Interraption:To show that the system \displaystyle\dot{{r}}

The coordinate system can be either the Cartesian system or the polar system. In case of the polar system, a point is determined by a distance and angle from a reference point while as in Cartesian coordinate system each point is determined by a pair of numerical coordinates.
It is given that ${\displaystyle x=r\cos \theta \text{and}y=r\sin \theta }$
Square and add the above equations.
${\displaystyle x^2+y^2=r^2{\cos }^2\theta +r^2{\sin }^2\theta }$
${\displaystyle x^2+y^2=r^2({\cos }^2\theta +{\sin }^2\theta )}$
Substitute ${\displaystyle ({\cos }^2\theta +{\sin }^2\theta )=1}$ so that
${\displaystyle x^2+y^2=r^2}$
Now,
${\displaystyle x=r\cos \theta }$
Differentiate.
${\displaystyle \dot{x}=\dot{r}\theta -r\theta \sin \theta }$
Substitute ${\displaystyle \dot{r}=r(1-r^2),\cos \theta =\frac{x}{r}\text{and}y\text{for}r\sin \theta }$ into the above equation.
${\displaystyle \dot{x}=\left(\frac{x\left(r(1-r^2)\right)}{r}\right)-y}$
${\displaystyle \dot{x}=x(1-r^2)-y}$
Substitute ${\displaystyle r^2=x^2+y^2}$
${\displaystyle \dot{x}=x(1-x^2-y^2)-y}$
${\displaystyle \dot{x}=x-y-x(x^2+y^2)}$
Thus, it is shown that ${\displaystyle \dot{x}=x-y-x(x^2+y^2)}$
Now,
${\displaystyle y=\sin \theta }$
Differentiate.
${\displaystyle \dot{y}=\dot{r}\sin \theta +r\theta \cos \theta }$
Substitute ${\displaystyle \dot{r}=r(1-r^2),\sin \theta =\frac{y}{r},\dot{\theta }=1\text{and}x=r\cos \theta }$ into the above equation.
${\displaystyle \dot{y}=\left(\frac{y\left(r(1-r^2)\right)}{r}\right)+x}$
${\displaystyle \dot{y}=y(1-r^2)+x}$
But ${\displaystyle r^2=x^2+y^2}$
${\displaystyle \dot{y}=x+y-y(x^2+y^2)}$
Therefore, it is shown that ${\displaystyle \dot{y}=x+y-y(x^2+y^2)}$
Answer:
For system

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