PP1, PP2 - Polar conversions and curve sketching

PP1, PP2 - Polar conversions and curve sketching

Locate points by a radial coordinate and an angle, then use those coordinates to describe curves. PP1 connects polar and Cartesian coordinates; PP2 uses symmetry, zeros and the sign of the radial coordinate to sketch circles, heart-shaped curves and petal-shaped curves accurately.

A radial coordinate and a direction

For r=2cos⁡2θr=2\cos2\theta, why does part of the curve lie below the horizontal axis while θ\theta runs through angles pointing upwards? The answer is a negative radial coordinate: it sends you backwards along the indicated direction.

You already know

Angles are measured in radians. A displacement of length dd at angle θ\theta has horizontal component dcos⁡θd\cos\theta and vertical component dsin⁡θd\sin\theta. Exact trigonometric values and quadrant signs determine both components.

Start with (r,θ)=(4,π/6)(r,\theta)=(4,\pi/6). Turn anticlockwise through π/6\pi/6, then travel 44 units along that ray. For (−4,π/6)(-4,\pi/6), make the same turn but travel 44 units in the opposite direction. These are different points.

Polar coordinates

The pole is the fixed origin OO. The initial line is the fixed ray from which angles are measured, normally the positive Cartesian xx-axis. In polar coordinates (r,θ)(r,\theta), θ\theta gives a direction and the signed radial coordinate rr gives a displacement along it: forwards if r>0r>0, backwards if r<0r<0. The distance from the pole is ∣r∣|r|. If r=0r=0, the point is the pole, whatever θ\theta is.

Thus rr is not the Cartesian xx-coordinate, and it is a distance only when non-negative. In particular, a negative rr is allowed; it does not mean a negative distance.

Polar to Cartesian coordinates

With the same origin and the positive xx-axis as initial line,

x=rcos⁡θ,y=rsin⁡θ,x2+y2=r2x=r\cos\theta,\qquad y=r\sin\theta,\qquad x^2+y^2=r^2

These relations hold for positive, zero and negative rr.

Not in the formula booklet: learn it.

Calculate both coordinates using the signed value of rr. Do not move the point a second time after the calculation: the minus sign has already done that.

Worked example 1: a positive radial coordinate

Question. Find the Cartesian coordinates of (4,π/6)(4,\pi/6).

Plan. Substitute r=4r=4 and the exact sine and cosine of π/6\pi/6.

x=4cos⁡π6=4×32=23horizontal componenty=4sin⁡π6=4×12=2vertical component\begin{aligned} x&=4\cos\frac{\pi}{6}=4\times\frac{\sqrt3}{2}=2\sqrt3 &&\text{horizontal component} \\ y&=4\sin\frac{\pi}{6}=4\times\frac12=2 &&\text{vertical component} \end{aligned}

Why this step? The first polar coordinate is a displacement along a ray, so it must be resolved into two Cartesian components.

Answer. (x,y)=(23,2)(x,y)=(2\sqrt3,2).

Check. x2+y2=12+4=16=r2x^2+y^2=12+4=16=r^2, and the point is in the first quadrant.

Change only rr from 44 to −4-4. Both Cartesian components change sign; the indicated angle stays π/6\pi/6.

Worked example 2: a negative radial coordinate

Question. Find the Cartesian coordinates of (−4,π/6)(-4,\pi/6).

Plan. Use the same trigonometric values and multiply both by the negative radial coordinate.

x=−4cos⁡π6=−23negative horizontal componenty=−4sin⁡π6=−2negative vertical component\begin{aligned} x&=-4\cos\frac{\pi}{6}=-2\sqrt3 &&\text{negative horizontal component} \\ y&=-4\sin\frac{\pi}{6}=-2 &&\text{negative vertical component} \end{aligned}

Why this step? Multiplying by a negative rr reverses the whole displacement, rather than just its horizontal component.

Answer. (x,y)=(−23,−2)(x,y)=(-2\sqrt3,-2). An alternative polar representation is (4,7π/6)(4,7\pi/6).

Check. The distance is still 44, and adding π\pi to the indicated direction takes you to the third quadrant.

The two points are opposite ends of a diameter through the pole.

