Unit 9
Parametric Equations, Polar Coordinates, and Vector-Valued Functions
Notes
Parametric equations
For $x = x(t)$ and $y = y(t)$:
- Slope: $\displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ (where $\displaystyle \frac{dx}{dt} \ne 0$).
- Second derivative: $\displaystyle \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$.
- Arc length from $t = a$ to $b$: $\displaystyle \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$.
Vector-valued functions and motion in the plane
Position $\vec r(t) = \langle x(t), y(t) \rangle$.
- Velocity $\vec v(t) = \langle x'(t), y'(t) \rangle$; acceleration $\vec a(t) = \langle x''(t), y''(t) \rangle$.
- Speed $\displaystyle = |\vec v(t)| = \sqrt{(x')^2 + (y')^2}$.
- Distance traveled on $[a, b]$: $\displaystyle \int_a^b |\vec v(t)|\,dt$.
- Position from velocity: $\displaystyle x(b) = x(a) + \int_a^b x'(t)\,dt$, and the same for $y$.
Polar curves
For $r = f(\theta)$, with $x = r\cos\theta$, $y = r\sin\theta$:
- Slope: $\displaystyle \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.
- Area inside the curve from $\alpha$ to $\beta$: $\displaystyle \frac{1}{2}\int_\alpha^\beta r^2\,d\theta$.
- Area between two polar curves: $\displaystyle \frac{1}{2}\int_\alpha^\beta \left(r_{outer}^2 - r_{inner}^2\right)d\theta$.
- $\displaystyle \frac{dr}{d\theta} > 0$ with $r > 0$ means the curve is moving away from the origin.
Topics in this unit
Each topic has its own short notes and practice.
- 9.1Defining and Differentiating Parametric Equations
- 9.2Second Derivatives of Parametric Equations
- 9.3Finding Arc Lengths of Curves Given by Parametric Equations
- 9.4Defining and Differentiating Vector-Valued Functions
- 9.5Integrating Vector-Valued Functions
- 9.6Solving Motion Problems Using Parametric and Vector-Valued Functions
- 9.7Defining Polar Coordinates and Differentiating in Polar Form
- 9.8Finding the Area of a Polar Region or the Area Bounded by a Single Polar Curve
- 9.9Finding the Area of the Region Bounded by Two Polar Curves
Practice
$\displaystyle \frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2} = 3$ at $t = 2$.
$\displaystyle \sqrt{3^2 + 4^2} = 5$.
Polar area is $\displaystyle \frac{1}{2}\int r^2\,d\theta$. (It equals $\pi$, a circle of radius 1.)
$(x')^2 + (y')^2 = \sin^2 t + \cos^2 t = 1$, so the length is $\pi$, half a unit circle.
$\displaystyle x = 1 + \int_0^3 2t\,dt = 10$ and $\displaystyle y = 2 + \int_0^3 1\,dt = 5$.
$\displaystyle \frac{dy}{dx} = \frac{3t}{2}$. Its $t$-derivative is $\displaystyle \frac{3}{2}$; divide by $\displaystyle \frac{dx}{dt} = 2t$.
A decreasing positive $r$ means the distance to the origin is shrinking.
Multiply by $r$: $r^2 = 4r\cos\theta$, so $x^2 + y^2 = 4x$, a circle of radius 2 centered at $(2, 0)$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A particle moves in the plane with position $\langle x(t), y(t) \rangle$, where $x'(t) = 2t - 4$ and $y'(t) = 3$, and it starts at $(0, 1)$ at $t = 0$.
(a) Find the speed of the particle at $t = 4$.
Show the answer and scoring guide
$\vec v(4) = \langle 4, 3 \rangle$, speed $\displaystyle = \sqrt{16 + 9} = 5$.
Check off the points you earned:
(b) Find the position of the particle at $t = 4$.
Show the answer and scoring guide
$\displaystyle x(4) = 0 + \int_0^4 (2t - 4)\,dt = 16 - 16 = 0$; $y(4) = 1 + 12 = 13$. Position $(0, 13)$.
Check off the points you earned:
(c) Write an integral for the total distance the particle travels from $t = 0$ to $t = 4$, and evaluate it with a calculator.
Show the answer and scoring guide
$\displaystyle \int_0^4 \sqrt{(2t - 4)^2 + 9}\,dt \approx 14.944$.
Check off the points you earned: