Find the curvature of the curve \( \vec{r}(t)=\langle 2 \cos (3 t), 2 \sin (3 t), 2 t\rangle \) at the point \( t=0 \) Give your answer to two decimal places

To find the curvature of the curve \(\vec{r}(t)=\langle 2 \cos (3 t), 2 \sin (3 t), 2 t\rangle\) at the point \(t=0\), we can use the formula for the curvature \(\kappa\) given by: \[ \kappa = \frac{\|\vec{r}'(t) \times \vec{r}''(t)\|}{\|\vec{r}'(t)\|^3} \] First, we need to calculate the first and second derivatives of \(\vec{r}(t)\). 1. **First Derivative**: \[ \vec{r}'(t) = \frac{d}{dt}\langle 2 \cos(3t), 2 \sin(3t), 2t \rangle = \langle -6 \sin(3t), 6 \cos(3t), 2 \rangle \] 2. **Second Derivative**: \[ \vec{r}''(t) = \frac{d}{dt}\langle -6 \sin(3t), 6 \cos(3t), 2 \rangle = \langle -18 \cos(3t), -18 \sin(3t), 0 \rangle \] Next, we evaluate these derivatives at \(t=0\): - \(\vec{r}'(0) = \langle -6 \sin(0), 6 \cos(0), 2 \rangle = \langle 0, 6, 2 \rangle\) - \(\vec{r}''(0) = \langle -18 \cos(0), -18 \sin(0), 0 \rangle = \langle -18, 0, 0 \rangle\) Now, we calculate the cross product \(\vec{r}'(0) \times \vec{r}''(0)\): \[ \vec{r}'(0) \times \vec{r}''(0) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 6 & 2 \\ -18 & 0 & 0 \end{vmatrix} = \hat{i}(6 \cdot 0 - 2 \cdot (-18)) - \hat{j}(0 \cdot 0 - 2 \cdot (-18)) + \hat{k}(0 \cdot 0 - 6 \cdot (-18)) \] \[ = \hat{i}(0 + 36) - \hat{j}(0 + 36) + \hat{k}(0 + 108) = \langle 36, -36, 108 \rangle \] Now we calculate the magnitude of the cross product: \[ \|\vec{r}'(0) \times \vec{r}''(0)\| = \sqrt{36^2 + (-36)^2 + 108^2} = \sqrt{1296 + 1296 + 11664} = \sqrt{14256} = 2\sqrt{3564} = 2 \cdot 6\sqrt{99} = 12\sqrt{99} \] Next, we need to compute \(\|\vec{r}'(0)\|\): \[ \|\vec{r}'(0)\| = \sqrt{0^2 + 6^2 + 2^2} = \sqrt{0 + 36 + 4} = \sqrt{40} = 2\sqrt{10} \] Finally, we can substitute these values into the curvature formula: \[ \kappa = \frac{12\sqrt{99}}{(2\sqrt{10})^3} = \frac{12\sqrt{99}}{8\sqrt{1000}} = \frac{12\sqrt{99}}{80\sqrt{10}} = \frac{3\sqrt{99}}{20\sqrt{10}} \] Now approximate this value to two decimal places. We calculate: \[ \sqrt{99} \approx 9.95, \quad \sqrt{10} \approx 3.16 \] Thus: \[ \kappa \approx \frac{3 \cdot 9.95}{20 \cdot 3.16} \approx \frac{29.85}{63.2} \approx 0.472 \] Therefore, the curvature of the curve at the point \(t=