Lesson Explainer: Polar Coordinates Mathematics • Higher Education

In this explainer, we will learn how to define and plot points given in polar coordinates and convert between the Cartesian and polar coordinates of a point.

When we think about points in a plane, we usually think of Cartesian coordinates as this is the most prevalent coordinate system. In particular, the Cartesian coordinates of a point are used in linear motion where specifying the axis of motion is simple and where the motion will take a linear path to a particular location.

Cartesian coordinates, in two dimensions, define a position as the linear displacement from the origin in two mutually perpendicular axes. The origin is the point where the axes intersect, and the points on the plane are specified by a pair of numbers (π‘₯,𝑦). This coordinate system also allows both positive and negative directions, relative to the origin, and each coordinate set defines a unique point in space.

Recall that when using Cartesian coordinates, we mark a point by how far along the horizontal axis it is (left or right), denoted by π‘₯, and how far along the vertical axis it is (up or down), denoted by 𝑦, relative to the origin.

In Cartesian coordinates, any point in space can be defined by a unique set of coordinates given by the pair of numbers (π‘₯,𝑦). However, there are other ways of representing the position of a point in the plane using a coordinate pair; we will explore one such way known as polar coordinates. These coordinates define a position in space using a combination of radial and angular units and a vector is specified by a straight-line displacement from the origin and the angle from the positive π‘₯-axis. These are known as the radial and angular coordinates (π‘Ÿ,πœƒ), which indicate the displacement from the origin and the angular direction.

Polar coordinates provide us with an alternative way of plotting points and drawing graphs; we can often express complicated graphs using simple polar functions. For example, π‘Ÿ=1 represents all points of displacement of one from the origin, which is the unit circle centered at the origin. In Cartesian coordinates, this is described by the curve π‘₯+𝑦=1.

Polar coordinates are naturally used in nonlinear motion, for example, if the motion involves a circular path. This makes polar coordinates useful in calculating the equations of motion for a lot of mechanical systems. It also has other real-world applications, such as in plan position indicators in radars and in gravitational fields the characteristics of a microphone and guiding industrial robots in various production applications, to name a few.

In polar coordinates, we mark a point by its displacement from the origin, denoted by π‘Ÿ, and its angle from the positive π‘₯-axis, denoted by πœƒ.

These are equivalent ways of defining the same point.

The modulus of the radial coordinate, |π‘Ÿ|, is equal to the length or distance from the origin. By the Pythagorean theorem, the distance between two points (π‘₯,𝑦) and (π‘Ž,𝑏) is given by 𝐿=(π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘).

Thus, for any points (π‘₯,𝑦) and the origin (π‘Ž,𝑏)=(0,0), this distance is given by 𝐿=√π‘₯+𝑦.

The radial coordinate is defined as π‘Ÿβ‰‘Β±πΏ.

If π‘Ÿ is negative, it means that the point lies in the quadrant on the opposite side of the pole. We also note that the modulus of the radial coordinate is equal to the length since |π‘Ÿ|=𝐿 and 𝐿>0.

Since we can form a right triangle from the coordinates, we express the sides of the triangle in terms of sinπœƒ and cosπœƒ.

This also allows us to express the Cartesian coordinates (π‘₯,𝑦) in terms of (π‘Ÿ,πœƒ).

Definition: Converting from Polar Coordinates to Cartesian Coordinates

The Cartesian coordinates (π‘₯,𝑦) can be written in terms of polar coordinates (π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

So, if we are given a point in polar coordinates, the displacement from the origin π‘Ÿ, and the angle πœƒ, we can determine the point in Cartesian coordinates, π‘₯ and 𝑦, using these equations.

As an example, let’s convert a point from polar coordinates, in radians, to Cartesian coordinates, for an acute angle located in the first quadrant.

Example 1: Finding the Cartesian Coordinates of a Point Given in Polar Coordinates

Find the Cartesian coordinates of the point ο€»1,πœ‹4.

Answer

In this example, we want to determine the Cartesian coordinates for a particular polar coordinate ο€»1,πœ‹4 located in the first quadrant, since πœ‹4βˆˆο€»0,πœ‹2.

Recall that polar coordinates define a point according to the displacement from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

For any given point in polar coordinates (π‘Ÿ,πœƒ) , we can find the equivalent point in Cartesian coordinates, (π‘₯,𝑦), using the formulae π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

Thus, substituting the radial coordinate, π‘Ÿ=1, and the angular coordinate, πœƒ=πœ‹4, we have π‘₯=1Γ—ο€»πœ‹4=√22,cos and 𝑦=1Γ—ο€»πœ‹4=√22.sin

Therefore, the given point in Cartesian coordinates is ο€Ώβˆš22,√22.

The counterclockwise angle is taken to be positive, while the clockwise angle is negative. In previous examples and diagrams, the angular components have been acute, since the points were in the first quadrant. If a point is in another quadrant, as shown in the diagram below, its angle is not acute.

In fact, although we derived the equations for the polar coordinates π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒcossin from acute angles ο€»0β‰€πœƒ<πœ‹2, we know that they remain true for any angle πœƒ.

Let’s consider an example where we convert a point that lies in the second quadrant with a nonacute angle from polar coordinates, in degrees, to Cartesian coordinates.

