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.