7.3 Polar coordinates

Calculus volume 2 Page 1 / 16

Locate points in a plane by using polar coordinates.
Convert points between rectangular and polar coordinates.
Sketch polar curves from given equations.
Convert equations between rectangular and polar coordinates.
Identify symmetry in polar curves and equations.

The rectangular coordinate system (or Cartesian plane) provides a means of mapping points to ordered pairs and ordered pairs to points. This is called a one-to-one mapping from points in the plane to ordered pairs. The polar coordinate system provides an alternative method of mapping points to ordered pairs. In this section we see that in some circumstances, polar coordinates can be more useful than rectangular coordinates.

Defining polar coordinates

To find the coordinates of a point in the polar coordinate system, consider [link] . The point $P$ has Cartesian coordinates $(x, y) .$ The line segment connecting the origin to the point $P$ measures the distance from the origin to $P$ and has length $r .$ The angle between the positive $x$ -axis and the line segment has measure $θ .$ This observation suggests a natural correspondence between the coordinate pair $(x, y)$ and the values $r$ and $θ .$ This correspondence is the basis of the polar coordinate system . Note that every point in the Cartesian plane has two values (hence the term ordered pair ) associated with it. In the polar coordinate system, each point also two values associated with it: $r$ and $θ .$

A point P(x, y) is given in the first quadrant with lines drawn to indicate its x and y values. There is a line from the origin to P(x, y) marked r and this line make an angle θ with the x axis. — An arbitrary point in the Cartesian plane.

Using right-triangle trigonometry, the following equations are true for the point $P :$

cos θ = \frac{x}{r} so x = r cos θ

sin θ = \frac{y}{r} so y = r sin θ .

Furthermore,

r^{2} = x^{2} + y^{2} and tan θ = \frac{y}{x} .

Each point $(x, y)$ in the Cartesian coordinate system can therefore be represented as an ordered pair $(r, θ)$ in the polar coordinate system. The first coordinate is called the radial coordinate and the second coordinate is called the angular coordinate . Every point in the plane can be represented in this form.

Note that the equation $tan θ = y / x$ has an infinite number of solutions for any ordered pair $(x, y) .$ However, if we restrict the solutions to values between $0$ and $2 π$ then we can assign a unique solution to the quadrant in which the original point $(x, y)$ is located. Then the corresponding value of r is positive, so $r^{2} = x^{2} + y^{2} .$

Converting points between coordinate systems

Given a point $P$ in the plane with Cartesian coordinates $(x, y)$ and polar coordinates $(r, θ),$ the following conversion formulas hold true:

x = r cos θ and y = r sin θ,

r^{2} = x^{2} + y^{2} and tan θ = \frac{y}{x} .

These formulas can be used to convert from rectangular to polar or from polar to rectangular coordinates.

Converting between rectangular and polar coordinates

Convert each of the following points into polar coordinates.

$(1, 1)$
$(−3, 4)$
$(0, 3)$
$(5 \sqrt{3}, −5)$

Convert each of the following points into rectangular coordinates.

$(3, π / 3)$
$(2, 3 π / 2)$
$(6, −5 π / 6)$

Use $x = 1$ and $y = 1$ in [link] :

$\begin{matrix} \begin{matrix} r^{2} & = & x^{2} + y^{2} \\ = & 1^{2} + 1^{2} \\ r & = & \sqrt{2} \end{matrix} & and & \begin{matrix} tan θ & = & \frac{y}{x} \\ = & \frac{1}{1} = 1 \\ θ & = & \frac{π}{4} . \end{matrix} \end{matrix}$

Therefore this point can be represented as $(\sqrt{2}, \frac{π}{4})$ in polar coordinates.
Use $x = −3$ and $y = 4$ in [link] :

$\begin{matrix} \begin{matrix} r^{2} & = & x^{2} + y^{2} \\ = & {(−3)}^{2} + {(4)}^{2} \\ r & = & 5 \end{matrix} & and & \begin{matrix} tan θ & = & \frac{y}{x} \\ = & - \frac{4}{3} \\ θ & = & − arctan (\frac{4}{3}) \\ \approx & 2.21. \end{matrix} \end{matrix}$

Therefore this point can be represented as $(5, 2.21)$ in polar coordinates.
Use $x = 0$ and $y = 3$ in [link] :

$\begin{matrix} \begin{matrix} r^{2} & = & x^{2} + y^{2} \\ = & {(3)}^{2} + {(0)}^{2} \\ = & 9 + 0 \\ r & = & 3 \end{matrix} & and & \begin{matrix} tan θ & = & \frac{y}{x} \\ = & \frac{3}{0} . \end{matrix} \end{matrix}$

Direct application of the second equation leads to division by zero. Graphing the point $(0, 3)$ on the rectangular coordinate system reveals that the point is located on the positive y -axis. The angle between the positive x -axis and the positive y -axis is $\frac{π}{2} .$ Therefore this point can be represented as $(3, \frac{π}{2})$ in polar coordinates.
Use $x = 5 \sqrt{3}$ and $y = −5$ in [link] :

$\begin{matrix} \begin{matrix} r^{2} & = & x^{2} + y^{2} \\ = & {(5 \sqrt{3})}^{2} + {(−5)}^{2} \\ = & 75 + 25 \\ r & = & 10 \end{matrix} & and & \begin{matrix} tan θ & = & \frac{y}{x} \\ = & \frac{−5}{5 \sqrt{3}} = - \frac{\sqrt{3}}{3} \\ θ & = & - \frac{π}{6} . \end{matrix} \end{matrix}$

Therefore this point can be represented as $(10, - \frac{π}{6})$ in polar coordinates.
Use $r = 3$ and $θ = \frac{π}{3}$ in [link] :

$\begin{matrix} \begin{matrix} x & = & r cos θ \\ = & 3 cos (\frac{π}{3}) \\ = & 3 (\frac{1}{2}) = \frac{3}{2} \end{matrix} & and & \begin{matrix} y & = & r sin θ \\ = & 3 sin (\frac{π}{3}) \\ = & 3 (\frac{\sqrt{3}}{2}) = \frac{3 \sqrt{3}}{2} . \end{matrix} \end{matrix}$

Therefore this point can be represented as $(\frac{3}{2}, \frac{3 \sqrt{3}}{2})$ in rectangular coordinates.
Use $r = 2$ and $θ = \frac{3 π}{2}$ in [link] :

$\begin{matrix} \begin{matrix} x & = & r cos θ \\ = & 2 cos (\frac{3 π}{2}) \\ = & 2 (0) = 0 \end{matrix} & and & \begin{matrix} y & = & r sin θ \\ = & 2 sin (\frac{3 π}{2}) \\ = & 2 (−1) = −2. \end{matrix} \end{matrix}$

Therefore this point can be represented as $(0, −2)$ in rectangular coordinates.
Use $r = 6$ and $θ = - \frac{5 π}{6}$ in [link] :

$\begin{matrix} \begin{matrix} x & = & r cos θ \\ = & 6 cos (- \frac{5 π}{6}) \\ = & 6 (- \frac{\sqrt{3}}{2}) \\ = & −3 \sqrt{3} \end{matrix} & and & \begin{matrix} y & = & r sin θ \\ = & 6 sin (- \frac{5 π}{6}) \\ = & 6 (- \frac{1}{2}) \\ = & −3. \end{matrix} \end{matrix}$

Therefore this point can be represented as $(−3 \sqrt{3}, −3)$ in rectangular coordinates.

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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