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Let's solve both questions one by one:

### Question 1:
A particle is moving along the parabola \( x^2 = 4(y + 6) \), and at the point \( (4, -2) \), the rate of change of its y-coordinate \( \frac{dy}{dt} \) is 2 units per second. We need to find the rate of change of the x-coordinate \( \frac{dx}{dt} \) at this point.

**Steps:**
1. The equation of the parabola is:
   \[
   x^2 = 4(y + 6)
   \]
   Differentiating both sides with respect to time \( t \) (using implicit differentiation):
   \[
   2x \frac{dx}{dt} = 4 \frac{dy}{dt}
   \]

2. Plug in the known values:
   - At the point \( (x, y) = (4, -2) \), we have \( x = 4 \) and \( \frac{dy}{dt} = 2 \).
   
   Substituting into the differentiated equation:
   \[
   2(4) \frac{dx}{dt} = 4(2)
   \]
   Simplifying:
   \[
   8 \frac{dx}{dt} = 8 \quad \Rightarrow \quad \frac{dx}{dt} = 1
   \]

Thus, the rate of change of the x-coordinate is **1 unit per second**.

### Answer: \( \boxed{1} \).

---

### Question 2:
A heap of rubbish in the shape of a cube is being compacted, and the volume decreases at a rate of 7 cubic meters per minute. The volume of the cube is 8 cubic meters, and we are asked to find the rate of change of the edge length \( s \) of the cube when its volume is exactly 8 cubic meters.

**Steps:**
1. The volume \( V \) of a cube is given by:
   \[
   V = s^3
   \]
   Differentiating both sides with respect to time \( t \):
   \[
   \frac{dV}{dt} = 3s^2 \frac{ds}{dt}
   \]

2. We know that:
   - \( \frac{dV}{dt} = -7 \) cubic meters per minute (the volume is decreasing).
   - When the volume is 8 cubic meters, the edge length \( s \) is:
     \[
     s^3 = 8 \quad \Rightarrow \quad s = 2 \, \text{meters}.
     \]

   Substituting into the differentiated equation:
   \[
   -7 = 3(2^2) \frac{ds}{dt}
   \]
   Simplifying:
   \[
   -7 = 12 \frac{ds}{dt} \quad \Rightarrow \quad \frac{ds}{dt} = \frac{-7}{12}
   \]

Thus, the rate of change of the edge length is \( \boxed{-\frac{7}{12}} \) meters per minute.

---

Would you like further details or clarification on any of the steps? Here are 5 related questions to explore:

1. How would the calculation change if the particle were moving along a different parabola, such as \( x^2 = 8(y + 2) \)?
2. What if the rate of change of the y-coordinate in Question 1 were negative? How would that affect the result?
3. How do you approach related rates problems in general? What are some common strategies?
4. What would happen in Question 2 if the cube's volume were increasing instead of decreasing?
5. How does the edge length change rate vary as the cube gets smaller or larger?

**Tip**: In related rates problems, always identify what is constant and what is changing with respect to time, and be clear about the units involved!

1. A particle is moving along the parabola x^2 = 4(y + 6). As the particle passes through the point (4, -2), the rate of change of its y-coordinate is 2 units per second. How fast, in units per second, is the x-coordinate changing at this instant?

2. A heap of rubbish in the shape of a cube is being compacted into a smaller cube. Given that the volume decreases at a rate of 7 cubic meters per minute, find the rate of change of an edge, in meters per minute, of the cube when the volume is exactly 8 cubic meters.

This page provides solutions to related rates problems involving a particle moving along a parabola and a cube's volume decreasing over time. Learn how to use implicit differentiation and the relationship between volume and edge length for geometric shapes. Suitable for high school students in Grades 11-12.

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