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

Step 1 ：Given the curve $\overrightarrow{r}(t)=⟨4\cos (3t),4\sin (3t),4t⟩$

Step 2 ：The first derivative of the curve is ${\overrightarrow{r}}^{\prime }(t)=⟨-12\sin (3t),12\cos (3t),4⟩$

Step 3 ：The second derivative of the curve is ${\overrightarrow{r}}^{″}(t)=⟨-36\cos (3t),-36\sin (3t),0⟩$

Step 4 ：The cross product of ${\overrightarrow{r}}^{\prime }(t)$ and ${\overrightarrow{r}}^{″}(t)$ is ${\overrightarrow{r}}^{\prime }(t)\times {\overrightarrow{r}}^{″}(t)=⟨144\sin (3t),-144\cos (3t),432⟩$

Step 5 ：The norm of the cross product is $||{\overrightarrow{r}}^{\prime }(t)\times {\overrightarrow{r}}^{″}(t)||=\sqrt{20736\sin (3t{)}^2+20736\cos (3t{)}^2+186624}$

Step 6 ：The norm of the first derivative is $||{\overrightarrow{r}}^{\prime }(t)||=\sqrt{144\sin (3t{)}^2+144\cos (3t{)}^2+16}$

Step 7 ：The curvature of the curve is given by $\kappa (t)=\frac{||{\overrightarrow{r}}^{\prime }(t)\times {\overrightarrow{r}}^{″}(t)||}{||{\overrightarrow{r}}^{\prime }(t)|{|}^3}=\frac{\sqrt{20736\sin (3t{)}^2+20736\cos (3t{)}^2+186624}}{(144\sin (3t{)}^2+144\cos (3t{)}^2+16{)}^{3/2}}$

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

Step 9 ：Final Answer: The curvature of the curve $\overrightarrow{r}(t)=⟨4\cos (3t),4\sin (3t),4t⟩$ at the point $t=0$ is $\boxed{{\displaystyle 0.225}}$