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Trigonometry Particle's position

dwsmith

Well-known member
Feb 1, 2012
1,673
Relatively easy question but I haven't done anything like this in awhile.
$$
\mathbf{r}(t) = \hat{\mathbf{x}}R\cos(\omega t) + \hat{\mathbf{y}}R\sin(\omega t)
$$
The particle moves in a circle so I want to show that the position is given by the above.
I know at $t = 0$ the particle is on the x-axis.
Therefore, we know that at t = 0, $(R,0)$.
Obviously it has to be that but I guess there is a way to show this basically.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
re: Particles position

Consider the parametrization:

$\displaystyle x(t)=R\cos(\omega t)$

$\displaystyle y(t)=R\sin(\omega t)$

Square both and add to eliminate the parameter.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
re: Particles position

Consider the parametrization:

$\displaystyle x(t)=R\cos(\omega t)$

$\displaystyle y(t)=R\sin(\omega t)$

Square both and add to eliminate the parameter.
That will just be $x(t)^2 + y(t)^2 = R^2$.
What about x and y hat?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
re: Particles position

Aren't they orthogonal unit vectors? I'm sorry, I'm not familiar with that notation. I interpreted it as:

$\displaystyle \vec{r}(t)=R\langle \cos(\omega t),\sin(\omega t) \rangle$
 

dwsmith

Well-known member
Feb 1, 2012
1,673
re: Particles position

Aren't they orthogonal unit vectors? I'm sorry, I'm not familiar with that notation. I interpreted it as:

$\displaystyle \vec{r}(t)=R\langle \cos(\omega t),\sin(\omega t) \rangle$
They could be but it doesn't specify. All we know is that they are vectors.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
re: Particles position

They could be but it doesn't specify. All we know is that they are vectors.
I just looked through the book. They are unit vectors. What a dumb notation.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
re: Particles position

Good deal. I believe I have seen that notation a long time ago in a galaxy far away, but I wasn't sure. ;)