Plotting the trajectory of a particle in polar coordinates

Hi,

Unfortunately, I am not quite sure whether I have solved/plotted the following task correctly


I started by resolving the expression ##\phi=2 \pi t## to t so that I can represent ##\rho(t)## with ##\rho(\phi)##

The vector ##\vec{e}_r## was written in my lecture as follows ##\vec{e}_{\rho}## , ##\vec{e}_{\rho}= \left(\begin{array}{c} \cos(\phi) \\ \sin(\phi) \end{array}\right)##

Then the position vector is ##\vec{r}(\phi)=\rho(\phi) \vec{e}_{\rho}=(\rho_0+\rho_1 \sin^2(2 \phi) ) \left(\begin{array}{c} \cos(\phi) \\ \sin(\phi) \end{array}\right)##

Then I plotted this expression using Mathematica from ##0## to ##2 \pi##



Is that correct? Because doesn't the two different colors mean that I got the trajectory for two different particles instead of just one?

Note that the expression ##\rho = 0.1 + \sin^2(4\pi t)## is always greater than or equal to 0.1. So, the trajectory never gets closer than 0.1 to the origin.

The format for Mathematica's PolarPlot is



Here, ##r## is just your ##\rho## expressed as a function of ##\phi##, but without the ##\vec{e}_{\rho}= \left(\begin{array}{c} \cos(\phi) \\ \sin(\phi) \end{array}\right)##. By putting in ##\vec{e}_{\rho}= \left(\begin{array}{c} \cos(\phi) \\ \sin(\phi) \end{array}\right)##, you are essentially going to Cartesian coordinates where the x and y components are expressed in terms of the parameter ##\phi##. In that case, you could use Mathematica's ParametricPlot. However, I think using PolarPlot is simpler here.

In addition to what @TSny said, can I add this.

Lambda96 said:

Then the position vector is ##\vec{r}(\phi)=\rho(\phi) \vec{e}_{\rho}=(\rho_0+\rho_1 \sin^2(2 \phi) ) \left(\begin{array}{c} \cos(\phi) \\ \sin(\phi) \end{array}\right)##

Note that ##(\rho_0+\rho_1 \sin^2(2 \phi))## is a scalar quantity. So

##(\rho_0+\rho_1 \sin^2(2 \phi) )\left(\begin{array}{c} \cos(\phi) \\ \sin(\phi) \end{array}\right)##

is equal to

## \left(\begin{array}{c} \rho_0\cos(\phi)+\rho_1 \sin^2(2 \phi)\cos(\phi) \\ \rho_0\sin(\phi)+\rho_1 \sin^2(2 \phi)\sin(\phi) \end{array}\right)##

This doesn't match what you entered in Mathematica.

You might want practise/check by plotting a few simple curves of known shapes.

Thank you TSny and Steve4Physics for your help

I have now redone the plot and got the following:

Looks good.

If you want to use the command ParametricPlot, then the format is

So, for this problem we would have


Or, maybe less confusing,

Thank you TSny for your help and also thank you for showing me how to plot the curve with ParametricPlot Since Mathematica is relatively new to me, this helped me a lot.