Parametric, Polar & Vector Functions

Parametric Equations

$x = f(t)$, $y = g(t)$.

First derivative: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Second derivative: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$

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

Speed: $v = \sqrt{(dx/dt)^2 + (dy/dt)^2}$

Polar Coordinates

$r = f(\theta)$.

Conversion: $x = r\cos\theta$, $y = r\sin\theta$.

Slope in polar: $\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}$

Area enclosed: $A = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta$

Area between two polar curves: $A = \frac{1}{2}\int_\alpha^\beta (r_{\text{outer}}^2 - r_{\text{inner}}^2)\,d\theta$

Vector-Valued Functions

$\vec{r}(t) = \langle x(t), y(t) \rangle$.

Velocity: $\vec{v}(t) = \langle x'(t), y'(t) \rangle$.

Acceleration: $\vec{a}(t) = \langle x''(t), y''(t) \rangle$.

Speed: $|\vec{v}| = \sqrt{(x')^2 + (y')^2}$.

Displacement: $\int_{t_1}^{t_2} \vec{v}(t)\,dt$.

Total distance: $\int_{t_1}^{t_2} |\vec{v}(t)|\,dt$.

Problem: Find the area enclosed by $r = 2\cos\theta$.

Solution: This is a circle. It traces from $\theta = -\pi/2$ to $\theta = \pi/2$ (or equivalently $0$ to $\pi$).

$A = \frac{1}{2}\int_0^\pi (2\cos\theta)^2\,d\theta = 2\int_0^\pi \cos^2\theta\,d\theta = 2\int_0^\pi \frac{1+\cos 2\theta}{2}\,d\theta$ $= \int_0^\pi (1 + \cos 2\theta)\,d\theta = [\theta + \frac{\sin 2\theta}{2}]_0^\pi = \pi$

• Parametric: $dy/dx = (dy/dt)/(dx/dt)$; arc length uses the speed formula.
• Polar area: $A = \frac{1}{2}\int r^2\,d\theta$.
• Vectors: velocity is the derivative of position; speed is the magnitude of velocity.
• Total distance = integral of speed; displacement = integral of velocity.