Why can't the position vector change direction without changing its length? (What would that path look like?)Davidllerenav said:For the second one I really can't Imagine how the direction can change without changing the module.
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.Doc Al said:Why can't the position vector change direction without changing its length? (What would that path look like?)
If the position vector is ##\vec r##, write an expression for its modulus.Davidllerenav said:How can I make this analitically?
Since it is a one-dimension movement, it would be ##|\vec r|=\sqrt{(\vec r_x)^2} \vec i##, right?haruspex said:If the position vector is ##\vec r##, write an expression for its modulus.
Exactly!Davidllerenav said:I guess that it could look like a circle, right? Because the length wouldn't change.
Ok, but how do I write that analitically?Doc Al said:Exactly!
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}##.Davidllerenav said:Since it is a one-dimension movement, it would be ##|\vec r|=\sqrt{(\vec r_x)^2} \vec i##, right?
I the subscript means that it is the x component of the position vector.haruspex said:Not sure what you intend by the right hand side of that. Why the x subscript and the i?
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.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.
Follow haruspex's advice! ;)Davidllerenav said:Ok, but how do I write that analitically?
It is ##x^2+y^2=r^2##.Doc Al said:Follow haruspex's advice! ;)
Hint: What's the equation for a circle?
But the modulus is not specific to the x direction.Davidllerenav said:I the subscript means that it is the x component of the position vector.
Yes, but absolute values are not convenient for algebraic manipulation. What's another way to get rid of a square root in an equation?Davidllerenav said:To avoid the square root, I can just write it as absolute value, right?
Rasing it all to square.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?
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.Davidllerenav said:Rasing it all to square.
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?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.
No need to break it into components. Look at the form I wrote ##\vec r^2## in in post #8.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?
Sorry, but I really don't remember.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.
Think calculus.Davidllerenav said:Sorry, but I really don't remember.
I guess by the derivative? If it is a constant its derivative is 0.haruspex said:Think calculus.
Right.Davidllerenav said:I guess by the derivative? If it is a constant its derivative is 0.
With respect to time? ##\vec r.\vec r## would be 0. And ##\vec r^2## would be 0 too, right?haruspex said:Right.
Can you differentiate either ##\vec r^2## or ##\vec r.\vec r##?
No, it's constant; its derivative is zero. Can you differentiate it?Davidllerenav said:With respect to time? ##\vec r.\vec r## would be 0.
No, because it doesn't depend on anything.haruspex said:No, it's constant; its derivative is zero. Can you differentiate it?
##\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.Davidllerenav said:No, because it doesn't depend on anything.
It would be ##\dot{\vec r} \vec r * \dot{\vec r}## using the chain rule. But shouldn't I use the product 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.
Ah, yes - that's what I meant.Davidllerenav said:But shouldn't I use the product rule?
Ok, so it would be ##\dot{\vec r}\vec r + \vec r \dot{\vec r}##haruspex said:Ah, yes - that's what I meant.
Right! and since the dot product commutes, those two terms are the same. So simplify it.Davidllerenav said:Ok, so it would be ##\dot{\vec r}\vec r + \vec r \dot{\vec r}##
That tells me that they are perpendicular, right?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?
Right. And what do those two perpendicular vectors represent in terms of the motion?Davidllerenav said:That tells me that they are perpendicular, right?
I think that they represent the components of the vector, is it right?haruspex said:Right. And what do those two perpendicular vectors represent in terms of the motion?
No, they are two different vectors, ##\vec r## and ##\dot{\vec r}##. What does each represent in terms the motion?Davidllerenav said:I think that they represent the components of the vector, is it right?
The position and the velocity?haruspex said:No, they are two different vectors, ##\vec r## and ##\dot{\vec r}##. What does each represent in terms the motion?
Right. So we deduce that in this motion the position vector is perpendicular, always, to the velocity vector.Davidllerenav said:The position and the velocity?
But what does that mean?haruspex said:Right. So we deduce that in this motion the position vector is perpendicular, always, to the velocity vector.