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dk_ch
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1. The problem
if {\vec{V}=ky\hat{i}+kx\hat{j} find the equation of the path
if {\vec{V}=ky\hat{i}+kx\hat{j} find the equation of the path
dk_ch said:1. The problem
if ##\vec{V}=ky\hat{i}+kx\hat{j}## find the equation of the path
haruspex said:You need to show your attempt, or at least describe your thinking.
dk_ch said: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.dk_ch said: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
haruspex said: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.
Yes, much better than my way.dk_ch said: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.
The equation for the path of a vector with components ky and kx is:
x = ky
y = kx
This is the standard form for a vector with components ky and kx.
The value of k determines the magnitude of the vector. A larger value of k will result in a longer vector, while a smaller value of k will result in a shorter vector. The direction of the vector remains the same regardless of the value of k.
The term hat (denoted by ^) indicates that the following letter is a unit vector. In this case, i and j are unit vectors in the x and y directions, respectively.
Yes, the path of this vector can be graphed on a 2D coordinate plane. The vector components ky and kx can be plotted as points on the x and y axes, and the resulting path will be a straight line passing through the origin.
The path of this vector represents the displacement of an object in 2D space. In physics, displacement is defined as the change in position of an object from its initial position to its final position. Therefore, the path of this vector can be used to analyze the motion of an object in two dimensions.