[DIAGRAM: asset_name: A negative radial coordinate reverses the ray; asset_slug: b042_043_signed_radius; caption: The angle is the same, but changing the sign of r moves to the opposite side of the pole; description: Two equal-scale Cartesian panels use the initial line as the positive x-axis and mark a ray at pi over 6. The first shows r equals 4 at (2 square root 3,2). The second shows r equals minus 4 at (-2 square root 3,-2), opposite the indicated ray. Both show distance 4 from the pole.]
The angle is the same, but changing the sign of r moves to the opposite side of the pole

Watch out

Wrong: (−4,π/6)(-4,\pi/6) has distance −4-4 from the pole. Right: its distance is 44, and its position is opposite the direction π/6\pi/6. The signed value rr and the distance ∣r∣|r| are different quantities.

Keep the negative radial coordinate and change the reference angle for your turn.

Choose a radius and the correct quadrant

A point has many polar representations. You can add a full turn to its angle, or reverse the sign of rr and add a half-turn. Unless asked otherwise, choose r≥0r\ge0 when converting from Cartesian coordinates, then give θ\theta in the interval requested.

Cartesian to polar coordinates

For a point (x,y)≠(0,0)(x,y)\ne(0,0), choosing r>0r>0 gives

r=x2+y2,cos⁡θ=xr,sin⁡θ=yrr=\sqrt{x^2+y^2},\qquad \cos\theta=\frac{x}{r},\qquad \sin\theta=\frac{y}{r}

Choose the quadrant from both signs. When x≠0x\ne0, tan⁡θ=y/x\tan\theta=y/x alone does not identify the quadrant. For any integer kk,

(r,θ),(r,θ+2kπ),(−r,θ+(2k+1)π)(r,\theta),\quad (r,\theta+2k\pi),\quad (-r,\theta+(2k+1)\pi)

represent the same point. At the pole, r=0r=0 and the angle is arbitrary.

Not in the formula booklet: learn it.

Polar coordinates do not automatically use the principal-argument interval from complex numbers. Here the questions specify 0≤θ<2π0\le\theta<2\pi; use a different interval if it is requested.

Worked example 3: a first-quadrant point

Question. Find polar coordinates of (23,2)(2\sqrt3,2), with r>0r>0 and 0≤θ<2π0\le\theta<2\pi.

Plan. Find the distance, then identify the angle from its cosine and sine.

r=(23)2+22=4positive distancecos⁡θ=32,sin⁡θ=12divide coordinates by 4θ=π6first quadrant\begin{aligned} r&=\sqrt{(2\sqrt3)^2+2^2}=4 &&\text{positive distance} \\ \cos\theta&=\frac{\sqrt3}{2},\qquad \sin\theta=\frac12 &&\text{divide coordinates by }4 \\ \theta&=\frac{\pi}{6} &&\text{first quadrant} \end{aligned}

Why this step? Choosing r>0r>0 means the angle points towards the actual point, rather than towards its opposite.

Answer. (r,θ)=(4,π/6)(r,\theta)=(4,\pi/6).

Check. Substitution into (rcos⁡θ,rsin⁡θ)(r\cos\theta,r\sin\theta) returns (23,2)(2\sqrt3,2).

Change only the sign of xx. The distance is unchanged, but the point moves into the second quadrant.

Worked example 4: a second-quadrant point

Question. Find polar coordinates of (−23,2)(-2\sqrt3,2), with r>0r>0 and 0≤θ<2π0\le\theta<2\pi.

Plan. Keep the positive distance and choose the second-quadrant angle with reference angle π/6\pi/6.

r=(−23)2+22=4same squared coordinatescos⁡θ=−32,sin⁡θ=12second quadrantθ=π−π6=5π6correct direction\begin{aligned} r&=\sqrt{(-2\sqrt3)^2+2^2}=4 &&\text{same squared coordinates} \\ \cos\theta&=-\frac{\sqrt3}{2},\qquad \sin\theta=\frac12 &&\text{second quadrant} \\ \theta&=\pi-\frac{\pi}{6}=\frac{5\pi}{6} &&\text{correct direction} \end{aligned}

Why this step? arctan⁡(−1/3)=−π/6\arctan(-1/\sqrt3)=-\pi/6 would put the point in the fourth quadrant, with both signs wrong.

Answer. (r,θ)=(4,5π/6)(r,\theta)=(4,5\pi/6).

Check. 4cos⁡(5π/6)=−234\cos(5\pi/6)=-2\sqrt3 and 4sin⁡(5π/6)=24\sin(5\pi/6)=2.

