Motion in the Plane — AP® Calculus BC Notes

Motion in the plane brings together everything in Unit 9 so far: the position vector, its derivatives, its integrals and the length of a path. Every free-response question on a particle in the plane asks some mix of the same five things. This page lists them, shows the tool for each, and points out where points are usually lost.

From position to velocity, speed and distance

Start from the position vector $\mathbf{r}(t) = \left\langle x(t), y(t) \right\rangle$ and follow the arrows:

Quantity	Formula	Kind of answer
Velocity	$\mathbf{v}(t) = \left\langle x'(t), y'(t) \right\rangle$	a vector
Acceleration	$\mathbf{a}(t) = \left\langle x''(t), y''(t) \right\rangle$	a vector
Speed	$\left\\| \mathbf{v}(t) \right\\|$ $= \sqrt{(x'(t))^2 + (y'(t))^2}$	a number, at least 0
Displacement on $[a, b]$	$\int_a^b \mathbf{v}(t)\,dt$ $= \mathbf{r}(b) - \mathbf{r}(a)$	a vector
Position at $t = b$	$\mathbf{r}(b)$ $= \mathbf{r}(a)$ $+ \int_a^b \mathbf{v}(t)\,dt$	a point or vector
Distance traveled on $[a, b]$	$\int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$	a number, at least 0

The table mirrors straight-line motion from Units 4 and 8, where speed was $|v(t)|$ and distance $\int_a^b |v(t)|\,dt$ . In the plane, the absolute value becomes the magnitude of a vector.

The path r(t) = 2t, 4t minus t squared for t from 0 to 4, an arch from (0, 0) up to (4, 4) and down to (8, 0), traced left to right. A dashed gray position vector goes from the origin to (2, 3), the point at t = 1. From that point a blue velocity vector 2, 2 is drawn; its length, the speed, is 2 root 2. A vermillion displacement vector 8, 0 runs from the start to the end, length 8, while the distance along the path, the integral of speed, is about 11.83. — **Figure 1.** Position, velocity, displacement and distance on one path. LevlPrep original diagram.

The figure shows $\mathbf{r}(t) = \left\langle 2t, 4t - t^2 \right\rangle$ for $0 \le t \le 4$ . At $t = 1$ the particle is at $(2, 3)$ with $\mathbf{v}(1) = \left\langle 2, 2 \right\rangle$ , so its speed is $2\sqrt{2}$ . Over the whole trip the displacement is $\mathbf{r}(4) - \mathbf{r}(0) = \left\langle 8, 0 \right\rangle$ , length 8, but the particle flies along the arch, a distance of $\int_0^4 \sqrt{4 + (4 - 2t)^2}\,dt$ $\approx 11.83$ .

A calculator question, start to finish

Worked example 1. A particle moves with $x'(t) = 2\sin(t^2) + 1$ and $y'(t) = e^{-t} + t$ for $0 \le t \le 3$ , and $(x(0), y(0)) = (2, 5)$ .

Speed at $t = 1$ : $\sqrt{(x'(1))^2 + (y'(1))^2}$ $= \sqrt{(2\sin 1 + 1)^2 + (e^{-1} + 1)^2}$ $\approx 3.012$ .

Position at $t = 3$ : $x(3)$ $= 2$ $+ \int_0^3 x'(t)\,dt$ $\approx 2$ $+ 4.547$ $= 6.547$ and $y(3)$ $= 5$ $+ \int_0^3 y'(t)\,dt$ $\approx 5$ $+ 5.450$ $= 10.450$ .

Distance traveled: $\int_0^3 \sqrt{(x'(t))^2 + (y'(t))^2}\,dt$ $\approx 7.900$ .

Each answer shows the setup first, then the calculator value to three decimal places, with stored values and no early rounding.

A question may also give a speed and ask for the time: "when is the speed 3?" Then write the equation $\sqrt{(x'(t))^2 + (y'(t))^2}$ $= 3$ and solve it on the calculator. Evaluating the speed at some time does not answer it.

Reading direction and behavior

Many points come from reading signs, and each question has its own derivative:

"Is the particle moving left or right?" Read $\frac{dx}{dt}$ . "Up or down?" Read $\frac{dy}{dt}$ .
"Is the $y$ -coordinate decreasing?" Read $\frac{dy}{dt} < 0$ , not $\frac{dy}{dx}$ . A path can fall as $x$ increases ( $\frac{dy}{dx} < 0$ ) while $x$ itself is decreasing, so that $y$ rises in time.
"Is the particle at rest?" Both components of the velocity are 0.
"Where is the particle farthest left?" A candidates test on $x(t)$ : compare $x$ at the endpoints and where $x'(t)$ changes sign.

Worked example 2. A particle has $x(t) = t^3 - 3t$ and $y(t) = 2t^2 - t^4$ for $-2 \le t \le 2$ . For which $t$ is the particle moving left and down?

Left: $x'(t) = 3t^2 - 3 < 0$ , so $-1 < t < 1$ .

Down: $y'(t)$ $= 4t - 4t^3$ $= 4t(1 - t)(1 + t) < 0$ , so $-1 < t < 0$ or $t > 1$ .

Both: $-1 < t < 0$ .

Two particles

With two particles, label everything with its particle ( $x_A$ , $y_B$ ). Their paths cross where the curves share a point; they collide only if they are at that point at the same time. Set the $x$ -coordinates equal, solve for $t$ , and check that the $y$ -coordinates agree at that same $t$ .

Where points are lost

Slip	Fix
Speed as $\left\|\frac{dy}{dx}\right\|$ or as one component	$\sqrt{(x'(t))^2 + (y'(t))^2}$
Position without the initial value, or with it subtracted	$x(b)$ $= x(a)$ $+ \int_a^b x'(t)\,dt$
A missing $dt$ , or limits that do not match the start	Write the full integral with its differential and the $t$ -interval
" $y$ is decreasing" read from $\frac{dy}{dx}$	Read the sign of $\frac{dy}{dt}$
Velocity given when acceleration is asked, or components swapped	Check which derivative, and keep the order $\left\langle x, y \right\rangle$
Fewer than three decimal places	Store values; round only the final answer