Motion Problems
Parametric & Vector-Valued Functions

Master position, velocity, acceleration, and speed using parametric equations and vector-valued functions for the AP Calculus BC exam.

AP Calculus BC

What Are Parametric Equations?

Instead of writing \( y \) as a function of \( x \), we express both coordinates as functions of a third variable, the parameter \( t \) (often representing time):

\( x(t) \)

Horizontal position as a function of time

\( y(t) \)

Vertical position as a function of time

\( t \)

Parameter (usually time in motion problems)

Visualizing a Parametric Path

Consider a particle with position \( \mathbf{r}(t) = \langle t^2 - 1,\, 2t \rangle \). As \( t \) increases, the particle traces a parabolic path.

Vector-Valued Functions

A vector-valued function packages parametric equations into a single vector. In the plane:

Each derivative of \( \mathbf{r}(t) \) has a physical interpretation in motion:

1

Position: \( \mathbf{r}(t) = \langle x(t),\, y(t) \rangle \) tells you where the particle is.

2

Velocity: \( \mathbf{v}(t) = \mathbf{r}'(t) = \langle x'(t),\, y'(t) \rangle \) tells you how fast and in what direction.

3

Acceleration: \( \mathbf{a}(t) = \mathbf{v}'(t) = \langle x''(t),\, y''(t) \rangle \) tells you how velocity changes.

Core Motion Formulas

Velocity

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

Speed

\( \|\mathbf{v}(t)\| = \sqrt{[x'(t)]^2 + [y'(t)]^2} \)

Acceleration

\( \mathbf{a}(t) = \langle x''(t),\, y''(t) \rangle \)

Distance Traveled

\( \displaystyle\int_{a}^{b} \|\mathbf{v}(t)\|\, dt \)

Slope: \( \frac{dy}{dx} \) for Parametric Curves

Do not simply take \( \frac{y''(t)}{x''(t)} \). You must differentiate the first derivative with respect to \( t \), then divide by \( \frac{dx}{dt} \).

Speed vs. Velocity

	Velocity \( \mathbf{v}(t) \)	Speed \( \\|\mathbf{v}(t)\\| \)
Type	Vector	Scalar
Formula	\( \langle x'(t), y'(t) \rangle \)	\( \sqrt{[x'(t)]^2+[y'(t)]^2} \)
Can be negative?	Yes (components can)	No (always \( \geq 0 \))
Gives direction?	Yes	No
Units	m/s (with direction)	m/s (magnitude only)

How to Solve Motion Problems

1

Identify given information: position \( \mathbf{r}(t) \), velocity \( \mathbf{v}(t) \), or acceleration \( \mathbf{a}(t) \)?

2

Determine what's asked: Speed? Slope \( \frac{dy}{dx} \)? Distance? Position at a time?

3

Differentiate to go from position → velocity → acceleration. Integrate to go the other way.

4

Apply the correct formula: Speed = \( \|\mathbf{v}(t)\| \), Distance = \( \int_a^b \|\mathbf{v}(t)\|\, dt \).

5

Use initial conditions when integrating to find the constant of integration.

Example 1: Finding Velocity & Speed

A particle moves with position \( x(t) = 3t^2 \) and \( y(t) = 2t^3 \). Find the velocity and speed at \( t = 1 \).

1

Differentiate: \( x'(t) = 6t \), \( y'(t) = 6t^2 \)

2

Velocity at \( t=1 \): \( \mathbf{v}(1) = \langle 6(1),\, 6(1)^2 \rangle = \langle 6, 6 \rangle \)

3

Speed: \( \|\mathbf{v}(1)\| = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} \)

Example 2: Total Distance Traveled

A particle has velocity \( \mathbf{v}(t) = \langle 2t,\, 3 \rangle \). Find the total distance traveled from \( t = 0 \) to \( t = 4 \).

1

Speed: \( \|\mathbf{v}(t)\| = \sqrt{(2t)^2 + 3^2} = \sqrt{4t^2 + 9} \)

2

Set up integral: \( \displaystyle\int_0^4 \sqrt{4t^2 + 9}\, dt \)

3

Evaluate (by trig sub or calculator): \( \approx 19.978 \)

Example 3: Position from Velocity

A particle has \( \mathbf{v}(t) = \langle \cos t,\, e^t \rangle \) and initial position \( \mathbf{r}(0) = \langle 2, 1 \rangle \). Find \( \mathbf{r}(t) \).

