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explain step by step how the final solution was synthesizedd like that

Let's break down the given equation step by step and understand how the final expression is synthesized.

### The given equation:
\[
V_z = - \left( \frac{\rho g \sin \alpha}{\eta} \right) \frac{x^2}{2} + \left( \frac{\rho g \sin \alpha}{\eta} \right) \frac{\delta^2}{2} = \frac{\rho g \sin \alpha \delta^2}{2} \left[ 1 - \left( \frac{x}{\delta} \right)^2 \right]
\]

### Step-by-step explanation:

1. **First Term of \( V_z \)**:
   \[
   - \left( \frac{\rho g \sin \alpha}{\eta} \right) \frac{x^2}{2}
   \]
   This is the first part of the equation. It involves \( \rho \) (density), \( g \) (gravitational acceleration), \( \alpha \) (angle), \( \eta \) (viscosity), and \( x^2 \) divided by 2.

   The coefficient in front, \( \frac{\rho g \sin \alpha}{\eta} \), appears as a common factor in both terms of the equation. This suggests that these terms are related and might simplify together later.

2. **Second Term of \( V_z \)**:
   \[
   + \left( \frac{\rho g \sin \alpha}{\eta} \right) \frac{\delta^2}{2}
   \]
   This is the second term, where \( \delta^2 \) appears in place of \( x^2 \). It's structured the same as the first term, but instead of \( x \), we have \( \delta \), which likely represents some characteristic length (perhaps the boundary or limit of \( x \)).

3. **Setting the Sum Equal to the Final Term**:
   \[
   V_z = - \left( \frac{\rho g \sin \alpha}{\eta} \right) \frac{x^2}{2} + \left( \frac{\rho g \sin \alpha}{\eta} \right) \frac{\delta^2}{2}
   \]
   Notice how both terms have the common factor \( \frac{\rho g \sin \alpha}{\eta} \). Let's factor this term out from both expressions.

4. **Factoring Out Common Terms**:
   Factor out \( \frac{\rho g \sin \alpha}{\eta} \) and \( \frac{1}{2} \):
   \[
   V_z = \frac{\rho g \sin \alpha}{\eta} \cdot \frac{1}{2} \left( \delta^2 - x^2 \right)
   \]

5. **Rewriting the Expression**:
   Now, notice the expression \( \delta^2 - x^2 \). This can be factored as a difference of squares:
   \[
   \delta^2 - x^2 = \delta^2 \left( 1 - \left( \frac{x}{\delta} \right)^2 \right)
   \]

6. **Final Expression**:
   Substituting this factorization into the equation:
   \[
   V_z = \frac{\rho g \sin \alpha}{\eta} \cdot \frac{1}{2} \cdot \delta^2 \left( 1 - \left( \frac{x}{\delta} \right)^2 \right)
   \]
   This matches the final form provided:
   \[
   V_z = \frac{\rho g \sin \alpha \delta^2}{2} \left( 1 - \left( \frac{x}{\delta} \right)^2 \right)
   \]

### Summary:
- The original terms both contained the factor \( \frac{\rho g \sin \alpha}{\eta} \).
- Factoring out this common term, along with \( \frac{1}{2} \), allowed the expression to be simplified.
- The difference \( \delta^2 - x^2 \) was then rewritten as \( \delta^2 \left( 1 - \left( \frac{x}{\delta} \right)^2 \right) \), leading to the final result.

### Relative Questions:
1. What does each variable (e.g., \( \rho \), \( g \), \( \alpha \), \( \eta \)) represent physically in this equation?
2. How can we interpret the term \( 1 - \left( \frac{x}{\delta} \right)^2 \) geometrically?
3. Could the factorization approach used here apply to other types of equations, such as in kinematics?
4. How would the solution change if viscosity \( \eta \) varied with \( x \)?
5. What physical scenario could this equation model, such as fluid flow or potential energy?

### Tip:
When simplifying expressions, always look for common factors that can be factored out, as this often leads to significant simplifications.

Explain step by step how the final solution for V_z was synthesized.

Understand how the fluid velocity equation V_z is simplified, factoring in variables like fluid density, gravity, angle, and viscosity. This step-by-step explanation covers key concepts in fluid mechanics and differential equations.

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