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The problem asks how fast the distance of a particle from the origin is increasing. Let’s summarize and solve it step by step.

### Problem summary:
- A particle moves in the \(xy\)-plane with the following rates of change:
  - \( \frac{dx}{dt} = -1 \, \text{m/s} \)
  - \( \frac{dy}{dt} = 2 \, \text{m/s} \)
- We need to determine the rate at which the **distance from the origin** is changing when the particle is at the point \( (5, 12) \).

---

### Solution:

1. **Distance from origin formula**:
   \[
   r = \sqrt{x^2 + y^2}
   \]
   where \(r\) is the distance from the origin.

2. **Differentiate \(r\) with respect to \(t\)** (using the chain rule):
   \[
   \frac{dr}{dt} = \frac{x}{\sqrt{x^2 + y^2}} \frac{dx}{dt} + \frac{y}{\sqrt{x^2 + y^2}} \frac{dy}{dt}
   \]

3. **Substitute** the values at the given point \( (x, y) = (5, 12) \):
   \[
   r = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
   \]

4. **Use the given rates** \( \frac{dx}{dt} = -1 \, \text{m/s} \) and \( \frac{dy}{dt} = 2 \, \text{m/s} \). Now substitute everything into the equation:

   \[
   \frac{dr}{dt} = \frac{5}{13}(-1) + \frac{12}{13}(2)
   \]

5. **Simplify the equation**:
   \[
   \frac{dr}{dt} = -\frac{5}{13} + \frac{24}{13} = \frac{19}{13}
   \]

6. **Final answer**:
   \[
   \frac{dr}{dt} = \frac{19}{13} \, \text{m/s} \approx 1.46 \, \text{m/s}
   \]

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The rate at which the particle's distance from the origin is increasing is approximately **1.46 m/s**.

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Would you like further clarifications, or do you have any additional questions?

Here are five follow-up questions related to this topic:
1. How would the answer change if \( \frac{dx}{dt} \) were positive?
2. What does the sign of \( \frac{dx}{dt} \) or \( \frac{dy}{dt} \) tell us about the particle's movement?
3. How can we find the angle of the particle's velocity with respect to the x-axis?
4. What happens to \( \frac{dr}{dt} \) if the particle were at the origin?
5. How would the equation change if we were calculating speed instead of the rate of change of the distance?

**Tip**: Using the chain rule correctly is essential for related rates problems like these. Make sure to carefully identify what quantities are changing with time.

Koordinat dari suatu partikel yang bergerak di bidang-xy terdirensialkan dengan dx/dt = -1 m/s dan dy/dt = 2 m/s. Seberapa cepat jarak partikel dengan titik asal bertambah ketika partikel tersebut berada di titik (5,12)?

Learn how to determine the rate at which a particle's distance from the origin increases given dx/dt = -1 m/s and dy/dt = 2 m/s at the point (5,12). This solution applies the distance formula and the chain rule for derivatives to solve a related rates problem in calculus.

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