CP 7.1 - Converting between polar and Cartesian

CP 7.1 - Converting between polar and Cartesian

Describe a point by its distance and direction from a fixed origin, then convert between that description and Cartesian coordinates. You will choose angles in the required range and translate simple curve equations without losing their domain restrictions.

Distance and direction locate a point

Can the angles −π3-\frac{\pi}{3} and 5π3\frac{5\pi}{3} describe the same point? Yes, provided the distance is the same: the two directions differ by one complete turn. Polar coordinates use distance and direction rather than horizontal and vertical displacement.

You already know

From A-level Mathematics, radians measure directed angles, and sine and cosine give the vertical and horizontal components of a length. Their signs depend on the quadrant. One full turn is 2π2\pi radians.

Fix an origin OO, called the pole, and a horizontal ray pointing to the right, called the initial line. A point PP is located by moving distance rr from the pole in direction θ\theta. Positive angles turn anticlockwise from the initial line.

Polar coordinates

The polar coordinates of PP are (r,θ)(r,\theta), where r=OP≥0r=OP\ge0 is its distance from the pole and θ\theta is a directed angle from the initial line to OPOP. For example, (3,π2)(3,\frac{\pi}{2}) is three units directly above the pole. The pair (π2,3)(\frac{\pi}{2},3) describes a different point: the order matters.

For a non-zero point, replacing θ\theta by θ+2kπ\theta+2k\pi, for integer kk, changes the description but not the position. At the pole, r=0r=0; any angle leads to the same point, and there is no distinguished direction. That is a striking difference from Cartesian coordinates: one point can have infinitely many coordinate pairs.

[DIAGRAM: asset_name: A point located by distance and direction; asset_slug: 9fm0-cp45-polar-triangle; caption: The radius and angle give the horizontal and vertical components; description: Equal-scale Cartesian axes show the point P at minus 3 comma 3 root 3. Its radius from the pole O is labelled r equals 6 and its anticlockwise angle from the initial line is two pi over 3. Dashed projections show x equals r cos theta equals minus 3 and y equals r sin theta equals 3 root 3. The initial line is the positive horizontal ray.]
The radius and angle give the horizontal and vertical components

The picture uses a second-quadrant point deliberately: rr remains positive while its horizontal component is negative. A signed component is different from a distance.

Key idea

The radius tells you how far; the angle tells you which direction. An angle range chooses one representative of a direction, while the pole has no unique angle.

Check the coordinate order and the direction convention together.

From polar to Cartesian coordinates

Resolve the radius horizontally and vertically. The same component formulae work in every quadrant because sine and cosine already carry the correct signs.

Cartesian components of polar coordinates

For r≥0r\ge0 and real θ\theta,

x=rcos⁡θy=rsin⁡θx=r\cos\theta \qquad y=r\sin\theta

Angles are in radians. These relations are not separately printed among the polar formulae.

Not in the formula booklet: learn it.

The first example lies in the first quadrant, where both components are positive.

Worked example 1: a first-quadrant point

Question. Convert (6,π3)(6,\frac{\pi}{3}) to Cartesian coordinates.

Plan. Substitute the radius and angle into the two component formulae.

x=6cos⁡π3=6×12=3horizontal componenty=6sin⁡π3=6×32=33vertical component\begin{aligned} x&=6\cos\frac{\pi}{3}=6\times\frac12=3 && \text{horizontal component} \\ y&=6\sin\frac{\pi}{3}=6\times\frac{\sqrt{3}}2=3\sqrt{3} && \text{vertical component} \end{aligned}

Why this step? Cosine goes with the horizontal coordinate, since the angle is measured from the horizontal initial line.

Answer. (x,y)=(3,33)(x,y)=(3,3\sqrt{3}).

Check. The distance is 32+(33)2=36=6\sqrt{3^2+(3\sqrt{3})^2}=\sqrt{36}=6, and both coordinates are positive.

Change only the angle from π3\frac{\pi}{3} to 2π3\frac{2\pi}{3}. The point is reflected across the vertical axis, so only its horizontal sign should change.

Worked example 2: a second-quadrant point

Question. Convert (6,2π3)(6,\frac{2\pi}{3}) to Cartesian coordinates.

