We are given a problem where a particle moves along the curve $y = 2 \sin\left( \frac{\pi x}{2} \right)$, and we need to find the rate at which the distance from the particle to the origin is changing as it passes through the point $\left( \frac{1}{3}, 1 \right)$. We are also given that the x-coordinate is increasing at a rate of $\sqrt{10}$ cm/s at this instant.

Step-by-Step Solution:

1. Distance formula from the particle to the origin: The distance $D$ from the particle at $(x, y)$ to the origin $(0, 0)$ is given by: $D = \sqrt{x^2 + y^2}$

2. Differentiating the distance with respect to time: We apply the chain rule to differentiate the distance function $D$ with respect to time $t$: $\frac{dD}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$ Simplifying: $\frac{dD}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}}$

3. Finding $\frac{dy}{dt}$ using the curve equation: The particle moves along the curve $y = 2 \sin\left( \frac{\pi x}{2} \right)$, so we differentiate $y$ with respect to $t$: $\frac{dy}{dt} = 2 \cos\left( \frac{\pi x}{2} \right) \cdot \frac{\pi}{2} \cdot \frac{dx}{dt}$ Simplifying: $\frac{dy}{dt} = \pi \cos\left( \frac{\pi x}{2} \right) \cdot \frac{dx}{dt}$

4. Substituting values at the given point $\left( \frac{1}{3}, 1 \right)$:

   • At $x = \frac{1}{3}$, $y = 1$.
   • $\frac{dx}{dt} = \sqrt{10}$ cm/s.

   Now, we substitute $x = \frac{1}{3}$ into the equation for $\frac{dy}{dt}$: $\frac{dy}{dt} = \pi \cos\left( \frac{\pi}{6} \right) \cdot \sqrt{10}$ Using $\cos\left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}$: $\frac{dy}{dt} = \pi \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{10}$

5. Substituting into the rate of change of distance formula: Now we calculate $\frac{dD}{dt}$ at $x = \frac{1}{3}$, $y = 1$: $\frac{dD}{dt} = \frac{\left( \frac{1}{3} \cdot \sqrt{10} \right) + \left( 1 \cdot \pi \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{10} \right)}{\sqrt{\left( \frac{1}{3} \right)^2 + 1^2}}$ Simplifying the denominator: $\sqrt{\left( \frac{1}{3} \right)^2 + 1^2} = \sqrt{\frac{1}{9} + 1} = \sqrt{\frac{10}{9}} = \frac{\sqrt{10}}{3}$

   Now, substitute back into the equation: $\frac{dD}{dt} = \frac{\frac{1}{3} \sqrt{10} + \pi \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{10}}{\frac{\sqrt{10}}{3}}$ The $\sqrt{10}$ terms cancel out, so we get: $\frac{dD}{dt} = 3 \left( \frac{1}{3} + \pi \cdot \frac{\sqrt{3}}{2} \right)$

6. Final answer: Simplify the expression for the rate of change of the distance from the particle to the origin: $\frac{dD}{dt} = 1 + \frac{3\pi \sqrt{3}}{2} \ \text{cm/s}$

Thus, the rate at which the distance from the particle to the origin is changing is $1 + \frac{3\pi \sqrt{3}}{2}$ cm/s.

Would you like further details or have any questions?

Here are 5 related questions:

1. How does the distance between two points change with respect to time?
2. What is the physical interpretation of the derivative in this context?
3. How would the result change if the particle moved along a different curve?
4. What happens if the rate of change of $x$ is negative, indicating a decreasing x-coordinate?
5. Can the distance to the origin ever decrease under these conditions?

Tip: In related rates problems, always relate the variables explicitly using a geometric or physical relationship before differentiating with respect to time.