Maths › Polar coordinates › Polar curves
Polar curves
Describe a point by how far and in what direction instead of across and up, and curves that torment cartesian algebra collapse into one short equation in r and θ.
Builds on Radians, arcs and small angles and Circles.
IN THIS TOPIC
- Plot points from (r, θ) and convert both ways with x = r cos θ, y = r sin θ.
- Convert polar equations to cartesian form and recognise the result.
- Sketch the standard families, including circles, half-lines, cardioids and roses.
- Locate tangents parallel and perpendicular to the initial line.
COMMON MISCONCEPTION
Polar coordinates are only a novelty; anything they describe is just as easy in x and y.
How far, which way
A point's polar coordinates are (r, θ), the distance from the pole and the anticlockwise angle from the initial line. The dictionary between the systems is immediate from a right triangle:
The booklet has a polar area formula but no conversions, so carry these three yourself. The triangle rebuilds them in seconds if you sketch it.
WORKED EXAMPLE
A polar equation in cartesian clothing
Convert r = 4 cos θ to cartesian form and identify the curve.
Multiply both sides by r to get r² = 4r cos θ.
Substitute: x² + y² = 4x.
Complete the square: (x − 2)² + y² = 4, a circle of radius 2 centred at (2, 0).
Spot check with θ = π/3, which gives r = 2 and the point (1, √3). Then (1 − 2)² + 3 = 4.
The standard sketches
Sketching runs on a short repertoire. r = a is a circle round the pole, θ = α a half-line at fixed bearing, and r = a(1 + cos θ) is the cardioid, a heart shape swelling to 2a on the initial line and pinching to zero opposite. r = a cos 2θ draws a four-petal rose. Tabulating r at θ = 0, π/2, π and 3π/2 pins each shape in seconds, which is far quicker than any cartesian attack on the same curves.
The rose needs a convention stated. A point's r was defined as a distance, but r = a cos 2θ turns negative over part of the turn, and the reading for curves like it is that (−r, θ) means the point (r, θ + π), the same distance out in the opposite direction. Where cos 2θ ≥ 0, on −π/4 ≤ θ ≤ π/4 and on 3π/4 ≤ θ ≤ 5π/4, the curve draws two petals along the initial line, and the stretches where cos 2θ < 0 supply the other two at right angles to them. For an area, take one loop at a time, over an interval on which r ≥ 0.
WORKED EXAMPLE
Reading a cardioid's vital points
For r = 1 + cos θ, find r at θ = 0, π/2 and π, and describe the curve.
θ = 0 gives r = 2. θ = π/2 gives r = 1. θ = π gives r = 0.
The curve bulges to 2 rightwards, passes 1 unit above the pole, and closes onto the pole itself at the back, a heart lying on its side.
The pinch at the pole happens because r reaches exactly zero there. Curves r = a + b cos θ with a > b never touch the pole at all.
GUIDED PRACTICE
Cartesian to polar
Find polar coordinates for the cartesian point (−3, 3), with 0 ≤ θ < 2π.
Show the working
r = √(9 + 9) = 3√2.
The point sits in the second quadrant, so θ = π − π/4 = 3π/4.
So (r, θ) = (3√2, 3π/4). Quoting θ = arctan(−1) = −π/4 without checking the quadrant is the standard trap.
Tangents parallel and perpendicular to the initial line
This is the one part of the topic students reliably fluff, and the cure is to stop thinking in polar for a moment. Write x = r cos θ and y = r sin θ as functions of θ alone. A tangent parallel to the initial line needs dy/dθ = 0 with dx/dθ ≠ 0. A tangent perpendicular to it needs dx/dθ = 0 with dy/dθ ≠ 0. If both derivatives vanish at once, as at a cusp, the test is silent and the limiting direction has to be examined instead. Differentiate, factorise, solve for θ, and only then go back for the r values.
WORKED EXAMPLE
Where the cardioid runs flat
Find the points on r = 1 + cos θ where the tangent is parallel to the initial line.
y = (1 + cos θ)sin θ, so dy/dθ = cos θ + cos 2θ.
Using cos 2θ = 2cos²θ − 1, that factorises as (2 cos θ − 1)(cos θ + 1).
cos θ = 1/2 gives θ = ±π/3, and cos θ = −1 gives θ = π, which is the cusp and not a genuine tangent.
At θ = π/3, r = 3/2, so the points are (3/2, π/3) and (3/2, −π/3), symmetric about the initial line.
Run the same method on x for the perpendicular case. dx/dθ = −sin θ(1 + 2cos θ) here, so those tangents sit at θ = ±2π/3 along with the trivial θ = 0. Always check whether a solution is the cusp or the pole before quoting it as an answer.
ASSESSMENT FOCUS
- To convert an equation, manufacture r², r cos θ and r sin θ. Multiplying through by r is the usual first move.
- Check the quadrant before writing θ. arctan alone cannot tell (−3, 3) from (3, −3).
- Sketch from a table of r at θ = 0, π/2, π, 3π/2, and mark where r = 0.
- State the range of θ that traces the curve once; it is part of a complete answer.
- For tangents, differentiate r sin θ or r cos θ, not r. Setting dr/dθ = 0 answers a different question.
CHECK YOURSELF
Convert the polar point (2, π/6) to cartesian coordinates, exactly.
Show a hint
x = r cos θ, y = r sin θ.
Show the answer
x = 2 cos(π/6) = √3 and y = 2 sin(π/6) = 1, so the point is (√3, 1).
x = r cos θ and y = r sin θ. Build r² and r cos θ to convert equations.
Sketch polar curves from r at the four compass angles, marking any pass through the pole.
Tangents parallel to the initial line come from dy/dθ = 0 and perpendicular ones from dx/dθ = 0, each with the other derivative non-zero.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the polar curves questions page.
CHECK YOUR PROGRESS
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- Plot points from (r, θ) and convert both ways with x = r cos θ, y = r sin θ.
- Convert polar equations to cartesian form and recognise the result.
- Sketch the standard families, including circles, half-lines, cardioids and roses.
- Locate tangents parallel and perpendicular to the initial line.
Open the full revision checklist to see every objective in the course in one place.