Example 2: Finding the Cartesian Coordinates of a Point Given in Polar Coordinates

Given that the polar coordinates of point 𝐴 are (4,120)∘, find the Cartesian coordinates of 𝐴.

Answer

In this example, we want to determine the Cartesian coordinates for a particular polar coordinate (4,120)∘ located in the second quadrant.

Recall that polar coordinates define a point according to the displacement from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

For any given point in polar coordinates (π‘Ÿ,πœƒ) , we can find the equivalent point in Cartesian coordinates, (π‘₯,𝑦), using the formulae π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

Thus, substituting the radial coordinate, π‘Ÿ=4, and the angular coordinate, πœƒ=120∘, we have π‘₯=4120=βˆ’2,cos∘ and 𝑦=4120=2√3.sin∘

Therefore, the point 𝐴 in Cartesian coordinates is ο€»βˆ’2,2√3.

So far, we have seen examples of how to convert a point from polar coordinates to Cartesian coordinates, using trigonometry. But what if we wanted to do the reverse, that is, convert a point from Cartesian coordinates into polar coordinates?

Let’s begin by recalling the equations expressing the components of the Cartesian coordinates, π‘₯ and 𝑦, in terms of the components of the polar coordinates, π‘Ÿ and πœƒ: π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

Using these, we want to write the components of polar coordinates, π‘Ÿ and πœƒ, in terms of the components of the Cartesian coordinates, π‘₯ and 𝑦, no matter which quadrant the point lies in.

If we take the square of each of these and add them up, by using the Pythagorean identity, we can eliminate πœƒ and show that these satisfy π‘₯+𝑦=π‘Ÿπœƒ+π‘Ÿπœƒ=π‘Ÿο€Ίπœƒ+πœƒο†=π‘Ÿ.cossincossin

Also, when we divide the equation for 𝑦 by the equation for π‘₯, we can cancel the π‘Ÿ that appears, for π‘Ÿβ‰ 0, to obtain 𝑦π‘₯=π‘Ÿπœƒπ‘Ÿπœƒ=πœƒπœƒ=πœƒ.sincossincostan

We note that this only holds true for π‘₯β‰ 0. We have a special case when π‘₯=0, 𝑦=π‘Žβˆˆβ„ or (0,π‘Ž) in the Cartesian coordinates. For this, we have π‘Ÿπœƒ=0,cos which leads to πœƒ=πœ‹2, πœƒ=βˆ’πœ‹2, or π‘Ÿ=0. We can ignore the case π‘Ÿ=0, since this would imply 𝑦=0, which corresponds to the origin (0,0), which in polar coordinates is denoted by (0,πœƒ), for any angle πœƒ.

Thus, πœƒ=πœ‹2 or πœƒ=βˆ’πœ‹2, which correspond to the 𝑦-axis. These angles place the point on the 𝑦-axis and one possible representation for the radial coordinate (for π‘Ÿ>0) is equal to the absolute value of the 𝑦-coordinate, π‘Ÿ=|π‘Ž|. A representation of the polar coordinates is ο€»π‘Ž,πœ‹2, for π‘Ž>0, and ο€»βˆ’π‘Ž,βˆ’πœ‹2, for π‘Ž<0.

So, if π‘₯β‰ 0, we have the following equation to determine the angle πœƒ: tanπœƒ=𝑦π‘₯.

The range for the inverse tangent function is ο€»βˆ’πœ‹2,πœ‹2 when the domain of the tangent function is restricted to the same interval, known as a principal branch. This is to ensure that the tangent function is one to one so that the inverse tangent function evaluates to a single value, known as the principal value.

Thus, as long as πœƒβˆˆο€»βˆ’πœ‹2,πœ‹2, we can take the inverse tangent of both sides of the equation to obtain πœƒ=𝑦π‘₯.tan

The angular coordinates πœƒβˆˆο€»βˆ’πœ‹2,πœ‹2 correspond to the first and fourth quadrants, or the quadrants where π‘₯>0.

In the next example, we will convert a point from Cartesian coordinates to polar coordinates, in degrees.

Example 3: Converting Coordinates into Polar Coordinates

Convert (2,3) to polar coordinates. Give the angle in degrees and round your answer to one decimal place.

Answer

In this example, we want to determine the polar coordinates, in degrees, for the particular Cartesian coordinates.

Recall that polar coordinates define a point according to the displacement from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

The Cartesian coordinates (π‘₯,𝑦) can be expressed in terms of the polar coordinates (π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

Now, let’s find the polar coordinates of the point (2,3) by using the graphical representation directly with the definition. The radial coordinate is just the displacement from the origin to the point (2,3), which we can find by using the Pythagorean theorem on the right triangle, as shown the diagram. We want to find the hypotenuse of this triangle using π‘Ÿ=√2+3=√13=3.605551….

Since the point (2,3) is located in the first quadrant, the angular coordinate πœƒ in polar coordinates will be the positive counterclockwise angle from the positive π‘₯-axis, as shown in the diagram. We can form a right triangle with angle πœƒ and sides of lengths 2 and 3. Since πœƒ is an acute angle, we can write this in terms of the sides using the inverse tangent as πœƒ=ο€Ό32=56.309932….tan∘

We could have also arrived at this answer by using the fact that we can convert the Cartesian coordinate (π‘₯,𝑦) located in the first quadrant into the polar coordinates (π‘Ÿ,πœƒ) by using π‘Ÿ=√π‘₯+𝑦,πœƒ=𝑦π‘₯,0<πœƒ<90.∘∘tanfor

This gives the same radial and angular coordinate after substituting π‘₯=2 and 𝑦=3.

