Polar Coordinates - Comprehensive Notes

Polar Coordinates: Basics

• Key idea: In polar coordinates, a point is given by (r, θ), where r is the distance from the origin (the pole) and θ is the angle from the polar axis (positive x-axis).
• Core relationships to Cartesian:
  • x = r \cos \theta
  • y = r \sin \theta
  • r^2 = x^2 + y^2
  • \tan \theta = \frac{y}{x}
• Important nuance: r can be negative. When r < 0, the point lies in the direction opposite to θ, i.e., the angle effectively shifts by π.
• Practical note: Sketching polar graphs often involves recognizing circles from r = a cos θ or r = a sin θ, and lines from θ = constant.

Converting Between Polar and Cartesian

• From polar to Cartesian (given r, θ):
  • x = r \cos \theta, \quad y = r \sin \theta
  • Example: The polar point $(-2, \frac{5\pi}{6})$ →
  • x = -2 \cos \frac{5\pi}{6} = -2\left(-\frac{\sqrt{3}}{2}\right) = \sqrt{3}
  • y = -2 \sin \frac{5\pi}{6} = -2\left(\frac{1}{2}\right) = -1
  • Cartesian: $(\sqrt{3}, -1)$.
• From Cartesian to polar (given x, y):
  • r = \sqrt{x^2 + y^2} (with r ≥ 0 by convention)
  • \theta = \operatorname{atan2}(y, x) which places θ in $[0, 2\pi)$ (adjusting for the correct quadrant)
  • Example: Cartesian $(-3, -3)$ →
  • r = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}
  • \theta = \arctan(y/x) = \arctan(1) = \tfrac{\pi}{4} but in QIII, so \theta = \tfrac{5\pi}{4}
• Negative radius interpretation:
  • A point with $(r, θ)$ is equivalent to $(|r|, θ)$ if you add π to θ when r < 0, or keep r negative with a corresponding angle. Both representations refer to the same Cartesian point.
• Quick check example (from above): polar $(-2, \tfrac{5\pi}{6})$ equals Cartesian $(\sqrt{3}, -1)$.
• Practice: Given a Cartesian point $(-3, -3)$, the polar form with $r>0$ is $(3\sqrt{2}, \tfrac{5\pi}{4})$; a representation with $r<0$ could be $(\, -3\sqrt{2}, \tfrac{5\pi}{4} - \pi)$. The key is to ensure θ lies in $[0, 2\pi)$ when required.

Sketching Polar Graphs: Key Shapes

• Circle from $r = a \cos\theta$:
  • Circle of diameter $|a|$; center at $(a/2, 0)$, on the right if $a>0$ and on the left if $a<0$.
• Circle from $r = a \sin\theta$:
  • Circle of diameter $|a|$; center at $(0, a/2)$.
• Lines from $\theta = \text{constant}$:
  • A straight line through the origin at angle θ0; slope is $\tan(\theta_0)$.
• Practical visualization:
  • If you plug in a few θ-values for $r = a \cos\theta$ or $r = a \sin\theta$, you recover the circle through polar coordinates with the claimed center and radius.
• Desmos/graphing tip:
  • To graph a polar point in Desmos, input as Cartesian coordinates: $(x, y) = (r\cos\theta, r\sin\theta)$. If plotting with a function $r = f(\theta)$, use the parametric form $x(\theta) = f(\theta)\cos\theta$, $y(\theta) = f(\theta)\sin\theta$.

Polar–Cartesian Inverse Relations (and Practice)

• Important identities to memorize:

  • x = r \cos \theta, \quad y = r \sin \theta,

  r^2 = x^2 + y^2,

  \tan \theta = \frac{y}{x}

• Converting a Cartesian point to polar and choosing a representative with $0 \le \theta < 2\pi$:

  • Example: Cartesian $(-3, -3)$ gives $r = 3\sqrt{2}$ and $\theta = \tfrac{5\pi}{4}$ (in $[0, 2\pi)$).

