The velocity of a particle moving in the `x-y` plane is given by
`(dx)/(dt) = 8 pi sin 2 pi t and (dy)/(dt) = 5 pi sin 2 pi t`
where, `t = 0, x = 8 and y = 0`, the path of the particle is.

Options

A. a straight line
B. a elliopse
C. a circle
D. a parabola

Answer

To solve the problem, we need to find the path of the particle given its velocity components in the x-y plane. The velocity components are given as:

\[ \frac{dx}{dt} = 8 \pi \sin(2 \pi t) \]
\[ \frac{dy}{dt} = 5 \pi \sin(2 \pi t) \]

### Step 1: Set up the relationship between \(dy\) and \(dx\)

We can find the relationship between \(dy\) and \(dx\) by dividing the two equations:

\[ \frac{dy}{dt} = 5 \pi \sin(2 \pi t) \]
\[ \frac{dx}{dt} = 8 \pi \sin(2 \pi t) \]

Dividing these two equations gives:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{5 \pi \sin(2 \pi t)}{8 \pi \sin(2 \pi t)} \]

### Step 2: Simplify the expression

The \(\pi\) and \(\sin(2 \pi t)\) terms cancel out:

\[ \frac{dy}{dx} = \frac{5}{8} \]

### Step 3: Integrate the relationship

Now we can integrate both sides. This means we will integrate with respect to \(x\):

\[ dy = \frac{5}{8} dx \]

Integrating both sides gives:

\[ y = \frac{5}{8}x + C \]

### Step 4: Determine the constant of integration \(C\)

We know that at \(t = 0\), \(x = 8\) and \(y = 0\). We can use these values to find \(C\):

Substituting \(x = 8\) and \(y = 0\):

\[ 0 = \frac{5}{8}(8) + C \]
\[ 0 = 5 + C \implies C = -5 \]

### Step 5: Write the final equation of the path

Now substituting \(C\) back into the equation gives:

\[ y = \frac{5}{8}x - 5 \]

Rearranging this, we get:

\[ 5x - 8y - 40 = 0 \]

This represents a straight line in the x-y plane.

### Conclusion

The path of the particle is represented by the equation:

\[ 5x - 8y - 40 = 0 \]