# Precalculus

## Analytic Geometry # Polar Coordinates

Polar coordinates are defined as follows: we select some point O on the plane, which is called the pole, and a ray coming out of it and called the polar axis. In addition, we specify a unit of measure.

Suppose we're given a polar coordinate system and let M be an arbitrary point on the plane distinct from O. We denote by ρ the distance of the point M from the pole O, and by φ the angle by which you need to turn the polar axis counterclockwise to align with the ray OM.

The numbers ρ and φ are called the polar coordinates of the point M. The number ρ is called the polar radius, and the number φ is called the polar angle. The symbol $$M\left({\rho,\varphi}\right)$$ indicates that the point M has polar coordinates ρ and φ.

It is usually considered that ρ and φ change within the following limits:

$0 \le \rho \lt +\infty,\;\;0 \le \varphi \lt 2\pi.$

However, in some cases, angles greater than 2π are considered, as well as negative ones, that is, angles counted from the polar axis clockwise.

## Converting Between Polar and Cartesian Coordinates

Let the origin of the Cartesian coordinate system is at the pole and the positive $$x-$$axis coincides with the polar axis. Let point M have Cartesian coordinates $$x$$ and $$y$$ and polar coordinates $$\rho$$ and $$\varphi.$$ The conversion from polar coordinates of the point M to Cartesian (rectangular) ones is carried out by the formulas

$x = \rho\cos\varphi,\;\;y = \rho\sin\varphi.$

The expression of polar coordinates in terms of Cartesian ones follows from these formulas:

$\rho = \sqrt{x^2 + y^2},\;\;\tan\varphi = \frac{y}{x}.$

## Equations in Polar Coordinates

Any equation in the variables $$\rho$$ and $$\varphi$$ can be interpreted as an equation in polar coordinates. Usually the radius $$\rho$$ is considered as a function and the angle $$\varphi$$ as an independent variable. In explicit form, a polar function is given by the equation

$\rho = \rho\left({\varphi}\right).$

### Examples of some famous polar curves:

$\left.\begin{array}{l} \text{Cardioid:} & \rho = 1 + \cos\varphi\\ \text{Archimedean spiral:} & \rho = a\varphi\\ \text{Logarithmic spiral:} & \rho = ae^{b\varphi}\\ \text{Hyperbolic spiral:} & \rho = \frac{a}{\varphi}\\ \text{Lemniscate of Bernoulli:} & \rho = \sqrt{\cos2\varphi}\\ \text{Galileo's spiral:} & \rho = a\varphi^2\\ \text{Fermat's spiral:} & \rho = \sqrt{\varphi} \end{array} \right.$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Plot points given by polar coordinates

1. $$A\left({3,\frac{\pi}{4}}\right)$$
2. $$B\left({2,\frac{3\pi}{2}}\right)$$
3. $$C\left({5,\frac{5\pi}{4}}\right)$$
4. $$D\left({4,0}\right)$$
5. $$E\left({4,-\frac{5\pi}{4}}\right)$$

### Example 2

Given points in a Cartesian coordinate system:

1. $$A\left({0,2}\right)$$
2. $$B\left({-3,0}\right)$$
3. $$C\left({1,\sqrt{3}}\right)$$
4. $$D\left({\sqrt{5},-\sqrt{5}}\right)$$

Find their polar coordinates.

### Example 3

Given points in a polar coordinate system:

1. $$A\left({4,\frac{\pi}{4}}\right)$$
2. $$B\left({3,\frac{\pi}{2}}\right)$$
3. $$C\left({\sqrt{2},\frac{3\pi}{4}}\right)$$
4. $$D\left({5,\pi}\right)$$

Find their Cartesian (rectangular) coordinates.

### Example 4

The vertex $$O$$ of a triangle $$OAB$$ is at the pole and the other two are points $$A\left({5,\frac{\pi}{4}}\right)$$ and $$B\left({6,\frac{\pi}{12}}\right).$$ Find the area of the triangle.

