Question

How do you convert $r = 2 - 2\cos \left( \theta \right)$ into rectangular form?

Answer 1

In this problem, we have given an equation in the polar form. Here we are asked to convert the given polar form equation into the rectangular form. To convert the given equation into the equation into the rectangular form we need to use some identities and by using that identity we can get a required solution.

Formula used:
Perfect square trinomial rule: $\left( {{a^2} + {b^2}} \right) = {a^2} + 2ab + {b^2}$

Complete step by step solution:
The polar form equation is $r = 2 - 2\cos \left( \theta \right)$.
Now, the identities that we are going to use are,
$x = r\cos \theta$
$y = r\sin \theta$
Now just squaring and adding the $x$, $y$ values, we get
${x^2} = {r^2}{\cos ^2}\theta$ and ${y^2} = {r^2}{\sin ^2}\theta$
Then, ${x^2} + {y^2} = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta$
Also, ${r^2}$ is common in both terms, so take ${r^2}$ as common.
$\Rightarrow {x^2} + {y^2} = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)$…(i)
As we know that, ${\cos ^2}\theta + {\sin ^2}\theta = 1$.
Substitute the value of ${\cos ^2}\theta + {\sin ^2}\theta$ in equation (i), we get
$\Rightarrow {x^2} + {y^2} = {r^2}$
Now, find the $\cos \theta$ from $x = r\cos \theta$ and substitute its value in given polar form.
$x = r\cos \theta$
$\Rightarrow \cos \theta = \dfrac{x}{r}$
Substitute $\cos \theta = \dfrac{x}{r}$ in $r = 2 - 2\cos \left( \theta \right)$, we get
$r = 2 - 2\cos \left( \theta \right)$
$\Rightarrow r = 2 - \dfrac{{2x}}{r}$
$\Rightarrow {r^2} = 2r - 2x$…(ii)
Also, we know that
${x^2} + {y^2} = {r^2} \Rightarrow \sqrt {{x^2} + {y^2}} = r$…(iii)
Now, substitute the value of ${r^2}$ and $r$ from (iii) in equation (ii).
${r^2} = 2r - 2x$
$\Rightarrow \left( {{x^2} + {y^2}} \right) = 2\sqrt {{x^2} + {y^2}} - 2x$
$\Rightarrow \left( {{x^2} + {y^2}} \right) + 2x = 2\sqrt {{x^2} + {y^2}}$
Square both sides of the equation, we get
$\Rightarrow {\left( {\left( {{x^2} + {y^2}} \right) + 2x} \right)^2} = 4\left( {{x^2} + {y^2}} \right)$
Now, use algebraic identity $\left( {{a^2} + {b^2}} \right) = {a^2} + 2ab + {b^2}$ to solve the equation.
$\Rightarrow {\left( {{x^2} + {y^2}} \right)^2} + 4x\left( {{x^2} + {y^2}} \right) + 4{x^2} = 4{x^2} + 4{y^2}$
$\Rightarrow {\left( {{x^2} + {y^2}} \right)^2} + 4x\left( {{x^2} + {y^2}} \right) - 4{y^2} = 0$

Final solution: Hence, the rectangular form of the given polar equation is ${\left( {{x^2} + {y^2}} \right)^2} + 4x\left( {{x^2} + {y^2}} \right) - 4{y^2} = 0$.

Additional information:
In two dimensions, the Cartesian coordinate $\left( {x,y} \right)$ specifics the location of a point P in the plane. A polar coordinate system in a plane consists of a fixed point $O$, called the pole (or origin), and a ray emanating from the pole, called the polar axis. In such a coordinate system we can associate with each point $P$ in the plane a pair of polar coordinates $\left( {r,\theta } \right)$, where $r$ is the distance from $P$ to the pole and $\theta$ is an angle from the polar axis to the ray $OP$. The number $r$ is called the radical coordinate of $P$ and the number $\theta$ the angular coordinate (or polar angle) of $P$.

Note:
In this problem our aim is to convert the given polar coordinate equation which is in the form $\left( {r,\theta } \right)$ into rectangular coordinates, for this we followed some important steps.
Relationship between Polar and Rectangular Coordinates:
Frequently, it will be useful to superimpose a rectangular $xy$-coordinate system on top of a polar coordinate system, making the positive $x$-axis coincide with the polar axis. If this is done, then every point P will have both rectangular coordinates $\left( {x,y} \right)$ and polar coordinates $\left( {r,\theta } \right)$. These coordinates are related by the equations
$x = r\cos \theta$, $y = r\sin \theta$…(1)
These equations are well suited for finding $x$ and $y$ when $r$ and $\theta$ are known. However, to find $r$ and $\theta$ when $x$ and $y$ are known, it is preferable to use the identities ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and $\tan \theta = \sin \theta /\cos \theta$ to rewrite (1) as
${r^2} = {x^2} + {y^2}$, $\tan \theta = \dfrac{y}{x}$…(2)