Plan. Use the exact sine and cosine values in the second quadrant.

x=6cos⁡2π3=6(−12)=−3cosine is negativey=6sin⁡2π3=6×32=33sine is positive\begin{aligned} x&=6\cos\frac{2\pi}{3}=6\left(-\frac12\right)=-3 && \text{cosine is negative} \\ y&=6\sin\frac{2\pi}{3}=6\times\frac{\sqrt{3}}2=3\sqrt{3} && \text{sine is positive} \end{aligned}

Why this step? The radius is still 66; the quadrant changes the sign of the component, not the sign of the distance.

Answer. (x,y)=(−3,33)(x,y)=(-3,3\sqrt{3}), as in the diagram.

Check. Its distance is again 66, and a negative horizontal coordinate with positive vertical coordinate puts it in the required quadrant.

Sketching a direction before substituting takes a moment and catches many sign slips.

Watch out

Wrong: x=rsin⁡θx=r\sin\theta, y=rcos⁡θy=r\cos\theta. At θ=0\theta=0, this would place a point above the pole instead of on the initial line. Right: horizontal cosine, vertical sine.

Move the angle to the third quadrant while keeping the radius fixed.

From Cartesian to polar coordinates

Find the distance using Pythagoras, then choose the angle using both coordinates. The ratio y/xy/x alone cannot distinguish opposite quadrants.

Squaring and adding the component formulae gives x2+y2=r2(cos⁡2θ+sin⁡2θ)=r2x^2+y^2=r^2(\cos^2\theta+\sin^2\theta)=r^2. Dividing them gives the tangent ratio when x≠0x\ne0.

Recovering radius and direction

For a non-zero point (x,y)(x,y),

r=x2+y2tan⁡θ=yx(x≠0)r=\sqrt{x^2+y^2} \qquad \tan\theta=\frac{y}{x}\quad(x\ne0)

Choose θ\theta in the requested range and the correct quadrant. If x=0x=0, use the upward or downward axis direction directly.

Not in the formula booklet: learn it.

An inverse-tangent result is a starting angle, not automatically the final polar angle.

Worked example 3: choosing the second quadrant

Question. Find polar coordinates for (−2,23)(-2,2\sqrt{3}), with r≥0r\ge0 and 0≤θ<2π0\le\theta<2\pi.

Plan. Calculate the radius, find a reference angle and use the coordinate signs to select the quadrant.

r=(−2)2+(23)2=4Pythagorastan⁡θ=23−2=−3component ratioθ=π−π3=2π3second quadrant\begin{aligned} r&=\sqrt{(-2)^2+(2\sqrt{3})^2}=4 && \text{Pythagoras} \\ \tan\theta&=\frac{2\sqrt{3}}{-2}=-\sqrt{3} && \text{component ratio} \\ \theta&=\pi-\frac{\pi}{3}=\frac{2\pi}{3} && \text{second quadrant} \end{aligned}

Why this step? A calculator may return −π3-\frac{\pi}{3} for the inverse tangent, but that direction has positive xx. The given point has negative xx and positive yy.

Answer. (r,θ)=(4,2π3)(r,\theta)=(4,\frac{2\pi}{3}).

Check. 4cos⁡2π3=−24\cos\frac{2\pi}{3}=-2 and 4sin⁡2π3=234\sin\frac{2\pi}{3}=2\sqrt{3}, recovering both coordinates.

Change only the sign of the vertical coordinate. The point moves to the third quadrant, while its distance remains the same.

Worked example 4: choosing the third quadrant

Question. Find polar coordinates for (−2,−23)(-2,-2\sqrt{3}), with r≥0r\ge0 and 0≤θ<2π0\le\theta<2\pi.

Plan. Use the same distance and reference angle, then select the direction with both coordinates negative.

r=4+12=4unchanged distancetan⁡θ=−23−2=3positive tangent ratioθ=π+π3=4π3third quadrant\begin{aligned} r&=\sqrt{4+12}=4 && \text{unchanged distance} \\ \tan\theta&=\frac{-2\sqrt{3}}{-2}=\sqrt{3} && \text{positive tangent ratio} \\ \theta&=\pi+\frac{\pi}{3}=\frac{4\pi}{3} && \text{third quadrant} \end{aligned}

Why this step? A positive tangent occurs in both the first and third quadrants. The coordinate signs, rather than the tangent sign alone, decide which one is needed.

Answer. (r,θ)=(4,4π3)(r,\theta)=(4,\frac{4\pi}{3}).

Check. 4cos⁡4π3=−24\cos\frac{4\pi}{3}=-2, 4sin⁡4π3=−234\sin\frac{4\pi}{3}=-2\sqrt{3}.

