Calculus BC · CED Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions · 14 min read · Updated 2026-05-11

Solving motion problems using parametric and vector-valued functions — AP Calculus BC

AP Calculus BC · CED Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions · 14 min read

1. Position, Velocity, and Acceleration Vectors ★★☆☆☆ ⏱ 3 min

For a particle moving in the 2D plane, x and y coordinates change independently with time, so we represent position as a vector-valued function of time. All operations on motion vectors are done component-wise: we differentiate or integrate each coordinate separately.

Velocity is the first derivative of position, found by differentiating each component separately:

\vec{v}(t) = \vec{r}'(t) = \langle x'(t), y'(t) \rangle

Acceleration is the derivative of velocity (second derivative of position):

\vec{a}(t) = \vec{v}'(t) = \vec{r}''(t) = \langle x''(t), y''(t) \rangle

When working backwards from acceleration to velocity or position, we integrate component-wise, then use initial conditions to solve for constants of integration for each component.

📐 Worked Example

A particle moves in the plane with acceleration $\vec{a}(t) = \langle 2t, 6 \rangle$ for $t \geq 0$. At $t=0$, velocity is $\vec{v}(0) = \langle 4, 0 \rangle$ and position is $\vec{r}(0) = \langle 0, 10 \rangle$. Find the position function $\vec{r}(t)$.

Integrate acceleration component-wise to get velocity:
$\int 2t dt = t^2 + C_x; \int 6 dt = 6t + C_y \implies \vec{v}(t) = \langle t^2 + C_x, 6t + C_y \rangle$
Use initial velocity to solve for constants:
$v_x(0) = C_x = 4; v_y(0) = C_y = 0 \implies \vec{v}(t) = \langle t^2 + 4, 6t \rangle$
Integrate velocity component-wise to get position:
$\int (t^2 + 4) dt = \frac{1}{3}t^3 + 4t + D_x; \int 6t dt = 3t^2 + D_y \implies \vec{r}(t) = \langle \frac{1}{3}t^3 + 4t + D_x, 3t^2 + D_y \rangle$
Use initial position to solve for constants and get the final position function:
$x(0) = D_x = 0; y(0) = D_y = 10 \implies \vec{r}(t) = \langle \frac{1}{3}t^3 + 4t, 3t^2 + 10 \rangle$

2. Speed and Direction of Motion ★★☆☆☆ ⏱ 2 min

Velocity is a vector with both magnitude and direction. The magnitude of the velocity vector is called speed, a non-negative scalar quantity that describes how fast the particle is moving regardless of direction.

\text{speed} = |\vec{v}(t)| = \sqrt{\left(x'(t)\right)^2 + \left(y'(t)\right)^2}

To find direction of motion, check the sign of each velocity component: positive $x'(t)$ means rightward motion, negative means left; positive $y'(t)$ means upward motion, negative means down. The slope $\frac{y'(t)}{x'(t)}$ gives the slope of the tangent to the particle's path at time $t$.

📐 Worked Example

A particle has position $\vec{r}(t) = \langle \cos t, 2 \sin t \rangle$ for $t \geq 0$. Find the speed of the particle at $t = \frac{\pi}{3}$, and state whether it is moving left/right and up/down at this time.

Differentiate position component-wise to get velocity:
$x'(t) = -\sin t; y'(t) = 2 \cos t \implies \vec{v}\left(\frac{\pi}{3}\right) = \langle -\sin\left(\frac{\pi}{3}\right), 2 \cos\left(\frac{\pi}{3}\right) \rangle$
Evaluate components at $t = \frac{\pi}{3}$:
$\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}; \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \implies \vec{v}\left(\frac{\pi}{3}\right) = \langle -\frac{\sqrt{3}}{2}, 1 \rangle$
Calculate speed as the magnitude of velocity:
$|v| = \sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2 + (1)^2} = \sqrt{\frac{3}{4} + 1} = \frac{\sqrt{7}}{2}$
Check component signs for direction: $v_x = -\frac{\sqrt{3}}{2} < 0$ (moving left), $v_y = 1 > 0$ (moving up).

3. Displacement vs Total Distance Traveled ★★★☆☆ ⏱ 4 min

A key distinction repeatedly tested on the AP exam is between displacement (net change in position) and total distance traveled (total length of the path traversed by the particle).

\text{Total Distance} = \int_a^b |\vec{v}(t)| dt = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt

This formula is identical to the arc length formula for a parametric curve, which makes sense because we are calculating the length of the path traced by the particle over time.

4. Projectile Motion Modeling ★★★☆☆ ⏱ 3 min

Projectile motion (motion of an object acted on only by gravity after launch) is one of the most common real-world applications tested on the AP exam. For a standard coordinate system with origin at the launch point, x horizontal, y upward, acceleration is entirely due to gravity, so horizontal acceleration is always 0:

\vec{a}(t) = \langle 0, -g \rangle

where $g = 32 \text{ ft/s}^2$ for imperial units, and $g = 9.8 \text{ m/s}^2$ for metric units. If launched with initial speed $v_0$ at angle $\theta$ from the horizontal from initial height $h$, integrating gives the standard position function:

\vec{r}(t) = \langle (v_0 \cos \theta) t, h + (v_0 \sin \theta)t - \frac{1}{2} g t^2 \rangle

Common questions ask for maximum height, time of impact, or range (horizontal distance at impact), all solved using derivative and integral rules for vector motion. Maximum height occurs when vertical velocity equals 0.

Common Pitfalls

Why: Students confuse the definitions of displacement and total distance, thinking 'distance' only means net change.

Why: Students forget integration is done component-wise, so each integral produces a separate constant.

Why: Students confuse velocity (vector) with speed (scalar magnitude of velocity).

Why: Students mix up which axis gravity acts on.

Why: Students confuse position (where the particle is) with velocity (the direction it is moving).

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