How to Convert Between Polar & Rectangular Coordinates

Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

Symmetry in a math problem may suggest a change of coordinate systems. In this lesson we show how to convert between two common coordinate systems: polar and rectangular.

Converting Words Like Math

If you meet someone from a different culture it's good to welcome them in their language. The word 'welcome' is converted to other sounds and letters. It's like the same point but expressed differently. In math, it's like using polar coordinates or rectangular coordinates.

In this lesson we convert between polar and rectangular coordinates in each of the four quadrants where the rules may differ slightly. It's like the French 'bienvenue' not being quite the same as the Spanish 'bienvenida'.

Finding Locations in the Plane

A plane has two dimensions, which implies 'two values'. Usually, we use an x and a y value to locate a point in the plane. A special point in the plane is the origin where x and y are both zero. In the first quadrant, to get to a point from the origin we move to the right a specific distance x and then up a specific distance y. The point is labeled as (x,y). These are the rectangular coordinates of the point.

rectangular coordinates

Then again, why not go directly to the point along some radius r instead of taking the 'scenic route' along x and y?

radius r

We can do so, but we also need a direction before heading from the origin. Remember, a plane has two dimensions and the rectangular system needed two values. Just one value is not enough. In addition to the r we need an angle θ. This angle is the counter-clockwise rotation referenced to the x-axis. We label the point as (r, θ). These are the polar coordinates of the point.

polar coordinates

We've just described two equivalent ways to express the same point. You are 'welcome'. But what if we only have r and θ; can we find x and y? Sure! From trigonometry, sine of θ is the opposite side over the hypotenuse. The hypotenuse is r and the opposite side is y. We write sin(θ) = y/r. This means y = r sin(θ). How about x? Since cosine is the adjacent side over the hypotenuse and the adjacent side is x, we get x = r cos(θ).

Instead of knowing r and θ, what if we had x and y. Can we find r and θ from x and y? But of course! A right-angled triangle is formed by x, y and r. Using Pythagoras' theorem we have r^2 = x^2 + y^2. Then, r = square root of (x^2 + y^2). What about θ?

From trigonometry, tangent θ is the opposite side over the adjacent side. Thus, tan(θ) is y/x so θ = atan(y/x). Atan is the 'arc tangent' which is also written as the 'inverse tangent' tan-1.

In the 'quadrant I' figure, r = 2, y = 1, x = 1.732 and θ = 30o. Do our equations check out?

Locating a point in quadrant I
relating coordinates

  • x = r cos(θ) => r cos(θ) = 2 cos(30o) = 2 (.866) = 1.732 = x, CHECK!
  • y = r sin(θ) => r sin(θ) = 2 sin(30o) = 2 (.5) = 1 = y, CHECK!
  • r = (x2 + y2)1/2 => (1.7322 + 12)1/2 = 2 = r, CHECK!
  • θ = tan-1 (y/x) => tan-1 (1/1.732) = 30o = θ, CHECK!

All four equations check out! At least in the first quadrant, we can convert between polar and rectangular coordinates. What about the other quadrants?

The Other Quadrants

Quadrant II

Locating a point in quadrant II

In quadrant II, start at the origin but go left; meaning x is negative. Then go up; y is again positive. The radius r is still the shortest path from the origin to the point. The angle θ is still measured from the positive x-axis, x is still r cos(θ) and y is still r sin(θ). Are there any differences? Yes! We must be attentive to tan(θ) and tan-1 (y/x). Just like with languages, we must interpret the result based on where we are.

Some calculators have a 'tan2' function that allows for tangent-related calculations in all 4 quadrants. But we can also look at the angle for tangent as an angle from 0 to 90o. This is the angle we saw in the Pythagoras triangle. In the 'quadrant II' figure, this is the 'brown' angle.

In the 'quadrant II' figure, x = -1.732 , y = 1, r = 2 and θ = 150o. Do you see how θ is the counter-clockwise rotation from the positive x-axis?

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