of the vector-valued function $\vec{r}(t) = (\sin t, e^t + t, 2t^2)$. Use this formula to find the torsion at $t = 0$.

In this example, we will use the second variant of the torsion formula. First we must compute the first three derivatives of $\vec{r}(t)$. We have that $\vec{r'}(t) = (\cos t, e^t + 1, 4t)$, $\vec{r''}(t) = (-\sin t, e^t, 4)$ and $\vec{r'''}(t) = (-\cos t, e^t, 0)$. Now we will compute $( \vec{r'}(t) \times \vec{r''}(t) ) \cdot \vec{r'''}(t)$ as follows:

(4)

\begin{align} \begin{vmatrix}\cos t & e^t + 1 & 4t \\ -\sin t & e^t & 4 \\ -\cos t & e^t & 0 \end{vmatrix} = 4t \begin{vmatrix} -\sin t & e^t\\ -\cos t & e^t \end{vmatrix} - 4 \begin{vmatrix} \cos t & e^t + 1\\ -\cos t & e^t \end{vmatrix} \\ = 4t(-e^t \sin t + e^t \cos t) - 4(e^t \cos t + (e^t + 1) \cos t) \\ = 4t(-e^t \sin t + e^t \cos t) - 4(2e^t \cos t + \cos t) \\ = -4te^t \sin t + 4te^t \cos t - 8e^t \cos t - 4 \cos t \\ = -4te^t \sin t + (4t - 8)e^t \cos t - 4 \cos t \end{align}

Now we need to compute $\vec{r'}(t) \times \vec{r''}(t)$ as follows:

(5)

\begin{align} \quad \quad \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\\cos t & e^t + 1 & 4t \\ -\sin t & e^t & 4 \end{vmatrix} \\ = (4e^t + 4 - 4te^t) \vec{i} - (4 \cos t + 4t \sin t) \vec{j} + (e^t \cos t + e^t\sin t + \sin t) \vec{k} \\ \end{align}

Therefore the length/magnitude, $\| \vec{r'}(t) \times \vec{r''}(t) \|$ is:

(6)

\begin{align} \quad \quad \| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{(4e^t + 4 - 4te^t)^2 + (-4 \cos t - 4t \sin t)^2 + (e^t \cos t + e^t\sin t + \sin t)^2 } \end{align}

Thus $\| \vec{r'}(t) \times \vec{r''}(t) \|^2$ is:

(7)

\begin{align} \quad \quad \| \vec{r'}(t) \times \vec{r''}(t) \|^2 =\biggr \rvert (4e^t + 4 - 4te^t)^2 + (-4 \cos t - 4t \sin t)^2 + (e^t \cos t + e^t\sin t + \sin t)^2 \biggr \rvert \end{align}

Putting this all together now, we have that:

(8)

\begin{align} \tau (t) = \frac{-4te^t \sin t + (4t - 8)e^t \cos t - 4 \cos t}{\biggr \rvert (4e^t + 4 - 4te^t)^2 + (-4 \cos t - 4t \sin t)^2 + (e^t \cos t + e^t\sin t + \sin t)^2 \biggr \rvert} \end{align}

If we plug in $t = 0$ we get that $\tau (0) = \frac{-12}{81}$.