| | 2025.01.28

1: Parametric Equations / Polar Coordinates

Alternative coordinate systems for describing curves and regions.

Parametric Equations

Instead of $y = f(x)$ , we describe a curve using a parameter $t$ :

x = f(t), \quad y = g(t), \quad a \leq t \leq b

Circle of radius $r$ :

x = r\cos t, \quad y = r\sin t, \quad 0 \leq t \leq 2\pi

Ellipse with semi-axes $a$ and $b$ :

x = a\cos t, \quad y = b\sin t, \quad 0 \leq t \leq 2\pi

Cycloid (point on a rolling wheel):

x = r(t - \sin t), \quad y = r(1 - \cos t)

If $x = f(t)$ and $y = g(t)$ , the slope is:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}

Second derivative:

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}

The length of a parametric curve from $t = a$ to $t = b$ :

L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt

A = \int_a^b y \, dx = \int_a^b g(t) f'(t) \, dt

A point is described by distance from origin and angle:

(r, \theta) \quad \text{where } r \geq 0, \quad 0 \leq \theta < 2\pi

Polar to Cartesian:

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

Cartesian to Polar:

r = \sqrt{x^2 + y^2}, \quad \tan\theta = \frac{y}{x}

Since $x = r\cos\theta$ and $y = r\sin\theta$ :

\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}

Area enclosed by $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$ :

A = \frac{1}{2}\int_\alpha^\beta r^2 \, d\theta = \frac{1}{2}\int_\alpha^\beta [f(\theta)]^2 \, d\theta

Intuition: A thin wedge has area $\frac{1}{2}r^2 \, d\theta$ (like a pizza slice).

L = \int_\alpha^\beta \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta

Concept	Cartesian	Parametric	Polar
Curve	$y = f(x)$	$x = f(t), y = g(t)$	$r = f(\theta)$
Slope	$\frac{dy}{dx}$	$\frac{dy/dt}{dx/dt}$	$\frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$
Arc length	$\int\sqrt{1 + (dy/dx)^2}\,dx$	$\int\sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$	$\int\sqrt{r^2 + (dr/d\theta)^2}\,d\theta$
Area	$\int y\,dx$	$\int g(t)f'(t)\,dt$	$\frac{1}{2}\int r^2\,d\theta$