Trajectory of a particle when its position vector changes

• Davidllerenav
In summary: I think that the derivative would be ##2 \vec r## if I use the product rule.With respect to time? ##\vec r.\vec r## would be 0. And ##\vec r^2## would be 0 too,... but it should be different, right? I think that the derivative would be ##2 \vec r## if I use the product rule.Yes, so what can you say about ##\vec r^2##?Yes, so what can you say about ##\vec r^2##?In summary, the position vector of a particle changes only by its module or direction. The trajectory of the particle's movement can be a line or a circle
Davidllerenav

Homework Statement

The position vector of a particle changes:
1. Only by its module.
2. Only by its direction.
What can be said about the trayectory of the movement of the particle? Obtain the answer analitically.

None.

The Attempt at a Solution

I think that the trayectory would be a line, at least for the first one. For the second one I really can't Imagine how the direction can change without changing the module. How can I make this analitically?

Davidllerenav said:
For the second one I really can't Imagine how the direction can change without changing the module.
Why can't the position vector change direction without changing its length? (What would that path look like?)

Doc Al said:
Why can't the position vector change direction without changing its length? (What would that path look like?)
I didn't said it can't, but I can't imagine it. I guess that it could look like a circle, right? Because the length wouldn't change.

Davidllerenav said:
How can I make this analitically?
If the position vector is ##\vec r##, write an expression for its modulus.

haruspex said:
If the position vector is ##\vec r##, write an expression for its modulus.
Since it is a one-dimension movement, it would be ##|\vec r|=\sqrt{(\vec r_x)^2} \vec i##, right?

Davidllerenav said:
I guess that it could look like a circle, right? Because the length wouldn't change.
Exactly!

Doc Al said:
Exactly!
Ok, but how do I write that analitically?

Davidllerenav said:
Since it is a one-dimension movement, it would be ##|\vec r|=\sqrt{(\vec r_x)^2} \vec i##, right?
Not sure what you intend by the right hand side of that. Why the x subscript and the i? It is just ##\sqrt{\vec r^2}=\sqrt{\vec r.\vec r}##.
Now write it in a form that avoids the square root, and think how you can express analytically that it is constant.

haruspex said:
Not sure what you intend by the right hand side of that. Why the x subscript and the i?
I the subscript means that it is the x component of the position vector.
haruspex said:
It is just ##\sqrt{\vec r^2}=\sqrt{\vec r.\vec r}##.
Now write it in a form that avoids the square root, and think how you can express analytically that it is constant.
To avoid the square root, I can just write it as absolute value, right? I think that it would be a constant because it is a number, not a vector, so it won't change.

Davidllerenav said:
Ok, but how do I write that analitically?

Hint: What's the equation for a circle?

Doc Al said:

Hint: What's the equation for a circle?
It is ##x^2+y^2=r^2##.

Last edited:
Davidllerenav said:
I the subscript means that it is the x component of the position vector.
But the modulus is not specific to the x direction.
Davidllerenav said:
To avoid the square root, I can just write it as absolute value, right?
Yes, but absolute values are not convenient for algebraic manipulation. What's another way to get rid of a square root in an equation?

haruspex said:
But the modulus is not specific to the x direction.

Yes, but absolute values are not convenient for algebraic manipulation. What's another way to get rid of a square root in an equation?
Rasing it all to square.

Davidllerenav said:
Rasing it all to square.
Right, so write an equation that relates the vector to the square of its modulus, and consider how you can express analytically that the modulus is constant.

haruspex said:
Right, so write an equation that relates the vector to the square of its modulus, and consider how you can express analytically that the modulus is constant.
Ok, so I end up with ##||\vec r||^2=\vec r^2## then the vector would be ##\vec r=(\vec r^2 cos(\theta), \vec r^2 sin(\theta))##, right?

Davidllerenav said:
Ok, so I end up with ##||\vec r||^2=\vec r^2## then the vector would be ##\vec r=(\vec r^2 cos(\theta), \vec r^2 sin(\theta))##, right?
No need to break it into components. Look at the form I wrote ##\vec r^2## in in post #8.
And you haven't answered my question about how, in algebra, one usually expresses the fact that a 'variable' is constant.

haruspex said:
No need to break it into components. Look at the form I wrote ##\vec r^2## in in post #8.
And you haven't answered my question about how, in algebra, one usually expresses the fact that a 'variable' is constant.
Sorry, but I really don't remember.

Davidllerenav said:
Sorry, but I really don't remember.
Think calculus.

haruspex said:
Think calculus.
I guess by the derivative? If it is a constant its derivative is 0.

Davidllerenav said:
I guess by the derivative? If it is a constant its derivative is 0.
Right.
Can you differentiate either ##\vec r^2## or ##\vec r.\vec r##?

haruspex said:
Right.
Can you differentiate either ##\vec r^2## or ##\vec r.\vec r##?
With respect to time? ##\vec r.\vec r## would be 0. And ##\vec r^2## would be 0 too, right?

Davidllerenav said:
With respect to time? ##\vec r.\vec r## would be 0.
No, it's constant; its derivative is zero. Can you differentiate it?

haruspex said:
No, it's constant; its derivative is zero. Can you differentiate it?
No, because it doesn't depend on anything.

Davidllerenav said:
No, because it doesn't depend on anything.
##\vec r## depends on time. Its derivative is ##\dot{\vec r}##. So what is the derivative of ##\vec r.\vec r##? Use the chain rule.

haruspex said:
##\vec r## depends on time. Its derivative is ##\dot{\vec r}##. So what is the derivative of ##\vec r.\vec r##? Use the chain rule.
It would be ##\dot{\vec r} \vec r * \dot{\vec r}## using the chain rule. But shouldn't I use the product rule?

Davidllerenav said:
But shouldn't I use the product rule?
Ah, yes - that's what I meant.

haruspex said:
Ah, yes - that's what I meant.
Ok, so it would be ##\dot{\vec r}\vec r + \vec r \dot{\vec r}##

Davidllerenav said:
Ok, so it would be ##\dot{\vec r}\vec r + \vec r \dot{\vec r}##
Right! and since the dot product commutes, those two terms are the same. So simplify it.
If the dot product of two vectors is zero, what does that tell you about the two vectors, geometrically?

haruspex said:
Right! and since the dot product commutes, those two terms are the same. So simplify it.
If the dot product of two vectors is zero, what does that tell you about the two vectors, geometrically?
That tells me that they are perpendicular, right?

Davidllerenav said:
That tells me that they are perpendicular, right?
Right. And what do those two perpendicular vectors represent in terms of the motion?

haruspex said:
Right. And what do those two perpendicular vectors represent in terms of the motion?
I think that they represent the components of the vector, is it right?

Davidllerenav said:
I think that they represent the components of the vector, is it right?
No, they are two different vectors, ##\vec r## and ##\dot{\vec r}##. What does each represent in terms the motion?

haruspex said:
No, they are two different vectors, ##\vec r## and ##\dot{\vec r}##. What does each represent in terms the motion?
The position and the velocity?

Davidllerenav said:
The position and the velocity?
Right. So we deduce that in this motion the position vector is perpendicular, always, to the velocity vector.

haruspex said:
Right. So we deduce that in this motion the position vector is perpendicular, always, to the velocity vector.
But what does that mean?

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