• Quick check of consistency:

  • The polar point $(r, \theta) = (\sqrt{18}, \tfrac{5\pi}{4})$ produces Cartesian $(x, y) = (r\cos\theta, r\sin\theta) = ( -3, -3 )$ after substitution.

• When converting from Cartesian to polar, be mindful of quadrant; you may also represent the same point with a negative $r$ and a different angle.

Slope of a Polar Curve (dy/dx)

• If a polar curve is given by $r = r(\theta)$, then with
  • x = r(\theta) \cos \theta, \quad y = r(\theta) \sin \theta
  • and with $r' = \dfrac{dr}{d\theta}$,
• Then the derivatives with respect to $\theta$ are:
  • \frac{dy}{d\theta} = r'(\theta) \sin\theta + r(\theta) \cos\theta
  • \frac{dx}{d\theta} = r'(\theta) \cos\theta - r(\theta) \sin\theta
• Therefore the slope of the tangent is:
  • \frac{dy}{dx} = \frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}} = \frac{r'(\theta) \sin\theta + r(\theta) \cos\theta}{r'(\theta) \cos\theta - r(\theta) \sin\theta}
• Example (conceptual, not fully simplified here): find the slope of the curve $r(\theta) = 1 + \sin\theta$ at $\theta = \tfrac{\pi}{4}$.
  • $r = 1 + \sin(\tfrac{\pi}{4}) = 1 + \tfrac{\sqrt{2}}{2}$
  • $r' = \cos(\tfrac{\pi}{4}) = \tfrac{\sqrt{2}}{2}$
  • $\sin\theta = \cos\theta = \tfrac{\sqrt{2}}{2}$
  • Compute $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta}$ using the formulas above, then take their ratio.
  • Result (numerical): approximately $-2.414$ (negative slope), illustrating the need to evaluate with both $r$ and $r'$.
• Takeaways:
  • The slope is well-defined except where $\dfrac{dx}{d\theta} = 0$ (vertical tangent) or $\dfrac{dy}{d\theta} = 0$ (horizontal tangent).

Area Inside a Polar Curve

• General formula (polar area):
  • A = \frac{1}{2} \int{\theta1}^{\theta_2} [r(\theta)]^2 \; d\theta
• Rationale: A small sector with radius $r$ and angle $d\theta$ has area approximately (\tfrac{1}{2} r^2 d\theta).
• Key steps:
  • Determine the angular interval that sweeps the region.
  • Integrate $[r(\theta)]^2$ over that interval and multiply by (1/2).
• Example 1: Area in the first and fourth quadrants bounded by $r = 3$ (a circle of radius 3)
  • angular span from $-\tfrac{\pi}{4}$ to $\tfrac{\pi}{4}$
  • A = \tfrac{1}{2} \int{-\pi/4}^{\pi/4} 3^2 \; d\theta = \tfrac{9}{2} \left[ \theta \right]{-\pi/4}^{\pi/4} = \tfrac{9}{2} \cdot \tfrac{\pi}{2} = \frac{9\pi}{4}
• Example 2: One petal of $r = \sin(2\theta)$
  • A rose with 4 petals; one petal occurs for $\theta$ from $0$ to $\tfrac{\pi}{2}$ (where $r \ge 0$)
  • A{\text{petal}} = \tfrac{1}{2} \int{0}^{\pi/2} \sin^2(2\theta) \; d\theta
  • Using $\sin^2 u = \frac{1 - \cos 2u}{2}$ with $u = 2\theta$ gives
  • A{\text{petal}} = \frac{1}{2} \cdot \frac{1}{2} \int{0}^{\pi/2} (1 - \cos 4\theta) \, d\theta = \frac{1}{4} \Big[ \theta - \frac{\sin 4\theta}{4} \Big]_{0}^{\pi/2} = \frac{\pi}{8} \approx 0.3927
• Example 3: Area between curves $r = 2\cos\theta$ and $r = \sqrt{3}$
  • Intersections satisfy $2\cos\theta = \sqrt{3}$ => $\cos\theta = \sqrt{3}/2$ => $\theta = \pm \pi/6$ (within $[-\pi/2, \pi/2]$ where both are nonnegative)
  • For $\theta \in [-\pi/6, \pi/6]$, the outer radius is $R(\theta) = 2\cos\theta$ and the inner radius is $r(\theta) = \sqrt{3}$
  • Area
  • A = \frac{1}{2} \int_{-\pi/6}^{\pi/6} \left[(2\cos\theta)^2 - (\sqrt{3})^2\right] d\theta
  • Simplify: $(2\cos\theta)^2 - 3 = 4\cos^2\theta - 3 = 2(1+\cos 2\theta) - 3 = -1 + 2\cos 2\theta$
  • A = \frac{1}{2} \int{-\pi/6}^{\pi/6} [-1 + 2\cos(2\theta)]\, d\theta = \frac{1}{2}\left[ -\theta + \sin(2\theta)\right]{-\pi/6}^{\pi/6} = -\frac{\pi}{6} + \frac{\sqrt{3}}{2} \approx 0.342
  • Note: one can double a symmetric half to simplify the calculation, if preferred.
• Quick tips for area problems:
  • Always identify the correct radial function(s) over the interval(s) of interest.
  • If curves intersect, split the integral at intersection angles and subtract as needed when one curve lies outside another.

