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To find the tangent line and normal line to the function \( g(t) = (4\sin t + 4\cos t)^2 \) at \( t = \frac{\pi}{2} \), we need to follow these steps:

1. **Find \( g(t) \) and \( g'(t) \) at \( t = \frac{\pi}{2} \)**:
   - Evaluate \( g(t) \) at \( t = \frac{\pi}{2} \).
   - Compute the derivative \( g'(t) \) and evaluate it at \( t = \frac{\pi}{2} \) for the slope of the tangent line.

2. **Equation of the Tangent Line**:
   - Use the point-slope form of the line equation: \( y - g(t_0) = g'(t_0)(t - t_0) \), where \( t_0 = \frac{\pi}{2} \).

3. **Equation of the Normal Line**:
   - The slope of the normal line is the negative reciprocal of the slope of the tangent line.

### Step 1: Evaluate \( g(t) \) at \( t = \frac{\pi}{2} \)

Given:
\[
g(t) = (4\sin t + 4\cos t)^2
\]

At \( t = \frac{\pi}{2} \):
\[
\sin\left(\frac{\pi}{2}\right) = 1, \quad \cos\left(\frac{\pi}{2}\right) = 0
\]

Thus,
\[
g\left(\frac{\pi}{2}\right) = (4 \cdot 1 + 4 \cdot 0)^2 = 16
\]

### Step 2: Compute \( g'(t) \)

Use the chain rule:
\[
g(t) = (4\sin t + 4\cos t)^2
\]
Let \( u = 4\sin t + 4\cos t \), then \( g(t) = u^2 \).

The derivative is:
\[
g'(t) = 2u \cdot \frac{du}{dt}
\]

Now, find \( \frac{du}{dt} \):
\[
u = 4\sin t + 4\cos t \implies \frac{du}{dt} = 4\cos t - 4\sin t
\]

Thus,
\[
g'(t) = 2(4\sin t + 4\cos t)(4\cos t - 4\sin t)
\]

Evaluate \( g'(t) \) at \( t = \frac{\pi}{2} \):
\[
g'\left(\frac{\pi}{2}\right) = 2(4\cdot 1 + 4\cdot 0)(4\cdot 0 - 4\cdot 1) = 2(4)(-4) = -32
\]

### Step 3: Equation of the Tangent Line

Using the point \( \left(\frac{\pi}{2}, 16\right) \) and slope \( m = -32 \):
\[
y - 16 = -32(t - \frac{\pi}{2})
\]
So, the tangent line equation is:
\[
y = -32t + 16 + 32\frac{\pi}{2}
\]
\[
y = -32t + 16 + 16\pi
\]

### Step 4: Equation of the Normal Line

The slope of the normal line is the negative reciprocal of \(-32\), which is \( \frac{1}{32} \).

Thus, the normal line equation is:
\[
y - 16 = \frac{1}{32}(t - \frac{\pi}{2})
\]
So,
\[
y = \frac{1}{32}t - \frac{\pi}{64} + 16
\]

### Summary
- **Tangent line equation**: \( y = -32t + 16 + 16\pi \)
- **Normal line equation**: \( y = \frac{1}{32}t - \frac{\pi}{64} + 16 \)

Would you like further details or have any questions?

### Further Exploration
1. How do we derive the equation of a tangent line in parametric form?
2. What is the significance of the negative reciprocal in finding the normal line?
3. How can the derivative \( g'(t) \) be interpreted geometrically?
4. What happens to the tangent and normal lines if we choose a different point \( t \)?
5. Can we generalize this approach to three-dimensional functions?

### Tip
Always check if the function simplifies before differentiating to avoid complex calculations.

Find the tangent line and normal line to the function \( g(t) = (4\sin t + 4\cos t)^2 \) at \( t = \frac{\pi}{2} \).

Explore how to find the tangent and normal lines for the function \( g(t) = (4\sin t + 4\cos t)^2 \) at \( t = \frac{\pi}{2} \). This calculus problem covers key concepts such as trigonometric derivatives and point-slope form, suitable for high school students in Grades 11-12.

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