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AP Calculus BC 9.6 Solving Motion Problems Using Parametric and Vector- Valued Functions Study Notes - New Syllabus

AP Calculus BC 9.6 Solving Motion Problems Using Parametric and Vector- Valued Functions Study Notes- New syllabus

AP Calculus BC 9.6 Solving Motion Problems Using Parametric and Vector- Valued Functions Study Notes – AP Calculus BC-  per latest AP Calculus BC Syllabus.

LEARNING OBJECTIVE

  • Solving an initial value problem allows us to determine an expression for the position of a particle moving in the plane.

Key Concepts: 

  • Solving Motion Problems Using Parametric and Vector-Valued Functions

AP Calculus BC-Concise Summary Notes- All Topics

Solving Motion Problems Using Parametric and Vector-Valued Functions

Solving Motion Problems Using Parametric and Vector-Valued Functions

Motion in a plane can be described using either parametric equations or vector-valued functions. These methods allow us to model movement in two or three dimensions and to compute position, velocity, and acceleration as functions of time.

Position Functions

  • In parametric form: $ x = f(t), \quad y = g(t) $ Here, \(t\) represents time, \(x\) and \(y\) describe the coordinates of the moving object.
  • In vector-valued form: $ \mathbf{r}(t) = \langle f(t), g(t) \rangle \quad \text{(2D)}, \quad \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle \quad \text{(3D)} $ The vector \(\mathbf{r}(t)\) gives the position of the particle at time \(t\).

Velocity Functions

  • Velocity is the time derivative of position: $ \mathbf{v}(t) = \mathbf{r}'(t) = \left\langle \dfrac{dx}{dt}, \dfrac{dy}{dt} \right\rangle $
  • The magnitude of velocity is called speed: $ \text{Speed} = |\mathbf{v}(t)| = \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2} $
  • Velocity gives both the speed and the direction of motion.

Acceleration Functions

  • Acceleration is the time derivative of velocity: $ \mathbf{a}(t) = \mathbf{v}'(t) = \left\langle \dfrac{d^2x}{dt^2}, \dfrac{d^2y}{dt^2} \right\rangle $
  • Acceleration describes how the velocity changes over time.

Speed, Distance, and Displacement

  • Displacement Vector: $ \mathbf{D} = \mathbf{r}(t_2) – \mathbf{r}(t_1) $ Represents the change in position from \(t_1\) to \(t_2\).
  • Total Distance Traveled: $ \text{Distance} = \int_{t_1}^{t_2} |\mathbf{v}(t)| \, dt $ This measures the total path length regardless of direction.
  • Speed at a Given Time: $ |\mathbf{v}(t)| = \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2} $

Motion in Three Dimensions

  • In 3D, position is given by: $ \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle $
  • Velocity and acceleration are: $ \mathbf{v}(t) = \langle x'(t), y'(t), z'(t) \rangle, \quad \mathbf{a}(t) = \langle x”(t), y”(t), z”(t) \rangle $
  • All formulas for speed, displacement, and distance apply similarly in three dimensions.

Tangential and Normal Components of Acceleration

  • Acceleration can be decomposed into two perpendicular components: $ a_T = \frac{d}{dt}|\mathbf{v}(t)| \quad \text{(tangential acceleration)} $ $ a_N = \sqrt{|\mathbf{a}(t)|^2 – a_T^2} \quad \text{(normal acceleration)} $
  • \(a_T\) changes the speed of the particle, \(a_N\) changes its direction.

 Unit Tangent and Normal Vectors

  • Unit Tangent Vector: $ \mathbf{T}(t) = \frac{\mathbf{v}(t)}{|\mathbf{v}(t)|} $ Represents the direction of motion.
  • Unit Normal Vector: $ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} $ Points towards the center of curvature of the path.

Practical Uses

  • Modeling projectile motion
  • Tracking satellites or spacecraft trajectories
  • Simulating car motion on curved roads
  • Analyzing roller coaster paths
  • Studying fluid particle motion in physics

Example

The position of a particle is given by \(\mathbf{r}(t) = \langle t^{2}, \; 3t, \; 2t^{3} \rangle\) m. Find the velocity vector and the speed at \(t = 2\) s.

▶️ Answer/Explanation

Velocity:

$ \mathbf{v}(t) = \langle 2t, \; 3, \; 6t^{2} \rangle $

At \(t = 2\): \(\mathbf{v}(2) = \langle 4, \; 3, \; 24 \rangle\)

Speed:

$ |\mathbf{v}(2)| = \sqrt{4^{2} + 3^{2} + 24^{2}} = \sqrt{16 + 9 + 576} = \sqrt{601} $

So, \(\mathbf{v}(2) = \langle 4, 3, 24 \rangle\) m/s, speed = \(\sqrt{601} \ \text{m/s}\).

Example

The position of a particle in the plane is \(\mathbf{r}(t) = \langle \sin t, \; \cos t \rangle\) m, where \(t\) is in seconds. Find the distance traveled from \(t = 0\) to \(t = \pi\).

▶️ Answer/Explanation

Velocity:

$ \mathbf{v}(t) = \langle \cos t, \; -\sin t \rangle $

Speed:

$ |\mathbf{v}(t)| = \sqrt{\cos^{2} t + \sin^{2} t} = 1 $

Distance:

$ \text{Distance} = \int_{0}^{\pi} 1 \, dt = \pi \ \text{m} $

Example

A particle’s position is \(\mathbf{r}(t) = \langle t, \; t^{2} \rangle\) m. Find \(a_{T}\) and \(a_{N}\) at \(t = 1\).

▶️ Answer/Explanation

Velocity:

$ \mathbf{v}(t) = \langle 1, \; 2t \rangle $

Acceleration:

$ \mathbf{a}(t) = \langle 0, \; 2 \rangle $

Speed at \(t = 1\):

$ |\mathbf{v}(1)| = \sqrt{1^{2} + (2)^{2}} = \sqrt{5} $

Tangential component:

$ a_{T} = \frac{\mathbf{v} \cdot \mathbf{a}}{|\mathbf{v}|} = \frac{1 \cdot 0 + 2 \cdot 2}{\sqrt{5}} = \frac{4}{\sqrt{5}} $

Normal component:

$ a_{N} = \frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|} = \frac{|(1)(2) – (2)(0)|}{\sqrt{5}} = \frac{2}{\sqrt{5}} $

Thus, \(a_{T} = \frac{4}{\sqrt{5}}\) m/s² and \(a_{N} = \frac{2}{\sqrt{5}}\) m/s².

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