A parametric curve describes position using a parameter $t$: $(x(t), y(t))$. This lets us represent curves that aren't functions of $x$ (like circles, spirals, and loops) and naturally model motion over time.
Given $x = x(t)$, $y = y(t)$ with $\dot{x} = dx/dt$ and $\dot{y} = dy/dt$:
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\dot{y}}{\dot{x}}$$
$$\frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left[\dfrac{dy}{dx}\right]}{dx/dt}$$
Arc length from $t = a$ to $t = b$: $$L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$$
$x = 3\cos t$, $y = 3\sin t$ for $t \in [0, 2\pi]$ traces a circle of radius 3.
Slope: $\dfrac{dy}{dx} = \dfrac{3\cos t}{-3\sin t} = -\cot t$
At $t = \pi/4$: slope $= -\cot(\pi/4) = -1$. Point: $\left(\tfrac{3}{\sqrt{2}}, \tfrac{3}{\sqrt{2}}\right)$.
Arc length (full circle): $L = \int_0^{2\pi}\sqrt{9\sin^2 t + 9\cos^2 t}\,dt = \int_0^{2\pi} 3\,dt = 6\pi$ ✓
For the cycloid $x = t - \sin t$, $y = 1 - \cos t$:
$\dot{x} = 1 - \cos t$, $\dot{y} = \sin t$
Horizontal tangents ($\dot{y} = 0$, $\dot{x} \neq 0$): $\sin t = 0 \Rightarrow t = \pi, 3\pi, \ldots$ (top of arch)
Vertical tangents ($\dot{x} = 0$, $\dot{y} \neq 0$): $\cos t = 1 \Rightarrow t = 0, 2\pi, \ldots$ (cusps at ground)
For $x = t^2$, $y = t^3 - 3t$, find $\dfrac{dy}{dx}$ and determine where the curve has a horizontal tangent.
AP Exam Tip: Parametric arc length is a common FRQ topic. Remember to square both derivatives before adding under the radical: $\sqrt{(dx/dt)^2 + (dy/dt)^2}$. The exam may ask for arc length over a specific interval — set up the integral and evaluate with a calculator.
Parametric curves: a circle and a cycloid. Adjust the parameter $t$ (slider) to trace the curve and see the velocity vector $(\dot{x}, \dot{y})$ at each point.
Figure 10.1 — Parametric Curve: Circle $x=3\cos t$, $y=3\sin t$ with Tangent Vector
Polar coordinates $(r, \theta)$ describe a point by its distance $r$ from the origin and angle $\theta$ from the positive $x$-axis. They're especially natural for curves with rotational symmetry.
$x = r\cos\theta \qquad y = r\sin\theta \qquad r^2 = x^2 + y^2 \qquad \tan\theta = y/x$
Slope of polar curve $r = f(\theta)$:
$$\frac{dy}{dx} = \frac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta} \quad \text{where } r' = \frac{dr}{d\theta}$$
Area enclosed by $r = f(\theta)$ from $\alpha$ to $\beta$:
$$A = \frac{1}{2}\int_\alpha^\beta r^2\,d\theta$$
Area between $r_{\text{outer}}$ and $r_{\text{inner}}$:
$$A = \frac{1}{2}\int_\alpha^\beta \left(r_{\text{outer}}^2 - r_{\text{inner}}^2\right)d\theta$$
Find the area of one petal of $r = \cos(2\theta)$.
The curve $r = \cos(2\theta)$ has 4 petals. One petal exists where $r \geq 0$ in $\left[-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right]$.
$$A = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \cos^2(2\theta)\,d\theta = \frac{1}{2}\int_{-\pi/4}^{\pi/4} \frac{1+\cos(4\theta)}{2}\,d\theta$$
$$= \frac{1}{4}\left[\theta + \frac{\sin(4\theta)}{4}\right]_{-\pi/4}^{\pi/4} = \frac{1}{4}\cdot\frac{\pi}{2} = \frac{\pi}{8}$$
Find the area enclosed by the cardioid $r = 1 + \cos\theta$ (integrate over $[0, 2\pi]$).
Polar curves: the 4-petal rose $r=\cos(2\theta)$ and the cardioid $r=1+\cos(\theta)$. Toggle each curve using the sliders.
Figure 10.2 — Polar Curves: Rose $r=\cos(2\theta)$ and Cardioid $r=1+\cos\theta$
A vector-valued function $\vec{r}(t) = \langle x(t), y(t) \rangle$ describes the position of a moving object. Its derivative gives velocity; the magnitude of velocity gives speed.
