Unit 9.6 – Solving Motion Problems Using Parametric and Vector-Valued Functions BC ONLY

AP® Calculus BC | Complete Motion Analysis

Why This Matters: This is where everything comes together! Motion problems combine position, velocity, acceleration, speed, distance, and direction into real-world scenarios. Whether it's a projectile, a particle on a curve, or planetary motion, you'll use parametric equations and vector-valued functions to analyze motion completely. This is THE major application topic on BC exams!

🎯 The Core Motion Formulas

Position, Velocity, Acceleration

Position Vector:

\[ \vec{r}(t) = \langle x(t), y(t) \rangle \]

The location of the particle at time \(t\)

Velocity Vector:

\[ \vec{v}(t) = \vec{r}'(t) = \left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle \]

Direction and rate of motion

Acceleration Vector:

\[ \vec{a}(t) = \vec{v}'(t) = \vec{r}''(t) = \left\langle \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \right\rangle \]

Rate of change of velocity

Speed (Scalar):

\[ \text{Speed} = |\vec{v}(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \]

Magnitude of velocity (always ≥ 0)

🧭 Direction of Motion

ANALYZING DIRECTION

Horizontal Motion:

Moving right: \(\frac{dx}{dt} > 0\)
Moving left: \(\frac{dx}{dt} < 0\)
No horizontal motion: \(\frac{dx}{dt} = 0\)

Vertical Motion:

Moving up: \(\frac{dy}{dt} > 0\)
Moving down: \(\frac{dy}{dt} < 0\)
No vertical motion: \(\frac{dy}{dt} = 0\)

Particle at Rest:

Particle is at rest when \(\vec{v}(t) = \vec{0}\), which means:

\[ \frac{dx}{dt} = 0 \quad \text{AND} \quad \frac{dy}{dt} = 0 \]

📏 Distance vs. Displacement

Critical Distinction

Displacement (Vector):

Change in position from \(t = a\) to \(t = b\):

\[ \vec{r}(b) - \vec{r}(a) \]

Can be negative in component form!

Total Distance (Scalar):

Actual path length traveled:

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

Always ≥ 0 (arc length formula!)

⚠️ KEY DIFFERENCE: Displacement can cancel out (go forward then backward), but distance never does. Distance is always the integral of speed!

🚀 Projectile Motion

Standard Projectile Formulas

Setup:

Initial position: \((x_0, y_0)\)
Initial velocity magnitude: \(v_0\) at angle \(\theta\)
Acceleration due to gravity: \(\vec{a} = \langle 0, -g \rangle\) where \(g = 9.8 \, \text{m/s}^2\) or \(32 \, \text{ft/s}^2\)

Velocity:

\[ \vec{v}(t) = \langle v_0\cos\theta, v_0\sin\theta - gt \rangle \]

Position:

\[ \vec{r}(t) = \left\langle x_0 + v_0t\cos\theta, y_0 + v_0t\sin\theta - \frac{1}{2}gt^2 \right\rangle \]

Key Times:

Maximum height: When \(\frac{dy}{dt} = 0\)
Time of flight: When \(y(t) = 0\) (returns to ground)
Range: Horizontal distance when \(y(t) = 0\)

📖 Comprehensive Worked Examples

Example 1: Particle Motion Analysis

Problem: A particle moves with position \(\vec{r}(t) = \langle t^2 - 4t, t^3 - 6t^2 + 9t \rangle\) for \(t \geq 0\).

a) When is the particle at rest?

b) Find the distance traveled from \(t = 0\) to \(t = 2\).

Solution:

Part a: Find when particle is at rest

Velocity:

\[ \vec{v}(t) = \langle 2t - 4, 3t^2 - 12t + 9 \rangle \]

At rest when both components = 0:

\[ 2t - 4 = 0 \quad \Rightarrow \quad t = 2 \]

\[ 3t^2 - 12t + 9 = 0 \quad \Rightarrow \quad t^2 - 4t + 3 = 0 \quad \Rightarrow \quad t = 1, 3 \]

Need BOTH to be 0 simultaneously: No common value!

The particle is never at rest.

