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

In summary: Good job.In summary, the problem is to find the equation of the path if ##\vec{V}=ky\hat{i}+kx\hat{j}## and the conversation involves discussing different approaches to solving it. The final solution is obtained by finding the relationship between x and y and integrating it to get the equation ##y^2 = x^2 + constant##, where x and y are functions of t.
  • #1
dk_ch
44
0
1. The problem
if {\vec{V}=ky\hat{i}+kx\hat{j} find the equation of the path
 
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  • #2
dk_ch said:
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.
 
  • #3
What do "x" and "y" represent here? If they are two independent variables this is surface not a path.
 
  • #4
are you dealing with parametric equations?
 
  • #5
haruspex said:
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
 
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  • #6
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

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.
 
  • #7
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.
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.
 
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  • #8
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.

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.
 
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  • #9
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.
Yes, much better than my way.
 
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1. What is the equation for the path of a vector with components ky and kx?

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.

2. How does the value of k affect the path of the vector?

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.

3. What does the term hat indicate in the vector equation?

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.

4. Can the path of this vector be graphed on a 2D coordinate plane?

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.

5. How is the path of this vector related to the concept of motion in physics?

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.

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