amazon.co.ukPolar coordinates for Edexcel 9FM0 Core Pure 2 and OCR MEI: converting between polar and cartesian form, sketching cardioids, limacons and rose curves, and finding enclosed areas with the half r-squared integral - revision with worked examples and mark-scheme logic.
The polar system. A point P is fixed by two numbers: r, its distance from the origin O (called the pole), and theta, the angle measured anticlockwise from the positive x-axis, the initial line. The link with cartesian coordinates is x = r cos theta, y = r sin theta; going the other way, r2 = x2 + y2 and tan theta = y/x. Because tan theta fixes theta only up to a multiple of pi, always settle the quadrant from the signs of x and y before quoting a final angle, and give theta in the range the question specifies (usually -pi < theta <= pi or 0 <= theta < 2pi), in radians.
Worked example (cartesian to polar). Express (-1, root3) in polar form. r = root(1 + 3) = 2. The point (-1, root3) is in the second quadrant, and tan theta = -root3 has reference angle pi/3, so theta = pi - pi/3 = 2pi/3. Hence (-1, root3) = (2, 2pi/3). Check: 2cos(2pi/3) = -1 and 2sin(2pi/3) = root3 ✓. Quoting theta = pi/3 (the raw arctan) loses the accuracy mark - this quadrant check is the single most examined detail in conversions.
Non-uniqueness and negative r. Polar coordinates are not unique: (2, 2pi/3) is the same point as (2, -4pi/3) and, using a negative radius, (-2, -pi/3). The convention for negative r is to plot the point on the opposite ray: (-r, theta) means (r, theta + pi). This matters when sketching, because limacons take negative r values over some theta-intervals and those points trace an inner loop.
Polar equations to cartesian. Multiply through by a suitable power of r so that r cos theta and r sin theta appear, then substitute. Worked example: r = 4cos theta becomes r2 = 4r cos theta, i.e. x2 + y2 = 4x, so (x - 2)2 + y2 = 4 - a circle of centre (2, 0) and radius 2. Two substitutions do all the work: r cos theta to x, r sin theta to y, r2 to x2 + y2. If an equation involves only r (for example r = 3), squaring gives x2 + y2 = 9 with no loss here, but in general squaring can introduce spurious points, so convert with the smallest multiplication that works.
Keep radians throughout and leave exact surds and pi in answers unless a decimal is demanded. Sketch the point roughly before working out theta: the sketch settles the quadrant instantly and protects you from the arctan trap. If a question asks for coordinates in the range 0 <= theta < 2pi, convert whatever you get into that range - marks are attached to the specified range, not to any valid description. Points on the vertical axis have theta = pi/2 or 3pi/2 directly, since tan theta = y/x cannot be formed when x = 0.
A reliable sketching routine. For any polar equation r = f(theta), build a table of r at standard angles (0, pi/6, pi/4, pi/3, pi/2, and onwards), then exploit symmetry: if only cos theta appears the curve is symmetric about the initial line (theta to -theta leaves r unchanged); if only sin theta appears it is symmetric about the line theta = pi/2. Mark three things on every sketch: the maximum value of r and where it occurs, every point where the curve passes through the pole (solve f(theta) = 0), and the tangent at the pole - when r = 0 at theta = theta0, the tangent there is the half-line theta = theta0.
The standard families. These shapes are examinable by name:
| Equation | Curve |
|---|---|
| r = 2a cos theta | Circle, centre (a, 0), radius a (r = 2a sin theta: centre (0, a)) |
| r = a(1 + cos theta) | Cardioid, cusp at the pole, max r = 2a on the initial line; -cos points left, +sin points up |
| r = a + b cos theta | Limacon; a >= b: no inner loop (dimple or oval); a < b: inner loop, pole where cos theta = -a/b |
| r = a cos n theta | Rose; n odd gives n petals, n even gives 2n petals; petal tips at distance a |
Worked example (rose). Sketch r = 2cos 3theta. The curve passes through the pole when 3theta = ±pi/2, i.e. theta = ±pi/6, and again every half-turn. One petal is traced for -pi/6 <= theta <= pi/6, centred on the initial line with tip at (2, 0). Since cos 3(theta + 2pi/3) = cos 3theta, the pattern repeats every 2pi/3, giving three petals with tips at (2, 0), (2, 2pi/3) and (2, 4pi/3). Odd n always yields n petals; r = 2cos 4theta would give eight.
