Unit 7, lesson 86

Converting Between Polar and Rectangular Coordinates

How to move points and equations between polar and rectangular form, how to choose the angle in the right quadrant, and how to check that a conversion neither adds nor loses points.

  • Theory part 171
  • Practice part 172
  • Polar Coordinates

A point does not move when you change the way you describe it. The point 33 units right and 33 units up is also the point 323\sqrt{2} units from the origin in the direction 45∘45^\circ. Some questions are easier in one description and some in the other, so you need a reliable way to translate. The translation is right triangle trigonometry plus one careful look at the quadrant.

Key terms

Polar to rectangular formulas

Definition Put the pole at the origin and the polar axis along the positive xx-axis. If a point has polar coordinates (r,θ)(r, \theta), its rectangular coordinates are

x=rcos⁡θ,y=rsin⁡θ.x = r\cos\theta, \qquad y = r\sin\theta.

These hold for every real rr, including r=0r = 0 and r<0r < 0. For r>0r > 0 the point is rr times the point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) of the unit circle. For r<0r < 0 the point is also (−r,θ+π)(-r, \theta + \pi), and −rcos⁡(θ+π)=rcos⁡θ-r\cos(\theta + \pi) = r\cos\theta and −rsin⁡(θ+π)=rsin⁡θ-r\sin(\theta + \pi) = r\sin\theta, so the same formulas come out.

In plain words Drop a perpendicular from the point to the xx-axis. You get a right triangle whose hypotenuse is the distance ∣r∣\lvert r \rvert, and its legs, with signs, are xx and yy.

Example For the polar point (−4,π3)(-4, \frac{\pi}{3}),

x=−4cos⁡π3=−2,y=−4sin⁡π3=−23.\begin{aligned} x &= -4\cos\frac{\pi}{3} = -2, \\[0.8em] y &= -4\sin\frac{\pi}{3} = -2\sqrt{3}. \end{aligned}

The point is in the third quadrant, which is what a negative radius with a first quadrant angle should give.

Radius from x and y

Definition If (r,θ)(r, \theta) is any polar name of the point (x,y)(x, y), then

x2+y2=r2cos⁡2θ+r2sin⁡2θ=r2.\begin{aligned} x^2 + y^2 &= r^2\cos^2\theta + r^2\sin^2\theta \\[0.4em] &= r^2. \end{aligned}

So ∣r∣=x2+y2\lvert r \rvert = \sqrt{x^2 + y^2}, and when a positive radius is wanted, r=x2+y2r = \sqrt{x^2 + y^2} for every point other than the origin.

In plain words This is the Pythagorean theorem. The radius is the distance from the origin, up to its sign.

Example The point (−8,6)(-8, 6) has r2=64+36=100r^2 = 64 + 36 = 100, so it is 1010 units from the origin. Its polar names use r=10r = 10 or r=−10r = -10.

Choosing the angle with atan2

Definition For a point (x,y)(x, y) with x≠0x \ne 0, every polar angle with a positive radius satisfies tan⁡θ=y/x\tan\theta = y/x. That equation alone does not settle the angle. It has two solutions in [0,2π)[0, 2\pi), one pointing at (x,y)(x, y) and one pointing at the opposite point (−x,−y)(-x, -y), and arctan⁡(y/x)\arctan(y/x) always lies in (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), so it only ever points into the right half of the plane. The two-input function atan2⁡(y,x)\operatorname{atan2}(y, x) fixes this. For (x,y)≠(0,0)(x, y) \ne (0, 0) it is the unique angle θ\theta in (−π,π](-\pi, \pi] with

cos⁡θ=xx2+y2,sin⁡θ=yx2+y2,\begin{aligned} \cos\theta &= \frac{x}{\sqrt{x^2 + y^2}}, \\[0.8em] \sin\theta &= \frac{y}{\sqrt{x^2 + y^2}}, \end{aligned}

