Lesson 4-3: Parametric & Polar Systems

Master parametric curves, polar coordinates, derivatives, arc length, and coordinate conversions with applications to complex curves.

Learning Objectives

Parametric Derivatives

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

\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{1}{dx/dt}

Polar Conversions

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

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

x = f(t), \quad y = g(t)

Where t is the parameter and f(t), g(t) are differentiable functions.

\text{Slope} = \frac{dy}{dx} = \frac{g'(t)}{f'(t)}

\text{Point} = (f(t_0), g(t_0))

r = a\cos\theta \quad \text{(Circle)}

r = a(1 + \cos\theta) \quad \text{(Cardioid)}

r = a\cos(n\theta) \quad \text{(Rose curve)}

\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}

Where r' = dr/dθ

Find the tangent line to the curve $x = t - \sin t, y = 1 - \cos t$ at $t = \pi$ .

Solution:

x'(t) = 1 - \cos t, \quad y'(t) = \sin t

\frac{dy}{dx} = \frac{\sin t}{1 - \cos t}

\text{At } t = \pi: \text{ slope } = 0, \text{ point } = (\pi, 2)

\text{Tangent: } y = 2

Convert the polar equation $r = 2\cos\theta$ to rectangular form.

Solution:

r = 2\cos\theta \Rightarrow r^2 = 2r\cos\theta

x^2 + y^2 = 2x

x^2 - 2x + y^2 = 0 \Rightarrow (x-1)^2 + y^2 = 1

\text{Circle: center }(1,0), \text{ radius } 1

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

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

Find the slope of the tangent to $x = t^2, y = t^3$ at $t = 2$ .

\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2} = 3

Convert $r = 4\sin\theta$ to rectangular form.

x^2 + (y-2)^2 = 4

This completes Unit 4. Return to Unit 4 Overview or continue to the next unit.