Finding the Orbit of a Particle with Varying Position Vector

AI Thread Summary
The position vector of a particle is given as a function of time, leading to a derived velocity vector through differentiation. When the constants b and c are equal, the particle's motion is circular with constant tangential speed. If b is greater than c, the motion becomes elliptical, with the larger value representing the semi-major axis and the smaller the semi-minor axis. The provided position vector aligns with the general form of an ellipse's equation. Understanding these relationships is crucial for accurately describing the particle's orbit.
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I'm given the position vector as a function of time for a particle (b, c and ω are constants):

\vec{r(t)} = \hat{x} b \cos(ωt) + \hat{y} c \sin(ωt)

To obtain it's velocity i differentiate \vec{r(t)} with respect to time and i obtain:

\vec{v(t)} = -\hat{x} ωb \sin(ωt) + \hat{y} ωc \cos(ωt)

Now i have to describe the orbit of this particle. I'm quite clear that if b=c the orbit is perfectly circular with constant tangential speed. But if b≠c (let's say b>c) is the motion elliptical with ±b as the semi-major axis?
Thanks.
 
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d2x said:
I'm given the position vector as a function of time for a particle (b, c and ω are constants):

\vec{r(t)} = \hat{x} b \cos(ωt) + \hat{y} c \sin(ωt)

To obtain it's velocity i differentiate \vec{r(t)} with respect to time and i obtain:

\vec{v(t)} = -\hat{x} ωb \sin(ωt) + \hat{y} ωc \cos(ωt)

Now i have to describe the orbit of this particle. I'm quite clear that if b=c the orbit is perfectly circular with constant tangential speed. But if b≠c (let's say b>c) is the motion elliptical with ±b as the semi-major axis?
Thanks.

Yes, the larger value will determine the semi-major axis, the smaller will determine the semi-minor axis of an elliptical trajectory. Your expression for r(t) is one form of the equation for an ellipse.
 
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