Polar to Cartesian Equation (Another View)

• Example: Convert $r = \cos\theta + \sin\theta$ to Cartesian.
  • Multiply both sides by $r$: $r^2 = r\cos\theta + r\sin\theta$.
  • Replace with Cartesian variables: $x^2 + y^2 = x + y$.
  • Complete the square to see the circle form:
  • $x^2 - x + y^2 - y = 0$ → $(x - \tfrac{1}{2})^2 + (y - \tfrac{1}{2})^2 = \tfrac{1}{2}$.
• Key takeaway: The polar form $r = f(\theta)$ and the Cartesian form can describe the same curve; sometimes one form is easier to analyze for geometry (center, radius) and integration.

Practice Notes and Takeaways

• Core identities to memorize:
  • x = r \cos \theta, \quad y = r \sin \theta, \
    r^2 = x^2 + y^2, \
    \tan \theta = \frac{y}{x}.
• When converting Cartesian to polar, ensure the angle θ lies in the requested interval (often $[0, 2\pi)$).
• When plotting polar curves, remember the sign of r affects the quadrant of the plotted point. You may prefer to adjust to a positive r with a shifted angle for clarity, especially when interpreting graphs.
• Real-world relevance: Polar coordinates are natural for problems with radial symmetry, circular motion, or angular measurements from a fixed axis; they complement Cartesian views and can simplify area, length, and intersection analyses.
• Desmos/graphing tip: Represent a polar curve parametrically via $x(\theta) = r(\theta) \cos \theta$, $y(\theta) = r(\theta) \sin \theta$ to visualize it accurately.

Connection to Prior Topics

• Builds on pre-calc/trigonometry ideas (sine/cosine relationships, unit circle geometry).
• Uses calculus concepts (area via integration, slope via derivatives) in a coordinate system best suited to rotational symmetry.
• Bridges geometric intuition (circles, lines) with algebraic manipulation (substituting $x = r\cos\theta$ and $y = r\sin\theta$).

Quick Summary of Key Formulas (for quick review)

• Cartesian from polar:
  • x = r \cos \theta, \quad y = r \sin \theta, \
    r^2 = x^2 + y^2, \
    \tan \theta = \frac{y}{x}
• Polar from Cartesian:
  • r = \sqrt{x^2 + y^2}, \quad \theta = \operatorname{atan2}(y, x) \in [0, 2\pi)
• Slope of polar curve:
  • Let $r' = \dfrac{dr}{d\theta}$; then
  • \frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}
• Area in polar coordinates:
  • A = \tfrac{1}{2} \int{\theta1}^{\theta_2} [r(\theta)]^2 \; d\theta

End of Notes