# Convert to a rectangular equation? r + rsintheta = 1

## Answer is ${x}^{2} + 2 y = 1$ Thanks in advance

##### 3 Answers
May 1, 2018

$r + r \sin \theta = 1$

becomes

${x}^{2} + 2 y = 1$

#### Explanation:

We know

${r}^{2} = {x}^{2} + {y}^{2}$

$x = r \cos \theta$

$y = r \sin \theta$

so

$r + r \sin \theta = 1$

becomes

$\setminus \sqrt{{x}^{2} + {y}^{2}} + y = 1$

$\setminus \sqrt{{x}^{2} + {y}^{2}} = 1 - y$

${x}^{2} + {y}^{2} = 1 - 2 y + {y}^{2}$

${x}^{2} + 2 y = 1$

The only iffy step is the squaring of the square root. Usually for polar equations we allow negative $r$, and if so the squaring doesn't introduce a new part.

May 1, 2018

Procedure in explanation.

#### Explanation:

To convert from polar to rectangular, we may use the following substitutions: x=rcosθ
y=rsinθ
${r}^{2} = {x}^{2} + {y}^{2}$
tanθ=y/x
Using 1 and 3,
$\sqrt{{x}^{2} + {y}^{2}} + y = 1$
Square the equation. Using the expansion of ${\left(a + b\right)}^{2}$
${x}^{2} + {y}^{2} + {y}^{2} + 2 y \sqrt{{x}^{2} + {y}^{2}} = 1$
$\implies {x}^{2} + 2 {y}^{2} + 2 y \sqrt{{x}^{2} + {y}^{2}} = 1$
$\implies {x}^{2} + 2 y \left(y + \sqrt{{x}^{2} + {y}^{2}}\right) = 1$

Notice that the coefficient of 2y is 1.(See the first equation I wrote using 1 and 3)
So ${x}^{2} + 2 y = 1$

Hope this helps!

May 1, 2018

${x}^{2} - 2 y = 1$

#### Explanation:

$r + r \sin \theta = 1$

We need to convert from polar to rectangular form.

We know that:
$x = r \cos \theta$

$y = r \sin \theta$

and

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

$- - - - - - - - - - - - - - - - - -$

We can substitute in these values for $\textcolor{red}{r}$ and $\textcolor{red}{r \sin \theta}$:
$\textcolor{red}{\sqrt{{x}^{2} + {y}^{2}} + y} = 1$

Subtract $\textcolor{red}{y}$ from both sides of the equation:
$\sqrt{{x}^{2} + {y}^{2}} + y \quad \textcolor{red}{- \quad y} = 1 \quad \textcolor{red}{- \quad y}$

$\sqrt{{x}^{2} + {y}^{2}} = 1 - y$

Square both sides of the equation:
${\left(\sqrt{{x}^{2} + {y}^{2}}\right)}^{\textcolor{red}{2}} = {\left(1 - y\right)}^{\textcolor{red}{2}}$

${x}^{2} + {y}^{2} = 1 - 2 y + {y}^{2}$

Subtract $\textcolor{red}{{y}^{2}}$ from both sides of the equation so they cancel:

${x}^{2} + \cancel{{y}^{2} \quad \textcolor{red}{- \quad {y}^{2}}} = 1 - 2 y + \cancel{{y}^{2} \quad \textcolor{red}{- \quad {y}^{2}}}$

${x}^{2} = 1 - 2 y$

Add $\textcolor{red}{2 y}$ to both sides of the equation to get the final answer in rectangular form:
${x}^{2} - 2 y = 1$

Hope this helps!