The interval also matters below the horizontal axis, where a negative angle may need a full turn added.

Watch out

Wrong: report θ=−π/6\theta=-\pi/6 when the question requires 0≤θ<2π0\le\theta<2\pi. Right: use the equivalent direction 11π/611\pi/6. Equivalent points can still fail a stated coordinate convention.

Change the imaginary-direction sign from the last example so that the point moves below the horizontal axis.

Convert the equation without losing the pole

The same coordinate relations convert equations. Substitute for every xx and yy, simplify using trigonometric identities, and check the pole separately whenever you divide by rr.

For example, the Cartesian equation x2+y2=4xx^2+y^2=4x becomes r2=4rcos⁡θr^2=4r\cos\theta. If r≠0r\ne0, division gives r=4cos⁡θr=4\cos\theta. The original curve also contains the pole, so keep it: the polar equation reaches it when cos⁡θ=0\cos\theta=0.

Worked example 5: a circle through the pole

Question. Convert r=4cos⁡θr=4\cos\theta, for −π/2≤θ≤π/2-\pi/2\le\theta\le\pi/2, into a Cartesian equation and identify the curve.

Plan. Multiply by rr to create r2r^2 and rcos⁡θr\cos\theta, then complete the square.

r2=4rcos⁡θmultiply by rx2+y2=4xuse the coordinate relations(x−2)2+y2=4complete the square\begin{aligned} r^2&=4r\cos\theta &&\text{multiply by }r \\ x^2+y^2&=4x &&\text{use the coordinate relations} \\ (x-2)^2+y^2&=4 &&\text{complete the square} \end{aligned}

Why this step? cos⁡θ\cos\theta alone is not xx; the product rcos⁡θr\cos\theta is. Multiplying first produces the correct Cartesian expression.

Answer. The circle with centre (2,0)(2,0) and radius 22, traced once over the given interval.

Check. At θ=0\theta=0, r=4r=4 gives (4,0)(4,0). At both endpoints, r=0r=0 gives the pole, also on the circle.

Change only cosine to sine, rotating the circle to lie above the initial line. The interval changes to trace that circle once with r≥0r\ge0.

Worked example 6: the same circle turned upwards

Question. Convert r=4sin⁡θr=4\sin\theta, for 0≤θ≤π0\le\theta\le\pi, into a Cartesian equation and identify the curve.

Plan. Multiply by rr, substitute y=rsin⁡θy=r\sin\theta, and complete the square in yy.

r2=4rsin⁡θmultiply by rx2+y2=4yconvert both sidesx2+(y−2)2=4complete the square\begin{aligned} r^2&=4r\sin\theta &&\text{multiply by }r \\ x^2+y^2&=4y &&\text{convert both sides} \\ x^2+(y-2)^2&=4 &&\text{complete the square} \end{aligned}

Why this step? The sine term gives the vertical coordinate, so the circle's centre moves onto the yy-axis.

Answer. The circle with centre (0,2)(0,2) and radius 22, traced once over the stated interval.

Check. θ=π/2\theta=\pi/2 gives r=4r=4, the point (0,4)(0,4), while θ=0\theta=0 and π\pi give the pole.

A circle through the pole needs only a single trigonometric term in polar form, even though its centre is away from the pole. On the first curve, (r,θ)=(4,0)(r,\theta)=(4,0) is 44 units from the pole but only 22 units from the circle's centre. Those two lengths measure different things.

[DIAGRAM: asset_name: Polar equations of circles through the pole; asset_slug: b042_043_polar_circles; caption: r equals 4 cosine theta gives a circle centred on the initial line; replacing cosine by sine turns it upwards; description: Two equal-scale panels show the circles centred at (2,0) and (0,2), each with radius 2 and passing through the pole. The points furthest from the pole are (4,0) and (0,4). The first panel gives the interval minus pi over 2 to pi over 2 and the second zero to pi, each tracing once with non-negative r.]
r equals 4 cosine theta gives a circle centred on the initial line

Watch out

Wrong: cancel r2r^2 from an equation and conclude that the pole is absent. Right: cancellation assumes r≠0r\ne0, so check (x,y)=(0,0)(x,y)=(0,0) in the original equation separately. Also avoid introducing an extra branch by squaring without checking the original equation and its angle range.

The next conversion uses a double-angle identity. Keep an explicit note about the point where cancellation is unavailable.

Continue this lesson.

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