# Elliptical motion in polar coordinates

• lorenz0
lorenz0
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:
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.

Much better.

lorenz0
dextercioby said:
Much better.
Thanks!

Sorry, there's an error in the ##\frac{dr}{dt}## calculation. The derivative of ##\cos## carries a minus.

lorenz0
dextercioby said:
Sorry, there's an error in the ##\frac{dr}{dt}## calculation. The derivative of ##\cos## carries a minus.
Corrected. Thanks again.

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?

lorenz0
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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