AP Calculus BC 9.4 Defining and Differentiating Vector- Valued Functions Study Notes - New Syllabus
AP Calculus BC 9.4 Defining and Differentiating Vector- Valued Functions Study Notes- New syllabus
AP Calculus BC 9.4 Defining and Differentiating Vector- Valued Functions Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Derivatives allow us to solve real-world problems involving rates of change.
Key Concepts:
- Defining and Differentiating Vector-Valued Functions
Defining and Differentiating Vector-Valued Functions
Defining and Differentiating Vector-Valued Functions
A vector-valued function is a function whose output is a vector rather than a scalar. In two or three dimensions, the position of a moving object can be described as a vector function of time:
$ \mathbf{r}(t) = \langle x(t), \, y(t) \rangle \quad \text{(2D)},$
$\mathbf{r}(t) = \langle x(t), \, y(t), \, z(t) \rangle \quad \text{(3D)} $
Here, \( x(t) \), \( y(t) \), and \( z(t) \) are scalar functions of the parameter \( t \) (often time), representing the components of the vector along each axis.
Geometric Interpretation:
- The vector \( \mathbf{r}(t) \) represents the position of a particle at time \( t \).
- The derivative \( \mathbf{r}'(t) \) represents the velocity vector, indicating both speed and direction of motion.
- The second derivative \( \mathbf{r}”(t) \) represents the acceleration vector.
Definition of the Derivative:
The derivative of a vector-valued function is defined componentwise. If $ \mathbf{r}(t) = \langle x(t), \, y(t), \, z(t) \rangle $ then $ \mathbf{r}'(t) = \langle x'(t), \, y'(t), \, z'(t) \rangle $
The derivative measures the instantaneous rate of change of the position vector with respect to \( t \).
Magnitude of the Velocity (Speed):
The speed at time \( t \) is the magnitude of the velocity vector:
$ \text{Speed} = \left\| \mathbf{r}'(t) \right\| = \sqrt{ \left( x'(t) \right)^{2} + \left( y'(t) \right)^{2} + \left( z'(t) \right)^{2} } $
Key Notes:
- Differentiation rules for vector-valued functions follow the same rules as scalar calculus — derivatives are computed componentwise.
- Common rules include:
- Sum Rule: \( \dfrac{d}{dt} \left[ \mathbf{u}(t) + \mathbf{v}(t) \right] = \mathbf{u}'(t) + \mathbf{v}'(t) \)
- Scalar Multiple Rule: \( \dfrac{d}{dt} \left[ c \, \mathbf{u}(t) \right] = c \, \mathbf{u}'(t) \)
- Product Rule (scalar × vector): \( \dfrac{d}{dt} \left[ f(t) \, \mathbf{u}(t) \right] = f'(t) \, \mathbf{u}(t) + f(t) \, \mathbf{u}'(t) \)
- Dot Product Rule: \( \dfrac{d}{dt} \left[ \mathbf{u} \cdot \mathbf{v} \right] = \mathbf{u}’ \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}’ \)
- Cross Product Rule: \( \dfrac{d}{dt} \left[ \mathbf{u} \times \mathbf{v} \right] = \mathbf{u}’ \times \mathbf{v} + \mathbf{u} \times \mathbf{v}’ \)
- The concepts of tangent lines, curvature, and motion analysis can be extended naturally to vector-valued functions.
Example
Let \( \mathbf{r}(t) = \langle t^{2}, \, 3t, \, e^{t} \rangle \). Find the velocity, speed, and acceleration at \( t = 1 \).
▶️ Answer/Explanation
Velocity: $ \mathbf{r}'(t) = \langle 2t, \, 3, \, e^{t} \rangle $ At \( t = 1 \): $ \mathbf{v}(1) = \langle 2, \, 3, \, e \rangle $
Speed: $ \|\mathbf{v}(1)\| = \sqrt{ 2^{2} + 3^{2} + e^{2} } = \sqrt{4 + 9 + e^{2}} = \sqrt{13 + e^{2}} $
Acceleration: $ \mathbf{r}”(t) = \langle 2, \, 0, \, e^{t} \rangle $ At \( t = 1 \): $ \mathbf{a}(1) = \langle 2, \, 0, \, e \rangle $
Example
A particle moves so that \( \mathbf{r}(t) = \langle \cos t, \, \sin t, \, t \rangle \). Find its velocity, speed, and acceleration at \( t = \dfrac{\pi}{2} \).
▶️ Answer/Explanation
Velocity: $ \mathbf{r}'(t) = \langle -\sin t, \, \cos t, \, 1 \rangle $ At \( t = \dfrac{\pi}{2} \): $ \mathbf{v}\left( \dfrac{\pi}{2} \right) = \langle -1, \, 0, \, 1 \rangle $
Speed: $ \left\| \mathbf{v} \left( \dfrac{\pi}{2} \right) \right\| = \sqrt{ (-1)^{2} + 0^{2} + 1^{2} } = \sqrt{2} $
Acceleration: $ \mathbf{r}”(t) = \langle -\cos t, \, -\sin t, \, 0 \rangle $ At \( t = \dfrac{\pi}{2} \): $ \mathbf{a}\left( \dfrac{\pi}{2} \right) = \langle 0, \, -1, \, 0 \rangle $