(1 point) Find the unit tangent vector at the indicated point of the vector function \[ \begin{array}{l} \mathbf{r}(t)=e^{9 t} \cos (t) \mathbf{i}+e^{9 t} \sin (t) \mathbf{j}+e^{9 t} \mathbf{k} \\ \mathbf{T}(\pi / 2)=\langle\square, \square, \square \end{array} \]

Step 1 ：Given the vector function \(\mathbf{r}(t)=e^{9 t} \cos (t) \mathbf{i}+e^{9 t} \sin (t) \mathbf{j}+e^{9 t} \mathbf{k}\), we are asked to find the unit tangent vector \(\mathbf{T}(\pi / 2)\).

Step 2 ：The unit tangent vector \(\mathbf{T}(t)\) at a point on a curve given by a vector function \(\mathbf{r}(t)\) is given by the derivative of \(\mathbf{r}(t)\) divided by the magnitude of the derivative of \(\mathbf{r}(t)\).

Step 3 ：First, we find the derivative of \(\mathbf{r}(t)\), which is \(\mathbf{r}'(t) = [-e^{9t}\sin(t) + 9e^{9t}\cos(t), 9e^{9t}\sin(t) + e^{9t}\cos(t), 9e^{9t}]\).

Step 4 ：Next, we find the magnitude of \(\mathbf{r}'(t)\), which is \(\|\mathbf{r}'(t)\| = \sqrt{(-e^{9t}\sin(t) + 9e^{9t}\cos(t))^2 + (9e^{9t}\sin(t) + e^{9t}\cos(t))^2 + 81e^{18t}}\).

Step 5 ：Then, we divide \(\mathbf{r}'(t)\) by its magnitude to get the unit tangent vector \(\mathbf{T}(t)\), which is \(\mathbf{T}(t) = \left[\frac{-e^{9t}\sin(t) + 9e^{9t}\cos(t)}{\sqrt{(-e^{9t}\sin(t) + 9e^{9t}\cos(t))^2 + (9e^{9t}\sin(t) + e^{9t}\cos(t))^2 + 81e^{18t}}}, \frac{9e^{9t}\sin(t) + e^{9t}\cos(t)}{\sqrt{(-e^{9t}\sin(t) + 9e^{9t}\cos(t))^2 + (9e^{9t}\sin(t) + e^{9t}\cos(t))^2 + 81e^{18t}}}, \frac{9e^{9t}}{\sqrt{(-e^{9t}\sin(t) + 9e^{9t}\cos(t))^2 + (9e^{9t}\sin(t) + e^{9t}\cos(t))^2 + 81e^{18t}}}\right]\).

Step 6 ：Finally, we substitute \(t = \pi/2\) into \(\mathbf{T}(t)\) to get the unit tangent vector at that point, which is \(\mathbf{T}(\pi / 2)=\left\langle-\frac{\sqrt{163}}{163}, \frac{9\sqrt{163}}{163}, \frac{9\sqrt{163}}{163}\right\rangle\).

Step 7 ：\(\boxed{\mathbf{T}(\pi / 2)=\left\langle-\frac{\sqrt{163}}{163}, \frac{9\sqrt{163}}{163}, \frac{9\sqrt{163}}{163}\right\rangle}\)