Polar Coordinates - Worksheets, Questions and Revision

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A-Level · Core Pure Mathematics (Further Maths)

FP.CP6 Polar Coordinates

EDEXCEL 9FM0 · Calculator allowed · about 130 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Throughout this question, polar coordinates (r, θ) are given relative to the initial line and the pole O, with r ≥ 0 unless stated otherwise.

(a)The point A has Cartesian coordinates (3, -3sqrt(3)). Find the polar coordinates of A, giving r exactly and θ in radians such that -π < θ ≤ π.(3)

(b)A curve C has polar equation r = 6 cos(θ), for -π/2 ≤ θ ≤ π/2. Show that the Cartesian equation of C can be written as (x - 3)² + y² = 9, and describe C geometrically.(4)

(Total for Question 1 is 7 marks)

A curve has polar equation r = 5 + 3 cos(θ), for -π < θ ≤ π.

(a)State, with a reason, the symmetry of the curve.(2)

(b)Complete the table below, giving values to 3 significant figures where appropriate, and hence sketch the curve for -π < θ ≤ π.
θ: 0, π/6, π/3, π/2, 2pi/3, 5pi/6, π
r: 8, ?, 6.5, 5, ?, ?, 2(5)

(Total for Question 2 is 7 marks)

A curve C has polar equation r = 8 sin(θ), for 0 ≤ θ ≤ π.

(a)Show that the Cartesian equation of C can be written as x² + (y - 4)² = 16.(3)

(b)Hence state the centre and radius of C.(2)

(c)Using A = (1/2) x integral from θ=0 to θ=π of r² dtheta, show that the area enclosed by C is 16pi, and verify this agrees with your answer to part (b).(4)

(Total for Question 3 is 9 marks)

A cardioid C has polar equation r = 3(1 - cos(θ)), for 0 ≤ θ ≤ 2pi.

(a)State the value of r when θ = 0 and when θ = π, and state the symmetry of C, giving a reason.(3)

(b)Find the exact area enclosed by C.(6)

(Total for Question 4 is 9 marks)

An Archimedean spiral has polar equation r = 2 θ, for 0 ≤ θ ≤ 2pi (θ measured in radians).

(a)Complete the table of values of r.
θ: 0, π/2, π, 3pi/2, 2pi
r: 0, ?, ?, ?, ?(2)

(b)Find the exact area enclosed between the curve and the initial line for 0 ≤ θ ≤ 2pi (the area swept out in one complete revolution).(5)

(c)A second spiral has equation r = k θ for the same interval 0 ≤ θ ≤ 2pi. Given that the area enclosed by this spiral (defined as in part (b)) is exactly twice the area found in part (b), find the value of k.(3)

(Total for Question 5 is 10 marks)

The cardioid C, with polar equation r = 3(1 - cos(θ)) for 0 ≤ θ < 2pi, is the curve from Question 4. Let x = r cos(θ) and y = r sin(θ).

(a)Show that dy/dtheta = 3cos(θ) - 3cos(2theta).(3)

(b)Hence find the exact coordinates of the two points on C, other than the pole, at which the tangent is parallel to the initial line.(7)

(Total for Question 6 is 10 marks)

A rose curve has polar equation r = 4 cos(3theta).

(a)State the number of petals of the curve and the maximum value of r.(2)

(b)One petal is traced as θ ranges from -π/6 to π/6. Show that the area of this petal is 4pi/3.(5)

(c)Hence find the total area enclosed by all three petals of the curve.(2)

(Total for Question 7 is 9 marks)

Using the curve C from Question 1(b), with r = 6 cos(θ) for -π/2 ≤ θ ≤ π/2, let x = r cos(θ) and y = r sin(θ).

(a)Show that x = 6cos²(θ) and y = 3sin(2theta), and hence find dx/dtheta and dy/dtheta in terms of θ.(4)

(b)Find the coordinates of the points on C at which (i) the tangent is parallel to the initial line, (ii) the tangent is perpendicular to the initial line.(6)

(Total for Question 8 is 10 marks)

A limacon has polar equation r = 1 + 2cos(θ), for 0 ≤ θ < 2pi. Since the constant term (1) is less than the coefficient of cos(θ) (2), the curve has an inner loop.

(a)Find the two values of θ, 0 ≤ θ < 2pi, for which r = 0, and hence state the range of values of θ for which the curve traces the inner loop.(3)

(b)Find the exact area enclosed by the inner loop.(6)

(Total for Question 9 is 9 marks)

A lemniscate has polar equation r² = 4cos(2theta).

