Question

A particle is moving with a velocity $\vec v = K\left( {y\hat i + x\hat j} \right)$, where $K$ is a constant. The general equation for its path is :
(A) $xy = {\text{constant}}$
(B) ${y^2} = {x^2}{\text{ + constant}}$
(C) $y = {x^2}{\text{ + constant}}$
(D) ${y^2} = x{\text{ + constant}}$

Answer 1

Hint: From the given equation of velocity we can write the x-component of the velocity as the differentiation of the x-component of position with respect to time. Similarly for the y-component we find the differentiation of the y-component of position with respect to time. From there we find the differential equation in terms of x and y. Solving that we get the equation of path.

Formula Used: In this solution we will be using the following formula,
$\Rightarrow \int {ada} = \dfrac{{{a^2}}}{2} + C$
where $a$ is any variable and $C$ is the constant of integration.

Complete step by step answer
In this question we are given the velocity as, $\vec v = K\left( {y\hat i + x\hat j} \right)$. From here we can see that the x-component of velocity is the coefficient of $\hat i$, that is $K_y$. Now the x-component of velocity can also be written as the x-component of position differentiated with respect to time. Therefore, we get,
$\Rightarrow \dfrac{{dx}}{{dt}} = K_y$
Similarly, the y-component of velocity is the coefficient of $\hat j$, that is $K_x$. So this too can be written as,
$\Rightarrow \dfrac{{dy}}{{dt}} = K_x$
Now taking the ratio of the second equation to the first equation we get,
$\Rightarrow \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}} = \dfrac{{K_x}}{{K_y}}$
We can cancel out the $dt$ and the $K$. So we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{y}$
Now, on doing cross multiplication we get,
$\Rightarrow ydy = xdx$
Now this is the differential equation and we can solve it to get the equation of the path. Therefore we integrate it on both the sides,
$\Rightarrow \int {ydy} = \int {xdx} $
Therefore we get,
$\Rightarrow \dfrac{{{y^2}}}{2} = \dfrac{{{x^2}}}{2} + C$
here $C$ is the constant of integration. We can cancel $\dfrac{1}{2}$ from both the sides of the equation. Therefore, we have,
$\Rightarrow {y^2} = {x^2} + 2C$
Since $C$ is a constant, so $2C$ is also a constant. Therefore we can write this as,
$\Rightarrow {y^2} = {x^2}{\text{ + constant}}$
Hence option (B) is correct.

Note
When a body moves from one point to another, it does that by describing a geometric line in space. This line is the trajectory of the particle and the position vector of the particle with respect to time gives the trajectory equation or the equation for the path of the particle.