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Edexcel A-level Further Maths revision notes

Polar coordinates

Section CP-7
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
3 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section CP-7

Checked against Edexcel 9FM0 section CP-7. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

In the exam: Formulae booklet provided · calculator allowed in every paper

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CP-7.1

Understand and use polar coordinates and be able to convert between polar and Cartesian coordinates.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Polar coordinates (r,θ)(r,\theta) locate a point at directed distance rr from the pole, with θ\theta measured anticlockwise from the initial line.
  • Convert using x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta; in reverse, use r2=x2+y2r^2=x^2+y^2 and determine θ\theta from tan⁡θ=y/x\tan\theta=y/x with a quadrant check.
  • For a polar curve, multiplying by rr can expose the substitutions rcos⁡θ=xr\cos\theta=x, rsin⁡θ=yr\sin\theta=y and r2=x2+y2r^2=x^2+y^2.
  • Coordinates are not unique: (r,θ)(r,\theta), (r,θ+2π)(r,\theta+2\pi) and (−r,θ+π)(-r,\theta+\pi) represent the same point.
  • Examiners expect an angle in the requested interval and a Cartesian equation simplified enough to identify the locus.
A polar point resolved into its Cartesian horizontal and vertical components.
Worked example

Convert r=6cos⁡θ−4sin⁡θr=6\cos\theta-4\sin\theta to Cartesian form and identify the curve.

  1. 1.Multiply by rr: r2=6rcos⁡θ−4rsin⁡θr^2=6r\cos\theta-4r\sin\theta.
  2. 2.Substitute to obtain x2+y2=6x−4yx^2+y^2=6x-4y.
  3. 3.Complete the squares: (x−3)2+(y+2)2=13(x-3)^2+(y+2)^2=13.

Answer: The curve is a circle with centre (3,−2)(3,-2) and radius 13\sqrt{13}.

Common mistakes

  • Don't use arctan⁡(y/x)\arctan(y/x) without correcting the angle to the point's quadrant.
  • Don't replace rr by x2+y2x^2+y^2 instead of replacing r2r^2 by x2+y2x^2+y^2.
  • Don't treat a negative value of rr as invalid rather than reversing the direction.

Exam tip

When an interval for θ\theta is given, state the quadrant before selecting the inverse-tangent value.

Tier 1 · Easy

ORIGINAL

1.

Convert the polar coordinates (4,π/6)(4,\pi/6) to Cartesian coordinates.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Express the Cartesian point (−3,33)(-3,3\sqrt3) in polar form, taking r>0r>0 and 0≤θ<2π0\leq\theta<2\pi.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1.

Convert the polar curve r=4cos⁡θ+2sin⁡θr=4\cos\theta+2\sin\theta to Cartesian form and identify the curve.

(5)

(Total for Question 1 is 5 marks)

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CP-7.2

Sketch curves with r given as a function of theta, including use of trigonometric functions.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A polar sketch should be built from zeros and extreme values of rr, symmetry, and values on the initial line.
  • A negative radius places the point in the direction opposite to θ\theta, so discarding negative values can remove a loop or petal.
  • The specification includes straight lines such as r=psec⁡(α−θ)r=p\sec(\alpha-\theta), circles, spirals, cardioids, limacons, rose curves and r2=a2cos⁡2θr^2=a^2\cos2\theta.
  • For tangents, write x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta: a tangent parallel to the initial line normally has dy/dθ=0dy/d\theta=0, while a perpendicular tangent normally has dx/dθ=0dx/d\theta=0.
  • Examiners expect key angles, pole crossings, symmetry and maximum radii to be labelled rather than an unsupported calculator sketch.
The four-petalled shape of the rose curve r=acos⁡4θr=a\cos4\theta (under r≥0r\geq0), with its symmetry and angular directions labelled.
Worked example

Sketch r=3(1+cos⁡θ)r=3(1+\cos\theta) for 0≤θ≤2π0\leq\theta\leq2\pi, labelling the pole crossing and maximum radius.