Thus, to one decimal place, the polar coordinates are (3.6,56.3).∘

In the previous example, we converted a point in Cartesian coordinates that lies in the first quadrant into polar coordinates. As seen in this example, we can compute the angular coordinate of a point by using πœƒ=𝑦π‘₯,tan in the first and fourth quadrants. However, this is no longer the case if the point lies in the second or third quadrant.

For the second and third quadrants, a value of πœ‹, in radians, or 180∘, in degrees, must be added to or subtracted from the angle πœƒ, to adjust the angular coordinate so that the point lies in the correct quadrant. This does not affect the tangent function itself since we have the identities tantantanπœƒ=(πœƒ+180)=(πœƒβˆ’180),∘∘ or, more generally, tantanπœƒ=(πœƒ+180π‘˜),π‘˜βˆˆβ„€.∘

To see this, consider a point in Cartesian coordinates that lies in the second quadrant, (π‘₯,𝑦)=(βˆ’π‘Ž,𝑏), with π‘Ž>0 and 𝑏>0.

The angular coordinate πœƒ for this point in polar coordinates will be the positive counterclockwise angle from the positive π‘₯-axis and angle 𝛼 is measured from the negative π‘₯-axis, as shown in the diagram. We can form a right triangle with angle 𝛼 and sides of lengths π‘Ž and 𝑏. Since 𝛼 is an acute angle, we can write this in terms of the sides using the inverse tangent as 𝛼=ο€½π‘π‘Žο‰.tan

We also have 𝛼+πœƒ=180∘, and substituting angle 𝛼, we can rearrange this to find πœƒ=βˆ’π›Ό+180=βˆ’ο€½π‘π‘Žο‰+180=ο€½π‘βˆ’π‘Žο‰+180,∘∘∘tantan where, for the last equality, we used the fact that the tangent function, and hence the inverse tangent function, is an odd function. Since we have (π‘₯,𝑦)=(βˆ’π‘Ž,𝑏), this is equivalent to πœƒ=𝑦π‘₯+180.tan∘

In the next example, we will find the polar coordinates of a particular point that lies in the second quadrant.

Example 4: Converting Coordinates into Polar Coordinates

Convert (βˆ’2,5) to polar coordinates. Give the angle in radians and round your answer to two decimal places.

Answer

In this example, we want to determine the polar coordinates (π‘Ÿ,πœƒ), in radians, for the particular Cartesian coordinates (βˆ’2,5).

Recall that polar coordinates define a point according to the displacement from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

The Cartesian coordinates (π‘₯,𝑦) can be expressed in terms of the polar coordinates (π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

Now, let’s find the polar coordinates of the point (βˆ’2,5) by using the graphical representation directly with the definition. The radial coordinate is just the displacement from the origin to the point (βˆ’2,5), which we can find from using the Pythagorean theorem on the right triangle, as shown the diagram. We want to find the hypotenuse of this triangle using π‘Ÿ=√2+5=√29=5.385164….

Since the point (βˆ’2,5) is located in the second quadrant, the angular coordinate πœƒ in polar coordinates will be the positive counterclockwise angle from the positive π‘₯-axis and angle 𝛼 is measured from the negative π‘₯-axis, as shown in the diagram. We can form a right triangle with angle 𝛼 and sides of lengths 2 and 5. Since 𝛼 is an acute angle, we can write this in terms of the sides using the inverse tangent as 𝛼=ο€Ό52.tan

We also have 𝛼+πœƒ=πœ‹, and substituting the angle 𝛼, we can rearrange this to find πœƒ=βˆ’π›Ό+πœ‹=βˆ’ο€Ό52+πœ‹=1.951302….tan

We could have also arrived at this answer by using the fact that we can convert the Cartesian coordinate (π‘₯,𝑦) located in the second quadrant into the polar coordinates (π‘Ÿ,πœƒ) by using π‘Ÿ=√π‘₯+𝑦,πœƒ=𝑦π‘₯+πœ‹,πœ‹2<πœƒ<πœ‹.tanfor

This gives the same radial and angular coordinate after substituting π‘₯=βˆ’2 and 𝑦=5.

Thus, the polar coordinates rounded to two decimal places is given by (5.39,1.95).

For the third quadrant we can show, in a similar way, that we have to subtract 180∘ from tanοŠ±οŠ§ο€»π‘¦π‘₯ to get the angular coordinate πœƒ in the correct quadrant.

Polar coordinates are not unique and there are many ways of representing the same point. As an example, let’s determine the polar coordinates of the point (1,1) in Cartesian coordinates.

The radial coordinate π‘Ÿ is the displacement from the origin, which we can determine using the Pythagorean theorem on the right triangle with sides of length 1 and angle πœƒ. In particular, π‘Ÿ=√1+1=√2, which is a particular choice for the radial coordinates with π‘Ÿ>0, but we could also use π‘Ÿ=βˆ’βˆš2.

There are many ways to express the angular coordinate πœƒ. One is the positive counterclockwise angle from the positive π‘₯-axis, which from the right triangle gives us the tangent in terms of the ratio of the opposite and adjacent side: tanπœƒ=11=1.

