Question

A particle is moving with a velocity $\vec V = {y^2}xi - {x^2}yj$. The general equation for its path is-
A. ${y^2} - {x^2} = {\text{constant}}$
B. ${y^2} + {x^2} = {\text{constant}}$
C. $xy = {\text{constant}}$
D. $y + x = {\text{constant}}$

Answer 1

As we know that, in this question equation of velocity is given and velocity is equal to rate of change of displacement. For finding the general equation for a path we just need to rewrite the velocity equation in its components, so that finding the equation will be easy.

Complete step by step solution:
Particle is moving with the velocity, $\vec V = {y^2}xi - {x^2}yj$
$\vec V = \dfrac{{dx}}{{dt}}i - \dfrac{{dy}}{{dt}}j$
Now, compare the above two equations-
Here, $\dfrac{{dx}}{{dt}} = {y^2}x$ ------ (1)
$\dfrac{{dy}}{{dt}} = - {x^2}y$ ----- (2)
As we know, velocity is the rate of change of displacement. So, we have substituted the values of ${y^2}x$ as $\dfrac{{dx}}{{dt}}$ and $- {x^2}y$ as $\dfrac{{dy}}{{dt}}$ by using the equation (1) and equation (2).

Divide equation (2) by equation (1),
$\dfrac{{dy}}{{dx}} = \dfrac{{ - {x^2}y}}{{{y^2}x}}$
$\Rightarrow\dfrac{{dy}}{{dx}} = \dfrac{{ - x}}{y}$
From, this equation, we can write-
$y.dy = - x.dx$
We can take the $- x.dx$ on the left hand side and then Integrate the resultant equation,
We get- $\dfrac{{{y^2}}}{2} + \dfrac{{{x^2}}}{2} = {\text{constant}}$
Now take L.C.M of the above equation, we get-
$\therefore{y^2} + {x^2} = {\text{constant}}$

So, option B is correct.

Note: Velocity is rate of change of position with respect to a frame of reference. It is a vector quantity, as it has magnitude as well as direction. Velocity of an object can be zero. Its SI unit is $m/\sec$ .In this question, while comparing the equations, remember to take negative sign, $\dfrac{{dy}}{{dt}} = - {x^2}y$. Also when we integrate any quantities, one constant term also comes after the integration along with the quantities.