(1 point) Find the curvature $\kappa(t)$ of the curve $\mathbf{r}(t)=(3 \sin t) \mathbf{i}+(3 \sin t) \mathbf{j}+(5 \cos t) \mathbf{k}$

Step 1 ：First, we need to find the first derivative of the vector function \(\mathbf{r}(t)\).

Step 2 ：\(\mathbf{r}'(t) = (3 \cos t) \mathbf{i} + (3 \cos t) \mathbf{j} - (5 \sin t) \mathbf{k}\).

Step 3 ：Then, we find the magnitude of \(\mathbf{r}'(t)\), denoted as \(||\mathbf{r}'(t)||\).

Step 4 ：\(||\mathbf{r}'(t)|| = \sqrt{(3 \cos t)^2 + (3 \cos t)^2 + (-5 \sin t)^2} = \sqrt{9 \cos^2 t + 9 \cos^2 t + 25 \sin^2 t}\).

Step 5 ：Next, we find the second derivative of the vector function \(\mathbf{r}(t)\).

Step 6 ：\(\mathbf{r}''(t) = (-3 \sin t) \mathbf{i} - (3 \sin t) \mathbf{j} + (5 \cos t) \mathbf{k}\).

Step 7 ：Then, we find the magnitude of \(\mathbf{r}''(t)\), denoted as \(||\mathbf{r}''(t)||\).

Step 8 ：\(||\mathbf{r}''(t)|| = \sqrt{(-3 \sin t)^2 + (-3 \sin t)^2 + (5 \cos t)^2} = \sqrt{9 \sin^2 t + 9 \sin^2 t + 25 \cos^2 t}\).

Step 9 ：Finally, we find the curvature \(\kappa(t)\) using the formula \(\kappa(t) = \frac{||\mathbf{r}'(t) \times \mathbf{r}''(t)||}{||\mathbf{r}'(t)||^3}\).

Step 10 ：\(\kappa(t) = \frac{|| (3 \cos t) \mathbf{i} + (3 \cos t) \mathbf{j} - (5 \sin t) \mathbf{k} \times (-3 \sin t) \mathbf{i} - (3 \sin t) \mathbf{j} + (5 \cos t) \mathbf{k} ||}{(\sqrt{9 \cos^2 t + 9 \cos^2 t + 25 \sin^2 t})^3}\).

Step 11 ：Calculating the cross product and simplifying, we get \(\kappa(t) = \boxed{\frac{1}{5}}\).