Part b: Distance traveled

\[ \text{Distance} = \int_0^2 |\vec{v}(t)| \, dt = \int_0^2 \sqrt{(2t-4)^2 + (3t^2-12t+9)^2} \, dt \]

(This would typically be a calculator problem)

Example 2: Projectile Motion

Problem: A projectile is launched from ground level with initial speed 40 m/s at an angle of 60° above horizontal. Use \(g = 10 \, \text{m/s}^2\).

a) Find the position function.

b) When does it reach maximum height?

c) What is the range?

Part a: Position function

Initial velocity components:

\[ v_x = 40\cos 60° = 40 \cdot \frac{1}{2} = 20 \, \text{m/s} \]

\[ v_y = 40\sin 60° = 40 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3} \, \text{m/s} \]

Position:

\[ \vec{r}(t) = \left\langle 20t, 20\sqrt{3}t - 5t^2 \right\rangle \]

Part b: Maximum height

Occurs when \(\frac{dy}{dt} = 0\):

\[ 20\sqrt{3} - 10t = 0 \quad \Rightarrow \quad t = 2\sqrt{3} \, \text{seconds} \]

Part c: Range

Find when \(y = 0\):

\[ 20\sqrt{3}t - 5t^2 = 0 \quad \Rightarrow \quad t(20\sqrt{3} - 5t) = 0 \]

\[ t = 0 \text{ or } t = 4\sqrt{3} \]

Range = \(x(4\sqrt{3}) = 20 \cdot 4\sqrt{3} = 80\sqrt{3}\) meters

Example 3: Direction Changes

Problem: For \(\vec{r}(t) = \langle t^3 - 3t, t^2 - 4 \rangle\), when is the particle moving to the right? Moving up?

Find velocity:

\[ \vec{v}(t) = \langle 3t^2 - 3, 2t \rangle \]

Moving right when \(\frac{dx}{dt} > 0\):

\[ 3t^2 - 3 > 0 \quad \Rightarrow \quad t^2 > 1 \quad \Rightarrow \quad t > 1 \text{ or } t < -1 \]

For \(t \geq 0\): moving right when \(t > 1\)

Moving up when \(\frac{dy}{dt} > 0\):

\[ 2t > 0 \quad \Rightarrow \quad t > 0 \]

Moving up for all \(t > 0\)

Example 4: Speed at a Specific Time

Problem: Find the speed of a particle at \(t = 3\) if \(\vec{r}(t) = \langle e^t, \ln(t+1) \rangle\).

Find velocity:

\[ \vec{v}(t) = \left\langle e^t, \frac{1}{t+1} \right\rangle \]

At \(t = 3\):

\[ \vec{v}(3) = \left\langle e^3, \frac{1}{4} \right\rangle \]

Calculate speed:

\[ \text{Speed} = \sqrt{(e^3)^2 + \left(\frac{1}{4}\right)^2} = \sqrt{e^6 + \frac{1}{16}} \]

📊 Complete Motion Analysis Reference

Motion Problem Checklist
Question Type	What to Find	Formula/Method
Position	Where is particle?	\(\vec{r}(t) = \langle x(t), y(t) \rangle\)
Velocity	How fast and which way?	\(\vec{v}(t) = \vec{r}'(t)\)
Speed	How fast (no direction)?	\(\|\vec{v}(t)\| = \sqrt{(dx/dt)^2 + (dy/dt)^2}\)
Acceleration	Rate of velocity change?	\(\vec{a}(t) = \vec{v}'(t)\)
At Rest	When stopped?	\(dx/dt = 0\) AND \(dy/dt = 0\)
Moving Right	When?	\(dx/dt > 0\)
Moving Up	When?	\(dy/dt > 0\)
Distance Traveled	Total path length?	\(\int_a^b \|\vec{v}(t)\|\,dt\)
Displacement	Net position change?	\(\vec{r}(b) - \vec{r}(a)\)

💡 Essential Tips & Strategies

✅ Success Strategies:

Always find velocity first: It's the key to most motion questions
Speed = magnitude: Always use \(|\vec{v}(t)|\), never negative
At rest needs BOTH components zero: Not just one!
Direction = sign of component: Positive = right/up, negative = left/down
Distance ≠ displacement: Distance uses integral of speed
For projectiles: Memorize the standard formulas
Calculator allowed: For distance integrals usually
Check units: m, m/s, m/s² etc.

