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01 Polar Coordinates and Conic Sections

Polar Coordinates

In polar coordinate system, we locate a point with reference to:

  1. pole a fixed point (usually fixed at the origin)
  2. initial ray a fixed line, passing through the pole (usually \(+x\) axis)

Let

  • \(r\) - directed distance of the point from pole
    • \(r > 0\) forward
    • \(r < 0\) backward
  • \(\theta\) - directed angle of radius vector from the initial ray
    • \(\theta < 0\) anti-clockwise
    • \(\theta > 0\) clockwise
  • \(P(r, \theta)\) - corresponding point

Polar

Circle Through Pole

\[ r = \pm a, \quad 0 \le \theta \le 2 \pi \]

represents a circle with center @pole and radius \(a\). Sign can be either, because it is the same circle traversed in the opposite direction

Straight line through pole

\[ \theta = \theta_0, \quad - \infty < r < \infty \]

IDK

\(r\) \(\theta\) Diagram
const const point
const inequality arc
inequality const straight line segment
inequality inequality region

Cartesian \(\iff\) Polar

Consider the point \(P(x, y) \iff P(r, \theta)\)

\[ \begin{aligned} x &= r \cos\theta \\ y &= r \sin\theta \\ r^2 &= x^2 + y^2 \\ \theta &= \tan^{-1} \left( \frac y x \right) \end{aligned} \]

Symmetry

Let \(r = f(\theta)\) be a polar curve

X-axis

\(P(r, \theta)\) and \(P'(r, - \theta)\) lie on same graph

Symmetry about Vary theta \(P(r, \theta)\) lies on the same graph as or \(P(r, \theta)\) lies on the same graph as
X-axis \(0 \le \theta \le \pi\) \(P'(r, -\theta)\) \(P'(-r, \pi -\theta)\)
Y-axis \(\frac{-\pi} 2 \le \theta \le \frac \pi 2\) \(P'(-r, -\theta)\) \(P'(r, \pi -\theta)\)
Origin \(0 \le \theta \le \frac \pi 2\) \(P'(-r, \theta)\) \(P'(r, \pi + \theta)\)

Shapes

Limacon

\[ r = a \pm b \cos\theta \\ \text{ or } \\ r = a \pm b \sin\theta \]
\(\frac a b\) Type
\(<1\) inner loop
\(=1\) cardioid
\(>1\) outer loop

Roses

\[ \begin{aligned} r &= a \cos(n\theta) \\ &\text{ or } \\ r &= a \sin(n\theta) \\ \text{No of petals } N &= \begin{cases} n, & n = \text{odd} \\ 2n, & n = \text{even} \end{cases} \\ \text{Axis of first petal } \theta &= \begin{cases} 0 & r = a \textcolor{orange}{\cos}(n \theta) \\ \dfrac \pi {2n} & r = a \textcolor{orange}{\sin} (n \theta) \end{cases} \\ \text{Length of petals} &= a \\ \text{Angular Gap between axes of petals} &= \frac{2 \pi}{N} \end{aligned} \]

Lemmiscates

\[ r^2 = a \cos\theta \\ \text{ or } \\ r^2 = a \sin\theta \\ \]

Straight Line

\[ r \cos(\theta-\theta_0) = r_0 \]
  • \(P(r, \theta)\) is any point on given line
  • \(P_0(r_0, \theta_0)\) is foot of \(\perp\)r from the pole

Circle

\[ r^2 + {r_0}^2 - 2 r r_0 \cos(\theta - \theta_0) = a^2 \\ \]
  • \(P(r, \theta)\) is any point on circle
  • \(P_0(r_0, \theta_0)\) is center of circle
  • \(a\) is radius

Radius passing through pole

\[ r_0 = a\\ r = 2a cos(\theta - \theta_0) \]

Center lies on axis

Center at \(r\)
\((a,0)\) \(2a \cos \theta\)
\((-a,0)\) \(-2a \cos \theta\)
\((a, \frac \pi 2)\) \(2a \sin \theta\)
\((a, -\frac \pi 2)\) \(-2a \sin \theta\)

Area under curve

For a polar curve \(r = f(\theta), \alpha \le \theta \le \beta\)

\[ A = \frac12 \int\limits_{\theta = \alpha}^\beta r^2 \cdot d\theta \]

For area bounded by the curves \(r_1 = f_1(\theta), r_2 = f_2(\theta), \alpha \le \theta \le \beta\) such that \(r_1 < r_2\)

\[ A = \frac12 \int\limits_{\theta = \alpha}^\beta {r_2}^2 - {r_1}^2 \cdot d\theta \]

Length of curve

For a curve \(r = f(\theta), \alpha \le \theta \le \beta\) traversed exactly once from \(\theta = \alpha \to \beta\)

\[ L = \int\limits_{\theta = \alpha}^\beta \sqrt{ r^2 + (r')^2 } \cdot d\theta \qquad \left[ r' = \frac{dr}{d \theta} \right] \]

Conic Sections

Let

  • \(P(r, \theta)\) be any point on the conic section with focus at origin
  • \(e = \dfrac{ \text{Distance bw focii} }{ \text{Distance bw vertices} }\)
Directrix \(r\)
\(x = a\) \(\frac{ke}{1 + e \cos\theta}\)
\(x = -a\) \(\frac{ke}{1 - e \cos\theta}\)
\(y = a\) \(\frac{ke}{1 + e \sin\theta}\)
\(y = -a\) \(\frac{ke}{1 - e \sin\theta}\)

Shapes

\(e\) Shape
\(0 < e < 1\) Ellipse
\(e = 1\) Parabola
\(e > 1\) Hyperbola

For ellipse,

\[ k = a \left[ \frac 1 e - e \right] \]
Last Updated: 2023-01-25 ; Contributors: AhmedThahir

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