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To solve this problem, let's break it down step-by-step.

### Given Information
1. The particle is moving along the curve defined by:
   \[
   y = 5 \sqrt{3x + 3}
   \]
2. As the particle passes through the point \( (2, 15) \), the \( x \)-coordinate increases at a rate of 4 units per second, i.e., \(\frac{dx}{dt} = 4\).
3. We need to find the rate of change of the distance from the particle to the origin at this instant.

### Solution
1. **Find the Distance Formula**: Let the distance from the particle at point \((x, y)\) to the origin \((0, 0)\) be \( D \). Using the distance formula, we have:
   \[
   D = \sqrt{x^2 + y^2}
   \]

2. **Differentiate with Respect to Time \( t \)**: We want to find \(\frac{dD}{dt}\). Differentiate both sides of the distance formula with respect to \(t\):
   \[
   \frac{dD}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x \cdot \frac{dx}{dt} + \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2y \cdot \frac{dy}{dt}
   \]
   Simplifying, we get:
   \[
   \frac{dD}{dt} = \frac{x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt}}{\sqrt{x^2 + y^2}}
   \]

3. **Calculate \( \frac{dy}{dt} \) at \( x = 2 \)**:
   Since \(y = 5 \sqrt{3x + 3}\), differentiate \( y \) with respect to \( x \) to find \(\frac{dy}{dx}\):
   \[
   \frac{dy}{dx} = 5 \cdot \frac{1}{2\sqrt{3x + 3}} \cdot 3 = \frac{15}{2\sqrt{3x + 3}}
   \]
   Now substitute \( x = 2 \) into this expression to get the value of \(\frac{dy}{dx}\) at that point:
   \[
   \frac{dy}{dx} = \frac{15}{2\sqrt{3 \cdot 2 + 3}} = \frac{15}{2\sqrt{9}} = \frac{15}{2 \cdot 3} = \frac{15}{6} = 2.5
   \]
   Since \(\frac{dx}{dt} = 4\), we can find \(\frac{dy}{dt}\) using the chain rule:
   \[
   \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = 2.5 \cdot 4 = 10
   \]

4. **Substitute Values into \(\frac{dD}{dt}\)**:
   Now, we know \( x = 2 \), \( y = 15 \), \( \frac{dx}{dt} = 4 \), and \( \frac{dy}{dt} = 10 \). Substitute these values into the expression for \(\frac{dD}{dt}\):
   \[
   \frac{dD}{dt} = \frac{2 \cdot 4 + 15 \cdot 10}{\sqrt{2^2 + 15^2}}
   \]
   Calculate the components:
   - \( 2 \cdot 4 = 8 \)
   - \( 15 \cdot 10 = 150 \)
   - \( 2^2 + 15^2 = 4 + 225 = 229 \)
   
   So,
   \[
   \frac{dD}{dt} = \frac{8 + 150}{\sqrt{229}} = \frac{158}{\sqrt{229}}
   \]

5. **Simplify the Answer**:
   Thus, the rate of change of the distance from the particle to the origin at this instant is:
   \[
   \frac{dD}{dt} = \frac{158}{\sqrt{229}} \approx 10.45 \text{ units per second}
   \]

Would you like further clarification on any step?

### Related Questions
1. How would the rate of change of distance be affected if \(\frac{dx}{dt}\) was different?
2. What is the physical interpretation of \(\frac{dD}{dt}\) in this context?
3. How would this process change if we were given a different curve equation?
4. Can we generalize this approach to find rates of change of distances for other moving particles?
5. How does the curve’s slope at a point influence the rate of change of distance from the origin?

**Tip**: Always verify units and interpret each term’s significance in related rate problems to ensure clarity in the result.

A particle is moving along the curve y = 5√(3x + 3). As the particle passes through the point (2,15), its x-coordinate increases at a rate of 4 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Calculate the rate of change of distance from a particle to the origin as it moves along the curve y = 5√(3x + 3) and passes through (2,15). This problem involves related rates and the application of the chain rule in calculus.

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