How to Compare Distances Using Polar Coordinates

Comparing distances using polar coordinates can be straightforward once you understand the basics of the polar coordinate system. Let’s break it down step-by-step.

Understanding Polar Coordinates

In the polar coordinate system, a point in the plane is represented by two values: the radius (r) and the angle (θ). The radius is the distance from the origin to the point, and the angle is measured from the positive x-axis.

For example, a point with polar coordinates (r, θ) is located r units away from the origin at an angle θ from the positive x-axis.

Converting Polar Coordinates to Cartesian Coordinates

Sometimes, it is easier to compare distances by converting polar coordinates to Cartesian coordinates (x, y). The conversion formulas are:
$x = r times text{cos}(theta)$
$y = r times text{sin}(theta)$

Distance Formula in Cartesian Coordinates

Once converted to Cartesian coordinates, you can use the Euclidean distance formula to find the distance between two points (x1, y1) and (x2, y2):
$d = sqrt{(x2 – x1)^2 + (y2 – y1)^2}$

Direct Comparison in Polar Coordinates

However, you can also compare distances directly in polar coordinates. The distance between two points (r1, θ1) and (r2, θ2) is given by:
$d = sqrt{r1^2 + r2^2 – 2r1r2 cos(θ2 – θ1)}$

Example

Let’s compare the distances between two points in polar coordinates: (3, 45°) and (4, 60°).

  1. Convert to Cartesian Coordinates:

    • For (3, 45°):
      $x1 = 3 cos(45°) = 3 times frac{sqrt{2}}{2} = frac{3sqrt{2}}{2}$
      $y1 = 3 sin(45°) = 3 times frac{sqrt{2}}{2} = frac{3sqrt{2}}{2}$
    • For (4, 60°):
      $x2 = 4 cos(60°) = 4 times frac{1}{2} = 2$
      $y2 = 4 sin(60°) = 4 times frac{sqrt{3}}{2} = 2sqrt{3}$
  2. Calculate the Distance:
    $d = sqrt{(2 – frac{3sqrt{2}}{2})^2 + (2sqrt{3} – frac{3sqrt{2}}{2})^2}$

  3. Simplify:
    After simplifying, you would get the distance between the two points.

Alternatively, using the direct polar distance formula:
$d = sqrt{3^2 + 4^2 – 2 times 3 times 4 cos(60° – 45°)}$
$d = sqrt{9 + 16 – 24 cos(15°)}$
$d = sqrt{25 – 24 times cos(15°)}$
$d = sqrt{25 – 24 times 0.9659}$
$d = sqrt{25 – 23.1816}$
$d = sqrt{1.8184}$
$d ≈ 1.35$

Conclusion

Comparing distances using polar coordinates can be done either by converting to Cartesian coordinates or directly using the polar distance formula. Both methods provide accurate results, but using polar coordinates directly can often be more efficient.

Citations

  1. 1. Khan Academy – Polar Coordinates
  2. 2. Math is Fun – Polar Coordinates
  3. 3. Wolfram MathWorld – Polar Coordinates

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