Question

A particle moves in x-y plane such that its position vector varies with time as $\vec{r}=\left( 2\sin 3t \right)\hat{i}+2\left( 1-\cos 3t \right)\hat{j}$. Find the equation of the trajectory of the particle.

Answer 1

In the cartesian coordinate system, we specify a point in terms of the x and y coordinate. The position vector in the cartesian coordinate system is defined as, $\vec{r}=x\hat{i}+y\hat{j}$. Comparing it with the given position vector and using the laws of trigonometry, we can find the required equation of motion of the particle.

Complete answer:
Position vector can be defined as a straight line from a reference point to a body which is moving and used to describe the position of the point relative to the reference point. As the body moves, the position vector can change its magnitude or direction or both magnitude and the direction.
The position vector of the particle moving in the x-y plane is given by the equation,
$\vec{r}=\left( 2\sin 3t \right)\hat{i}+2\left( 1-\cos 3t \right)\hat{j}$
The position vector of a particle in the x-y plane is generally given by the equation,
$\vec{r}=x\hat{i}+y\hat{j}$
Comparing the above two vectors we can write,
$\begin{aligned} & x=2\sin 3t \\\ & y=2\left( 1-\cos 3t \right) \\\ \end{aligned}$
From the above we can write that,
$\begin{aligned} & \sin 3t=\dfrac{x}{2} \\\ & \cos 3t=1-\dfrac{y}{2} \\\ \end{aligned}$
One of the fundamental trigonometric identities is,
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
So, the summation of the square of the above terms will give,
${{\sin }^{2}}3t+{{\cos }^{2}}3t=1$
Putting the values obtained in the above expression we get that,
$\begin{aligned} & {{\left( \dfrac{x}{2} \right)}^{2}}+{{\left( 1-\dfrac{y}{2} \right)}^{2}}=1 \\\ & \dfrac{{{x}^{2}}}{4}+1-y+\dfrac{{{y}^{2}}}{4}=1 \\\ & \dfrac{{{x}^{2}}}{4}+\dfrac{{{y}^{2}}}{4}-y=0 \\\ & {{x}^{2}}+{{y}^{2}}-4y=0 \\\ \end{aligned}$
So, the equation of trajectory of the particle in motion in the x-y plane is given by the equation
${{x}^{2}}+{{y}^{2}}-4y=0$.

Note:
In the cartesian coordinate system we take (0,0) as the reference point and measure the required position or the position vector of the particles. We also have a 3 dimensional cartesian coordinate system where the position is described by three points.
We should remember the trigonometric identities to solve this type of question.