4.09a - Using polar coordinates

4.09a - Using polar coordinates

Describe a point by its distance and direction from a fixed point. You will convert points and curve equations between polar and cartesian coordinates, keeping the angle range and the convention that distance is never negative.

Distance and direction

Why does the equation r=4cos⁡θr = 4\cos\theta describe a circle through the origin, rather than a wave? The cosine gives the distance along a rotating ray, not the height above a horizontal axis. By the end of the lesson you will be able to uncover that circle algebraically.

Start with a point reached by turning through π3\frac{\pi}{3} anticlockwise from a fixed horizontal ray, then travelling 6 units. This description fixes a point just as precisely as giving its horizontal and vertical coordinates.

[DIAGRAM: asset_name: Distance and direction of a point; asset_slug: h245_polar_point; caption: A point can be located by a distance r and an angle theta, or by its horizontal and vertical coordinates; description: Equal-scale cartesian axes with positive x-axis labelled initial line and origin labelled pole O. A blue ray reaches P at cartesian coordinates (3, 3 root 3), with r = 6 marked alongside. A sage angle arc is labelled theta = pi over 3. Dashed projections label x = 3 and y = 3 root 3. A right panel states polar coordinates [6, pi over 3] and cartesian coordinates (3, 3 root 3)]
A point can be located by a distance r and an angle theta, or by its horizontal and vertical coordinates

You already know

A full turn is 2π2\pi radians. Sine and cosine give the vertical and horizontal coordinates of a point on the unit circle, including their signs in every quadrant.

Polar coordinates use the same angles but allow the distance from the centre to vary.

Polar coordinates

The pole is a fixed point OO, and the initial line is a fixed ray from OO. The polar coordinates [r,θ][r, \theta] locate a point a distance rr from OO, at angle θ\theta from the initial line. Anticlockwise angles are positive; clockwise angles are negative. We use r≥0r \ge 0. For example, [3,π][3, \pi] is 3 units along the ray opposite the initial line. A pair with r=−3r = -3 is not allowed under this convention.

Unless told otherwise, take the pole as the cartesian origin and the initial line as the positive xx-axis. Write the radius first: [r,θ][r, \theta], not [θ,r][\theta, r].

A point does not have a unique polar description until an angle interval is chosen. The pairs [2,π2][2, \frac{\pi}{2}] and [2,5π2][2, \frac{5\pi}{2}] locate the same point, because their angles differ by a full turn. Typical intervals are 0≤θ<2π0 \le \theta < 2\pi or −π<θ≤π-\pi < \theta \le \pi. Follow the interval in the question. A negative angle means a clockwise turn; after turning far enough, you can finish above or below the initial line. At the pole, r=0r = 0 and every angle locates the same point; there is no unique direction from OO to itself.

Key idea

The radius is a distance, so r≥0r \ge 0. Angles may be negative or exceed one full turn, but equivalent angles differ by an integer multiple of 2π2\pi.

A negative cartesian coordinate and a negative polar radius mean different things: a point to the left of OO still has a positive distance from OO.

From polar to cartesian coordinates

Project the radius onto the horizontal and vertical axes to turn distance and direction into xx and yy.

For [6,π3][6, \frac{\pi}{3}], the horizontal projection is 6cos⁡π3=36\cos\frac{\pi}{3} = 3, and the vertical projection is 6sin⁡π3=336\sin\frac{\pi}{3} = 3\sqrt{3}. The same projection rule works in every quadrant: sine and cosine supply the signs.

Polar to cartesian conversion

When the pole is the origin and the initial line is the positive xx-axis,

x=rcos⁡θy=rsin⁡θx = r\cos\theta \qquad y = r\sin\theta

where r≥0r \ge 0.
Not in the formula booklet: learn it.

Calculate the two coordinates separately; this keeps their signs visible.

Worked example 1: a point in the first quadrant

Question. Find the exact cartesian coordinates of [10,π6][10, \frac{\pi}{6}].

Plan. Substitute the radius and angle into the horizontal and vertical projection formulae.

x=10cos⁡π6=10×32horizontal projection=53simplifyy=10sin⁡π6=10×12=5vertical projection\begin{aligned} x &= 10\cos\tfrac{\pi}{6} = 10\times\tfrac{\sqrt{3}}{2} && \text{horizontal projection} \\ &= 5\sqrt{3} && \text{simplify} \\ y &= 10\sin\tfrac{\pi}{6} = 10\times\tfrac12 = 5 && \text{vertical projection} \end{aligned}

Why this step? The radius is the hypotenuse, so multiplying by cosine and sine gives its two components.