### Example 5

Transform the polar coordinate equation to Cartesian coordinates: $\rho = 2\sin\varphi.$

### Example 6

Transform the Cartesian coordinate equation to polar coordinates: $x^2 + y^2 = 4x.$

### Example 1.

Plot points given by polar coordinates

1. $$A\left({3,\frac{\pi}{4}}\right)$$
2. $$B\left({2,\frac{3\pi}{2}}\right)$$
3. $$C\left({5,\frac{5\pi}{4}}\right)$$
4. $$D\left({4,0}\right)$$
5. $$E\left({4,-\frac{5\pi}{4}}\right)$$

Solution.

To plot a point in polar coordinates, we can start with the angle $$\varphi$$ which is measured from the polar axis. A positive angle goes counterclockwise. Respectively, a negative angle goes clockwise. The distance $$\rho$$ is counted out from the pole along the ray determined by the angle $$\varphi$$ (the terminal side of the angle). The points are shown in figure below.

### Example 2.

Given points in a Cartesian coordinate system:

1. $$A\left({0,2}\right)$$
2. $$B\left({-3,0}\right)$$
3. $$C\left({1,\sqrt{3}}\right)$$
4. $$D\left({\sqrt{5},-\sqrt{5}}\right)$$

Find their polar coordinates.

Solution.

For each point we calculate the polar radius $$\rho$$ and angle $$\varphi$$ using the formulas

$\rho = \sqrt{x^2 + y^2},\;\;\tan\varphi = \frac{y}{x}.$
1. $$A\left({0,2}\right)$$
$\rho = \sqrt{0^2 + 2^2} = 2,\;\tan\varphi = \frac{2}{0} = \infty, \Rightarrow \varphi = \frac{\pi}{2};$
$A\left({0,2}\right) \mapsto A\left({2,\frac{\pi}{2}}\right);$
2. $$B\left({-3,0}\right)$$
$\rho = \sqrt{\left({-3}\right)^2 + 0^2} = 3,\;\tan\varphi = \frac{0}{-3} = 0, \Rightarrow \varphi = 0;$
$B\left({-3,0}\right) \mapsto B\left({3,0}\right);$
3. $$C\left({1,\sqrt{3}}\right)$$
$\rho = \sqrt{1^2 + \left({\sqrt{3}}\right)^2} = 2,\;\tan\varphi = \frac{\sqrt{3}}{1} = \sqrt{3}, \Rightarrow \varphi = \frac{\pi}{3};$
$C\left({1,\sqrt{3}}\right) \mapsto C\left({2,\frac{\pi}{3}}\right);$
4. $$D\left({\sqrt{5},-\sqrt{5}}\right)$$
$\rho = \sqrt{\left({\sqrt{5}}\right)^2 + \left({-\sqrt{5}}\right)^2} = \sqrt{10},\;\tan\varphi = \frac{-\sqrt{5}}{\sqrt{5}} = -1, \Rightarrow \varphi = \frac{3\pi}{4};$
$D\left({\sqrt{5},-\sqrt{5}}\right) \mapsto D\left({\sqrt{10},\frac{3\pi}{4}}\right).$

### Example 3.

Given points in a polar coordinate system:

1. $$A\left({4,\frac{\pi}{4}}\right)$$
2. $$B\left({3,\frac{\pi}{2}}\right)$$
3. $$C\left({\sqrt{2},\frac{3\pi}{4}}\right)$$
4. $$D\left({5,\pi}\right)$$

Find their Cartesian (rectangular) coordinates.

Solution.

