Questions on polar curves

Question 1

Convert the polar point (4, 2π/3) to cartesian coordinates, exactly.

Accepted 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.

Question 2

Find polar coordinates (r, θ) for the cartesian point (1, −1), with −π < θ ≤ π.

Accepted 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.

Question 3

A point has polar coordinates (−3, π/4). Using x = r cos θ, y = r sin θ directly with r = −3, find its exact cartesian coordinates, and confirm the same point is (3, π/4 + π).

Accepted Answer

x = −3 cos(π/4) = −3√2/2 and y = −3 sin(π/4) = −3√2/2, so the point is (−3√2/2, −3√2/2). M1 for substituting r = −3 directly, A1 for both coordinates. Using the sign convention (−r, θ) = (r, θ + π), the same point is (3, π/4 + π) = (3, 5π/4): x = 3 cos(5π/4) = −3√2/2 and y = 3 sin(5π/4) = −3√2/2, agreeing exactly. B1 for the check.

Question 4

Convert the cartesian equation y = √3x to polar form.

Accepted Answer

tan θ = y/x = √3, so θ = π/3. M1 for tan θ = y/x, A1 for θ = π/3. Since r may take any real value, positive or negative, this single equation θ = π/3 already covers the whole line: r > 0 traces the ray into the first quadrant, and r < 0, reinterpreted as (r, π/3 + π), traces the opposite ray into the third quadrant. B1 for the line being covered fully with no extra equation needed.

Question 5

Convert r = 6 sin θ to cartesian form and identify the curve.

Accepted 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.

Question 6

Convert r = 2 sec θ to cartesian form and describe the curve.

Accepted 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.

Question 7

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.

Accepted 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.

Question 8

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.

Accepted 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.

Question 9

Convert the cartesian equation x² + y² = 4x + 6y to polar form, giving r in terms of θ, and hence state the cartesian centre and radius of the circle.

Accepted Answer

Substituting x² + y² = r², 4x = 4r cos θ and 6y = 6r sin θ: r² = 4r cos θ + 6r sin θ. Dividing by r, away from the pole: r = 4 cos θ + 6 sin θ. M1 for the substitution, A1 for dividing by r to reach this form. Completing the square in the original equation, (x − 2)² + (y − 3)² = 13, so the centre is (2, 3) and the radius is √13. M1 for completing the square, A1 for the centre and radius.

Question 10

The curve r = 2 sin 2θ is a rose. Find every value of θ in [0, 2π) for which r = 0, and state how many petals the curve has.

Accepted Answer

r = 0 when sin 2θ = 0, that is 2θ = 0, π, 2π, 3π, so θ = 0, π/2, π, 3π/2. M1 for sin 2θ = 0, A1 for the four values of θ. These four zeros split the full turn into four arcs, and because the coefficient 2 inside sin 2θ is even, the arcs where r is negative trace petals in new positions rather than retracing petals already drawn, so the curve has 4 petals, twice the coefficient. B1 for the count with the reasoning.

Question 11

For the spiral r = 2θ, θ ≥ 0, find r at θ = 0, π/2, π and 2π, and describe how the curve behaves as θ increases.

Accepted Answer

r = 0, π, 2π and 4π respectively, growing steadily. M1 for the four values, A1 for all correct. Unlike a circle or a cardioid, r grows without bound as θ increases: the curve starts at the pole and winds anticlockwise, moving further from the pole on every turn rather than retracing a closed curve. B1 for the description.

Question 12

The diagram shows one curve drawn on a polar grid. The curve is one of r = 4 cos θ, r = 4 sin θ, r = 2 + 2 cos θ and r = 4 cos 2θ. Use values of r read from the diagram to determine which of the four it is, and convert that equation to cartesian form, describing the curve it represents.