Since the point lies in the first quadrant and πœƒ is an acute angle in the diagram, we can find the angle directly from the inverse tangent: πœƒ=(1)=45.tan∘

Thus, a polar coordinate to describe the point (1,1) is ο€»βˆš2,45ο‡βˆ˜.

Another polar coordinate can be found if we use the negative clockwise angle from the positive π‘₯-axis which would give an equivalent point as ο€»βˆš2,45βˆ’360=ο€»βˆš2,βˆ’315ο‡βˆ˜βˆ˜βˆ˜. In fact, if we make a full revolution from this point, in either the clockwise or counterclockwise direction, we return back to the same point. So, another representation would be ο€»βˆš2,45+360=ο€»βˆš2,405ο‡βˆ˜βˆ˜βˆ˜.

This shows a key difference when using polar coordinates, as it allows for an infinite number of coordinate pairs to describe any given point. This is because we can add any integer multiple of a full revolution (360∘ or 2πœ‹) to the angular coordinate πœƒ to get an equivalent point in polar coordinates. This follows because the trigonometric functions, which are used to define polar coordinates, are themselves periodic.

The radial coordinate can also be positive or negative, and when negative radial coordinates are used, the angular coordinate places the location in the opposite quadrant from the intended point, though typically we keep the radial coordinate positive and modify the angle accordingly by adding or subtracting πœ‹ or 180∘ from πœƒ to place the location in the opposite quadrant.

These equivalence conditions can be summarized as follows.

Definition: Periodicity Condition for Polar Coordinates

If (π‘Ÿ,πœƒ) describes polar coordinates, then we can express the equivalent polar coordinates as (π‘Ÿ,πœƒ)=(π‘Ÿ,πœƒ+2πœ‹π‘›)()=(π‘Ÿ,πœƒ+360𝑛)(),radiansdegrees∘ and (βˆ’π‘Ÿ,πœƒ)=(π‘Ÿ,πœƒ+(2𝑛+1)πœ‹)()=(π‘Ÿ,πœƒ+180(2𝑛+1))(),radiansdegrees∘ for any π‘›βˆˆβ„€.

In other words, we add an even integer multiple of πœ‹ (or 180∘) for positive radial coordinates and an odd integer multiple of πœ‹ (or 180∘) for negative radial coordinates. For the second condition, we usually consider the points as (βˆ’π‘Ÿ,πœƒ)=(π‘Ÿ,πœƒΒ±πœ‹).

Now, let’s look at an example where we have to find multiple equivalent representations of a polar coordinate.

Example 5: Multiple Representations of Polar Coordinates

Which of the ordered pairs (4,βˆ’30)∘, (4,330)∘, (4,390)∘, and (4,βˆ’390)∘ does not describe the position of point 𝐡 in the diagram?

Answer

In this example, we want to find the equivalent representations of the same polar coordinate, in degrees, and determine which of the given points does not describe the position of that point.

Recall that polar coordinates define a point according to the displacement from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

These coordinates are not unique, since we can add any integer multiple of a full revolution (360∘) to the angular coordinate πœƒ to get an equivalent point in polar coordinates. In particular, (π‘Ÿ,πœƒ)=(π‘Ÿ,πœƒ+360𝑛),π‘›βˆˆβ„€.∘

Another way to have an equivalent representation is using a negative radius which places the point in the opposite quadrant and is equivalent to adding or subtracting half a revolution (180∘); taking the periodicity condition into account, this can be written as (π‘Ÿ,πœƒ)=(βˆ’π‘Ÿ,πœƒ+180+360𝑛),π‘›βˆˆβ„€.∘∘

From the diagram, we can read the polar coordinates of 𝐡, which has a radial coordinate 4 and angular coordinate βˆ’30∘ and thus the polar coordinate (4,βˆ’30)∘.

Since the angular coordinate is periodic by 360∘, we can add and subtract a full revolution to find equivalent representations; in particular, πœƒ=βˆ’30+360=330,∘∘∘ and πœƒ=βˆ’30βˆ’360=βˆ’390.∘∘∘

Therefore, two equivalent representations are the polar coordinates (4,330)∘ and (4,βˆ’390)∘.

The remaining point (4,390)∘ lies in the first quadrant, but the angle given is outside the range βˆ’180<πœƒβ‰€180∘∘ and we can find the angle within this range by subtracting 360∘ from the angle to write the point in a standardized form. Hence, the equivalent point is (4,390)=(4,390βˆ’360)=(4,30).∘∘∘∘

While the radial coordinate is the same, this is clearly different from point 𝑃, since it is in a completely different quadrant from the angular coordinate. We can also plot this point in relation to point 𝐡.

Another way to check whether two points in polar coordinates are coincident or not is by calculating the distance between the two points. If the distance is zero, then the two points are coincident, while a nonzero value indicates that they are not.

Hence, the ordered pair that does not describe point 𝐡 in the diagram is (4,390).∘

Thus, due to these equivalences in writing the polar coordinates, in order to express polar coordinates in a standardized form with π‘Ÿ>0 and βˆ’πœ‹<πœƒβ‰€πœ‹ in radians or βˆ’180<πœƒβ‰€180∘∘ in degrees, we may have to adjust the value of the angular coordinate πœƒ.