If the question instead specified −π<θ≤π-\pi<\theta\le\pi, the second answer would use −2π3-\frac{2\pi}{3}. Polar coordinates do not impose one universal angle interval: follow the interval in the question.

Watch out

Wrong: θ=arctan⁡(y/x)\theta=\arctan(y/x) without checking the point. Right: find a reference angle and use the signs of both xx and yy, then enforce the requested range. At (0,0)(0,0), do not calculate 0/00/0: state r=0r=0 and no unique direction.

The opening directions now have a concrete position: at radius 44, both −π3-\frac{\pi}{3} and 5π3\frac{5\pi}{3} give (2,−23)(2,-2\sqrt{3}).

Converting curve equations

Use x=rcos⁡θx=r\cos\theta, y=rsin⁡θy=r\sin\theta and r2=x2+y2r^2=x^2+y^2 to translate a whole equation. Keep track of restrictions when multiplying, dividing or squaring.

For example, x2+y2=9x^2+y^2=9 becomes r2=9r^2=9, hence r=3r=3 because r≥0r\ge0. A circle centred at the pole has a constant radius: its equation becomes simpler in polar coordinates. The opposite direction can reveal a circle hidden inside a trigonometric polar equation.

Worked example 5: a circle touching the pole

Question. Convert r=6cos⁡θr=6\cos\theta to Cartesian form and identify the curve.

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

r2=6rcos⁡θmultiply by rx2+y2=6xCartesian substitutions(x−3)2+y2=9complete the square\begin{aligned} r^2&=6r\cos\theta && \text{multiply by }r \\ x^2+y^2&=6x && \text{Cartesian substitutions} \\ (x-3)^2+y^2&=9 && \text{complete the square} \end{aligned}

Why this step? Multiplication by rr creates exactly the combinations with Cartesian meanings. Dividing by rr instead would require a separate check at the pole.

Answer. The curve is a circle with centre (3,0)(3,0) and radius 33. Its polar points use angles with cos⁡θ≥0\cos\theta\ge0.

Check. The circle passes through (0,0)(0,0) and (6,0)(6,0). Both occur in the polar description: r=0r=0 at θ=π2\theta=\frac{\pi}{2}, and r=6r=6 at θ=0\theta=0.

Change only cosine to sine. This changes which Cartesian coordinate appears and turns the circle's position through a right angle.

Worked example 6: the circle moves above the pole

Question. Convert r=6sin⁡θr=6\sin\theta to Cartesian form and identify the curve.

Plan. Multiply by rr, replace rsin⁡θr\sin\theta by yy, then complete the square.

r2=6rsin⁡θmultiply by rx2+y2=6ysubstitutex2+(y−3)2=9complete the square\begin{aligned} r^2&=6r\sin\theta && \text{multiply by }r \\ x^2+y^2&=6y && \text{substitute} \\ x^2+(y-3)^2&=9 && \text{complete the square} \end{aligned}

Why this step? The angle still starts at the horizontal initial line, so the sine component is yy. It is the curve that changes, not the coordinate convention.

Answer. A circle with centre (0,3)(0,3) and radius 33, using polar angles with sin⁡θ≥0\sin\theta\ge0.

Check. At θ=π2\theta=\frac{\pi}{2}, r=6r=6, giving (0,6)(0,6); at θ=0\theta=0, the point is the pole.

The component relations are not printed in the formula booklet, and these particular circle conversions are not printed either. Reconstruct them rather than memorising each separate case.

Watch out

Wrong: cancel rr from r2=6rcos⁡θr^2=6r\cos\theta and forget the case r=0r=0. Right: discuss the pole whenever a division by rr is used. Also, a restricted angle interval may describe only an arc of the Cartesian circle.

The next circle has the same form with a different scale.

Quick Recap

  • Polar coordinates use distance r≥0r\ge0 and an angle from the initial line; the origin is the pole.
  • Convert to Cartesian form with x=rcos⁡θx=r\cos\theta, y=rsin⁡θy=r\sin\theta.
  • Recover r=x2+y2r=\sqrt{x^2+y^2}, then choose the angle using the quadrant and the required range.
  • For equations, use the component relations and r2=x2+y2r^2=x^2+y^2; preserve restrictions and check the pole before cancelling rr.

The opening angles differ by 2π2\pi, so they give the same direction and, at equal radius, the same point.