1

Integrate velocity: \( x(t) = \int \cos t\, dt = \sin t + C_1 \), \( y(t) = \int e^t\, dt = e^t + C_2 \)

2

Apply initial conditions: \( x(0) = \sin 0 + C_1 = 2 \Rightarrow C_1 = 2 \). And \( y(0) = e^0 + C_2 = 1 \Rightarrow C_2 = 0 \).

3

Result: \( \mathbf{r}(t) = \langle \sin t + 2,\, e^t \rangle \)

Displacement vs. Distance

Total Distance

\( \displaystyle\int_a^b \|\mathbf{v}(t)\|\, dt \)
Always \( \geq 0 \), measures path length

Displacement

\( \mathbf{r}(b) - \mathbf{r}(a) \)
A vector from start to end, can be negative in components

When Is a Particle Speeding Up?

A particle's speed is increasing when the velocity and acceleration vectors are working "together." Formally:

The dot product \( \mathbf{v} \cdot \mathbf{a} = x'(t)\,x''(t) + y'(t)\,y''(t) \) measures how much acceleration is in the direction of motion.

Position via Definite Integrals

If you know the velocity and an initial position, you can find position at any time \( t \) using integration:

This is the accumulation form of the Fundamental Theorem. The AP exam often gives velocity functions and an initial position, then asks for position at a later time.

Common Mistakes to Avoid

✕ Confusing Speed and Velocity

Speed is \( \|\mathbf{v}(t)\| \), a scalar. Velocity is a vector. Don't forget the square root for speed.

✕ Wrong \( \frac{d^2 y}{dx^2} \)

The second derivative is NOT \( \frac{y''(t)}{x''(t)} \). You must use \( \frac{\frac{d}{dt}[dy/dx]}{dx/dt} \).

✕ Distance = Displacement

Distance uses \( \int\|\mathbf{v}\|\, dt \). Displacement is \( \mathbf{r}(b)-\mathbf{r}(a) \). They are not the same!

Summary of Key Formulas

Quantity	Formula	Type
Position	\( \mathbf{r}(t) = \langle x(t), y(t) \rangle \)	Vector
Velocity	\( \mathbf{v}(t) = \langle x'(t), y'(t) \rangle \)	Vector
Speed	\( \sqrt{[x'(t)]^2+[y'(t)]^2} \)	Scalar
Acceleration	\( \mathbf{a}(t) = \langle x''(t), y''(t) \rangle \)	Vector
Slope	\( \frac{dy}{dx} = \frac{y'(t)}{x'(t)} \)	Scalar
Distance	\( \int_a^b \\|\mathbf{v}(t)\\|\, dt \)	Scalar

Worked practice: Practice Problems, Problem 4 — applied

Work through this original lesson problem, then reveal the solution and illustrations.

A drone follows the path x(t) = 4t and y(t) = 12t − 4.9t² (meters, with t in seconds) for t ≥ 0 until it hits the ground. (a) Find the velocity vector v(t). (b) Determine the time when the drone reaches its maximum height. (c) Set up, but do not evaluate, an integral that gives the total distance the drone travels from launch until it returns to y = 0.

Show answer and explanation

Point 1 (1 pt): Velocity vector. x′(t) = 4, y′(t) = 12 − 9.8t. So v(t) = ⟨4, 12 − 9.8t⟩. Point 2 (1 pt): Maximum height occurs when y′(t) = 0. Set 12 − 9.8t = 0, giving t = 12/9.8 = 60/49 ≈ 1.224 seconds. Point 3 (1 pt): Verify this is a maximum. Since y″(t) = −9.8 < 0, the critical point is indeed a maximum. Point 4 (1 pt): Find the time the drone returns to y = 0. Set y(t) = 12t − 4.9t² = t(12 − 4.9t) = 0, giving t = 0 (launch) and t = 12/4.9 = 120/49 ≈ 2.449 seconds. Point 5 (1 pt): Correct arc length integral setup. Distance = ∫₀^(120/49) √[(4)² + (12 − 9.8t)²] dt = ∫₀^(120/49) √[16 + (12 − 9.8t)²] dt. This integral represents the total path length of the drone's parabolic trajectory.

Ballistic trajectory with same-scale velocity arrows (4,12), (4,0), and (4,−12) at launch, peak and return height. — In the idealized unpowered ballistic phase, horizontal velocity stays constant while vertical velocity reaches zero at the apex.