The conventions we use take the counterclockwise angle as positive and clockwise as negative. For the first and second quadrants, the angle is in the positive counterclockwise direction from the positive π‘₯-axis, while for the third and fourth quadrants, the angle is in the negative clockwise direction from the positive π‘₯-axis.

We can summarize what we have covered so far in a definition, which can be used to convert a point from one coordinate system to another in any quadrant.

Definition: Converting from Cartesian Coordinates to Polar Coordinates

One possible representation of the polar coordinates (π‘Ÿ,πœƒ) with π‘Ÿβ‰₯0 can be expressed in terms of Cartesian coordinates (π‘₯,𝑦) as π‘Ÿ=√π‘₯+𝑦,πœƒ=⎧βŽͺβŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺβŽͺβŽ©ο€»π‘¦π‘₯π‘₯>0,𝑦π‘₯+(180πœ‹)π‘₯<0,𝑦>0,𝑦π‘₯ο‡βˆ’(180πœ‹)π‘₯<0,𝑦<0,90πœ‹2π‘₯=0,𝑦>0,βˆ’90βˆ’πœ‹2π‘₯=0,𝑦<0,∘∘∘∘tanfortanorfortanorfororfororfor for βˆ’πœ‹<πœƒβ‰€πœ‹ in radians or βˆ’180<πœƒβ‰€180∘∘ in degrees.

The origin (0,0) in Cartesian coordinates, with π‘₯=0 and 𝑦=0, has a special case where the polar coordinates can be represented by π‘Ÿ=0 or (0,πœƒ) for any angle πœƒ.

This information can be communicated efficiently using the following diagram.

Another way to express polar coordinates for all quadrants is if we instead define Μ‚π‘Ÿ=ο­π‘Ÿπ‘₯>0,βˆ’π‘Ÿπ‘₯<0,Μ‚πœƒ=𝑦π‘₯,π‘₯β‰ 0,ififtanforall with polar coordinates ο€ΊΜ‚π‘Ÿ,Μ‚πœƒο†. In other words, we can define the angular coordinate for all π‘₯β‰ 0 by the inverse tangent function without having to add or subtract πœ‹ or 180∘, but instead by changing the sign of the radial coordinate for specific quadrants. In particular, for the first and fourth quadrants where π‘₯>0, we have Μ‚π‘Ÿ=π‘Ÿ, and for the second and third quadrants where π‘₯<0, we have Μ‚π‘Ÿ=βˆ’π‘Ÿ. This works because of the property ο€ΊΜ‚π‘Ÿ,Μ‚πœƒο†=ο€Ίβˆ’π‘Ÿ,Μ‚πœƒο†=ο€Ίπ‘Ÿ,Μ‚πœƒΒ±πœ‹ο†, where Μ‚πœƒΒ±πœ‹=𝑦π‘₯ο‡Β±πœ‹=πœƒ,tan which is equivalent to the definition of πœƒ for the second and third quadrant (π‘₯<0), where we have to add or subtract πœ‹ from the inverse tangent.

As an example, let’s consider the points (4,3), (βˆ’4,3), (βˆ’4,βˆ’3), and (4,βˆ’3) in Cartesian coordinates, where each point is located in the different quadrants, as shown in the graph. We want to determine the polar coordinates of these points in a standardized way and in degrees with βˆ’180<πœƒβ‰€180∘∘.

The radial coordinate π‘Ÿ for these points in polar coordinates will be the same, since π‘Ÿ=√4+3=(βˆ’4)+3=(βˆ’4)+(βˆ’3)=4+(βˆ’3)=5, and we take π‘Ÿ>0 for the standardized form.

The difference will be with the angular coordinate πœƒ, since this will determine the direction and hence which quadrant the point in polar coordinates will lie in.

The point (4,3) is in the first quadrant and we can determine the angular coordinate from the general formula as πœƒ=ο€Ό34=36.869897645….tan∘

As expected, this angle is acute, as 0β‰€πœƒ<90∘∘, which places the point in the first quadrant, and it is positive, which represents the counterclockwise direction from the positive π‘₯-axis.

The point (βˆ’4,3) is in the second quadrant and the angular coordinate is πœƒ=ο€Ό3βˆ’4+180=βˆ’ο€Ό34+180=βˆ’36.869897645…+180=143.13010235….tantan∘∘∘∘∘

As expected, we have 90<πœƒ<180∘∘, which places the point in the second quadrant, and it is positive, which represents the counterclockwise direction from the positive π‘₯-axis.

The point (βˆ’4,βˆ’3) is in the third quadrant and the angular coordinate is πœƒ=ο€Όβˆ’3βˆ’4οˆβˆ’180=ο€Ό34οˆβˆ’180=36.869897645β€¦βˆ’180=βˆ’143.13010235….tantan∘∘∘∘∘

As expected, we have βˆ’180<πœƒ<βˆ’90∘∘, which places the point in the third quadrant, and it is negative, which represents the clockwise direction from the positive π‘₯-axis.

Finally, the point (4,βˆ’3) is in the fourth quadrant and the angular coordinate is πœƒ=ο€Όβˆ’34=βˆ’ο€Ό34=βˆ’36.869897645….tantan∘

As expected, we have βˆ’90<πœƒ<0∘, which places the point in the fourth quadrant, and it is negative, which represents the clockwise direction from the positive π‘₯-axis.

