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The Polar Coordinate System

Calculus 2 · Chapter 9: Parametric Equations & Polar CoordinatesSection 9.4 · lesson 48 of 549 min · free

← Arc Length and Surface Area in Parametric FormNext lesson: Polar Curves →

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In this lesson

Locating a point by distance and direction.

  • Read (r,\theta) as range and bearing
  • Handle the infinitely many names of one point
  • Convert polar to Cartesian directly
  • Convert Cartesian to polar with a quadrant check

Transcript

Everything said in the video, section by section (1,278 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 9.4

The Polar Coordinate System

An air traffic controller doesn't describe an aircraft as ten miles west and six miles south. They give a range and a bearing. A distance from the antenna, and a direction. Radar, sonar, and a great many physical problems are naturally posed that way. And the Cartesian grid — left-right, up-down — fits them awkwardly. Polar coordinates locate a point by exactly those two numbers instead. The distance r from a fixed point called the pole, which is the origin. And the angle theta, measured counter-clockwise from a fixed ray — the positive x-axis.

The Idea

Range and Bearing, Not Left and Up

One feature of polar coordinates surprises people, so let's address it up front. The radius r is allowed to be negative. And a negative radius has a precise meaning. The pair minus r, theta means: face along the ray at angle theta. Then walk backwards a distance absolute r. So minus two, forty-five degrees does not land in the first quadrant. It lands in the third. That convention might seem like an unnecessary complication. But it lets a single equation trace curves that would otherwise need several. You'll see why in the next section, with roses and cardioids.

Non-Uniqueness

Every Point Has Infinitely Many Names

Now a genuinely important consequence of the polar system. Every point other than the pole has infinitely many polar representations. If r comma theta locates a point P, then so does every pair of two forms. First: r, theta plus three hundred sixty n. For any integer n. Second: minus r, theta plus one eighty, plus three hundred sixty n. In radians those read r, theta plus two pi n. And minus r, theta plus pi plus two pi n. And the pole itself is a special case. It's written zero comma theta, for any angle theta whatsoever. Since zero distance is reached regardless of which direction you face. Two distinct mechanisms generate those duplicates. Adding a full revolution returns the ray to where it started. So theta and theta plus three sixty point identically. Separately, negating r while rotating a half turn reverses direction twice. And the two reversals cancel. Let's see it. The point P has polar coordinates two comma forty-five degrees. Produce three more names. Add a full revolution: two comma four hundred five degrees. Subtract one: two comma minus three fifteen. That represents rotating clockwise by three fifteen, rather than counter-clockwise by forty-five. A longer route to the same ray. Use a negative radius: minus two, comma two twenty-five. And you can verify all four by converting to Cartesian form. Each one reduces to root two, root two. All four are correct names for a single point.

Polar to Cartesian

The Easy Direction

Converting from polar to Cartesian is the easy direction. Drop a perpendicular from the point to the x-axis. The resulting right triangle has hypotenuse r, and angle theta at the origin. So basic trigonometry reads off both coordinates directly. x equals r cosine theta. And y equals r sine theta. No cautions attach to this direction. Substitute and evaluate. And the formulas remain valid for negative r — the trig functions handle the sign automatically. Let's do the radar example. An aircraft appears at range twelve nautical miles, bearing two hundred ten degrees. x equals twelve cosine two ten. Cosine of two ten is minus root three over two. So x is minus six root three. y equals twelve sine two ten. Sine of two ten is minus one half. So y is minus six. The Cartesian position is minus six root three, minus six. Approximately minus ten point three nine, minus six. So the aircraft lies about ten point four nautical miles west, and six nautical miles south, of the antenna. Both coordinates negative. Consistent with a third-quadrant bearing.

Cartesian to Polar

The Direction That Needs Care

The reverse conversion demands more care. Solving for r involves a square root. And solving for theta involves an inverse tangent. And both of those operations lose information about sign. The relationships are: r squared equals x squared plus y squared. And tangent theta equals y over x, provided x isn't zero. Taking r as the positive square root gives the principal representation. But the angle requires a quadrant check. And this is where students lose marks. The arctangent function returns values only between minus ninety and ninety degrees. Which covers Quadrants One and Four. Only. So if the point lies in Quadrant Two or Three, you must add one hundred eighty degrees to the arctangent result. And if x equals zero, the point lies on the y-axis. Theta is ninety degrees when y is positive, and two seventy when y is negative. Look at the picture, because it makes the necessity vivid. Four points, one in each quadrant, all at equal distance from the origin. Every one of them satisfies absolute y over x equals one. So the arctangent alone cannot distinguish them. It returns only forty-five degrees, or minus forty-five. It's the signs of x and y that identify the quadrant. And thereby the correct angle. Which is why the quadrant check is a required step. Not an optional refinement.

