AP Calculus BC

Unit 9

Parametric Equations, Polar Coordinates, and Vector-Valued Functions

9 topics · 35 practice questions · Free response

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.

Practice

For $x = t^2$, $y = t^3$, what is $\displaystyle \frac{dy}{dx}$ at $t = 2$?
A particle has $\vec v(t) = \langle 3, 4t \rangle$. What is its speed at $t = 1$?
Which integral gives the area inside one full loop of $r = 2\sin\theta$ ($0 \le \theta \le \pi$)?
For $x = \cos t$, $y = \sin t$, which gives the arc length from $t = 0$ to $t = \pi$?
A particle at $(1, 2)$ when $t = 0$ has $\vec v(t) = \langle 2t, 1 \rangle$. Where is it at $t = 3$?
For $x = t^2$, $y = t^3$, what is $\displaystyle \frac{d^2y}{dx^2}$?
On an interval where $r = f(\theta) > 0$ and $\displaystyle \frac{dr}{d\theta} < 0$, the curve is:
Convert $r = 4\cos\theta$ to rectangular form.

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: