The question aims to find the rate of **change** in **distance** of the **particle** from the **origin** as it moves along the given **curve** and its **movement increases.**

The background concepts needed for this question include basic **calculus,** which includes **derivatives** and calculating **distance** by using **distance formula** and some **trigonometric ratios.**

## Expert Answer

The given information about the question is given as:

\[ Curve\ y\ =\ 2 \sin(\pi \frac{x} {2}) \]

\[ A\ Point\ on\ the\ Curve\ ,\ p\ =\ (1/3, 1) \]

\[ Rate\ of\ Change\ of\ in\ x-coordinate\ \dfrac{dx}{dt} = \sqrt{10} cm/s \]

To calculate the **rate of change** in **distance,** we can use the **distance formula.** The **distance** from the **origin** to the **particle** is given as:

\[ S = \sqrt{(x – 0)^2 + (y – 0)^2} \]

\[ S = \sqrt{ x^2 + y^2 } \]

Taking the **derivative** of the **distance** $S$ with respect to **time** $t$ to calculate the **rate of change** in **distance,** we get:

\[ \dfrac{ dS }{ dt } = \dfrac{d}{ dt } \sqrt{ x^2 + y^2 } \]

To successfully calculate this **derivative,** we will use the **chain rule** as:

\[ \dfrac{ dS }{ dt } = \dfrac{d}{ d(x^2 + y^2) } (\sqrt{ x^2 + y^2 }) \times \dfrac{ d(x^2 + y^2)}{ dt } \]

Solving the **derivative,** we get:

\[ \dfrac{ dS }{ dt } = \dfrac{1}{ 2 \sqrt{ x^2 + y^2 }} . \Big[ 2x \dfrac{ dx }{ dt } + 2y \dfrac{ dy }{ dt } \Big] \hspace{0.4in} (1) \]

To solve this equation, we need the value of $\dfrac{ dy }{ dt }$. We can calculate its value by **derivating** the equation of the given **curve.** The equation of the curve is given as:

\[ y = 2 \sin (\pi \dfrac{x}{2}) \]

Taking the **derivative** of the **curve** $y$ with respect to **time** $t$, we get:

\[ \dfrac{ dy }{ dt } = \dfrac{d}{ dt } 2 \sin (\pi \dfrac{x}{2}) \]

Solving the equation, we get:

\[ \dfrac{ dy }{ dt } = \pi \cos ( \pi \dfrac{x}{2}) \times \dfrac{ dx }{ dt } \]

Substituting the values, we get:

\[ \dfrac{ dy }{ dt } = \pi \cos ( \pi (\dfrac{\frac{1}{3}}{2} )) \times \sqrt{10} \]

Solving it, we get:

\[ \dfrac{ dy }{ dt } = \dfrac{ \pi }{ 2 } \sqrt{30} \]

Substituting the values in equation $(1)$, we get:

\[ \dfrac{ dS }{ dt } = \dfrac{1}{2 \sqrt{ (\dfrac{1}{3})^2 + (1)^2 }} . \Big[ 2 (\dfrac{1}{3}) \sqrt{10} + 2 (1) (\dfrac{ \pi } {2} \sqrt{30}) \Big] \]

Solving the equation, we get:

\[ \dfrac{ dS }{ dt } = 9.2 cm/s \]

## Numerical Result

The **rate of change** of **distance** from the **origin** of the **particle** moving along the **curve** is calculated to be:

\[ \dfrac{ dS }{ dt } = 9.2 cm/s \]

## Example

Find the **distance** of a **particle** moving along the **curve** $y$ from the **origin** to the **point** $(3, 4)$.

The **distance formula** is given as:

\[ S = \sqrt{ (x – x’)^2 + (y – y’)^2 } \]

Here, the given **coordinates** are:

\[ (x, y) = (3, 4) \]

\[ (x’, y’) = (0, 0) \]

Substituting the values, we get:

\[ S = \sqrt{ (3 – 0)^2 + (4 – 0)^2 } \]

\[ S = \sqrt{ 3^2 + 4^2 } \]

\[ S = \sqrt{ 25 } \]

\[ S = 5 units \]

The **distance** of the **particle** from the **origin** to the **point** given on the **curve** is $25$.