  1. 1.The equation is unchanged by θ↦−θ\theta\mapsto-\theta, so the curve is symmetric about the initial line.
  2. 2.r=0r=0 at θ=π\theta=\pi, while r=6r=6 at θ=0\theta=0.
  3. 3.At θ=π/2\theta=\pi/2 and 3π/23\pi/2, r=3r=3; join the points as one cardioid with its cusp at the pole.

Answer: A right-facing cardioid, symmetric about the initial line, with cusp at the pole and furthest point (6,0)(6,0).

Common mistakes

  • Don't plot a negative radius in the direction θ\theta instead of the opposite direction.
  • Don't call dy/dθ=0dy/d\theta=0 sufficient for a horizontal tangent without checking that dx/dθ≠0dx/d\theta\ne0.
  • Don't sketch a rose petal at every zero of rr rather than using extrema to locate petal axes.

Exam tip

For a sketch question, find zeros and extrema of rr and label them on the curve before joining smoothly.

Tier 1 · Easy

ORIGINAL

1.

Sketch the polar curve r=3r=3, labelling its key geometric features.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1.

Sketch the polar curve r=2cos⁡θr=2\cos\theta. State its Cartesian equation, centre and radius.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

For 0≤θ<2π0\leq\theta<2\pi and taking r≥0r\geq0, sketch the complete polar curve r=4sin⁡2θr=4\sin2\theta, stating the intervals of θ\theta on which r≥0r\geq0 and labelling the directions and lengths of all petals and the values of θ\theta at which the curve passes through the pole.

(5)

(Total for Question 1 is 5 marks)

CP-7.3

Find the area enclosed by a polar curve.

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The polar area swept while θ\theta runs from α\alpha to β\beta is A=12∫αβr2 dθA=\dfrac12\int_\alpha^\beta r^2\,d\theta. The limits must describe the required region exactly; find them from pole crossings, intersections or symmetry, because one complete loop may occupy only part of a full revolution.
  • For a region between curves over the same angles, use 12∫(router2−rinner2) dθ\dfrac12\int(r_{\text{outer}}^2-r_{\text{inner}}^2)\,d\theta, checking which radius is larger throughout.
  • Trigonometric identities are often needed before integration.
  • Examiners expect a sketch or clear limit justification, the factor 12\tfrac12, correct squaring of rr, and an exact value unless a decimal is requested.
  • A symmetric multiplier is valid only when the chosen sector is not already the whole region.
Worked example

Find the exact area of one petal of r=4cos⁡(2θ)r=4\cos(2\theta).

  1. 1.The petal centred on the initial line is traced from θ=−π/4\theta=-\pi/4 to θ=π/4\theta=\pi/4.
  2. 2.A=12∫−π/4π/416cos⁡2(2θ) dθA=\dfrac12\int_{-\pi/4}^{\pi/4}16\cos^2(2\theta)\,d\theta.
  3. 3.Using cos⁡2(2θ)=12(1+cos⁡4θ)\cos^2(2\theta)=\dfrac12(1+\cos4\theta) gives A=4[θ+14sin⁡4θ]−π/4π/4=2πA=4[\theta+\tfrac14\sin4\theta]_{-\pi/4}^{\pi/4}=2\pi.

Answer: The area of one petal is 2π2\pi square units.

Common mistakes

  • Don't integrate rr rather than r2r^2 in the polar area formula.
  • Don't omit the factor 12\tfrac12 from A=12∫r2 dθA=\tfrac12\int r^2\,d\theta.
  • Don't use 0≤θ≤2π0\leq\theta\leq2\pi for one loop, because this counts the same region repeatedly.

Exam tip

State how the intersection or pole-crossing equations produce the angular limits before evaluating the integral.

Tier 1 · Easy

ORIGINAL

1.

Find the exact area of the sector enclosed by r=2r=2 and the rays θ=0\theta=0 and θ=π/3\theta=\pi/3.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

1.

Determine the exact area enclosed by the loop r=3sin⁡θr=3\sin\theta.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

1.

The curves r=2r=2 and r=4cos⁡θr=4\cos\theta enclose a region that lies inside r=4cos⁡θr=4\cos\theta but outside r=2r=2. Find its exact area.

(7)

(Total for Question 1 is 7 marks)

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