Chapter 10, Problem 32RE

To determine

To find:The equivalent polar equation of the Cartesian equation: $x^2+y^2-2y=0$ .

Answer to Problem 32RE

The equivalent polar equation of the Cartesian equation: $x^2+y^2-2y=0$ is $r=2\sin \theta$ .

Explanation of Solution

Given information:

The Cartesian equation of the curve is $x^2+y^2-2y=0$ .

Formula used:

In the Polar co-ordinate system, points are identified by $\left(r,\theta \right)$ where $r$ is the distance of the point to the origin and $\theta$ is the directional angle or the angle from the initial ray which can be considered as the positive $x$ -axis.

Polar-Rectangular Conversion formula:

  $x=r\cos \theta$

  $y=r\sin \theta$

  $r=\sqrt{x^2+y^2}$

  $\tan \theta =\frac{y}{x}$

Calculation:

Let $\left(r,\theta \right)$ be the equivalent polar coordinates of the point $\left(x,y\right)$ where $r$ is the distance of the point to the origin and $\theta$ is the directional angle or the angle from the initial ray which can be considered as the positive $x$ -axis.

Apply the Polar-Rectangular conversion formula and substitute $x=r\cos \theta$ , $y=r\sin \theta$ in the given Cartesian equation $x^2+y^2-2y=0$ to obtain the equivalent polar equation as:

  $\begin{array}{c}{\left(r\cos \theta \right)}^2+{\left(r\sin \theta \right)}^2-2\left(r\sin \theta \right)=0\\ r^2{\cos }^2\theta +r^2{\sin }^2\theta -2r\sin \theta =0\\ r^2\left({\cos }^2\theta +{\sin }^2\theta \right)-2r\sin \theta =0\\ r^2\cdot 1-2r\sin \theta =0\end{array}$

Simplify further to obtain:

  $\begin{array}{c}{r}^2-2r\sin \theta =0\\ r\left(r-2\sin \theta \right)=0\end{array}$

Either $r=0$ or $r-2\sin \theta =0$

Since $r=\sqrt{x^2+y^2}$ which cannot be equal to $0$ ,so,

  $\begin{array}{l}r-2\sin \theta =0\\ r=2\sin \theta \end{array}$

Thus, the equivalent polar equation of the Cartesian equation: $x^2+y^2-2y=0$ is $r=2\sin \theta$ .