Worked Example 3

A Quadrant II Conversion

Let's work a Quadrant Two conversion carefully. Convert the Cartesian point minus three, three, to principal polar coordinates. Step one. Compute r. r squared equals minus three squared, plus three squared. Nine plus nine. Eighteen. Take the positive square root and simplify. Root eighteen is root of nine times two. Which is three root two. About four point two four. Step two. Compute the arctangent. Tangent theta equals three over minus three. Which is minus one. And arctangent of minus one is minus forty-five degrees. Step three. Apply the quadrant check. The point minus three, three has x negative and y positive. So it's in Quadrant Two. But minus forty-five degrees lies in Quadrant Four. So the arctangent has returned the wrong ray. Adding one eighty corrects it. Theta equals minus forty-five, plus one eighty. One hundred thirty-five degrees. Step four. Verify by converting back. x equals three root two, cosine one thirty-five. Cosine of one thirty-five is minus root two over two. So x is three root two, times minus root two over two. Which is minus three. Correct. And y works out to three. Correct. The original point is recovered. So the principal polar representation is three root two, one hundred thirty-five degrees. Now — had the quadrant check been skipped, the answer three root two, minus forty-five, would have named the point three, minus three. A reflection through the origin. In the wrong half of the plane entirely.

Key Takeaways

What to Carry Forward

Let's compare the two conversions. Polar to Cartesian: x equals r cosine theta, y equals r sine theta. No cautions. Substitute directly. And it's valid for negative r. Cartesian to polar: r equals root x squared plus y squared, theta equals arctangent y over x. Add one eighty in Quadrants Two and Three. And handle x equals zero separately. That asymmetry is worth noting. Moving to Cartesian coordinates is mechanical. Moving to polar coordinates requires a decision — based on the signs of x and y. Five takeaways. Polar coordinates give distance and direction. A natural fit for radar, sonar, and anything with circular symmetry. Every point but the pole has infinitely many names. From adding revolutions, and from negating r with a half-turn. And the pole is zero comma theta, for any theta. Polar to Cartesian is mechanical, and valid even for negative r. Cartesian to polar needs a quadrant check. Arctangent covers only Quadrants One and Four, so add one eighty in Two and Three. And always verify by converting back. Skipping the check names a point reflected through the origin. Next up is section nine point five — Polar Curves. Where a single equation traces roses, cardioids and spirals.

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FAQ

Questions students ask us

About The Polar Coordinate System

What do polar coordinates (r,\theta) mean?
Polar coordinates locate a point by its distance r from the pole (the origin) and its angle \theta, measured counterclockwise from the positive x-axis, much like the range and bearing used in radar. The radius may be negative: (-r,\theta) means face along the ray at angle \theta, then walk backward a distance |r|. For example, (-2, 45^\circ) lands in the third quadrant, not the first.
Why does a point have infinitely many polar coordinate representations?
A point has infinitely many representations because adding full revolutions, or negating r while turning a half revolution, leaves the location unchanged. If (r,\theta) names a point, so do (r,\theta+2\pi n) and (-r,\theta+\pi+2\pi n) for any integer n. For example, (2,45^\circ), (2,405^\circ), (2,-315^\circ), and (-2,225^\circ) all name (\sqrt{2},\sqrt{2}). The pole is (0,\theta) for any \theta.
How do you convert polar coordinates to Cartesian coordinates?
Use x = r\cos\theta and y = r\sin\theta, which come from a right triangle with hypotenuse r and angle \theta at the origin. The formulas work even for negative r. For example, a radar contact at range 12 and bearing 210^\circ gives x = 12\cos 210^\circ = -6\sqrt{3} and y = 12\sin 210^\circ = -6. That places the point at about (-10.39, -6), in the third quadrant.
Why is \theta = \arctan(y/x) sometimes wrong when converting to polar coordinates?
Arctangent only returns angles between -90^\circ and 90^\circ, so it gives the wrong ray for points in Quadrants II and III. In those quadrants, add 180^\circ. For (-3,3), r = \sqrt{18} = 3\sqrt{2} and \arctan(-1) = -45^\circ, but the correct angle is 135^\circ. Skipping this check names the point reflected through the origin. If x = 0, use \theta = 90^\circ when y > 0 and \theta = 270^\circ when y < 0.

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