To solve the given problem, let's break it down step by step:

Problem Breakdown:

1. Curve Equation: $y = 4\sqrt{2x} + 3$
2. Point of Interest: $(3, 12)$
3. Rate of Change of $x$: $\frac{dx}{dt} = 2 \, \text{units/second}$
4. Goal: Find the rate of change of the distance from the particle to the origin, i.e., $\frac{dD}{dt}$, where $D$ is the distance to the origin.

Distance Formula:

The distance $D$ from the particle at $(x, y)$ to the origin is: $D = \sqrt{x^2 + y^2}$

Taking the derivative with respect to time $t$, we use the chain rule: $\frac{dD}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$

This simplifies to: $\frac{dD}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{\sqrt{x^2 + y^2}}$

Step 1: Evaluate $y$ and $\frac{dy}{dx}$

Given $y = 4\sqrt{2x} + 3$, compute $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{4}{\sqrt{2x}} \cdot \frac{d}{dx}(2x) = \frac{4}{\sqrt{2x}} \cdot \sqrt{2} = \frac{4}{\sqrt{x}}$

Next, find $\frac{dy}{dt}$: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = \frac{4}{\sqrt{x}} \cdot \frac{dx}{dt}$

At $x = 3$: $\frac{dy}{dt} = \frac{4}{\sqrt{3}} \cdot 2 = \frac{8}{\sqrt{3}}$

Step 2: Evaluate $D$ at $(x, y) = (3, 12)$

$D = \sqrt{x^2 + y^2} = \sqrt{3^2 + 12^2} = \sqrt{9 + 144} = \sqrt{153}$

Step 3: Evaluate $\frac{dD}{dt}$

Substitute values into the formula: $\frac{dD}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{\sqrt{x^2 + y^2}}$

At $(x, y) = (3, 12)$: $\frac{dD}{dt} = \frac{3(2) + 12\left(\frac{8}{\sqrt{3}}\right)}{\sqrt{153}}$

Simplify: $\frac{dD}{dt} = \frac{6 + \frac{96}{\sqrt{3}}}{\sqrt{153}}$

Rationalize $\frac{96}{\sqrt{3}}$: $\frac{96}{\sqrt{3}} = \frac{96\sqrt{3}}{3} = 32\sqrt{3}$

Thus: $\frac{dD}{dt} = \frac{6 + 32\sqrt{3}}{\sqrt{153}}$

Final Answer:

$\frac{dD}{dt} = \frac{6 + 32\sqrt{3}}{\sqrt{153}} \, \text{units/second}.$

Would you like further clarification or numerical approximation? Here are related questions to explore:

1. How does the chain rule apply to this type of problem?
2. What if the rate of change of $y$ was constant instead of $x$?
3. Can the distance rate of change ever be negative?
4. What happens when the particle is closest to the origin?
5. How does changing $\frac{dx}{dt}$ impact $\frac{dD}{dt}$?

Tip: Always check units to ensure your rates are consistent!