Cartesian and polar coordinate system we have discussed earlier, see here ». In this article we will discuss about the conversion between them.[mathjax]

## Polar coordinate System

In polar coordinates, we determine position of a *Point* or any *Geometric* *Element* on a plane by radical distance $r$ from the pole $0$, and rotational angle $θ$ from the polar axis. In cartesian coordinates, we specify any position by left-right, up-down distance $(x, y)$ from the origin $(0, 0)$.

let $P$ is a point in polar coordinates, $P(r, θ)$. if we superposition the polar axis on a Cartesian plane, then the position of $P$ in Cartesian coordinates, $P(x, y)$. Therefore, according to trigonometric ratio, $sin θ=\frac{y}{r}$, $cos θ=\frac{x}{r}$, $tan θ=\frac{y}{x}$.

## Relation Between Cartesian and Polar Coordinates

According to Pythagorean-theorem, for any *right-triangle*, $r^{2}=x^{2}+y^{2}$. $r$ is the hypotenuse of right-triangle. $x$ is called Adjacent, and $y$ is the Opposite.

$tan θ=\frac{opposite}{adjacent}$, Or, $tan θ=\frac{y}{x}$.

Therefore, in our case, $r$ is the radical distance is polar-coordinates, $θ$, is the angle, and $x, y$ is the cartesian coordinate of the point, $P (x, y)$. So the basic relationship between polar and cartesian coordinates,

$x=r cos θ, \quad\text{} y=r sinθ$,

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

$tan θ=\frac{y}{x}$.

## Converting between rectangular and Polar – Equations

As-we-know the Relationship between Cartesian, and Polar Coordinate-system, therefore conversion between them is now easy. Let see some examples.

## Converting rectangular to polar – Equations

**Example 1**: Given an equation, $x^{2}+y^{2}=1$, transform it to Polar form.

* Solution*: We know, $r^{2}=x^{2}+y^{2}$, so, $r^{2}=1, \text{Or, }r=±1$. Hence, this is the polar form of given rectangular equation.

**Example 2**: Given rectangular Equation $y=2x+1$, convert it to polar equation.

** Solution**: Similarly, from above relationship, $x= r cos θ, \quad\text{}y=r sin θ$. Substitute these value to given equation, $r sin θ=2(r cos θ)+1,$ ⇒ $r sin θ-2r cos θ=1.$ Taking $r$ out, $r(sin θ-2cos θ)=1,$ finally,

$$r=\frac{1}{sin θ-2cos θ}.$$

This is the polar form of given rectangular equation.

**Example 3**: Given rectangular equation, $y=\frac{3}{x},$ convert it to polar equation.

* Solution*: Substitute, $x= r cos θ, \quad\text{}y=r sin θ,$ to above equation gives,

$$r sin θ=\frac{3}{r cos θ}, ⇒ r^{2}=\frac{3}{cos θ. sin θ}.$$

Similarly, This is the solution.

## Converting polar to rectangular – Equations

**Example 1**: Given Polar Equation, $r=3cos θ,$ convert it to rectangular form.

* Solution*: multiplying $r$ both side of the equation, $r.r = 3r.cos θ$.

As we know, $x= r cos θ\text{, }r=x^{2}+y^{2},$ therefore, $x^{2}+y^{2}=3x$. Finally, $x^{2}+y^{2}-3x=0$, is the rectangular form of the polar equation.

**Example 2**: Polar equation, $r^{2}=-3sec θ$, transform it to rectangular equation.

** Solution**: $r^{2}=-\frac{3}{cos θ},$ or, $\frac{1}{r}.r^{2}=-\frac{3}{r. cos θ}$. Or, $r=\frac{3}{x},$ because, $x=r.cos θ.$ Again, $r=\sqrt{x^{2}+y^{2}}$. Therefore, the rectangular equation, $\sqrt{x^{2}+y^{2}}=\frac{3}{x},$ or, $x^{2}+y^{2}=\frac{9}{x^{2}}.$

This is the end of discussion on cartesian or rectangular and polar coordinate system. And equation-conversion between them.