(a)State the two ranges of values of θ, for 0 ≤ θ < 2pi, for which the curve is defined (i.e. r² ≥ 0), describing the right-hand and left-hand loops.(2)

(b)Find the exact area enclosed by one loop of the curve.(5)

(c)Find the values of θ at which the curve has a tangent at the pole for the right-hand loop, and briefly interpret this result.(3)

(Total for Question 10 is 10 marks)

A region is bounded by the curve r = f(θ) between θ = α and θ = β (α < β), and the two half-lines θ = α and θ = β.

(a)By considering the region as the limit of a sum of thin circular sectors, prove that the area A of the region is given by A = (1/2) x integral from α to β of r² dtheta.(5)

(Total for Question 11 is 5 marks)

A circle C1 has polar equation r = 3/2. A cardioid C2 has polar equation r = 1 + cos(θ), for -π < θ ≤ π. The two curves intersect and enclose a common region R, which is symmetrical about the initial line.

(a)Find the value of θ in the interval 0 ≤ θ ≤ π at which C1 and C2 intersect.(3)

(b)Sketch C1 and C2 on the same diagram, for -π < θ ≤ π, showing clearly the region R that lies inside both curves.(3)

(c)Show that, for 0 ≤ θ ≤ π/3, the boundary of R nearer to the pole is C1, and that for π/3 ≤ θ ≤ π, the boundary of R nearer to the pole is C2.(2)

(d)Hence show that the area of R is 7pi/4 - (9sqrt(3))/8.(8)

(Total for Question 12 is 16 marks)

Mark scheme · FP.CP6 Polar Coordinates

Question 1

(a) M1 r² = 3² + (-3sqrt(3))² = 9 + 27 = 36 leading to r = 6
(a) M1 tan(θ) = y/x = -√3, reference angle π/3
(a) A1 θ = -π/3 (fourth quadrant, since x > 0, y < 0), cao
(a) Answer: (r, θ) = (6, -π/3)
(b) M1 multiply both sides by r: r² = 6r cos(θ)
(b) M1 use r² = x² + y² and r cos(θ) = x to obtain x² + y² = 6x
(b) A1 complete the square: (x - 3)² + y² = 9, cso (answer printed)
(b) B1 C is a circle with centre (3, 0) and radius 3
(b) Answer: (x-3)² + y² = 9; circle, centre (3,0), radius 3

Question 2

(a) B1 the curve is symmetrical about the initial line (θ = 0)
(a) B1 valid reason: cos(-θ) = cos(θ), so r(θ) = r(-θ) for all θ
(a) Answer: Symmetrical about the initial line, since r is an even function of θ
(b) M1 substitute θ = π/6, 2pi/3, 5pi/6 into r = 5 + 3cos(θ)
(b) A1 r(π/6) = 5 + (3sqrt(3))/2 = 7.60 (3 s.f.)
(b) A1 r(2pi/3) = 3.5 and r(5pi/6) = 5 - (3sqrt(3))/2 = 2.40 (3 s.f.)
(b) B1 smooth closed curve symmetrical about the initial line, with a single dimple (no inner loop, since 5 > 3), max r = 8 at θ = 0 and min r = 2 at θ = π
(b) Answer: r(π/6) = 7.60, r(2pi/3) = 3.5, r(5pi/6) = 2.40 (all 3 s.f. where relevant); dimpled limacon, symmetric about the initial line, ranging from r=8 at θ=0 to r=2 at θ=π

Question 3

(a) M1 multiply by r: r² = 8r sin(θ)
(a) M1 substitute r² = x²+y² and r sin(θ) = y: x² + y² = 8y
(a) A1 complete the square: x² + (y-4)² = 16, cso (answer printed)
(a) Answer: x² + (y-4)² = 16
(b) B1 centre (0, 4)
(b) B1 radius 4
(b) Answer: Centre (0,4), radius 4
(c) M1 Area = (1/2) x integral from 0 to π of (8sin(θ))² dtheta = 32 x integral from 0 to π of sin²(θ) dtheta
(c) M1 use sin²(θ) = (1 - cos(2theta))/2 to write the integrand as 16(1-cos(2theta))
(c) A1 integrate: 16[θ - sin(2theta)/2] evaluated from 0 to π = 16pi, cso
(c) B1 verification: this matches π x radius² = π x 4² = 16pi from part (b)
(c) Answer: Area = 16pi

