Finding the Curvature of a Space Curve

Find the curvature \(\kappa\) of the space curve \(\mathbf{r}(t)=t\mathbf{i}+t^{2}\mathbf{j}+t^{3}\mathbf{k}\).

Solution To use formula (4), we need to find \(\mathbf{r}^{\prime} (t),\) \(\left\Vert \mathbf{r}^{\prime} (t) \right\Vert,\) \(\mathbf{r}^{\prime \prime} (t),\) \(\mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t),\) and \(\left\Vert \mathbf{r}^{\prime} (t) \times \mathbf{r}^{\prime \prime} (t) \right\Vert.\) \begin{eqnarray*} \mathbf{r}^{\prime} (t) &=&\mathbf{i}+2t\mathbf{j}+3t^{2}\mathbf{k}\qquad \left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;=\;\sqrt{ 1+4t^{2}+9t^{4}}\\ \mathbf{r}^{\prime \prime} (t) &=&2\mathbf{j}+6t\mathbf{k} \end{eqnarray*} \begin{eqnarray*} \mathbf{r}^{\prime} (t)\times \mathbf{r}^{\prime \prime} (t)&\;=\;&\left\vert \begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] 1 & 2t & 3t^{2}\\ 0 & 2 & 6t \end{array} \right\vert\;=\;6t^{2}\mathbf{i}-6t\mathbf{j}+2\mathbf{k} \\[6pt] \left\Vert \mathbf{r}^{\prime} (t)\times \mathbf{r}^{\prime \prime} (t)\right\Vert &\;=\;&\sqrt{36t^{4}+36t^{2}+4}=2\sqrt{9t^{4}+9t^{2}+1} \end{eqnarray*}

Now we use formula (4). The curvature \(\kappa\) is \begin{equation*} \kappa\;=\;\dfrac{\left\Vert \mathbf{r}^{\prime} (t)\times \mathbf{r}^{\prime \prime} (t)\right\Vert }{\left\Vert \mathbf{r}^{\prime} (t) \right\Vert ^{3}}=\frac{2\sqrt{ 9t^{4}+9t^{2}+1}}{(1+4t^{2}+9t^{4})^{3/2}} \end{equation*}