Thus, a possible representation of the polar coordinates, accurate to two decimal places, of the points in Cartesian coordinates is given by (π‘₯,𝑦)⟢(π‘Ÿ,πœƒ),(4,3)⟢(5,36.87),(βˆ’4,3)⟢(5,143.13),(βˆ’4,βˆ’3)⟢(5,βˆ’143.13),(4,βˆ’3)⟢(5,βˆ’36.87).∘∘∘∘

We could have also used a negative radius to represent a point in the opposite quadrant as (βˆ’π‘Ÿ,πœƒ)=(π‘Ÿ,πœƒΒ±180)∘; that is, (5,βˆ’143.13)∘ is equivalent to (βˆ’5,36.87)∘, and (5,βˆ’36.87)∘ is equivalent to (βˆ’5,143.13)∘.

Now, let’s consider an example where we convert a point, in the fourth quadrant, from Cartesian coordinates to polar coordinates, in radians.

Example 6: Converting Coordinates into Polar Coordinates

Represent the point with Cartesian coordinates (1,βˆ’1) in terms of polar coordinates.

  1. ο€»βˆš2,πœ‹4
  2. ο€»βˆš2,βˆ’πœ‹4
  3. ο€»βˆ’βˆš2,βˆ’πœ‹4
  4. ο€»2,βˆ’πœ‹4
  5. ο€»βˆ’2,πœ‹4

Answer

In this example, we want to determine the polar coordinates (π‘Ÿ,πœƒ), in radians, for the particular Cartesian coordinates (1,βˆ’1).

Recall that polar coordinates define a point according to the displacement from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ. We use the conventions where angle πœƒ is the positive counterclockwise angle in the first and second quadrants and the negative clockwise angle in the third and fourth quadrants, and all the options are given in this form. We also restrict the angles to βˆ’πœ‹<πœƒβ‰€πœ‹, in order to write them in a standardized form.

The Cartesian coordinates (π‘₯,𝑦) can be expressed in terms of the polar coordinates (π‘Ÿ,πœƒ) as π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin

The polar coordinates (π‘Ÿ,πœƒ) are not unique and there are equivalent ways of describing the same point, since the trigonometric functions used to define them are periodic.

Now, let’s find the polar coordinates of the point (1,βˆ’1) in a standardized form by using the definition. The radial coordinate is just the displacement from the origin to the point (1,βˆ’1), which we can find by using the Pythagorean theorem on the right triangle, as shown the diagram. We want to find the hypotenuse of this triangle using π‘Ÿ=√1+1=√2.

Since the point (1,βˆ’1) is located in the fourth quadrant, the angular coordinate πœƒ in polar coordinates will be the negative clockwise angle from the positive π‘₯-axis, as shown in the diagram. The positive measure of this angle is 𝛼 and thus πœƒ=βˆ’π›Ό.

We can form a right triangle with angle 𝛼 and sides of length 1. Since 𝛼 is an acute angle, we can write this in terms of the sides using the inverse tangent as 𝛼=ο€Ό11=πœ‹4.tan

Thus, we have πœƒ=βˆ’πœ‹4. We could have also arrived at this answer by using the fact that we can convert the Cartesian coordinate (π‘₯,𝑦) located in the fourth quadrant into the polar coordinates (π‘Ÿ,πœƒ) by using π‘Ÿ=√π‘₯+𝑦,πœƒ=𝑦π‘₯,βˆ’πœ‹2<πœƒ<0.tanfor

This gives the same radial and angular coordinate after substituting π‘₯=1 and 𝑦=βˆ’1.

Thus, the polar coordinates are ο€»βˆš2,βˆ’πœ‹4.

This is option B.

We can also use a polar grid for polar coordinates, which is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane. The first coordinate is radius π‘Ÿ or the length of the line segment from the pole and angle πœƒ indicates the direction. Consider the point ο€»2,πœ‹4 in polar coordinates as shown in the diagram with the polar grid.

This grid can also be used in real-world applications in navigation. For example, if a sailboat encounters rough weather over 12 km from the port and is blown off course by a strong wind, we can use this polar grid with polar coordinates to indicate the location of the sailboat to the coast guard.

Next, let’s look at how we can use a graph with a polar grid to determine the polar coordinates of a particular point.

Example 7: Graphing Polar Coordinates

Consider the points plotted on the graph.

Write down the polar coordinates of 𝐢, giving the angle πœƒ in the range βˆ’πœ‹<πœƒβ‰€πœ‹.

Answer

In this example, we want to determine the polar coordinates of a particular point, specified on a polar grid.

Recall that polar coordinates define a point according to the displacement from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ. A polar grid for polar coordinates is represented as a series of concentric circles radiating out from the pole, or the origin of the coordinate plane.

From the graph with the polar grid, each circle increases its radius by 0.5 from the last, starting from π‘Ÿ=0.5 and ending at π‘Ÿ=2.5. In other words, the radius is 0.5π‘˜ for π‘˜=1,2,3,4,5, with the π‘˜th concentric circle starting from the one closest to the origin.

The point 𝐢 lies in the fourth quadrant and on the second concentric circle that has radius 0.5Γ—2=1.

Each segment or ray represents an angle of πœ‹12, starting from πœƒ=0 and going all around the circle.

The point 𝐢 is in between the ray of argument 5πœ‹3 and that of 11πœ‹6 as marked on the graph.

