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A particle moves along the curve y=2 sin(pi x/2). As the particle passes through the point (1/3, 1), its x-coordinate increases at a rate of sqrt{10} cm/s. How fast is the distance from the particle to the origin changing at this instant?

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 $25$.

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