Question 4

(a) B1 r = 0 when θ = 0 (C has a cusp at the pole)
(a) B1 r = 6 when θ = π (the point furthest from the pole)
(a) B1 symmetrical about the initial line, since cos(-θ) = cos(θ)
(a) Answer: r(0) = 0, r(π) = 6; symmetrical about the initial line
(b) M1 Area = (1/2) x integral from 0 to 2pi of [3(1-cos(θ))]² dtheta
(b) M1 expand: (1-cos(θ))² = 1 - 2cos(θ) + cos²(θ)
(b) M1 use cos²(θ) = (1+cos(2theta))/2 throughout the integrand
(b) A1 integral from 0 to 2pi of 1 dtheta = 2pi, integral of cos(θ) dtheta = 0, integral of cos²(θ) dtheta = π
(b) M1 combine: Area = (9/2)(2pi - 0 + π) = (9/2)(3pi)
(b) A1 Area = 27pi/2, cao
(b) Answer: Area = 27pi/2 (= 13.5pi, approx 42.4 square units)

Question 5

(a) B1 r(π/2) = π and r(π) = 2pi
(a) B1 r(3pi/2) = 3pi and r(2pi) = 4pi
(a) Answer: r = π, 2pi, 3pi, 4pi at θ = π/2, π, 3pi/2, 2pi respectively
(b) M1 Area = (1/2) x integral from 0 to 2pi of (2theta)² dtheta
(b) M1 = 2 x integral from 0 to 2pi of θ² dtheta
(b) M1 integrate: 2[θ³/3] from 0 to 2pi
(b) A1 substitute limits: 2 x (2pi)³/3 = 2 x 8pi³/3
(b) A1 Area = 16pi³/3, cao
(b) Answer: Area = 16pi³/3 (approx 165 square units, 3 s.f.)
(c) M1 new area = (k²/2) x integral from 0 to 2pi of θ² dtheta = (k²/2)(8pi³/3) = 4k² π³/3
(c) M1 set 4k² π³/3 = 2 x 16pi³/3, giving k² = 8
(c) A1 k = 2sqrt(2), cao (positive root taken)
(c) Answer: k = 2sqrt(2) (approx 2.83)

Question 6

(a) M1 write y = 3(1-cos(θ))sin(θ) = 3sin(θ) - 3sin(θ)cos(θ)
(a) M1 use sin(θ)cos(θ) = (1/2)sin(2theta), so y = 3sin(θ) - (3/2)sin(2theta)
(a) A1 differentiate: dy/dtheta = 3cos(θ) - 3cos(2theta), cso
(a) Answer: dy/dtheta = 3cos(θ) - 3cos(2theta)
(b) M1 set dy/dtheta = 0: cos(θ) = cos(2theta)
(b) M1 use cos(2theta) = 2cos²(θ) - 1 to form 2cos²(θ) - cos(θ) - 1 = 0
(b) A1 factorise (2cos(θ)+1)(cos(θ)-1) = 0, giving cos(θ) = -1/2 (rejecting cos(θ)=1, θ=0, the cusp at the pole)
(b) A1 θ = 2pi/3 and θ = 4pi/3
(b) M1 find r = 3(1-(-1/2)) = 9/2 at both values
(b) A1 point 1: (x,y) = (-9/4, 9sqrt(3)/4), cao
(b) B1ft point 2: (x,y) = (-9/4, -9sqrt(3)/4), ft from point 1 (mirror image under the symmetry of part (a) reasoning)
(b) Answer: (-9/4, 9sqrt(3)/4) and (-9/4, -9sqrt(3)/4)

Question 7

(a) B1 3 petals (the coefficient of θ, 3, is odd, so the curve has 3 petals)
(a) B1 maximum r = 4
(a) Answer: 3 petals, maximum r = 4
(b) M1 Area = (1/2) x integral from -π/6 to π/6 of 16cos²(3theta) dtheta
(b) M1 use cos²(3theta) = (1+cos(6theta))/2 to write the integrand as 4(1+cos(6theta))
(b) M1 integrate: 4[θ + sin(6theta)/6]
(b) A1 substitute limits: 4[(π/6 + 0) - (-π/6 + 0)] = 4 x π/3
(b) A1 Area = 4pi/3, cso (answer printed)
(b) Answer: Area of one petal = 4pi/3
(c) M1 total area = 3 x (4pi/3), using that the 3 petals do not overlap and the curve retraces the same petals over π ≤ θ < 2pi (since n is odd)
(c) A1 total area = 4pi, cao
(c) Answer: Total area = 4pi (approx 12.6 square units)