The point 𝐢 is on the ray of argument 12ο€Ό5πœ‹3+11πœ‹6=7πœ‹4.

But this is the counterclockwise angle from the positive π‘₯-axis. Since we require πœƒ to be in the range βˆ’πœ‹<πœƒβ‰€πœ‹ and the angle is taken clockwise for this quadrant, the negative direction from the positive π‘₯-axis, we have to subtract 2πœ‹ to get πœƒ=7πœ‹4βˆ’2πœ‹=βˆ’πœ‹4.

Equivalently, we can also find the angle graphically, since 𝐢 has 3 segments of angle πœ‹12 in the clockwise direction and will be negative: πœƒ=βˆ’3Γ—πœ‹12=βˆ’πœ‹4.

Thus, the polar coordinates of 𝐢 are given by ο€»1,βˆ’πœ‹4.

For any two points in polar coordinates, we can also calculate the distance between the points in polar coordinates using the following definition.

Definition: Distance between Two Points in Polar Coordinates

The distance between two points (π‘Ÿ,πœƒ) and (π‘Ÿ,πœƒ) in polar coordinates is given by 𝑑=ο„π‘Ÿ+π‘Ÿβˆ’2π‘Ÿπ‘Ÿ(πœƒβˆ’πœƒ).cos

We can derive this from the distance between two points (π‘₯βˆ’π‘¦) and (π‘₯,𝑦) in Cartesian coordinates, given by 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦).

We can use this to find the distance between two points (π‘Ÿ,πœƒ) and (π‘Ÿ,πœƒ) in polar coordinates by converting the Cartesian to polar using π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ,cossin for 𝑖=1,2. Thus, the distance in polar coordinates is 𝑑=(π‘₯βˆ’π‘₯)+(π‘¦βˆ’π‘¦)=(π‘Ÿπœƒβˆ’π‘Ÿπœƒ)+(π‘Ÿπœƒβˆ’π‘Ÿπœƒ)=ο„ο€Ήπ‘Ÿπœƒβˆ’2π‘Ÿπ‘Ÿπœƒπœƒ+π‘Ÿπœƒο…+ο€Ίπ‘Ÿπœƒβˆ’2π‘Ÿπ‘Ÿπœƒπœƒ+π‘Ÿπœƒο†=ο„π‘Ÿο€Ίπœƒ+πœƒο†+π‘Ÿο€Ίπœƒ+πœƒο†βˆ’2π‘Ÿπ‘Ÿ(πœƒπœƒ+πœƒπœƒ).coscossinsincoscoscoscossinsinsinsincossincossincoscossinsin

Now, by applying the Pythagorean identity and the angle difference formula for cosine given by coscoscossinsin(πœƒβˆ’πœƒ)=πœƒπœƒ+πœƒπœƒ, we can rewrite the expression for the distance as 𝑑=ο„π‘Ÿ+π‘Ÿβˆ’2π‘Ÿπ‘Ÿ(πœƒβˆ’πœƒ).cos

Returning to the example with the sailboat, suppose the locations of the port and the sailboat are (12,60)∘ and (9,210)∘, respectively, as depicted on the diagram. To specify the polar coordinates in a standardized form, we can modify the angle of the sailboat, as the second angle, 210∘, is not in the range βˆ’180<πœƒβ‰€180∘∘. The polar coordinate (9,210)∘ is the same as (9,210βˆ’360)=(9,βˆ’150)∘∘∘.

Thus, the sailboat is located at (9,βˆ’150)∘ and the port is located at (12,60)∘. We can calculate the distance between the port and the sailboat if we substitute π‘Ÿ=12, πœƒ=60∘ and π‘Ÿ=9, πœƒ=βˆ’150∘: 𝑑=√12+9βˆ’2Γ—12Γ—9(60+150)=√225βˆ’216210=225+108√3.∘∘∘coscos

We should note that for this distance formula, we do not actually need to write the polar coordinates in a standardized form first as it works for any angle πœƒ, since the cosine function is periodic. Using the coordinates (9,210)∘ for the sailboat and (12,60)∘ for the port directly, we can substitute π‘Ÿ=12, πœƒ=60∘ and π‘Ÿ=9, πœƒ=210∘ to obtain the same result: 𝑑=√12+9βˆ’2Γ—12Γ—9(60βˆ’210)=√225βˆ’216(βˆ’150)=225+108√3.∘∘∘coscos

Thus, the distance between the port and the sailboat, accurate to two decimal places, is given by 𝑑=20.30.km

In the next example, we will look at how to determine the distance between two points in polar coordinates.

Example 8: Finding the Distance between Two Polar Coordinates

Find the distance between the polar coordinates (2,πœ‹) and ο€Ό3,βˆ’3πœ‹4. Give your answer accurate to two decimal places.

Answer

In this example, we want to find the distance between two polar coordinates, expressed in radians.

Recall that polar coordinates define a point according to the displacement from the origin, denoted by π‘Ÿ, and the angular direction from the positive π‘₯-axis, denoted by πœƒ.