Ballistic motion over 0≤t≤120/49. The speed integral gives path length about 18.46 m; endpoint displacement has magnitude about 9.80 m.

Question 1 of 5

A particle moves with position \( x(t) = t^3 \) and \( y(t) = t^2 \). What is the speed of the particle at \( t = 2 \)?

First, \( x'(t)=3t^2 \) and \( y'(t)=2t \). At \( t=2 \): \( x'(2)=12 \), \( y'(2)=4 \). Speed \( = \sqrt{12^2+4^2} = \sqrt{144+16} = \sqrt{160} = 4\sqrt{10} \). Choice A gives the velocity vector, not the speed scalar.

Question 2 of 5

For the parametric curve \( x(t) = \sin t \), \( y(t) = \cos(2t) \), what is \( \frac{dy}{dx} \) at \( t = \frac{\pi}{6} \)?

We need \( \frac{dy/dt}{dx/dt} \). Here \( \frac{dy}{dt} = -2\sin(2t) \) and \( \frac{dx}{dt} = \cos t \). At \( t = \pi/6 \): \( \frac{dy}{dt} = -2\sin(\pi/3) = -2 \cdot \frac{\sqrt{3}}{2} = -\sqrt{3} \), and \( \frac{dx}{dt} = \cos(\pi/6) = \frac{\sqrt{3}}{2} \). So \( \frac{dy}{dx} = \frac{-\sqrt{3}}{\sqrt{3}/2} = -\sqrt{3} \cdot \frac{2}{\sqrt{3}} = -2 \).

Question 3 of 5

A particle has velocity \( \mathbf{v}(t) = \langle 4t, 3 \rangle \) and position \( \mathbf{r}(0) = \langle 1, -2 \rangle \). What is \( \mathbf{r}(2) \)?

Integrate: \( x(t) = \int 4t\, dt = 2t^2 + C_1 \). \( x(0) = 1 \Rightarrow C_1 = 1 \). So \( x(t) = 2t^2 + 1 \). \( y(t) = \int 3\, dt = 3t + C_2 \). \( y(0) = -2 \Rightarrow C_2 = -2 \). So \( y(t) = 3t - 2 \). At \( t = 2 \): \( x(2) = 2(4)+1 = 9 \), \( y(2) = 6-2 = 4 \). Thus \( \mathbf{r}(2) = \langle 9, 4 \rangle \).

Question 4 of 5

Which integral gives the total distance traveled by a particle with velocity \( \mathbf{v}(t) = \langle e^t, t^2 \rangle \) from \( t = 0 \) to \( t = 3 \)?

Total distance = \( \int_a^b \|\mathbf{v}(t)\|\, dt = \int_0^3 \sqrt{(e^t)^2 + (t^2)^2}\, dt = \int_0^3 \sqrt{e^{2t}+t^4}\, dt \). Choice A integrates a vector (gives displacement). Choice C adds components instead of using the Pythagorean theorem. Choice D takes the magnitude of the displacement vector.

Question 5 of 5

A particle has \( \mathbf{v}(t) = \langle 2, -6t \rangle \) and \( \mathbf{a}(t) = \langle 0, -6 \rangle \). At \( t = 1 \), is the particle speeding up or slowing down?

At \( t=1 \): \( \mathbf{v}(1) = \langle 2, -6 \rangle \) and \( \mathbf{a}(1) = \langle 0, -6 \rangle \). The dot product is \( (2)(0) + (-6)(-6) = 0 + 36 = 36 > 0 \). Since \( \mathbf{v} \cdot \mathbf{a} > 0 \), acceleration has a positive component in the direction of motion, so the particle is speeding up.

Motion ProblemsParametric & Vector-Valued Functions

What Are Parametric Equations?

Visualizing a Parametric Path

Vector-Valued Functions

Core Motion Formulas

Slope: \( \frac{dy}{dx} \) for Parametric Curves

Speed vs. Velocity

How to Solve Motion Problems

Example 1: Finding Velocity & Speed

Example 2: Total Distance Traveled

Example 3: Position from Velocity

Displacement vs. Distance

When Is a Particle Speeding Up?

Position via Definite Integrals

Common Mistakes to Avoid

Summary of Key Formulas

Worked practice: Practice Problems, Problem 4 — applied

Question 1 of 5

Question 2 of 5

Question 3 of 5

Question 4 of 5

Question 5 of 5

Motion Problems
Parametric & Vector-Valued Functions