Math Core

In AP Calculus AB you studied a particle moving back and forth along a line. Now the particle can move anywhere in the plane. The ideas are the same (velocity is the derivative of position, acceleration is the derivative of velocity), but each of them is now a vector, and speed and distance need the Pythagorean theorem.

Position, velocity and acceleration vectors

If a particle's position at time $t$ is $\vec{r}(t) = \langle x(t), y(t) \rangle$ , then:

Motion in the plane

\begin{aligned} \text{velocity:} \quad & \vec{v}(t) = \vec{r}\,'(t) = \langle x'(t),\, y'(t) \rangle \\ \text{acceleration:} \quad & \vec{a}(t) = \vec{v}\,'(t) = \langle x''(t),\, y''(t) \rangle \\ \text{speed:} \quad & |\vec{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2} \end{aligned}

Speed is the length of the velocity vector. It is a number, never negative, with no direction.

The velocity vector points in the direction of motion, tangent to the path. Each component tells you about motion in one direction:

$x'(t) > 0$ : moving right; $x'(t) < 0$ : moving left.
$y'(t) > 0$ : moving up; $y'(t) < 0$ : moving down.
The particle is at rest only when $x'(t) = 0$ and $y'(t) = 0$ at the same time.

Worked example: Velocity, speed and acceleration at an instant

A particle moves with position $\vec{r}(t) = \langle t^3 - 3t,\; t^2 \rangle$ . Find its velocity, speed and acceleration at $t = 2$ , and the time interval on which it moves to the left.

$\vec{v}(t) = \langle 3t^2 - 3,\; 2t \rangle$ , so $\vec{v}(2) = \langle 9, 4 \rangle$ .

Speed: $|\vec{v}(2)| = \sqrt{81 + 16} = \sqrt{97} \approx 9.849$ .

$\vec{a}(t) = \langle 6t,\; 2 \rangle$ , so $\vec{a}(2) = \langle 12, 2 \rangle$ .

Moving left means $x'(t) = 3t^2 - 3 < 0$ , which happens for $-1 < t < 1$ . The particle is never at rest, since $3t^2 - 3 = 0$ needs $t = \pm 1$ while $2t = 0$ needs $t = 0$ .

The path of r(t) = ⟨t³ − 3t, t²⟩ for −2 ≤ t ≤ 2. At t = 2 the particle is at (2, 4), moving along the dashed tangent line in the direction of v(2) = ⟨9, 4⟩.Open in grapher →

Position from velocity

Going the other way, you integrate. If you know $\vec{v}(t)$ and the position at one time, integrate each component and use the known position to find the constants. Equivalently, use the Fundamental Theorem on each coordinate:

x(b) = x(a) + \int_a^b x'(t)\, dt, \qquad y(b) = y(a) + \int_a^b y'(t)\, dt.

This form is the one to use on the calculator section, when the velocity components can't be integrated by hand.

Worked example: Finding a later position

A particle has velocity $\vec{v}(t) = \langle 3t^2,\; 2t + 1 \rangle$ and is at $(4, -1)$ when $t = 1$ . Where is it at $t = 2$ ?

\begin{aligned} x(2) &= 4 + \int_1^2 3t^2\, dt = 4 + (8 - 1) = 11 \\ y(2) &= -1 + \int_1^2 (2t + 1)\, dt = -1 + (t^2 + t)\Big|_1^2 = -1 + (6 - 2) = 3 \end{aligned}

The particle is at $(11, 3)$ .

Distance traveled versus displacement

Two different questions sound alike:

Displacement from $t = a$ to $t = b$ is the change in position, the vector $\displaystyle\int_a^b \vec{v}(t)\, dt = \vec{r}(b) - \vec{r}(a)$ . Its length is the straight-line distance between the starting and ending points.
Total distance traveled is the length of the path actually followed, which is the integral of speed:

\text{distance} = \int_a^b |\vec{v}(t)|\, dt = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\, dt.

That second formula is the arc length formula from earlier in the unit, now read as "speed times time, added up."

Worked example: Distance and displacement

A particle moves with $\vec{v}(t) = \langle 2t,\; t^2 - 1 \rangle$ for $0 \le t \le 3$ . Find the total distance traveled and the magnitude of the displacement.

Speed: $(2t)^2 + (t^2 - 1)^2 = 4t^2 + t^4 - 2t^2 + 1 = t^4 + 2t^2 + 1 = (t^2 + 1)^2$ , so $|\vec{v}(t)| = t^2 + 1$ .

Distance: $\displaystyle\int_0^3 (t^2 + 1)\, dt = 9 + 3 = 12$ .

Displacement: $\left\langle \displaystyle\int_0^3 2t\, dt,\; \int_0^3 (t^2 - 1)\, dt \right\rangle = \langle 9, 6 \rangle$ , with length $\sqrt{81 + 36} = \sqrt{117} \approx 10.817$ .

The path is longer than the straight line between its endpoints, as it must be.

Common mistake

Speed is not $x'(t) + y'(t)$ , and distance is not $|x(b) - x(a)| + |y(b) - y(a)|$ . Speed is the length of the velocity vector, so it needs the square root of the sum of squares. When a question says "speed" or "total distance," reach for $\sqrt{(x')^2 + (y')^2}$ .

A calculator-active problem

Worked example: Numerical distance traveled

A particle moves with $\vec{v}(t) = \left\langle \cos(t^2),\; e^{0.5t} \right\rangle$ . Find the total distance it travels from $t = 0$ to $t = 3$ .

The speed is $\sqrt{\cos^2(t^2) + e^{t}}$ (because $\left(e^{0.5t}\right)^2 = e^t$ ). With a calculator,

\int_0^3 \sqrt{\cos^2(t^2) + e^{t}}\, dt \approx 7.411.

Tip

On the AP exam, round to three decimal places, and store intermediate results in the calculator rather than retyping rounded values. Early rounding is a common way to lose the answer point.

Practice

Practice 1

A particle has position $\vec{r}(t) = \langle t^2 + 1,\; 3t - t^2 \rangle$ . Find its velocity vector at $t = 2$ . Enter it as $(a, b)$ .

Enter a point like (2, -3)

Practice 2

For the particle in the previous problem, what is its speed at $t = 2$ ? Give an exact answer or a decimal to three places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

A particle moves with position $\vec{r}(t) = \langle t^3,\; 4\sin t \rangle$ . What is its acceleration vector at $t = \dfrac{\pi}{2}$ ?

Practice 4

A particle has velocity $\vec{v}(t) = \langle 4t - 1,\; 3t^2 \rangle$ and is at $(2, -1)$ when $t = 0$ . Find its position at $t = 2$ . Enter it as $(x, y)$ .

Enter a point like (2, -3)

Practice 5

A particle moves with position $\vec{r}(t) = \langle t^3 - 3t,\; 4t - t^2 \rangle$ for $t \ge 0$ . On which interval is the particle moving to the left and upward?

Practice 6

A particle moves with velocity $\vec{v}(t) = \langle 6t,\; 3t^2 - 3 \rangle$ . Find the total distance it travels from $t = 0$ to $t = 2$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Calculator allowed. A particle moves with velocity $\vec{v}(t) = \left\langle e^{\sin t},\; t\cos t \right\rangle$ . It is at $(1, 0)$ when $t = 0$ . Find the total distance the particle travels from $t = 0$ to $t = 2$ , to three decimal places.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.