Maths › Polar coordinates › Polar curves
Polar curves
A polar equation gives a signed radial coordinate as a function of angle. Determine a complete tracing interval, test symmetry, find zeros and extrema, and calculate enough intermediate values to identify the curve. For tangents, use dx/dθ and dy/dθ; if both vanish, evaluate the limiting gradient rather than discarding the point.
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
Every point has exactly one pair of polar coordinates.
Polar representations are not unique. Adding 2π to θ names the same point, the pole is (0, θ) for every θ, and in a polar curve equation a negative value is read by (−r, θ) = (r, θ + π), which is how r = a cos 2θ traces four petals. A question that fixes an interval such as 0 ≤ θ < 2π is fixing which representation to quote.
Polar coordinates and the sign convention for r
A point's polar coordinates are (r, θ), the distance from the pole and the anticlockwise angle from the initial line. The conversions follow from one right-angled triangle:
Check the current formula booklet for your awarding organisation; whether or not these are supplied, you must know when they apply and be able to derive them from the triangle.
A point may be represented by (r, θ) with r ≥ 0; in a polar curve equation, a negative value is interpreted by (−r, θ) = (r, θ + π). Polar representations are not unique: adding 2π to θ gives the same point, and the pole is (0, θ) for every θ.
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).
Check with θ = π/3, which gives r = 2 and the point (1, √3). Then (1 − 2)² + 3 = 4.
Sketching the standard curves
To sketch a polar curve, determine a complete tracing interval, test symmetry, find the zeros and extrema of r, and calculate enough intermediate values to identify the shape. r = a is a circle centred on the pole, θ = α a half-line from the pole, and r = a(1 + cos θ) is the cardioid, with r = 2a on the initial line and r = 0 in the opposite direction. r = a cos 2θ is a four-petal rose.
The rose requires the sign convention. r = a cos 2θ is negative over part of the turn, and (−r, θ) denotes the point (r, θ + π). Where cos 2θ ≥ 0, on −π/4 ≤ θ ≤ π/4 and on 3π/4 ≤ θ ≤ 5π/4, the curve traces two petals along the initial line; the intervals where cos 2θ < 0 give the other two, perpendicular to them. For an area, integrate over one loop at a time, on an interval where r ≥ 0.
WORKED EXAMPLE
Reading values off a cardioid
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 reaches r = 2 along the initial line, passes 1 unit above the pole, and returns to the pole at θ = π.
The cusp at the pole occurs because r = 0 there. A curve r = a + b cos θ with a > b > 0 has r > 0 throughout and does not reach the pole.
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). The calculator value arctan(−1) = −π/4 is in the fourth quadrant, so the quadrant must be checked before quoting θ.
Tangents parallel and perpendicular to the initial line
Work in cartesian components. Write x = r cos θ and y = r sin θ as functions of θ alone. A tangent parallel to the initial line requires dy/dθ = 0 with dx/dθ ≠ 0; a tangent perpendicular to it requires dx/dθ = 0 with dy/dθ ≠ 0. If both derivatives vanish, as at a cusp, evaluate the limiting gradient (dy/dθ)/(dx/dθ) instead. Differentiate, factorise, solve for θ, then find the corresponding values of r.
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, the cusp, where dx/dθ also vanishes and the test above does not apply.
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.
At θ = π both derivatives vanish. Taking the limiting ratio (dy/dθ)/(dx/dθ), or expanding near π, gives gradient 0; the cusp has tangent y = 0.
Apply the same method to x for the perpendicular case. Here dx/dθ = −sin θ(1 + 2cos θ), giving tangents at θ = ±2π/3 and at θ = 0. Check whether a solution is a cusp or the pole before quoting it.
Describe the drawing
The cardioid with five marked points. Amber dots at theta equals plus and minus pi over 3 carry dashed horizontal tangents, where dy by d theta vanishes; coral dots at theta equals 0 and plus and minus 2 pi over 3 carry dashed vertical tangents, where dx by d theta vanishes. At the cusp, theta equals pi, both derivatives vanish at once and neither test applies, so the pole point is marked grey and excluded.
ASSESSMENT FOCUS
- To convert an equation, produce r², r cos θ and r sin θ; multiplying through by r is usually the first step.
- Check the quadrant before writing θ: arctan alone does not distinguish (−3, 3) from (3, −3).
- Sketch from symmetry, the zeros and extrema of r, and enough intermediate values to fix the shape.
- State the interval of θ that traces the curve once; it is part of a complete answer.
- For tangents, differentiate r sin θ and r cos θ, not r; dr/dθ = 0 locates extreme values of r instead.
- If dx/dθ and dy/dθ both vanish, evaluate the limiting gradient (dy/dθ)/(dx/dθ) rather than discarding the point.
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; form r² and r cos θ to convert equations. In a curve equation, (−r, θ) denotes (r, θ + π).
Sketch a polar curve from a complete tracing interval, its symmetry, the zeros and extrema of r, and sufficient intermediate values.
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; if both vanish, take the limiting gradient.
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.
USED LATER IN
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.