Graphing in Polar Coordinates & Geometric Applications
§4.1 The Polar Coordinate Frame & Classical Curve Tracing
1. Coordinate Transformation Equations
In the plane $\mathbb{R}^2$, a point $P$ is determined by its directed distance $r$ from the origin (pole) and counterclockwise angle $\theta$ from the positive horizontal axis (polar axis):
2. Symmetry Criteria for Polar Curves $r = f(\theta)$
- Symmetry about Polar Axis (Horizontal Axis): The equation is unchanged when replacing $(r, \theta)$ with $(r, -\theta)$ or $(-r, \pi - \theta)$.
- Symmetry about Normal Axis $\theta = \pi/2$ (Vertical Axis): The equation is unchanged when replacing $(r, \theta)$ with $(r, \pi - \theta)$ or $(-r, -\theta)$.
- Symmetry about the Pole (Origin): The equation is unchanged when replacing $(r, \theta)$ with $(-r, \theta)$ or $(r, \theta + \pi)$.
3. Zoology of Classical Polar Curves
- Cardioids: $r = a(1 \pm \cos\theta)$ or $r = a(1 \pm \sin\theta)$ ($a > 0$). Heart-shaped curves passing through the pole with a cusp at $r = 0$.
- Limaçons: $r = a + b\cos\theta$. If $a < b$, it features an inner loop; if $a = b$, it is a cardioid; if $a > b$, it is dimpled or convex.
- Rose Curves: $r = a\cos(n\theta)$ or $r = a\sin(n\theta)$. If $n \in \mathbb{N}$ is odd, the curve has $n$ petals; if $n$ is even, the curve has $2n$ petals.
- Lemniscates: $r^2 = a^2\cos(2\theta)$. Figure-eight curves with nodal tangents at $\theta = \pm \pi/4$.
§4.2 Tangents to Polar Curves & Angle Between Radius Vector and Tangent
1. Slope of the Cartesian Tangent $\frac{dy}{dx}$
Treating $\theta$ as a parametric variable with $x(\theta) = r(\theta)\cos\theta$ and $y(\theta) = r(\theta)\sin\theta$:
2. The Polar Angle $\psi$ (Angle Between Radius Vector and Tangent)
Let $\phi$ be the inclination angle of the tangent line with the polar axis ($\tan\phi = \frac{dy}{dx}$) and $\theta$ be the vectorial angle of $P$. The angle $\psi = \phi - \theta$ between the radius vector $OP$ and the tangent line satisfies:
Substituting $\tan\phi = \frac{r' \sin\theta + r \cos\theta}{r' \cos\theta - r \sin\theta}$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$:
This remarkably elegant identity proves that the geometry of polar tangents depends solely on $r$ and its angular rate of change.
§4.3 Areas Enclosed by Polar Curves & Multi-Looped Sectors
1. Derivation of the Polar Area Element
Consider an infinitesimal sector between rays $\theta$ and $\theta + d\theta$ bounded by $r = f(\theta)$. Approximating this sector by a circular sector of radius $r$ and central angle $d\theta$:
Integrating from $\theta = \alpha$ to $\theta = \beta$ yields the total swept area:
2. Area Between Two Polar Curves
If $r_{\text{outer}}(\theta) \ge r_{\text{inner}}(\theta) \ge 0$ for $\theta \in [\alpha, \beta]$:
3. Critical Precaution: Multi-Loop & Overlapping Limits
For rose curves $r = a\cos(n\theta)$, a single petal is bounded between adjacent roots where $r = 0$. For $r = a\cos(2\theta)$, setting $2\theta = \pm \frac{\pi}{2} \implies \theta \in [-\frac{\pi}{4}, \frac{\pi}{4}]$ bounds exactly one petal. Integrating over $[0, 2\pi]$ directly without symmetry can count overlapping loops multiple times.
§4.4 Arc Length in Polar Coordinates
1. Analytical Derivation of $ds$
Recall $dx = (r'\cos\theta - r\sin\theta)d\theta$ and $dy = (r'\sin\theta + r\cos\theta)d\theta$. Squaring and summing:
2. Total Arc Length Integral
§4.5 Surfaces and Volumes of Revolution in Polar Coordinates
1. Revolution About the Polar Axis ($x$-axis)
When revolving about the polar axis ($y = 0$), the perpendicular distance to the rotation axis is $y = r\sin\theta$.
- Surface Area of Revolution: $$\mathbf{S = 2\pi \int_\alpha^\beta r\sin\theta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta}$$
- Volume of Revolution: Revolving differential triangular sectors $dA$ gives conical rings: $$\mathbf{V = \frac{2}{3}\pi \int_\alpha^\beta r^3 \sin\theta \, d\theta}$$
2. Revolution About the Normal Axis $\theta = \pi/2$ ($y$-axis)
The perpendicular distance is $x = r\cos\theta$:
Step-by-Step Solved Examination Problems
Comprehensive analytical derivations, multi-tier solutions (Foundational, Intermediate Exam, and Honors/Proof Challenge) with complete line-by-line verification.
For the cardioid $r = 2(1 + \cos\theta)$: (a) Find the angle $\psi$ between the radius vector and the tangent line at $\theta = \frac{\pi}{3}$. (b) Find the Cartesian slope $\frac{dy}{dx}$ of the tangent line at this point.
Part (a): Compute the Angle $\psi$
Differentiating $r = 2(1 + \cos\theta)$ with respect to $\theta$:
Using the polar tangent identity $\tan\psi = \frac{r}{dr/d\theta}$:
Thus, $\psi = \frac{\pi}{2} + \frac{\theta}{2}$. Evaluating at $\theta = \frac{\pi}{3}$:
Part (b): Compute the Cartesian Slope $\frac{dy}{dx}$
The inclination angle of the tangent is $\phi = \theta + \psi = \frac{\pi}{3} + \frac{2\pi}{3} = \pi$.
Therefore, the Cartesian slope is:
The tangent line to the cardioid at $\theta = \frac{\pi}{3}$ is perfectly horizontal.
$\psi = \frac{2\pi}{3}$ ($120^\circ$); \quad $\frac{dy}{dx} = 0$ (horizontal tangent)
Consider the four-petaled rose $r = 4\cos(2\theta)$. (a) Find the exact area of one complete petal. (b) Find the total area enclosed by all four petals.
Part (a): Area of One Petal
The tip of the petal on the positive polar axis occurs at $\theta = 0$ where $r = 4$. The boundaries of this petal occur where $r = 0$:
The area of this petal is:
Part (b): Total Area of All Four Petals
By fourfold rotational symmetry, the total enclosed area is:
$A_{\text{petal}} = 2\pi$; \quad $A_{\text{total}} = 8\pi$
For the complete cardioid $r = a(1 + \cos\theta)$ ($a > 0$): (a) Prove that its total perimeter is $L = 8a$. (b) Find the total surface area generated by revolving this cardioid about the initial polar axis.
Part (a): Total Perimeter of the Cardioid
We compute $r' = -a\sin\theta$. The arc length element is:
Using symmetry about the polar axis, integrate from $\theta = 0$ to $\pi$ where $\cos(\theta/2) \ge 0$:
Part (b): Surface Area Revolved About the Polar Axis
The surface area element is $dS = 2\pi y \, ds = 2\pi (r\sin\theta) ds$:
Using $1 + \cos\theta = 2\cos^2(\theta/2)$ and $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$:
Integrating over $\theta \in [0, \pi]$ with substitution $u = \cos(\theta/2) \implies du = -\frac{1}{2}\sin(\theta/2)d\theta$:
$L = 8a$; \quad $S = \frac{32}{5}\pi a^2$