To solve this problem, we'll use the chain rule and compute the rate of change of the distance from the particle to the origin.

Step 1: Define the distance formula

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

We want to find the rate of change of $D$, i.e., $\frac{dD}{dt}$, at the point where $(x, y) = (2, 15)$ and $\frac{dx}{dt} = 3$.

Step 2: Differentiate $D$ with respect to $t$

Differentiating $D$ with respect to $t$: $\frac{dD}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x\frac{dx}{dt} + 2y\frac{dy}{dt}.$ Simplify: $\frac{dD}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{\sqrt{x^2 + y^2}}.$

Step 3: Find $\frac{dy}{dt}$ using the chain rule

The equation of the curve is: $y = 5\sqrt{4x + 1}.$ Differentiate $y$ with respect to $t$: $\frac{dy}{dt} = \frac{d}{dt} \left( 5\sqrt{4x + 1} \right).$ Using the chain rule: $\frac{dy}{dt} = 5 \cdot \frac{1}{2\sqrt{4x + 1}} \cdot 4 \cdot \frac{dx}{dt}.$ Simplify: $\frac{dy}{dt} = \frac{10}{\sqrt{4x + 1}} \cdot \frac{dx}{dt}.$

Step 4: Evaluate $\frac{dy}{dt}$ at $x = 2$ and $\frac{dx}{dt} = 3$

At $x = 2$, substitute into $4x + 1$: $4x + 1 = 4(2) + 1 = 9.$ Thus: $\frac{dy}{dt} = \frac{10}{\sqrt{9}} \cdot 3 = \frac{10}{3} \cdot 3 = 10.$

Step 5: Substitute into the formula for $\frac{dD}{dt}$

At $(x, y) = (2, 15)$: $x^2 + y^2 = 2^2 + 15^2 = 4 + 225 = 229, \quad \sqrt{x^2 + y^2} = \sqrt{229}.$ Now substitute $x = 2$, $y = 15$, $\frac{dx}{dt} = 3$, and $\frac{dy}{dt} = 10$: $\frac{dD}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{\sqrt{x^2 + y^2}} = \frac{2(3) + 15(10)}{\sqrt{229}}.$ Simplify: $\frac{dD}{dt} = \frac{6 + 150}{\sqrt{229}} = \frac{156}{\sqrt{229}}.$

Final Answer:

$\boxed{\frac{dD}{dt} = \frac{156}{\sqrt{229}} \text{ units per second.}}$

Let me know if you'd like further clarifications or additional examples!

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Related Questions:

1. How does the chain rule apply in this problem?
2. What happens to $\frac{dD}{dt}$ if $\frac{dx}{dt}$ changes?
3. How would you find $\frac{dD}{dt}$ if the curve equation changes?
4. How does the distance rate change when $x$ or $y$ becomes negative?
5. Can we solve this using parametric equations instead?

Tip: When solving related rates problems, always identify which variables depend on time and use the chain rule systematically.