We calculate the Cartesian coordinates $$\left({x,y}\right)$$ by the formulas

$x = \rho\cos\varphi,\;\;y = \rho\sin\varphi.$
1. $$A\left({4,\frac{\pi}{4}}\right)$$
$x = 4\cos\frac{\pi}{4} = 4\cdot \frac{\sqrt{2}}{2} = 2\sqrt{2},\;y = 4\sin\frac{\pi}{4} = 4\cdot \frac{\sqrt{2}}{2} = 2\sqrt{2};$
$A\left({4,\frac{\pi}{4}}\right) \mapsto A\left({2\sqrt{2},2\sqrt{2}}\right);$
2. $$B\left({3,\frac{\pi}{2}}\right)$$
$x = 3\cos\frac{\pi}{2} = 0,\;y = 3\sin\frac{\pi}{2} = 3;$
$B\left({3,\frac{\pi}{2}}\right) \mapsto B\left({0,3}\right);$
3. $$C\left({\sqrt{2},\frac{3\pi}{4}}\right)$$
$x = \sqrt{2}\cos\frac{3\pi}{4} = \sqrt{2}\cdot \left({-\frac{\sqrt{2}}{2}}\right) = -1,\;y = \sqrt{2}\sin\frac{3\pi}{4} = \sqrt{2}\cdot \frac{\sqrt{2}}{2} = 1;$
$C\left({\sqrt{2},\frac{3\pi}{4}}\right) \mapsto C\left({-1,1}\right);$
4. $$D\left({5,\pi}\right)$$
$x = 5\cos\pi = 5\cdot \left({-1}\right) = -5,\;y = 5\sin\pi = 5\cdot 0 = 0;$
$D\left({5,\pi}\right) \mapsto D\left({-5,0}\right).$

### Example 4.

The vertex $$O$$ of a triangle $$OAB$$ is at the pole and the other two are points $$A\left({5,\frac{\pi}{4}}\right)$$ and $$B\left({6,\frac{\pi}{12}}\right).$$ Find the area of the triangle.

Solution.

Calculate the angle $$AOB:$$

$\angle AOB = \frac{\pi}{4} - \frac{\pi}{12} = \frac{3\pi - \pi}{12} = \frac{2\pi}{12} = \frac{\pi}{6}.$

The lengths of sides $$OA$$ and $$OB$$ are known: $$OA = 5,$$ $$OB = 6.$$ Then the area of the triangle can be calculated by the formula

$A = \frac{1}{2}OA \cdot OB \cdot \sin \left({AOB}\right).$

Substituting the values we get

$A = \frac{1}{2} \cdot 5 \cdot 6\sin \left({\frac{\pi}{6}}\right) = 15\sin \left({\frac{\pi}{6}}\right) = 15\cdot\frac{1}{2} = \frac{15}{2}.$

### Example 5.

Transform the polar coordinate equation to Cartesian coordinates: $\rho = 2\sin\varphi.$

Solution.

Multiply both sides of the equation by $$\rho:$$

$\rho = 2\sin\varphi, \Rightarrow \rho^2 = 2\rho\sin\varphi.$

This adds another root $$\rho = 0$$ to the equation, which, however, coincides with the solution $$\rho\left({\varphi = 0}\right) = 0$$ of the original equation. Now we can apply the conversion formulas

$\rho^2 = x^2 + y^2,\;x=\rho\sin\varphi.$

This yields:

$x^2 + y^2 = 2x.$

Let's do some algebra:

$x^2 + y^2 = 2x, \Rightarrow x^2 -2x + y^2 = 0, \Rightarrow \left({x^2 -2x + 1}\right) + y^2 = 1, \Rightarrow \left({x - 1}\right)^2 + y^2 = 1.$

We got the equation of a circle with center at point $$\left({1,0}\right)$$ and radius $$1.$$

### Example 6.

Transform the Cartesian coordinate equation to polar coordinates: $x^2 + y^2 = 4x.$

Solution.

Using the conversion formulas we get

$x^2 + y^2 = 4x, \Rightarrow \rho^2 = 4\rho\cos\varphi, \Rightarrow \rho\left({\rho - 4\cos\varphi}\right) = 0.$

Notice that here the root $$\rho = 0$$ coincides with the solution of another equation $$\rho - 4\cos\varphi = 0$$ since

$\rho\left({\varphi = \frac{\pi}{2}}\right) = 4\cos\frac{\pi}{2} = 4\cdot0 = 0.$

Therefore, cancelling by $$\rho,$$ we get the final equation in polar coordinates in the form

$\rho = 4\cos\varphi.$