Chapter 14.3, Problem 30WE

To determine

To find: The equivalent rectangular-coordinate equation for the given polar equation.

Answer to Problem 30WE

  $y^2-2y-1=0$

Explanation of Solution

Given information:

An equation in polar coordinate is given as $r\left(1-\sin \theta \right)=1$ .

Concepts used: Conversion formula for coordinate systems are given as-

1. From polar to rectangular:

  $x=r\cos \theta$

  $y=r\sin \theta$

2. From rectangular to polar:

  $r^2=x^2+y^2\Rightarrow r=\pm \sqrt{x^2+y^2}$

  $\cos \theta =\frac{x}{r}$ ; $\sin \theta =\frac{y}{r}$ ; $\tan \theta =\frac{x}{y}$ .

Calculation:

Given equation in polar-coordinate is $r\left(1-\sin \theta \right)=1$ .

Converting into equivalent rectangular-coordinate equation -

  $\begin{array}{l}r\left(1-\sin \theta \right)=1\\ \Rightarrow r-r\sin \theta =1\left\{\text{Multiplying by }r\text{ on both side}\right\}\\ \Rightarrow r=1+r\sin \theta \\ \text{Put }r=\sqrt{x^2+y^2},r\sin \theta =y\\ \Rightarrow \sqrt{x^2+y^2}=1+y\\ \Rightarrow x^2+y^2=1+y^2+2y\left\{\text{Taking square on both side}\right\}\\ \Rightarrow y^2-2y-1=0\end{array}$

Hence, desired equivalent rectangular-coordinate equation would be $y^2-2y-1=0$ .