and case by case it equals

atan2⁡(y,x)={arctan⁡(y/x)if x>0,arctan⁡(y/x)+πif x<0and y≥0,arctan⁡(y/x)−πif x<0and y<0,π/2if x=0and y>0,−π/2if x=0and y<0.\begin{gathered} \operatorname{atan2}(y, x) = \\[0.4em] \begin{cases} \arctan(y/x) & \text{if } x > 0, \\[0.4em] \arctan(y/x) + \pi & \text{if } x < 0 \\ & \text{and } y \ge 0, \\[0.4em] \arctan(y/x) - \pi & \text{if } x < 0 \\ & \text{and } y < 0, \\[0.4em] \pi/2 & \text{if } x = 0 \\ & \text{and } y > 0, \\[0.4em] -\pi/2 & \text{if } x = 0 \\ & \text{and } y < 0. \end{cases} \end{gathered}

It is undefined at the origin. Note the order of the inputs, yy first. If you want an angle in [0,2π)[0, 2\pi), add 2π2\pi to a negative result.

In plain words The tangent only sees the ratio y/xy/x, so it cannot tell (2,−2)(2, -2) from (−2,2)(-2, 2). Look at the signs of xx and yy to find the quadrant, then take the angle that points there.

Example Take the point (−2,2)(-2, 2). A calculator gives arctan⁡(−1)=−π4\arctan(-1) = -\frac{\pi}{4}, which points into the fourth quadrant. The point is in the second quadrant, with x<0x < 0 and y≥0y \ge 0, so

atan2⁡(2,−2)=−π4+π=3π4.\operatorname{atan2}(2, -2) = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}.

The careless name (22,−π4)(2\sqrt{2}, -\frac{\pi}{4}) describes (2,−2)(2, -2), the point on the other side of the origin.

Point conversion

Definition To convert a polar point (r,θ)(r, \theta) to rectangular form, compute x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta. To convert a rectangular point (x,y)≠(0,0)(x, y) \ne (0, 0) to polar form with a positive radius, take

r=x2+y2,θ=atan2⁡(y,x),\begin{aligned} r &= \sqrt{x^2 + y^2}, \\[0.4em] \theta &= \operatorname{atan2}(y, x), \end{aligned}

and add an integer multiple of 2π2\pi to θ\theta if another range is wanted. The origin converts to (0,θ)(0, \theta) with any θ\theta.

In plain words Polar to rectangular is a direct calculation. Rectangular to polar needs the distance and a quadrant check.

Example Start with the rectangular point (3,−33)(3, -3\sqrt{3}). Its radius is

r=9+27=6.r = \sqrt{9 + 27} = 6.

Since x>0x > 0, the angle is arctan⁡(−3)=−π3\arctan(-\sqrt{3}) = -\frac{\pi}{3}, which points into the fourth quadrant as it should. So the point is (6,−π3)(6, -\frac{\pi}{3}), or (6,5π3)(6, \frac{5\pi}{3}) if you want 0≤θ<2π0 \le \theta < 2\pi. Converting back, 6cos⁡(−π3)=36\cos(-\frac{\pi}{3}) = 3 and 6sin⁡(−π3)=−336\sin(-\frac{\pi}{3}) = -3\sqrt{3}.

Equation conversion

Definition A rectangular equation in xx and yy becomes a polar equation when you substitute x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta. A polar equation becomes rectangular when you rewrite it in terms of

rcos⁡θ=x,rsin⁡θ=y,r2=x2+y2,\begin{aligned} r\cos\theta &= x, \\[0.4em] r\sin\theta &= y, \\[0.4em] r^2 &= x^2 + y^2, \end{aligned}

often after multiplying both sides by rr. The graph of a polar equation is the set of points that have at least one polar name satisfying it. A conversion is correct when the two equations have exactly the same graph.

In plain words Substitute in one direction. In the other, hunt for rcos⁡θr\cos\theta, rsin⁡θr\sin\theta and r2r^2, and make them appear if they are not there yet.

Example The line x+y=4x + y = 4 becomes rcos⁡θ+rsin⁡θ=4r\cos\theta + r\sin\theta = 4, which gives

r=4cos⁡θ+sin⁡θ.r = \frac{4}{\cos\theta + \sin\theta}.

