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.