Worked example (limacon). Sketch r = 1 + 2cos theta. Here a = 1 < b = 2, so there is an inner loop. The pole is reached when cos theta = -1/2, at theta = 2pi/3 and 4pi/3; between these angles r is negative, tracing the small inner loop on the left. Maximum r = 3 at theta = 0. This exact question type - identify the loop, find the pole angles, quote the maximum - appears regularly, and the pole angles also become the limits in loop-area questions (see Section 3).
Sketch marks are typically B1 for overall shape, B1 for symmetry, B1 for pole behaviour and B1 for labelled key points such as (2a, 0) on a cardioid. Always label the initial line and mark r values on the axes. A table of values with the r = 0 rows included is acceptable working and banks partial credit even if the final sketch wobbles.
The area formula. The area swept out by the radius vector from theta = alpha to theta = beta is A = (1/2) integral of r2 d theta - given in the Edexcel 9FM0 formula booklet, but you must choose the correct limits yourself. Limits come from the region in the question, the pole (r = 0), or intersections with another curve (equate the two r expressions). For curves symmetric about the initial line, compute half and double. For a region between two curves use A = (1/2) integral of (r_outer2 - r_inner2) d theta over the interval where the outer curve encloses the region.
Worked example (full cardioid). Find the area enclosed by r = a(1 + cos theta). By symmetry about the initial line: A = 2 × (1/2) integral from 0 to pi of a2(1 + cos theta)2 d theta = a2 integral of (1 + 2cos theta + cos2 theta) d theta. Use cos2 theta = (1 + cos 2theta)/2: the integral evaluates to a2[theta + 2sin theta + theta/2 + sin 2theta/4] from 0 to pi = a2(pi + pi/2) = 3pi a2/2. Exact answers in terms of pi are expected.
Worked example (between two curves). Region inside r = 2(1 + cos theta) but outside the circle r = 2: intersections where 2(1 + cos theta) = 2, i.e. cos theta = 0, so theta = ±pi/2. Then A = (1/2) integral from -pi/2 to pi/2 of [4(1 + cos theta)2 - 4] d theta = (1/2) integral of (8cos theta + 4cos2 theta) d theta = (1/2)[8sin theta + 2theta + sin 2theta] from -pi/2 to pi/2 = (1/2)(16 + 2pi) = 8 + pi. The cos theta and sin 2theta terms are antisymmetric and vanish across the symmetric interval - a built-in accuracy check.
Tangents to polar curves. The polar gradient formula is not printed in the booklet - derive it: x = r cos theta and y = r sin theta, so dy/dx = (dy/d theta)/(dx/d theta) = (r' sin theta + r cos theta)/(r' cos theta - r sin theta), where r' = dr/d theta. Worked example: for r = a(1 + cos theta) at theta = 0: r' = -a sin theta, so dx/d theta = -a sin theta cos theta - a(1 + cos theta) sin theta = 0, while dy/d theta = -a sin2 theta + a(1 + cos theta) cos theta = 2a at theta = 0. Zero denominator with non-zero numerator: the tangent is vertical, the line x = 2a, at the furthest point (2a, 0). Also remember the tangent at the pole: r = 0 at theta = theta0 gives the line theta = theta0.
State the limits and justify them (from r = 0 or from an intersection) - method marks hang on the limits, not just the integration. When subtracting r2 values, put the outer curve first, and never integrate r instead of r2. For single petals, find the two angles where r = 0 and integrate between them. If both dx/d theta and dy/d theta are zero at a point, that is the tangent-at-the-pole case - use the half-line rule.