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=rcos⁡θx = r \cos \theta
    • y=rsin⁡θy = r \sin \theta
    • r2=x2+y2r^2 = x^2 + y^2
    • tan⁡θ=yx\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=rcos⁡θ,y=rsin⁡θx = r \cos \theta, \quad y = r \sin \theta
    • Example: The polar point $(-2, \frac{5\pi}{6})$ →
    • x=−2cos⁡5π6=−2(−32)=3x = -2 \cos \frac{5\pi}{6} = -2\left(-\frac{\sqrt{3}}{2}\right) = \sqrt{3}
    • y=−2sin⁡5π6=−2(12)=−1y = -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=x2+y2r = \sqrt{x^2 + y^2} (with r ≥ 0 by convention)
    • θ=atan2⁡(y,x)\theta = \operatorname{atan2}(y, x) which places θ in $[0, 2\pi)$ (adjusting for the correct quadrant)
    • Example: Cartesian $(-3, -3)$ →
    • r=(−3)2+(−3)2=18=32r = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}
    • θ=arctan⁡(y/x)=arctan⁡(1)=π4\theta = \arctan(y/x) = \arctan(1) = \tfrac{\pi}{4} but in QIII, so θ=5π4\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=rcos⁡θ,y=rsin⁡θ,</li></ul><p>r2=x2+y2,</p><p>tan⁡θ=yxx = r \cos \theta, \quad y = r \sin \theta, </li></ul> <p>r^2 = x^2 + y^2, </p> <p>\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(θ)cos⁡θ,y=r(θ)sin⁡θx = 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:
      • dydθ=r′(θ)sin⁡θ+r(θ)cos⁡θ\frac{dy}{d\theta} = r'(\theta) \sin\theta + r(\theta) \cos\theta
      • dxdθ=r′(θ)cos⁡θ−r(θ)sin⁡θ\frac{dx}{d\theta} = r'(\theta) \cos\theta - r(\theta) \sin\theta
    • Therefore the slope of the tangent is:
      • dydx=dydθdxdθ=r′(θ)sin⁡θ+r(θ)cos⁡θr′(θ)cos⁡θ−r(θ)sin⁡θ\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=12∫<em>θ</em>1θ2[r(θ)]2  dθA = \frac{1}{2} \int<em>{\theta</em>1}^{\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=12∫<em>−π/4π/432  dθ=92[θ]</em>−π/4π/4=92⋅π2=9π4A = \tfrac{1}{2} \int<em>{-\pi/4}^{\pi/4} 3^2 \; d\theta = \tfrac{9}{2} \left[ \theta \right]</em>{-\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<em>petal=12∫</em>0π/2sin⁡2(2θ)  dθA<em>{\text{petal}} = \tfrac{1}{2} \int</em>{0}^{\pi/2} \sin^2(2\theta) \; d\theta
      • Using $\sin^2 u = \frac{1 - \cos 2u}{2}$ with $u = 2\theta$ gives
      • A<em>petal=12⋅12∫</em>0π/2(1−cos⁡4θ) dθ=14[θ−sin⁡4θ4]0π/2=π8≈0.3927A<em>{\text{petal}} = \frac{1}{2} \cdot \frac{1}{2} \int</em>{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=12∫−π/6π/6[(2cos⁡θ)2−(3)2]dθA = \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=12∫<em>−π/6π/6[−1+2cos⁡(2θ)] dθ=12[−θ+sin⁡(2θ)]</em>−π/6π/6=−π6+32≈0.342A = \frac{1}{2} \int<em>{-\pi/6}^{\pi/6} [-1 + 2\cos(2\theta)]\, d\theta = \frac{1}{2}\left[ -\theta + \sin(2\theta)\right]</em>{-\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=x2+y2,θ=atan2⁡(y,x)∈[0,2π)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
      • dydx=r′sin⁡θ+rcos⁡θr′cos⁡θ−rsin⁡θ\frac{dy}{dx} = \frac{r' \sin \theta + r \cos \theta}{r' \cos \theta - r \sin \theta}
    • Area in polar coordinates:
      • A=12∫<em>θ</em>1θ2[r(θ)]2  dθA = \tfrac{1}{2} \int<em>{\theta</em>1}^{\theta_2} [r(\theta)]^2 \; d\theta

    End of Notes