Path of a Vector: $\vec{V}=ky\hat{i}+kx\hat{j}$

[dk_ch]

1. The problem
if {\vec{V}=ky\hat{i}+kx\hat{j} find the equation of the path

 

[haruspex]

1. The problem
if $\vec{V}=ky\hat{i}+kx\hat{j}$ find the equation of the path

You need to show your attempt, or at least describe your thinking.

 

[HallsofIvy]

What do "x" and "y" represent here? If they are two independent variables this is surface not a path.

 

[BiGyElLoWhAt]

are you dealing with parametric equations?

 

[dk_ch]

You need to show your attempt, or at least describe your thinking.

I wrote vel components separately as dx/dt and dy/dt . then the relations were integrated to have two equations relating x,y and t. Finally i eliminated t to get the equation. was i right?

i got eq (1) x = kyt +c1 and eq (2) y = kxt+c2
and final eq was (x-c1)/(y-c2)=y/x

 

[dauto]

I wrote vel components separately as dx/dt and dy/dt . then the relations were integrated to have two equations relating x,y and t. Finally i eliminated t to get the equation. was i right?

i got eq (1) x = yt +c1 and eq (2) y = xt+c2
and final eq was (x-c1)/(y-c2)=y/x

Well that solution is not correct is it? Just differentiate it to check with the original equation (making sure to use the product rule) and you will see that.

 

[haruspex]

I wrote vel components separately as dx/dt and dy/dt . then the relations were integrated to have two equations relating x,y and t. Finally i eliminated t to get the equation. was i right?

i got eq (1) x = yt +c1 and eq (2) y = xt+c2
and final eq was (x-c1)/(y-c2)=y/x

y and x are functions of t, so you cannot integrate y with respect to t and get yt.
I would start by getting out of vectors and represent the equation as two simultaneous scalar differential equations. Then you can eliminate one dependent variable to obtain a second order ODE.
There may be neater ways.

You will also need to known the position at t=0.

 

[dk_ch]

y and x are functions of t, so you cannot integrate y with respect to t and get yt.
I would start by getting out of vectors and represent the equation as two simultaneous scalar differential equations. Then you can eliminate one dependent variable to obtain a second order ODE.
There may be neater ways.

You will also need to known the position at t=0.

thanks , I had doubt .
May I do it as follows .
dx/dt =ky and dy/dt=kx hence dx/dy=y/x or xdx=ydy .now by integrating we can get the result as y^2=x^2+ constant. .

please confirm.

 

[haruspex]

thanks , I had doubt .
May I do it as follows .
dx/dt =ky and dy/dt=kx hence dx/dy=y/x or xdx=ydy .now by integrating we can get the result as y^2=x^2+ constant. .

please confirm.

Yes, much better than my way.