Going the other way, multiply r=6cos⁡θr = 6\cos\theta by rr to get r2=6rcos⁡θr^2 = 6r\cos\theta, which is x2+y2=6xx^2 + y^2 = 6x. Completing the square in xx gives

(x−3)2+y2=9,(x - 3)^2 + y^2 = 9,

the circle of radius 33 centered at (3,0)(3, 0).

Domain and tracing checks

Definition Some algebra steps change the set of points an equation describes. Multiplying both sides by rr can add the pole. Dividing both sides by an expression can lose the points where that expression is 00, and squaring both sides can add points where the two sides had opposite signs. The domain check asks where each formula is defined and whether such a step added or lost points. The pole needs its own test, since it lies on the graph of a polar equation exactly when some pair (0,θ)(0, \theta) satisfies the equation. The tracing check follows r=f(θ)r = f(\theta) as θ\theta increases and records which part of the curve is drawn, so you know an interval of θ\theta that draws the whole curve once.

In plain words After converting, ask two questions. Did any step add or remove a point, especially the origin? Which values of θ\theta actually draw the curve?

Example For r=6cos⁡θr = 6\cos\theta, multiplying by rr could have added the pole. It did not, because the pole is already on the curve, since θ=π2\theta = \frac{\pi}{2} gives r=0r = 0. A short table traces the curve.

θ\thetar=6cos⁡θr = 6\cos\theta(x,y)(x, y)
0066(6,0)(6, 0)
π4\frac{\pi}{4}323\sqrt{2}(3,3)(3, 3)
π2\frac{\pi}{2}00(0,0)(0, 0)
3π4\frac{3\pi}{4}−32-3\sqrt{2}(3,−3)(3, -3)
π\pi−6-6(6,0)(6, 0)

Every row satisfies (x−3)2+y2=9(x - 3)^2 + y^2 = 9. As θ\theta runs from 00 to π2\frac{\pi}{2} the point covers the upper half of the circle. From π2\frac{\pi}{2} to π\pi the radius is negative and the point covers the lower half. So 0≤θ<π0 \le \theta < \pi draws the circle once, and the next interval of length π\pi draws it again. For the line, the formula r=4/(cos⁡θ+sin⁡θ)r = 4/(\cos\theta + \sin\theta) is undefined where cos⁡θ+sin⁡θ=0\cos\theta + \sin\theta = 0, at θ=3π4+nπ\theta = \frac{3\pi}{4} + n\pi. Those angles point along y=−xy = -x, which is parallel to the line, so no point of the line is lost.

Check yourself

  1. Convert the polar point (−2,7π6)(-2, \frac{7\pi}{6}) to rectangular coordinates.
Show the answer

The conversion formulas give

x=−2cos⁡7π6=3,y=−2sin⁡7π6=1,\begin{aligned} x &= -2\cos\frac{7\pi}{6} = \sqrt{3}, \\[0.8em] y &= -2\sin\frac{7\pi}{6} = 1, \end{aligned}

so the point is (3,1)(\sqrt{3}, 1), in the first quadrant.

  1. Write the rectangular point (−3,−1)(-\sqrt{3}, -1) in polar form with r>0r > 0 and −π<θ≤π-\pi < \theta \le \pi.
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The radius is r=3+1=2r = \sqrt{3 + 1} = 2. Since x<0x < 0 and y<0y < 0, the point is in the third quadrant, so

θ=arctan⁡13−π=π6−π=−5π6.\begin{aligned} \theta &= \arctan\frac{1}{\sqrt{3}} - \pi \\[0.8em] &= \frac{\pi}{6} - \pi \\[0.8em] &= -\frac{5\pi}{6}. \end{aligned}

The point is (2,−5π6)(2, -\frac{5\pi}{6}).

  1. Convert x2+y2=6yx^2 + y^2 = 6y to polar form. Does dividing by rr lose the pole?
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Substituting gives r2=6rsin⁡θr^2 = 6r\sin\theta, so r=0r = 0 or r=6sin⁡θr = 6\sin\theta. The pole is already on r=6sin⁡θr = 6\sin\theta, because θ=0\theta = 0 gives r=0r = 0. Nothing is lost, and the polar form is r=6sin⁡θr = 6\sin\theta.

How this lesson connects

Where this leads in Calculus