Find the curvature of the curve $\vec{r}(t)=\langle 4 \cos (3 t), 4 \sin (3 t), 4 t\rangle$ at the point $t=0$ Give your answer to two decimal places

Step 1 ：Given the curve \(\vec{r}(t)=\langle 4 \cos (3 t), 4 \sin (3 t), 4 t\rangle\)

Step 2 ：The first derivative of the curve is \(\vec{r}'(t) = \langle -12\sin(3t), 12\cos(3t), 4 \rangle\)

Step 3 ：The second derivative of the curve is \(\vec{r}''(t) = \langle -36\cos(3t), -36\sin(3t), 0 \rangle\)

Step 4 ：The cross product of \(\vec{r}'(t)\) and \(\vec{r}''(t)\) is \(\vec{r}'(t) \times \vec{r}''(t) = \langle 144\sin(3t), -144\cos(3t), 432 \rangle\)

Step 5 ：The norm of the cross product is \(||\vec{r}'(t) \times \vec{r}''(t)|| = \sqrt{20736\sin(3t)^2 + 20736\cos(3t)^2 + 186624}\)

Step 6 ：The norm of the first derivative is \(||\vec{r}'(t)|| = \sqrt{144\sin(3t)^2 + 144\cos(3t)^2 + 16}\)

Step 7 ：The curvature of the curve is given by \(\kappa(t) = \frac{||\vec{r}'(t) \times \vec{r}''(t)||}{||\vec{r}'(t)||^3} = \frac{\sqrt{20736\sin(3t)^2 + 20736\cos(3t)^2 + 186624}}{(144\sin(3t)^2 + 144\cos(3t)^2 + 16)^{3/2}}\)

Step 8 ：Substituting \(t=0\) into the formula, we get \(\kappa(0) = \frac{9}{40}\)

Step 9 ：Final Answer: The curvature of the curve \(\vec{r}(t)=\langle 4 \cos (3 t), 4 \sin (3 t), 4 t\rangle\) at the point \(t=0\) is \(\boxed{0.225}\)