Question 8

(a) B1 x = r cos(θ) = 6cos(θ)cos(θ) = 6cos²(θ); y = r sin(θ) = 6cos(θ)sin(θ) = 3sin(2theta)
(a) M1 differentiate x: dx/dtheta = -12sin(θ)cos(θ)
(a) A1 dx/dtheta = -6sin(2theta), oe
(a) A1 dy/dtheta = 6cos(2theta)
(a) Answer: dx/dtheta = -6sin(2theta), dy/dtheta = 6cos(2theta)
(b) M1 (i) set dy/dtheta = 0: cos(2theta) = 0, giving θ = π/4 or θ = -π/4
(b) M1 (i) find r = 6cos(π/4) = 3sqrt(2) at each, then x = r cos(θ) = 3, y = r sin(θ) = ±3
(b) A1 (i) points (3, 3) and (3, -3), cao
(b) M1 (ii) set dx/dtheta = 0: sin(2theta) = 0, giving θ = 0 or θ = +-π/2
(b) A1 (ii) at θ = 0, r = 6, giving the point (6, 0)
(b) A1 (ii) at θ = +-π/2, r = 0, giving the pole (0, 0); both points have a vertical tangent
(b) Answer: (i) (3,3) and (3,-3); (ii) (6,0) and the pole (0,0)

Question 9

(a) M1 set 1 + 2cos(θ) = 0, giving cos(θ) = -1/2
(a) A1 θ = 2pi/3 and θ = 4pi/3, cao
(a) B1 the inner loop is traced for 2pi/3 ≤ θ ≤ 4pi/3, oe
(a) Answer: θ = 2pi/3, 4pi/3; inner loop traced for 2pi/3 ≤ θ ≤ 4pi/3
(b) M1 Area = (1/2) x integral from 2pi/3 to 4pi/3 of (1+2cos(θ))² dtheta
(b) M1 expand and use cos²(θ)=(1+cos(2theta))/2 to write integrand as 3 + 4cos(θ) + 2cos(2theta)
(b) M1 integrate: 3theta + 4sin(θ) + sin(2theta)
(b) A1 value at θ=4pi/3: 4pi - 2sqrt(3) + √3/2 = 4pi - (3sqrt(3))/2
(b) A1 value at θ=2pi/3: 2pi + 2sqrt(3) - √3/2 = 2pi + (3sqrt(3))/2
(b) A1 Area = (1/2)(2pi - 3sqrt(3)) = π - (3sqrt(3))/2, cao
(b) Answer: Area of inner loop = π - (3sqrt(3))/2 (approx 0.544 square units, 3 s.f.)

Question 10

(a) B1 right-hand loop: -π/4 ≤ θ ≤ π/4 (equivalently 0≤θ≤π/4 or 7pi/4≤θ<2pi)
(a) B1 left-hand loop: 3pi/4 ≤ θ ≤ 5pi/4
(a) Answer: -π/4 ≤ θ ≤ π/4 (right-hand loop) and 3pi/4 ≤ θ ≤ 5pi/4 (left-hand loop)
(b) M1 Area = (1/2) x integral from -π/4 to π/4 of 4cos(2theta) dtheta
(b) M1 = 2 x integral from -π/4 to π/4 of cos(2theta) dtheta
(b) M1 integrate: 2[sin(2theta)/2] = [sin(2theta)]
(b) A1 substitute limits: sin(π/2) - sin(-π/2) = 1-(-1) = 2
(b) A1 Area = 2, cao
(b) Answer: Area of one loop = 2 (square units)
(c) M1 set r = 0: cos(2theta) = 0
(c) A1 2theta = +-π/2, giving θ = π/4 and θ = -π/4
(c) B1 interpretation: the lines θ = π/4 and θ = -π/4 are tangent to the curve at the pole, i.e. the loop touches the origin tangentially along these two directions
(c) Answer: θ = π/4 and θ = -π/4

Question 11

(a) B1 divide the region into n thin sectors, each subtending an angle delta(θ) = (β - α)/n at the pole
(a) B1 the area of one such sector, of radius approximately r, is approximately (1/2) r² delta(θ), by the formula for the area of a circular sector
(a) B1 the total area is approximately the sum over all n sectors: A is approximately σ (1/2) r² delta(θ), for θ from α to β
(a) B1 as n tends to infinity, delta(θ) tends to 0, and the sum tends to the definite integral: A = (1/2) x integral from α to β of r² dtheta
(a) B1 cso: the approximation error in each sector's area tends to zero faster than delta(θ) as delta(θ) -> 0, so the limit gives the exact area, establishing the result
(a) Answer: A = (1/2) x integral from α to β of r² dtheta