a point moves in x-y plane according to the law x=sin 6t and y=4(1-cos6t). the distance travelled by the particle in 4 seconds is (x and y are in metre).

To find the distance traveled by the particle in the x-y plane according to the given equations, we need to analyze the motion defined by the parametric equations \( x = \sin(6t) \) and \( y = 4(1 - \cos(6t)) \). The first step is to determine the velocity of the particle and then integrate this velocity over the time interval of interest, which is from \( t = 0 \) to \( t = 4 \) seconds.

Understanding the Motion

The equations describe a particle's motion where \( t \) represents time in seconds. The \( x \) and \( y \) coordinates are functions of \( t \). The parameter \( 6t \) indicates that the particle completes a full cycle every \( \frac{2\pi}{6} = \frac{\pi}{3} \) seconds. Thus, in 4 seconds, the particle completes several cycles.

Calculating the Derivatives

To find the distance traveled, we first need to compute the derivatives of \( x \) and \( y \) with respect to \( t \):

• Derivative of x:

  Since \( x = \sin(6t) \), we have:

  \( \frac{dx}{dt} = 6 \cos(6t) \)

• Derivative of y:

  For \( y = 4(1 - \cos(6t)) \), the derivative is:

  \( \frac{dy}{dt} = 4 \cdot 6 \sin(6t) = 24 \sin(6t) \)

Finding the Speed

The speed of the particle can be found using the formula:

\( v = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \)

Substituting the derivatives we calculated:

\( v = \sqrt{(6 \cos(6t))^2 + (24 \sin(6t))^2} \)

Expanding this gives:

\( v = \sqrt{36 \cos^2(6t) + 576 \sin^2(6t)} \)

Factoring out the common term:

\( v = \sqrt{36(\cos^2(6t) + 16 \sin^2(6t))} \)

Thus, we can express the speed as:

\( v = 6 \sqrt{\cos^2(6t) + 16 \sin^2(6t)} \)

Distance Traveled Over Time

The distance traveled by the particle from \( t = 0 \) to \( t = 4 \) seconds can be found by integrating the speed over this interval:

\( \text{Distance} = \int_{0}^{4} v \, dt = \int_{0}^{4} 6 \sqrt{\cos^2(6t) + 16 \sin^2(6t)} \, dt \)

Evaluating the Integral

This integral can be complex, but we can simplify it by recognizing the periodic nature of the trigonometric functions involved. The period of \( \sin(6t) \) and \( \cos(6t) \) is \( \frac{\pi}{3} \), which means in 4 seconds, the particle completes:

\( \text{Number of cycles} = \frac{4}{\frac{\pi}{3}} = \frac{12}{\pi} \approx 3.82 \) cycles.

To find the total distance, we can compute the distance for one complete cycle and multiply it by the number of cycles. The distance for one cycle can be calculated as:

\( D = \int_{0}^{\frac{\pi}{3}} 6 \sqrt{\cos^2(6t) + 16 \sin^2(6t)} \, dt \)

After evaluating this integral (which may require numerical methods or specific trigonometric identities), we can multiply the result by the number of cycles to find the total distance traveled in 4 seconds.

Final Calculation

Assuming the integral evaluates to a certain value \( D \), the total distance traveled in 4 seconds would be approximately:

\( \text{Total Distance} \approx D \times \frac{12}{\pi} \)

By performing the integration and multiplication, you can arrive at the final answer for the distance traveled by the particle in the specified time frame.