Question

Convert \[r = 1 - \sin \theta \] into Cartesian form.

Answer 1

Hint:We know that the Cartesian form is also known as the rectangular form.To convert Polar coordinates $\left( {r,\theta } \right)$ to rectangular coordinates $\left( {x,y} \right)$, we have the equation:
$x = r\cos \theta \\
\Rightarrow y = r\sin \theta $
So by using the above equation and by using the process of substitution we can convert into \[r = 1 - \sin \theta \] the Cartesian form.

Complete step by step answer:
Given, $r = 1 - \sin \theta .....................\left( i \right)$.We know that rectangle coordinates are the Cartesian coordinates seen in the Cartesian plane which is represented by $\left( {x,y} \right)$ and polar coordinates give the position of a point in a plane by using the length $r$ and the angle made to the fixed point $\theta $, and is represented by $\left( {r,\theta } \right).$We know that (i) which is a polar coordinate is to be converted to a Cartesian coordinate. For that we can use the formula:
$x = r\cos \theta .................\left( {ii} \right) \\
\Rightarrow y = r\sin \theta ..................\left( {iii} \right) \\ $
Now we can find the value of $\sin \theta $ from equation (ii):
$\Rightarrow y = r\sin \theta \\
\Rightarrow \sin \theta = \dfrac{y}{r}......................\left( {iii} \right) \\ $
Substituting (iii) in (i) we can write:
\[\Rightarrow r = 1 - \sin \theta \\
\Rightarrow r = 1 - \dfrac{y}{r}......................\left( {iv} \right) \\ \]
Now we have to convert \[r = 1 - \sin \theta \] into Cartesian form and we have to express $r$ in the form of rectangular coordinates.
Such that we know from (ii) and (iii):
\[x = r\cos \theta \\
\Rightarrow y = r\sin \theta \\
\Rightarrow {x^2} + {y^2} = {\left( {r\cos \theta } \right)^2} + {\left( {r\sin \theta } \right)^2} \\
\Rightarrow {x^2} + {y^2} = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) \\
\Rightarrow {x^2} + {y^2} = {r^2} \\
\Rightarrow r = \sqrt {{x^2} + {y^2}} .......................\left( v \right) \\ \]
Now substituting the value of $r$ in (iv) we can write:
\[r = 1 - \dfrac{y}{r} \\
\Rightarrow\sqrt {{x^2} + {y^2}} = 1 - \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}......................\left( {vi} \right) \\ \]
On simplifying (vi) we can write:
\[\sqrt {{x^2} + {y^2}} = 1 - \dfrac{y}{{\sqrt {{x^2} + {y^2}} }} \\
\Rightarrow\sqrt {{x^2} + {y^2}} = \dfrac{{\sqrt {{x^2} + {y^2}} - y}}{{\sqrt {{x^2} + {y^2}} }} \\
\therefore{x^2} + {y^2} = \sqrt {{x^2} + {y^2}} - y.......................\left( {vii} \right) \\ \]
Therefore \[r = 1 - \sin \theta \] in Cartesian form can be written as: \[{x^2} + {y^2} = \sqrt {{x^2} + {y^2}} - y\].

Note:We know that to convert a polar coordinate $\left( {r,\theta } \right)$ to a rectangular coordinate $\left( {x,y} \right)$, we can use the formula:
$x = r\cos \theta \\
\Rightarrow y = r\sin \theta $
In a similar manner convert rectangular coordinate $\left( {x,y} \right)$ to a polar coordinate $\left( {r,\theta } \right)$, we can use the formula:
$r = \sqrt {\left( {{x^2} + {y^2}} \right)} \\
\Rightarrow\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) $
Also while choosing $\theta $ it’s better to choose it in radians since when $\theta $ is in radians the calculations become much easier.