Precalculus

Analytic Geometry

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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.

Definition of polar coordinates
Figure 1.

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.

Points in polar coordinate system
Figure 2.

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.

A triangle given vertices in polar coordinates
Figure 3.

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

\[\rho = 4\cos\varphi.\]