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: $\frac{dr}{dt} = \frac{19}{13} \, \text{m/s} \approx 1.46 \, \text{m/s}$

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

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.