Answer. (53,5)(5\sqrt{3}, 5)

Check. x2+y2=75+25=100=102x^2 + y^2 = 75 + 25 = 100 = 10^2, and both coordinates are positive, as the angle requires.

Worked example 2 changes only the quadrant, keeping the same radius and reference angle. The horizontal coordinate changes sign because cosine is negative in the second quadrant.

Worked example 2: a point in the second quadrant

Question. Find the exact cartesian coordinates of [10,5π6][10, \frac{5\pi}{6}].

Plan. Use the same projection formulae, with the exact sine and cosine of the given angle.

x=10cos⁡5π6=10×(−32)cosine is negative=−53simplifyy=10sin⁡5π6=10×12=5sine is positive\begin{aligned} x &= 10\cos\tfrac{5\pi}{6} = 10\times\left(-\tfrac{\sqrt{3}}{2}\right) && \text{cosine is negative} \\ &= -5\sqrt{3} && \text{simplify} \\ y &= 10\sin\tfrac{5\pi}{6} = 10\times\tfrac12 = 5 && \text{sine is positive} \end{aligned}

Why this step? A positive radius does not make both coordinates positive; the angle determines their signs.

Answer. (−53,5)(-5\sqrt{3}, 5)

Check. (−53)2+52=100(-5\sqrt{3})^2 + 5^2 = 100; the point is left of and above the pole.

Keeping the angle intact avoids a common sign error.

Watch out

Wrong: 10cos⁡5π6=5310\cos\frac{5\pi}{6} = 5\sqrt{3}. That uses the reference angle without its quadrant. Right: cos⁡5π6=−32\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}, so x=−53x = -5\sqrt{3}.

The next check keeps the second-quadrant angle and changes the radius.

From cartesian to polar coordinates

Find the distance first, then use the quadrant to select an angle in the requested interval.

Pythagoras gives the distance. Dividing the two projection formulae gives tan⁡θ=yx\tan\theta = \frac{y}{x} when x≠0x \ne 0. But tangent repeats after π\pi, so a calculator's inverse tangent alone cannot distinguish opposite quadrants.

Cartesian to polar conversion

For a point (x,y)(x, y),

r=x2+y2r = \sqrt{x^2 + y^2}

For a non-zero point with x≠0x \ne 0,

tan⁡θ=yx\tan\theta = \frac{y}{x}

Choose θ\theta using the quadrant and the required angle interval. If x=0x = 0, read the direction from the vertical axis instead.
Not in the formula booklet: learn it.

On the positive yy-axis use θ=π2\theta = \frac{\pi}{2}; on the negative yy-axis use 3π2\frac{3\pi}{2} or −π2-\frac{\pi}{2}, depending on the interval. At (0,0)(0, 0), give r=0r = 0 and recognise that the angle is arbitrary.

Worked example 3: choosing the second quadrant

Question. Find polar coordinates for (−2,23)(-2, 2\sqrt{3}), with 0≤θ<2π0 \le \theta < 2\pi.

Plan. Use Pythagoras for the radius and a reference angle for the direction, then choose the second quadrant.

r=(−2)2+(23)2=16=4distance from Otan⁡α=∣23−2∣=3acute reference angleα=π3exact tangentθ=π−π3=2π3x<0, y>0\begin{aligned} r &= \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{16} = 4 && \text{distance from } O \\ \tan\alpha &= \left|\frac{2\sqrt{3}}{-2}\right| = \sqrt{3} && \text{acute reference angle} \\ \alpha &= \tfrac{\pi}{3} && \text{exact tangent} \\ \theta &= \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3} && x < 0,\ y > 0 \end{aligned}

Why this step? The signs place the point in the second quadrant, so the angle must be between π2\frac{\pi}{2} and π\pi.

Answer. [4,2π3][4, \frac{2\pi}{3}]

Check. 4cos⁡2π3=−24\cos\frac{2\pi}{3} = -2 and 4sin⁡2π3=234\sin\frac{2\pi}{3} = 2\sqrt{3}.

Worked example 4 changes only the sign of yy. The distance and reference angle are unchanged, but the point now lies in the third quadrant.

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