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6.3. Polar Coordinates

In Section 1, we introduced rectangular coordinates (also known as Cartesian coordinates). These are the most commonly used coordinates, but there are many others. Perhaps the second most common set of coordinates is polar coordinates. In polar coordinates, a point P is represented by its distance r from the origin O and the angle θ formed by the ray from the O to P and the positive x-axis. All this is shown in the following diagram.

Figure 6.3.1: Polar Coordinates

To make a graph in polar coordinates, it is necessary to use * polar graph paper*. This is graph paper with concentric circles centered about a point in the middle of the page and radial lines eminating outward in all directions from the central point (the origin). Circles centered about the origin are as easy to graph with polar graph paper as vertical or horizontal lines are with rectangular graph paper. Also, as we will see in the following section, many curves are easy to graph with polar graph paper which are very difficult to graph with rectangular graph paper.

It is useful to know how to transform between rectangular and polar coordinates. First we go from polar to rectangular coordinates. It is easy to see from the following diagram that the rectangular coordinates x and y are given in terms of r and θ by the following equations:

**(6.3.2a)**x = r cos θ**(6.3.2b)**y= r sin θ

Figure 6.3.2: Coordinate Transformation Diagram

To go the other way, note that the radial coordinate r is just the distance from O to P, which by the Pythagorean Theorem is equal to √x^{2} + y ^{2}. We also see that the tangent of the angle θ is equal to y / x, since y is the opposite side of the right triangle in the lower-right and x is the adjacent side. Thus we have the following transformation law for going from rectangular to polar coordinates:

**(6.3.3a)**r = √x^{2}+ y^{2}**(6.3.3b)**tan θ = y / x (r > 0)

Note that Equation (6.3.3b) does not quite pin down θ, since tan θ remains unchanged if we add or subtract π from θ. In order to get around this ambiguity, it is also necessary to look at the signs of x and y. If y is positive, then we know that P lies in the upper half-plane (the region above the x-axis), whence θ is between 0 and π. On the other hand, if y is negative, then P lies in the lower half-plane (the region below the x-axis), whence θ is between π and 2π. What if y = 0? In this case, if x is positive, then P lies on the positive x-axis and we have θ = 0. On the other hand, if x is negative, then P lies on the negative x-axis and we have θ = π. The only other case to consider is the special case x = y = 0 (r = 0), corresponding to the origin. For this special point, θ is undefined.

What if x = 0 in (6.3.3b)? Then tan θ = y / x is undefined, implying that θ is either π/2 or 3π/2. Which one it is depends on the sign of y; if y is positive, then θ = π/2 and if y is negative, then θ = 3π/2.

**Example 1: **Convert the following points from polar to rectangular coordinates

- (a) (1, 0)
- (b) (2, π)
- (c) (5, π/2)
- (d) (7, 3π/2)
- (e) (√2, π/4)
- (f) (2, π/6)
- (g) (2, 3π/4)
- (h) (2, 5π/4)
- (i) (6, 5π/3)
- (j) (5, arctan(4/3))

**Solution: **To convert to rectangular coordinates we use equations (6.3.2).

- (a) We have r = 1 and θ = 0, whence x = 1 cos 0 = 1 and y = 1 sin 0 = 0. Thus the rectangular coordinate representation of this point is (1,0).
- (b) We have r = 2 and θ = π, whence x = 2 cos π = -2 and y = 2 sin π = 0. Thus the rectangular coordinate representation of this point is (-2,0).
- (c) We have r = 5 and θ = π/2, whence x = 5 cos π/2 = 0 and y = 5 sin π/2 = 5. Thus the rectangular coordinate representation of this point is (0,5).
- (d) We have r = 7 and θ = 3π/2, whence x = 7 cos 3π/2 = 0 and y = 7 sin 3π/2 = -7. Thus the rectangular coordinate representation of this point is (0,-7).
- (e) We have r = √2 and θ = π/4, whence x = √2 cos π/4 = (√2)(√2/2) = 1 and y = √2 sin π/4 = (√2)(√2/2) = 1. Thus the rectangular coordinate representation of this point is (1,1).
- (f) We have r = 2 and θ = π/6, whence x = 2 cos π/6 = (2)(√3/2) = √3 and y = 2 sin π/6 = (2)(1/2) = 1 . Thus the rectangular coordinate representation of this point is (√3,1).
- (g) We have r = 2 and θ = 3π/4, whence x = 2 cos 3π/4 = (2)(-√2/2) = - √2 and y = 2 sin 3π/4 = (2)(√2/2) = √2. Thus the rectangular coordinate representation of this point is (-√2, √2).
- (h) We have r = 2 and θ = 5π/4, whence x = 2 cos 5π/4 = (2)(-√2/2) = - √2 and y = 2 sin 5π/4 = (2)(-√2/2) = - √2. Thus the rectangular coordinate representation of this point is (-√2, -√2).
- (i) We have r = 6 and θ = 5π/3, whence x = 6 cos 5π/3 = (6)(1/2) = 3 and y = 6 sin 5π/3 = (6)(-√3/2) = - 3√3. Thus the rectangular coordinate representation of this point is (3, -√3).
- (j) We have r = 5 and θ = arctan(4/3), whence x = 5 cos(arctan(4/3)) and y = 5 sin(arctan(4/3)). How can we simplify these expressions? The trick is not to simplify them directly but rather to use properties of the inverse coordinate transformation, from rectangular to polar coordinates. Thus, for instance, we see that r = √x
^{2}+ y^{2}= 5 and tan θ = y / x = 4/3. Thus we must solve the simultaneous equations x^{2}+ y^{2}= 25 and y / x = 4/3 for x and y. Writing x = 3t and y = 4t yields x^{2}+ y^{2}= 25t^{2}= 25, whence t = ±1. This yields the solutions (3,4) and (-3,-4). Which one is correct? Note that θ = arctan(4/3) is between 0 and π/2, implying that x and y are both positive. Thus the rectangular coordinate representation of the point is (3,4).