The distance between two points (π‘Ÿ,πœƒ) and (π‘Ÿ,πœƒ) in polar coordinates is given by 𝑑=ο„π‘Ÿ+π‘Ÿβˆ’2π‘Ÿπ‘Ÿ(πœƒβˆ’πœƒ).cos

We can substitute π‘Ÿ=2, πœƒ=πœ‹οŠ§ and π‘Ÿ=3, πœƒ=βˆ’3πœ‹4: 𝑑=ο„ž2+3βˆ’2Γ—2Γ—3ο€Όπœ‹+3πœ‹4=ο„Ÿ2+3βˆ’2Γ—2Γ—3Γ—βˆš22=13βˆ’6√2=2.124786724….cos

Therefore, the distance accurate to two decimal places is 2.12.

Now, how can we tell whether two points (π‘Ÿ,πœƒ) and (π‘Ÿ,πœƒ) are coincident (i.e., they describe the same point)? From the periodicity conditions of equivalent points in polar coordinates, they are coincident when π‘Ÿ=π‘Ÿ,πœƒ=πœƒ+2π‘›πœ‹, or π‘Ÿ=βˆ’π‘Ÿ,πœƒ=πœƒ+(2𝑛+1)πœ‹, for any π‘›βˆˆβ„€. We can express both of these as a single condition: π‘Ÿ=(βˆ’1)π‘Ÿ,πœƒ=πœƒ+π‘˜πœ‹,οŠ§ο‡οŠ¨οŠ§οŠ¨ for any π‘˜βˆˆβ„€.

Another way to check if two points in polar coordinates are coincident or not is by calculating the distance between the two points. If the distance is zero, then the two points are coincident, while a nonzero value indicates that they are not. We can check this by using the distance formula in polar coordinates, by substituting π‘Ÿ=(βˆ’1)π‘ŸοŠ§ο‡οŠ¨ and πœƒ=πœƒ+π‘˜πœ‹οŠ§οŠ¨: 𝑑=ο„π‘Ÿ+π‘Ÿβˆ’2π‘Ÿπ‘Ÿ(πœƒβˆ’πœƒ)=ο„žο€Ί(βˆ’1)π‘Ÿο†+π‘Ÿβˆ’2(βˆ’1)π‘ŸΓ—π‘Ÿ(πœƒ+π‘˜πœ‹βˆ’πœƒ)=2π‘Ÿβˆ’2(βˆ’1)π‘Ÿ(π‘˜πœ‹).οŠ¨οŠ§οŠ¨οŠ¨οŠ§οŠ¨οŠ§οŠ¨ο‡οŠ¨οŠ¨οŠ¨οŠ¨ο‡οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨οŠ¨ο‡οŠ¨οŠ¨coscoscos

Since we have cos(π‘˜πœ‹)=(βˆ’1), the distance becomes 𝑑=2π‘Ÿβˆ’2(βˆ’1)π‘ŸΓ—(βˆ’1)=2π‘Ÿβˆ’2π‘Ÿ=0.οŠ¨οŠ¨ο‡οŠ¨οŠ¨ο‡οŠ¨οŠ¨οŠ¨οŠ¨

As expected, the distance between two coordinates that describe the same point is zero. For a nonzero distance, the two points would not describe the same point.

Key Points

  • A polar coordinate is of the form (π‘Ÿ,πœƒ), which denotes the displacement from the origin and the angle from the positive π‘₯-axis.
  • The conventions we use take the counterclockwise angle as positive and clockwise as negative. For the first and second quadrants, the angle πœƒ is in the positive counterclockwise direction from the positive π‘₯-axis, while for the third and fourth quadrants, the angle πœƒ is in the negative clockwise direction from the positive π‘₯-axis.
  • To convert from polar (π‘Ÿ,πœƒ) to Cartesian (π‘₯,𝑦), we use π‘₯=π‘Ÿπœƒ,𝑦=π‘Ÿπœƒ.cossin
  • To convert from Cartesian (π‘₯,𝑦) to polar (π‘Ÿ,πœƒ), one possible representation is given by π‘Ÿ=√π‘₯+𝑦,πœƒ=⎧βŽͺβŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺβŽͺβŽ©ο€»π‘¦π‘₯π‘₯>0,𝑦π‘₯+(180πœ‹)π‘₯<0,𝑦>0,𝑦π‘₯ο‡βˆ’(180πœ‹)π‘₯<0,𝑦<0,90πœ‹2π‘₯=0,𝑦>0,2703πœ‹2π‘₯=0,𝑦<0,∘∘∘∘tanfortanorfortanorfororfororfor for βˆ’πœ‹<πœƒβ‰€πœ‹ in radians or βˆ’180<πœƒβ‰€180∘∘ in degrees.
  • We can find equivalent polar coordinates by adding or subtracting any integer multiple of a full revolution (360∘ or 2πœ‹), or by using a negative radius which places the coordinate in the opposite quadrant, which is the same as adding or subtracting half a revolution (180∘ or πœ‹) from πœƒ. In particular, we have (π‘Ÿ,πœƒ)=(π‘Ÿ,πœƒ+2πœ‹π‘›)=(βˆ’π‘Ÿ,πœƒ+πœ‹+2πœ‹π‘›),π‘›βˆˆβ„€.
  • The distance between two points in polar coordinates (π‘Ÿ,πœƒ) and (π‘Ÿ,πœƒ) is given by 𝑑=ο„π‘Ÿ+π‘Ÿβˆ’2π‘Ÿπ‘Ÿ(πœƒβˆ’πœƒ).cos

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