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Polar curves questions
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.
7 original questions · 27 marks · the polar curves notes · Polar coordinates
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Convert the polar point (4, 2π/3) to cartesian coordinates, exactly.
Worked answer
x = 4 cos(2π/3) = −2 and y = 4 sin(2π/3) = 2√3, so the point is (−2, 2√3), in the second quadrant as an argument between π/2 and π requires. B1 B1 for the two coordinates.Find polar coordinates (r, θ) for the cartesian point (1, −1), with −π < θ ≤ π.
Worked answer
r = √2. The point sits in the fourth quadrant, so θ = −π/4. B1 for r, M1 for arctan with the quadrant checked, A1 for θ. Quoting +π/4 from arctan without checking the quadrant is the standard slip; the sketch settles it.Convert r = 6 sin θ to cartesian form and identify the curve.
Worked answer
Multiply by r: r² = 6r sin θ, so x² + y² = 6y. Completing the square: x² + (y − 3)² = 9, a circle of radius 3 centred at (0, 3), sitting on the pole. M1 for multiplying by r, A1 for x² + y² = 6y, M1 for completing the square, A1 for the centre and radius. r = a sin θ and r = a cos θ are always circles through the origin.Convert r = 2 sec θ to cartesian form and describe the curve.
Worked answer
r = 2 sec θ means r = 2/cos θ, so r cos θ = 2. Since x = r cos θ, the curve is the vertical line x = 2, two units to the right of the pole. M1 for r cos θ = 2, A1 for x = 2, B1 for describing the vertical line. A polar equation this short can still be a perfectly straight line.For the cardioid r = 3(1 + cos θ), find r at θ = 0, π/2 and π, and sketch the curve, marking those three points.
Worked answer
r = 6, 3 and 0 respectively. The curve bulges to (6, 0) on the initial line, passes three units above the pole, and pinches onto the pole itself at the back, a heart on its side traced once as θ makes a full turn. B1 B1 B1 for the three values of r, B1 for the sketch.The curve r = 2 cos 3θ is a rose. State how many petals it has, the maximum value of r, and the first positive θ at which the curve passes through the pole.
Worked answer
Three petals, each of maximum radius 2. The curve reaches the pole when cos 3θ = 0, first at 3θ = π/2, so θ = π/6. B1 for three petals, B1 for the maximum r, B1 for θ = π/6. An odd multiple of θ gives that many petals; an even multiple gives twice as many, since the negative-r halves land in new positions.The cardioid has equation r = 3(1 + cos θ) for 0 ≤ θ < 2π. Find the polar coordinates of the points on the curve that are furthest from the initial line, giving exact values.
Worked answer
Distance from the initial line is |y|, and y = r sin θ = 3(1 + cos θ) sin θ. Differentiating with the product rule gives dy/dθ = 3[(1 + cos θ)cos θ − sin²θ]. Replacing sin²θ by 1 − cos²θ gives 3[2cos²θ + cos θ − 1] = 3(2 cos θ − 1)(cos θ + 1). The stationary values are cos θ = 1/2 and cos θ = −1, that is θ = π/3, θ = π and, by symmetry, θ = 5π/3. At θ = π the curve is at the pole, so the greatest distance comes at θ = π/3, where r = 3(3/2) = 9/2 and y = (9/2)(√3/2) = 9√3/4 ≈ 3.90, and equally at its mirror image θ = 5π/3, the same distance below the line. On the stated interval the points are (9/2, π/3) and (9/2, 5π/3). M1 for y = r sin θ, A1 for the product, M1 for differentiating, A1 for dy/dθ, M1 for using sin²θ = 1 − cos²θ and factorising, A1 for cos θ = 1/2, A1 for r = 9/2, A1 for both θ = π/3 and θ = 5π/3. Maximising r instead gives θ = 0 and the wrong points entirely; the question asks about distance from the line, so r sin θ is the function to differentiate.
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