**Example 2: **Convert the following points from rectangular to polar coordinates

- (a) (5,0)
- (b) (1,1)
- (c) (0,2)
- (d) (-π, -π)
- (e) (1/2, √3/2)
- (f) (-√3, -1)
- (g) (-4,-3)
- (h) (5,-12)

**Solution: **To convert to polar coordinates we use equations (6.3.3).

- (a) We have x = 5 and y = 0, whence r = √5
^{2}+ 0^{2}= 5 and tan θ = 0 / 5 = 0, implying either θ = 0 or θ = π. To determine which one it is, we note that y = 0 and x > 0, which together imply θ = 0. Thus, the polar coordinate representation of this point is (5,0). (It is worth noting that the rectangular and polar representation of a point are both the same if and only if the point lies on the positive x-axis. We leave the proof of this fact as an exercise.) - (b) We have x = y = 1, whence r = √1
^{2}+ 1^{2}= √2^{}and tan θ = 1 / 1 = 1, implying either θ = π/4 or θ = 5π/4. Since y is positive, we see that θ = π/4 is the correct solution. Thus, the polar coordinate representation of this point is (√2, π/4). - (c) We have x = 0 and y = 2, whence r = √0
^{2}+ 2^{2}= 2 and tan θ = 2 / 0 is undefined. Since y is positive, this implies that θ = π/2. Thus, the polar coordinate representation of this point is (2, π/2). - (d) We have x = y = -π, whence r = √(-π)
^{2}+ (-π)^{2}= π√2^{}and tan θ = (-π) / (-π) = 1, implying either θ = π/4 or θ = 5π/4. Since y is negative, we see that θ = 5π/4 is the correct solution. Thus, the polar coordinate representation of this point is (π√2, 5π/4). - (e) We have x = 1/2 and y = √3/2, whence r = √(1/2)
^{2}+ (√3/2)^{2}= 1^{}and tan θ = (√3/2) / (1/2) = √3, implying either θ = π/3 or θ = 4π/3. Since y is positive, we see that θ = π/3 is the correct solution. Thus, the polar coordinate representation of this point is (1, π/3). - (f) We have x = -√3 and y = -1, whence r = √(-√3)
^{2}+ 1^{2}= 2^{}and tan θ = (-1) / (-√3) = √3/3, implying either θ = π/6 or θ = 7π/6. Since y is negative, we see that θ = 7π/6 is the correct solution. Thus, the polar coordinate representation of this point is (2, 7π/6). - (g) We have x = -4 and y = -3, whence r = √(-4)
^{2}+ (-3)^{2}= 5 and tan θ = 3/4, implying either θ = arctan(3/4) or θ = π + arctan(3/4). To see which solution is correct, we note that y is negative, implying that θ is between π and 2π. Since only the second solution satisfies this condition, we see that θ = π + arctan(3/4). Thus, the polar coordinate representation of this point is (5, π + arctan(3/4)). - (h) We have x = 5 and y = -12, whence r = √5
^{2}+ (-12)^{2}= 13 and tan θ = -12 / 5, implying either θ = π + arctan(-12/5) = π - arctan(12/5) or θ = 2π - arctan(12/5). To see which solution is correct, we note that y is negative, implying that θ is between π and 2π. Since only the second solution satisfies this condition, we see that θ = 2π - arctan(12/5). Thus, the polar coordinate representation of this point is (13, 2π - arctan(12/5)).

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