1.5 Polar coordinates

Syllabus
9231–2028–2029
Topic
1.5
Level
AS

Learning objectives

Convert the locus, not just its symbols

With the syllabus convention $r\ge0$,x=r\cos\theta,\qquad y=r\sin\theta,\qquad r^2=x^2+y^2,\qquad \tan\theta=\frac yx\ (x\ne0).

Cartesian to polar: replace x and y by rcosθr\cos\theta and rsinθr\sin\theta, use x2+y2=r2x^2+y^2=r^2, then simplify while retaining the stated angle and r0r\ge0 conditions. For x2+y2=4xx^2+y^2=4x, r2=4rcosθr^2=4r\cos\theta, giving r=4cosθr=4\cos\theta together with the pole r=0r=0; dividing by r without checking would hide that point.

Polar to Cartesian: replace $r\cos\theta$ by x, $r\sin\theta$ by y and $r^2$ by $x^2+y^2$. Forr=2(\cos\theta+\sin\theta),multiplybyrtoobtainmultiply by r to obtainx^2+y^2=2x+2y,oror(x-1)^2+(y-1)^2=2.

Do not introduce negative-r alternatives: this course uses r0r\ge0. Algebraic conversion can add or lose points when multiplying, squaring or dividing, so check the pole, domain and original equation after simplifying.

Build a polar sketch from angular events and radial extrema

Use the stated interval (normally 0θ<2π0\le\theta<2\pi or π<θπ-\pi<\theta\le\pi): test symmetry by replacing theta with θ-\theta, πθ\pi-\theta or θ+π\theta+\pi; solve r=0r=0 for pole visits; find intersections with the initial line at allowed angles representing that ray; solve dr/dθ=0dr/d\theta=0 and check endpoints for least/greatest r; then join only the admissible r0r\ge0 branches in increasing theta order.

At a pole value θ0\theta_0 where r0r\to0, the limiting angle gives the approach direction. Check whether the interval traces into and out of the pole, ends there, or meets it more than once; this distinguishes a crossing, cusp/loop contact or endpoint. Do not plot a formula-produced negative r on the opposite ray under this syllabus convention—restrict to where r is non-negative.

Feature for r=a(1+cosθ)r=a(1+\cos\theta), a>0a>0 Deduction
Symmetry r(θ)=r(θ)r(-\theta)=r(\theta), so symmetry about the initial line
Initial line θ=0\theta=0 gives (2a,0)(2a,0); the pole is also on every radial line
Pole r=0r=0 at θ=π\theta=\pi; both sides approach along the same line, forming the cardioid cusp
Radial extrema 0r2a0\le r\le2a; maximum 2a2a at θ=0\theta=0, minimum 0 at θ=π\theta=\pi

A polar sketch is not the Cartesian graph of r against theta. The final plane curve must show symmetry, labelled initial-line intersections, correct pole behaviour and least/greatest radial distances; dense plotting is not required.

Choose one traversal before integrating squared radius

A thin sector of angle $d\theta$ has area approximately $\tfrac12r^2d\theta$. Therefore, when $r=f(\theta)$ traces the intended boundary once from $\alpha$ to $\beta$,A=\frac12\int_{\alpha}^{\beta}r^2,d\theta.

Find the boundary angles from intersections, the initial line or r=0r=0. Check the sketch to ensure the interval covers the desired region once. Split at a pole or branch change when necessary. Between two curves on the same rays, use 12(router2rinner2)dθ\tfrac12\int(r_{outer}^2-r_{inner}^2)d\theta.

The upper half of $r=a(1+\cos\theta)$ is traced once for $0\le\theta\le\pi$. ThusA=\frac{a^2}{2}\int_0^\pi(1+\cos\theta)^2d\theta=\frac{a^2}{2}\int_0^\pi(1+2\cos\theta+\cos^2\theta)d\theta=\frac{3\pi a^2}{4}.

The integrand is r2r^2, not r, and the factor one-half is essential. Squaring makes area non-negative, but it does not prevent double counting: only the sketch and angular traversal establish correct limits.