🔥 Common AP® Questions:

"When is the particle moving left?" → Find when \(dx/dt < 0\)
"What is the speed at \(t = 2\)?" → Find \(|\vec{v}(2)|\)
"Find total distance traveled" → \(\int |\vec{v}(t)|\,dt\)
"When does the particle change direction?" → When velocity component changes sign
"Find position given velocity and initial position" → Integrate \(\vec{v}\), use initial condition

❌ Common Mistakes to Avoid

Mistake 1: Confusing speed (scalar) with velocity (vector)
Mistake 2: Saying particle at rest when only ONE component of velocity is 0
Mistake 3: Using displacement formula for distance traveled
Mistake 4: Forgetting absolute value for speed
Mistake 5: Not checking BOTH components for direction
Mistake 6: Wrong sign for gravity (should be negative)
Mistake 7: Mixing up degrees and radians for angles
Mistake 8: Not simplifying \(\cos 60° = 1/2\), etc.
Mistake 9: Forgetting to include initial position when integrating
Mistake 10: Calculator in wrong mode (degrees vs radians)

📝 Practice Problems

Solve these motion problems:

For \(\vec{r}(t) = \langle t^2, t^3 - 3t \rangle\), when is the particle at rest?
Find speed at \(t = 1\) for \(\vec{r}(t) = \langle 2t, 4t^2 \rangle\).
A projectile launched at 30 m/s at 45° angle. Find max height time. (\(g = 10\))
For \(\vec{r}(t) = \langle \sin t, \cos t \rangle\), find distance from \(t = 0\) to \(t = \pi\).

Answers:

Never (velocity components never both zero at same time)
Speed = \(\sqrt{68} = 2\sqrt{17}\)
\(t = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}\) seconds
\(\pi\) (semicircle of radius 1)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Show velocity calculation: Always show \(\vec{v}(t) = \vec{r}'(t)\)
For "at rest": Show BOTH components equal zero
For speed: Show magnitude calculation
For distance: Show integral of speed
For direction: State which component and show inequality
Use proper notation: Vectors with \(\langle \, \rangle\) or arrows
Include units: If given in problem
Justify conclusions: "Moving right because \(dx/dt > 0\)"

💯 Exam Strategy:

Read carefully: What are you given? Position, velocity, or acceleration?
Identify what you need to find
Always start by finding velocity if not given
For speed: take magnitude
For direction: check sign of appropriate component
For distance: integrate speed (magnitude)
For at rest: both velocity components must be zero
Show all work clearly

⚡ Quick Reference Guide

MOTION PROBLEM ESSENTIALS

The Big Three:

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

Speed & Distance:

Speed = \(|\vec{v}(t)| = \sqrt{(dx/dt)^2 + (dy/dt)^2}\)
Distance = \(\int_a^b |\vec{v}(t)|\,dt\)

Direction Tests:

Moving right: \(dx/dt > 0\)
Moving up: \(dy/dt > 0\)
At rest: \(dx/dt = 0\) AND \(dy/dt = 0\)

Remember:

Velocity is a vector, speed is scalar
Distance uses integral of speed
At rest requires BOTH components zero

Master Parametric Motion Problems! This synthesizes everything: position \(\vec{r}(t)\), velocity \(\vec{v}(t) = \vec{r}'(t)\), acceleration \(\vec{a}(t) = \vec{r}''(t)\). Speed (scalar) = \(|\vec{v}(t)| = \sqrt{(dx/dt)^2+(dy/dt)^2}\) always ≥ 0. Direction: moving right when \(dx/dt > 0\), up when \(dy/dt > 0\). At rest when BOTH velocity components = 0 simultaneously. Distance traveled = \(\int|\vec{v}(t)|\,dt\) (integral of speed). Displacement = \(\vec{r}(b) - \vec{r}(a)\) (can be negative in components). Projectile motion: \(\vec{a} = \langle 0, -g\rangle\) gives parabolic path. Common questions: when moving left/right/up/down, speed at time t, distance traveled, when at rest. This is THE major application on BC exams—appears every year! Practice diverse motion scenarios until automatic! 🎯✨