Elliptical motion in polar coordinates

AI Thread Summary
The discussion focuses on verifying a solution for elliptical motion in polar coordinates. The original poster presents their calculations, showing the trajectory as an ellipse centered at the origin with specified axes. Several corrections are made regarding the derivatives, particularly addressing errors in the calculations of dr/dt and the use of arctan. Participants emphasize the importance of maintaining accuracy in the mathematical expressions, especially concerning the presence of squared terms and the correct application of derivatives. The final consensus is that the corrections improve the clarity and correctness of the solution.
lorenz0
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Homework Statement
Given that ##x(t)=a\cos(\Omega t)## and ##y(t)=b\sin(\Omega t)##, identify the trajectory and then find the position and velocity vectors in polar coordinates.
Relevant Equations
##\vec{r}=r\hat{r}=\sqrt{x^2+y^2}\hat{r}##, ##\theta=\arctan\left(\frac{y}{x}\right)##, ##\vec{v}=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}=\frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}##
I think I have completed the exercise but since I have seldom used polar coordinates I would be grateful if someone would check out my work and tell me if I have done everything correctly. Thanks.
My solution follows.

Since ##\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1## it follows that the trajectory is an ellipse centered at the origin with axes ##a## and ##b.## Now, ##\vec{r}=r\hat{r}=\sqrt{x^2+y^2}\hat{r}=\sqrt{a^2\cos^2(\Omega t)+b^2\sin^2(\Omega t)}\hat{r}## and
##\theta=\arctan\left(\frac{y}{x}\right)=\arctan\left(\frac{b}{a}\tan(\Omega t)\right)## so, since ##\frac{dr}{dt}=\frac{-2a^2\cos(\Omega t)\sin(\Omega t)\Omega+2b^2\cos(\Omega t)\sin(\Omega t)\Omega}{2\sqrt{a^2\cos^2(\Omega t)+b^2\sin(\Omega t)}}=\frac{(b^2-a^2)\Omega\sin(2\Omega t)}{2r}## and ##\frac{d\theta}{dt}=\frac{1}{1+\left(\frac{b}{a}\tan(\Omega t)\right)^2}\cdot\frac{b}{a}\cdot \frac{1}{\cos^2(\Omega t)}\cdot\Omega=\frac{a^2\cos^2(\Omega t)}{a^2\cos^2(\Omega t)+b^2\sin^2(\Omega t)}\cdot\frac{b}{a}\cdot\frac{1}{\cos^2(\Omega t)}\cdot\Omega=\frac{ab\Omega}{r^2}## we have that
\begin{align*}
\vec{v}&=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}=\frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}\\ &=\frac{(b^2-a^2)\Omega\sin(2\Omega t)}{2r}\hat{r}+r\frac{ab\Omega}{r^2}\hat{\theta}\\
&=\frac{(b^2-a^2)\Omega\sin(2\Omega t)}{2r}\hat{r}+\frac{ab\Omega}{r}\hat{\theta}
\end{align*}
 
Last edited:
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Why do you have the squared tangent in the ##\theta (t)## calculation? (OP was edited).
 
dextercioby said:
Why do you have the squared tangent in the ##\theta (t)## calculation?
Typo. Fixed, thanks.
 
Please fix the whole calculation. There's no 2 anymore, since there's no square in the argument of ##\arctan ## (OP was again edited).
 
Last edited:
dextercioby said:
Please fix the whole calculation. There's no 2 anymore, since there's no square in the argument of ##\arctan ##.
Done.
 
dextercioby said:
Much better.
Thanks!
 
Sorry, there's an error in the ##\frac{dr}{dt}## calculation. The derivative of ##\cos## carries a minus.
 
dextercioby said:
Sorry, there's an error in the ##\frac{dr}{dt}## calculation. The derivative of ##\cos## carries a minus.
Corrected. Thanks again.
 
  • #10
Sorry, last check also in ##\frac{dr}{dt}## calculation. The 2 in the denominator stays there, if you use the 2 in the numerator to obtain the ##\sin## of double angle, right?
 
  • #11
dextercioby said:
Sorry, last check also in ##\frac{dr}{dt}## calculation. The 2 in the denominator stays there, if you use the 2 in the numerator to obtain the ##\sin## of double angle, right?
You are right.
 
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