Curvature at a Point on a Curve Examples 1

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 $\vec{r}(t) = (3t, 4 \sin t, 4 \cos t)$. Use this formula to find the curvature at $t = \pi$

We will use the third formula once again. First let's calculate the first and second derivatives of $\vec{r}(t)$. First, $\vec{r'}(t) = (3, 4 \cos t, -4 \sin t)$. Secondly, $\vec{r''}(t) = (0, - 4\sin t, -4 \cos t)$.

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

(1)

\begin{align} \quad \quad \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 3 & 4 \cos t & -4 \sin t\\ 0 & -4 \sin t & -4 \cos t \end{vmatrix} = \begin{vmatrix}4 \cos t & -4 \sin t\\ -4 \sin t & -4 \cos t \end{vmatrix} \vec{i} - \begin{vmatrix}3 & -4 \sin t\\ 0& -4 \cos t \end{vmatrix} \vec{j} + \begin{vmatrix} 3 & 4 \cos t\\ 0 & -4 \sin t \end{vmatrix} \vec{j} \\ = (-16 \cos ^2 t - 16 \sin^2 t, 12 \cos t, -12 \sin t) = (-16, 12 \cos t, -12 \sin t) \end{align}

The magnitude of $\vec{r'}(t) \times \vec{r''}(t)$ is $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{256 + 144 \cos ^2 t + 144 \sin ^2} = \sqrt{400} = 20$. Furthermore, the magnitude of $\vec{r'}(t)$ is $\| \vec{r'}(t) \| = \sqrt{9 + 16 \cos ^2 t + 16 \sin ^2 t} = \sqrt{25} = 5$. Therefore $\| \vec{r'}(t) \|^3 = 5^3 = 125$.

Putting this all together and we get that:

(2)

\begin{align} \kappa (t) = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3} = \frac{20}{125} = \frac{4}{25} \end{align}

We notice that the curvature of $\vec{r}(t)$ is constant, and so at $t = \pi$, $\kappa (\pi) = \frac{4}{25}$.

Example 2

Find the curvature of $\vec{r}(t) = t \vec{i} + t^2 \vec{j} + t^3 \vec{k}$. Use this formula to find the curvature at $t = 0$

We will use the third formula to calculate the curvature. We need to find the first and second derivatives of $\vec{r}(t)$. First, $\vec{r'}(t) = (1, 2t, 3t^2)$. Secondly, $\vec{r''}(t) = (0, 2, 6t)$.

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

(3)

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

The magnitude of $\vec{r'}(t) \times \vec{r''}(t)$ is $\| \vec{r'}(t) \times \vec{r''}(t) \| = \sqrt{(6t^2)^2 + (-6t)^2 + (2)^2} = \sqrt{36t^4 + 36t^2 + 4}$. Furthermore, the magnitude of $\vec{r'}(t)$ is $\| \vec{r'}(t) \| = \sqrt{(1)^2 + (2t)^2 + (3t^2)^2} = \sqrt{9t^4 + 4t^2 + 1}$ and so $\| \vec{r'}(t) \|^3 = \left ( \sqrt{9t^4 + 4t^2 + 1} \right)^3$.

Putting this all together and we get that:

(4)

\begin{align} \kappa (t) = \frac{\| \vec{r'}(t) \times \vec{r''}(t) \|}{\| \vec{r'}(t) \|^3} = \frac{\sqrt{36t^4 + 36t^2 + 4}}{\left ( \sqrt{9t^4 + 4t^2 + 1} \right)^3} \end{align}

Now plugging in $t = 0$ and we get that $\kappa (0) = \frac{\sqrt{4}}{\left ( \sqrt{1} \right )^3} = 2$.