Let's start by solving this step by step.

1. Find the unit tangent vector $\mathbf{T} \left( \frac{\pi}{2} \right)$: We are given the vector-valued function: $\mathbf{r}(t) = \langle 2 \cos(t), 2 \sin(t), 5 \sin^2(t) \rangle$ First, we compute the derivative $\mathbf{r}'(t)$ to find the tangent vector: $\mathbf{r}'(t) = \left\langle \frac{d}{dt}(2\cos(t)), \frac{d}{dt}(2\sin(t)), \frac{d}{dt}(5\sin^2(t)) \right\rangle$ Differentiating: $\mathbf{r}'(t) = \langle -2 \sin(t), 2 \cos(t), 10 \sin(t) \cos(t) \rangle$

   Now evaluate at $t = \frac{\pi}{2}$: $\mathbf{r}'\left( \frac{\pi}{2} \right) = \langle -2 \sin\left( \frac{\pi}{2} \right), 2 \cos\left( \frac{\pi}{2} \right), 10 \sin\left( \frac{\pi}{2} \right) \cos\left( \frac{\pi}{2} \right) \rangle$ Simplifying the trigonometric values: $\mathbf{r}'\left( \frac{\pi}{2} \right) = \langle -2, 0, 0 \rangle$

2. Find the unit tangent vector $\mathbf{T}\left( \frac{\pi}{2} \right)$: The unit tangent vector is given by: $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}$ The magnitude of $\mathbf{r}'\left( \frac{\pi}{2} \right)$ is: $|\mathbf{r}'\left( \frac{\pi}{2} \right)| = \sqrt{(-2)^2 + 0^2 + 0^2} = \sqrt{4} = 2$ Therefore, the unit tangent vector is: $\mathbf{T}\left( \frac{\pi}{2} \right) = \frac{1}{2} \langle -2, 0, 0 \rangle = \langle -1, 0, 0 \rangle$

3. Find the parametric equations of the tangent line: The parametric equation of a tangent line at a point $t = \frac{\pi}{2}$ is given by: $\mathbf{r}(t) = \mathbf{r}\left( \frac{\pi}{2} \right) + s \mathbf{T}\left( \frac{\pi}{2} \right)$ First, find $\mathbf{r}\left( \frac{\pi}{2} \right)$: $\mathbf{r}\left( \frac{\pi}{2} \right) = \langle 2 \cos\left( \frac{\pi}{2} \right), 2 \sin\left( \frac{\pi}{2} \right), 5 \sin^2\left( \frac{\pi}{2} \right) \rangle$ Simplifying: $\mathbf{r}\left( \frac{\pi}{2} \right) = \langle 0, 2, 5 \rangle$ Now, substitute $\mathbf{r}\left( \frac{\pi}{2} \right) = \langle 0, 2, 5 \rangle$ and $\mathbf{T}\left( \frac{\pi}{2} \right) = \langle -1, 0, 0 \rangle$: $\mathbf{r}(t) = \langle 0, 2, 5 \rangle + s \langle -1, 0, 0 \rangle = \langle -s, 2, 5 \rangle$

   Therefore, the parametric equations of the tangent line are: $x(t) = -s$ $y(t) = 2$ $z(t) = 5$

Let me know if you'd like further clarification or have any questions! Here are 5 questions you can explore related to this:

1. How do you compute the derivative of a vector-valued function in higher dimensions?
2. Why is finding the unit tangent vector important in calculus?
3. What is the geometric interpretation of the tangent vector?
4. How do you find parametric equations for lines in different contexts, such as curves in space?
5. What role does the magnitude of the tangent vector play in defining the unit tangent vector?

Tip: Always check the trigonometric values (e.g., $\sin$ and $\cos$) at special angles like $\frac{\pi}{2}$, $\pi$, etc., as they simplify many expressions.