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

Hint: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.