Given position vector $\vec{r}(t) = \langle x(t), y(t) \rangle$:
$\vec{v}(t) = \vec{r}'(t) = \langle x'(t), y'(t) \rangle$
$|\vec{v}(t)| = \sqrt{[x'(t)]^2 + [y'(t)]^2}$
$\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t)$
$\Delta\vec{r} = \displaystyle\int_a^b \vec{v}(t)\,dt = \langle \int x'dt,\, \int y'dt \rangle$
Total distance traveled from $t=a$ to $t=b$: $\displaystyle\int_a^b |\vec{v}(t)|\,dt$
A ball is launched with velocity $\vec{v}(0) = \langle 20, 40 \rangle$ ft/s from position $(0, 0)$. With gravity $g = 32$ ft/s², find position at $t = 2$ s.
Acceleration: $\vec{a} = \langle 0, -32 \rangle$
Velocity: $\vec{v}(t) = \int\vec{a}\,dt + \vec{v}(0) = \langle 20, 40-32t \rangle$
Position: $\vec{r}(t) = \int\vec{v}\,dt + \vec{r}(0) = \langle 20t, 40t - 16t^2 \rangle$
At $t=2$: $\vec{r}(2) = \langle 40, 80-64 \rangle = \langle 40, 16 \rangle$ ft.
Max height: when $y'(t) = 40-32t=0 \Rightarrow t=1.25$ s. $y_{\max} = 40(1.25) - 16(1.25)^2 = 25$ ft.
A particle has velocity $\vec{v}(t) = \langle 3t^2, 2t \rangle$ and starts at $(1, 2)$. Find its position at $t = 2$ and the total distance traveled on $[0,2]$.
AP Exam Tip: Vector FRQ problems often ask for: position at a given time (integrate velocity + use IC), total distance (integrate speed $|\vec{v}|$), or the time when the particle is at rest ($\vec{v} = \vec{0}$). Keep velocity and position as separate vectors — don't confuse $x(t)$ and $y(t)$ components.
Projectile motion as a vector-valued function. The red vector shows velocity at each point. Adjust $t$ with the slider to trace the path.
Figure 10.3 — Vector-Valued Function: Projectile $\vec{r}(t) = \langle 20t,\, 40t-16t^2 \rangle$
Find $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ for $x = t^2 + 1$, $y = t^3 - t$.
Find the arc length of the curve $x = e^t\cos t$, $y = e^t\sin t$ for $0 \leq t \leq \pi$.
Convert the polar equation $r = 4\sin\theta$ to Cartesian form. What curve is this?
Find the area inside the circle $r = 3$ and outside the cardioid $r = 1 + \cos\theta$.
A particle moves with $\vec{v}(t) = \langle \cos t, 2t \rangle$ and $\vec{r}(0) = \langle 0, 0 \rangle$. Find position at $t = \pi$ and total distance on $[0, \pi]$.
Find the area of one petal of the rose $r = 3\sin(3\theta)$.
Find the slope of the tangent to $r = 2 + \sin\theta$ at $\theta = \pi/2$.
(AP Style) A particle moves in the $xy$-plane with $x'(t) = t^2 - 3$ and $y'(t) = 2t - 4$, with $x(0) = 1$, $y(0) = 2$. At what times is the particle at rest? Find position at those times.
$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ — divide the $y$-derivative by the $x$-derivative with respect to parameter $t$.
$\dfrac{d^2y}{dx^2} = \dfrac{d}{dt}\!\left[\dfrac{dy}{dx}\right] \div \dfrac{dx}{dt}$ — used for concavity of parametric curves.
$L = \displaystyle\int_\alpha^\beta \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}\,dt$
$|\mathbf{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}$ — the magnitude of the velocity vector at time $t$.
$x = r\cos\theta$, $y = r\sin\theta$, $r^2 = x^2 + y^2$, $\tan\theta = y/x$
$\dfrac{dy}{dx} = \dfrac{r'\sin\theta + r\cos\theta}{r'\cos\theta - r\sin\theta}$ where $r' = dr/d\theta$
$A = \dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta$
$A = \dfrac{1}{2}\displaystyle\int_\alpha^\beta (r_{\text{outer}}^2 - r_{\text{inner}}^2)\,d\theta$
$\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 = |\mathbf{a}||\mathbf{b}|\cos\theta$. Vectors are perpendicular when $\mathbf{a}\cdot\mathbf{b}=0$.
$\mathbf{r}(t) \xrightarrow{d/dt} \mathbf{v}(t) \xrightarrow{d/dt} \mathbf{a}(t)$. Integrate backwards to find position from velocity.