Find the length of the curve $\vec{r}(t)=\langle 5 \cos (4 t), 5 \sin (4 t), 5 t\rangle$ for $-3 \leq t \leq 6$ Give your answer to two decimal places

Step 1 ：First, we need to find the derivative of the vector function \(\vec{r}(t)\). The derivative of \(\vec{r}(t) = \langle 5 \cos (4 t), 5 \sin (4 t), 5 t \rangle\) is \(\vec{r}'(t) = \langle -20 \sin (4 t), 20 \cos (4 t), 5 \rangle\).

Step 2 ：Next, we find the magnitude of the derivative \(\vec{r}'(t)\). The magnitude of \(\vec{r}'(t)\) is \(\sqrt{400 \sin^2 (4 t) + 400 \cos^2 (4 t) + 25}\).

Step 3 ：Finally, we integrate this magnitude from -3 to 6 to find the length of the curve. The length of the curve is \(\int_{-3}^{6} \sqrt{400 \sin^2 (4 t) + 400 \cos^2 (4 t) + 25} dt\).

Step 4 ：Calculating the integral, we find that the length of the curve is approximately 185.54.