Accepted Answer

Reading the curve where it meets a circle of the grid on a numbered ray, and writing the angles in radians: r = 4 at θ = 0, r = 2 at θ = 60° = π/3, and r = 0 at θ = 90° = π/2. B1 for the three values read from the diagram. Testing the candidates at θ = 0 first, r = 4 sin θ gives 0, not 4, so it goes. At θ = π/3 the survivors give 4 cos(π/3) = 2, 2 + 2cos(π/3) = 3 and 4 cos(2π/3) = −2. M1 for testing the candidates against the readings. Only r = 4 cos θ matches all three readings. A1 for the identification. Multiplying by r gives r² = 4r cos θ, and since r² = x² + y² and r cos θ = x, this is x² + y² = 4x. M1 for multiplying by r and substituting. Completing the square, (x − 2)² + y² = 4, a circle of centre (2, 0) and radius 2, passing through the pole with the segment of the initial line from 0 to 4 as a diameter. A1 for the cartesian equation and the description. Multiplying by r is the standard move: it converts r cos θ into x at the cost of introducing the pole as a possible extra solution, and here the pole is on the curve anyway, at θ = π/2.

Question 13

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.

Accepted 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.

Question 14

For the cardioid r = 1 + cos θ, find the values of θ, and the corresponding points, at which the tangent is parallel to the initial line. Then find the values of θ, and the points, at which the tangent is perpendicular to the initial line, excluding the pole.

Accepted Answer

y = (1 + cos θ) sin θ, so by the product rule dy/dθ = (1 + cos θ)cos θ − sin²θ, and replacing sin²θ by 1 − cos²θ gives dy/dθ = cos θ + cos 2θ. M1 for the product rule, A1 for dy/dθ = cos θ + cos 2θ. Using cos 2θ = 2cos²θ − 1, this is 2cos²θ + cos θ − 1, which factorises as (2 cos θ − 1)(cos θ + 1). M1 for the double angle and factorising. cos θ = 1/2 gives θ = ±π/3, where r = 3/2; cos θ = −1 gives θ = π, the cusp, where dx/dθ also vanishes, so this test does not apply there. The parallel tangents touch at (3/2, π/3) and (3/2, −π/3). A1 for both points.
x = (1 + cos θ) cos θ, so dx/dθ = −sin θ − 2 sin θ cos θ = −sin θ(1 + 2 cos θ). M1 for dx/dθ, A1 for dx/dθ = −sin θ(1 + 2 cos θ). This is zero when sin θ = 0, at θ = 0 or π, or when cos θ = −1/2, at θ = ±2π/3. At θ = π both derivatives vanish together, since it is the cusp, so that solution is excluded; the other three are genuine. M1 for solving and excluding the cusp. At θ = 0, r = 2; at θ = ±2π/3, r = 1/2. The perpendicular tangents touch at (2, 0), (1/2, 2π/3) and (1/2, −2π/3). A1 for all three points.

Question 15

The limaçon r = 1 + 2cos θ has an inner loop. Find the two values of θ in [0, 2π) at which r = 0, and hence state the range of θ that traces the inner loop. Find the least value of r on the inner loop, and the value of θ at which it occurs.

Accepted Answer

r = 0 when cos θ = −1/2, so θ = 2π/3 or θ = 4π/3. M1 for cos θ = −1/2, A1 for both values. Between these, for 2π/3 < θ < 4π/3, cos θ < −1/2 and r is negative: this is the range that traces the inner loop. B1 for the range. r = 1 + 2cos θ is least where cos θ is least, at θ = π, giving r = 1 + 2(−1) = −1. M1 for identifying θ = π, A1 for r = −1.

Question 16

The curve C has equation r² = 9cos 2θ. Find the two intervals of θ in [0, 2π) for which C does not exist, and find the greatest value of r, stating the value(s) of θ at which it occurs and the value of r at θ = π/4.

Accepted Answer

r² cannot be negative, so C exists only where cos 2θ ≥ 0. Solving cos 2θ < 0 gives 2θ between π/2 and 3π/2, or between 5π/2 and 7π/2, that is θ between π/4 and 3π/4, or between 5π/4 and 7π/4. M1 for cos 2θ ≥ 0 as the condition, A1 for both intervals where C does not exist. The greatest value of r² is 9, when cos 2θ = 1, that is θ = 0 or θ = π, giving the greatest value of r as 3, at θ = 0 and θ = π. M1 for cos 2θ = 1, A1 for r = 3 with both angles. At θ = π/4, cos 2θ = cos(π/2) = 0, so r² = 0 and r = 0 there, the pole itself: the curve just reaches the pole at the boundary of existence and does not cross it. A1 for r = 0 there.