To solve the problem of finding the curvature $\kappa(t)$ of the given vector-valued function $\mathbf{r}(t) = \langle 4\cos(3t), 4\sin(3t), 5t \rangle$ at $t = 0$, we follow these steps:

Formula for Curvature

The curvature $\kappa(t)$ of a vector-valued function $\mathbf{r}(t)$ is given by: $\kappa(t) = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}.$

Step 1: Compute $\mathbf{r}'(t)$

The derivative of $\mathbf{r}(t)$ is: $\mathbf{r}'(t) = \frac{d}{dt}\langle 4\cos(3t), 4\sin(3t), 5t \rangle = \langle -12\sin(3t), 12\cos(3t), 5 \rangle.$

At $t = 0$, we have: $\mathbf{r}'(0) = \langle -12\sin(0), 12\cos(0), 5 \rangle = \langle 0, 12, 5 \rangle.$

Step 2: Compute $\mathbf{r}''(t)$

The second derivative of $\mathbf{r}(t)$ is: $\mathbf{r}''(t) = \frac{d}{dt}\langle -12\sin(3t), 12\cos(3t), 5 \rangle = \langle -36\cos(3t), -36\sin(3t), 0 \rangle.$

At $t = 0$, we have: $\mathbf{r}''(0) = \langle -36\cos(0), -36\sin(0), 0 \rangle = \langle -36, 0, 0 \rangle.$

Step 3: Compute $\mathbf{r}'(t) \times \mathbf{r}''(t)$

The cross product $\mathbf{r}'(t) \times \mathbf{r}''(t)$ is calculated as: