Curvature at a Point on a Curve Examples 2

# Curvature at a Point on a Curve Examples 2

Recall that if $\vec{r}(t) = (x(t), y(t), z(t))$ is a vector-valued function that traces the smooth curve $C$, then the curvature of $C$ at the point $\vec{r}(t)$ can be given by either of the formulas:

• $\kappa (s) = \biggr \| \frac{d\hat{T}(s)}{ds} \biggr \| = \biggr \| \frac{d\hat{T}(s) / dt}{ds / dt} \biggr \|$ if we find an arc length parameterization of $\vec{r}(t)$ as $\vec{r}(s)$.
• $\kappa (t) = \frac{\| \hat{T'}(t) \|}{\| \vec{r'}(t) \|}$ if $\hat{T'}(t)$ is not too difficult to calculate.
• $\kappa (t) = \kappa (t) = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3}$.

## Example 1

Find the curvature of $\vec{r}(t) = (1 - t^3, t^3 + 2, -2t^3)$.

We will use the third formula to calculate $\kappa (t)$.

First we will calculate the first and second derivatives. Firstly, $\vec{r'}(t) = (-3t^2, 3t^2, -6t^2)$, and secondly, $\vec{r''}(t) = (-6t, 6t, -12t)$. We now need to take the cross product $\vec{r'}(t) \times \vec{r''}(t)$ as follows:

(1)
\begin{align} \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ -3t^2 & 3t^2 & -6t^2\\ -6t & 6t & -12t \end{vmatrix} = \begin{vmatrix} 3t^2& -6t^2\\ 6t& -12t \end{vmatrix} \vec{i} - \begin{vmatrix} -3t^2& -6t^2\\ -6t & -12t\end{vmatrix} \vec{j} + \begin{vmatrix} -3t^2& 3t^2\\ -6t & 6t \end{vmatrix} \vec{k} \\ = (-36t^3 + 36t^3) \vec{i} - (36t^3 - 36t^3) \vec{j} + (-18t^3 - 18t^3) \vec{j} = (0, 0, 0) \end{align}

Since we have arrived at the zero-vector, we know that $\| \vec{r'}(t) \times \vec{r''}(t) \| = 0$ and so $\kappa (t) = 0$. This implies that $\vec{r}(t)$ represents a line in $\mathbb{R}^3$.

## Example 2

Find the curvature of the elliptical helix $\vec{r}(t) = (2 \cos t, 3 \sin t, t^2)$. Use this formula to find the curvature at $t = \pi$

First we will calculate the first and second derivatives. Firstly, $\vec{r'}(t) = (-2 \sin t, 3 \cos t, 2t)$, and secondly, $\vec{r''}(t) = (-2 \cos t, -3 \sin t, 2)$. Now we need to take the cross product $\vec{r'}(t) \times \vec{r''}(t)$ as follows:

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

Now we need to calculate the magnitude of $\vec{r'}(t) \times \vec{r''}(t)$, that is $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{(6 \cos t + 6t \sin t)^2 + (4 \sin t - 4t \cos t)^2 + (6)^2} = \sqrt{(6 \cos t + 6t \sin t)^2 + (4 \sin t - 4t \cos t)^2 + 36}$.

Now we need to calculate $\| \vec{r'}(t) \| = \sqrt{(-2\sin t)^2 + (3 \cos t)^2 + (2t)^2} = \sqrt{4 \sin ^2 t + 9 \cos ^2 t + 4t^2} = \sqrt{5 \cos^2 t + 4t^2 + 4}$. Therefore $\| \vec{r'}(t) \|^3 = \left ( \sqrt{5 \cos^2 t + 4t^2 + 4} \right )^3$. Putting this all together and we have that:

(3)
\begin{align} \kappa (t) = \frac{\sqrt{(6 \cos t + 6t \sin t)^2 + (4 \sin t - 4t \cos t)^2 + 36}}{\left ( \sqrt{5 \cos^2 t + 4t^2 + 4} \right )^3} \end{align}

So for $t = \pi$ we have that $\kappa(\pi) = \frac{\sqrt{16\pi^2 +72}}{\left ( \sqrt{ 4 \pi ^2 + 9} \right)^3}$.