FP2.7 - Polar coordinates

Syllabus
2019
Topic
Level
A2

Interpret and sketch polar curves

A polar point (r,θ)(r,\theta) lies a distance r0r\ge0 from the pole in the direction measured by θ\theta from the initial line. Cartesian and polar descriptions are connected byx=rcosθ,y=rsinθ,r2=x2+y2.x=r\cos\theta,\qquad y=r\sin\theta,\qquad r^2=x^2+y^2.At the pole r=0r=0, the angle is not unique.

Polar equation Structure to recognise
θ=α\theta=\alpha Ray from the pole at angle α\alpha when r0r\ge0.
r=psec(αθ)r=p\sec(\alpha-\theta) Straight line xcosα+ysinα=px\cos\alpha+y\sin\alpha=p.
r=ar=a Circle centred at the pole, radius aa.
r=2acosθr=2a\cos\theta Circle centre (a,0)(a,0), radius aa.
r=kθr=k\theta Spiral whose radius changes linearly with angle.
r=a(1±cosθ)r=a(1\pm\cos\theta) Cardioid; the sign determines its orientation.
r=a(3+2cosθ)r=a(3+2\cos\theta) Limacon with ar5aa\le r\le5a when a>0a>0.
r=acos2θr=a\cos2\theta Rose-type curve; locate permitted intervals, zeros and extrema.
r2=a2cos2θr^2=a^2\cos2\theta Lemniscate; real points require cos2θ0\cos2\theta\ge0.

To sketch reliably, first state the permitted θ\theta interval. Test symmetry by replacing θ\theta with θ-\theta or πθ\pi-\theta, solve r=0r=0 for passages through the pole, find maximum and minimum rr, and calculate a small table at exact angles. Plot in increasing θ\theta so the order of loops or arcs is clear.

Cartesian conversion can confirm the shape. For example,r=2acosθr2=2arcosθ,r=2a\cos\theta\quad\Longrightarrow\quad r^2=2ar\cos\theta,so x2+y2=2ax(xa)2+y2=a2.x^2+y^2=2ax\quad\Longrightarrow\quad (x-a)^2+y^2=a^2.This proves that the polar curve is the circle described in the table.

Curves drawn over the same ray intersect where their admissible radii are equal. Solve r1(θ)=r2(θ)r_1(\theta)=r_2(\theta), retain angles in the stated range, and substitute back for the exact radius. The pole may be a shared point even when it arises from different angle values.

Under the stated convention r0r\ge0, discard angle intervals that make an unsquared radius negative unless the question explicitly defines a signed-radius plotting convention. Do not treat tanθ=y/x\tan\theta=y/x alone as a quadrant check, and do not assume one period traces every curve exactly once; use symmetry, zeros and the supplied interval.

Find polar areas and axis-parallel tangents

A narrow polar sector of angle dθd\theta has area approximately 12r2dθ\tfrac12r^2d\theta. Therefore the area swept once as θ\theta runs from α\alpha to β\beta isA=12αβr2dθ.A=\frac12\int_{\alpha}^{\beta}r^2\,d\theta.The limits must describe the actual boundary rays or curve intersections of the required region.

Step Check
Find boundary angles Solve intersections or use the stated rays; keep only admissible values.
Identify the swept region Check whether one integral gives it directly or whether a second polar area or a triangle must be subtracted.
Square before integrating Expand r2r^2 and use exact trigonometric identities.
Evaluate in radians Retain exact values of π\pi and radicals until the end.

For the upper half of the cardioid r=a(1+cosθ)r=a(1+\cos\theta), 0θπ0\le\theta\le\pi,A=a220π(1+2cosθ+cos2θ)dθ.A=\frac{a^2}{2}\int_0^\pi(1+2\cos\theta+\cos^2\theta)\,d\theta.Using cos2θ=12(1+cos2θ)\cos^2\theta=\tfrac12(1+\cos2\theta) givesA=a22(π+π2)=3πa24.A=\frac{a^2}{2}\left(\pi+\frac\pi2\right)=\frac{3\pi a^2}{4}.

dxdθ=rcosθrsinθ,dydθ=rsinθ+rcosθ,dydx=dy/dθdx/dθ\frac{dx}{d\theta}=r'\cos\theta-r\sin\theta,\qquad \frac{dy}{d\theta}=r'\sin\theta+r\cos\theta,\qquad \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}

Required tangent Condition
Parallel to the initial line dy/dθ=0dy/d\theta=0 and dx/dθ0dx/d\theta\ne0.
At right angles to the initial line dx/dθ=0dx/d\theta=0 and dy/dθ0dy/d\theta\ne0.

For r=2acosθr=2a\cos\theta, at θ=π/4\theta=\pi/4 we have r=2ar=\sqrt2a. Alsodydθ=2acos2θ=0,dxdθ=2asin2θ=2a0.\frac{dy}{d\theta}=2a\cos2\theta=0,\qquad \frac{dx}{d\theta}=-2a\sin2\theta=-2a\ne0.Thus the tangent is parallel to the initial line at (r,θ)=(2a,π/4)(r,\theta)=(\sqrt2a,\pi/4), the top point (a,a)(a,a) of the circle.

The area formula counts a region once only when the chosen interval traces its boundary once. If dx/dθdx/d\theta and dy/dθdy/d\theta are both zero, the simple horizontal/vertical test is inconclusive and the local curve needs separate analysis. Never divide by dx/dθdx/d\theta before checking whether it is zero, and use radians in integration.