A vector-valued function assigns a vector to each parameter value $t$, typically representing the position of a moving particle: $\mathbf{r}(t) = \langle x(t), y(t) \rangle$. The derivative gives velocity, and the second derivative gives acceleration.
🎯 In this section you will learn
- What vector-valued functions are and how to interpret them
- How to find the derivative of a vector-valued function
- Velocity, speed, and acceleration of a moving particle
- How to find the tangent vector and parametric equations of the tangent line
📌 Definition and Notation
$$ \mathbf{r}(t) = \langle x(t), y(t) \rangle = x(t)\mathbf{i} + y(t)\mathbf{j} $$
A vector-valued function representing position at time $t$
💡 Physical Interpretation
- $\mathbf{r}(t)$ = position vector of a particle at time $t$
- $\mathbf{v}(t) = \mathbf{r}'(t)$ = velocity vector
- $||\mathbf{v}(t)||$ = speed (scalar)
- $\mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t)$ = acceleration vector
📌 Derivatives of Vector-Valued Functions
$$ \mathbf{r}'(t) = \langle x'(t), y'(t) \rangle $$
$$ \mathbf{a}(t) = \mathbf{r}''(t) = \langle x''(t), y''(t) \rangle $$
Given $\mathbf{r}(t) = \langle t^2, t^3 \rangle$, find $\mathbf{v}(t)$ and $\mathbf{a}(t)$.
①
Velocity
$\mathbf{v}(t) = \mathbf{r}'(t) = \langle 2t, 3t^2 \rangle$
②
Acceleration
$\mathbf{a}(t) = \mathbf{v}'(t) = \langle 2, 6t \rangle$
For $\mathbf{r}(t) = \langle 2\cos t, 2\sin t \rangle$, find the speed at $t = \pi/4$.
①
Find velocity
$\mathbf{v}(t) = \langle -2\sin t, 2\cos t \rangle$
②
Speed formula
$||\mathbf{v}(t)|| = \sqrt{(-2\sin t)^2 + (2\cos t)^2} = \sqrt{4\sin^2 t + 4\cos^2 t} = \sqrt{4} = 2$
③
Speed is constant
At $t = \pi/4$, speed $= 2$
📌 Tangent Vectors and Tangent Lines
$$ \text{Tangent vector at } t = \mathbf{r}'(t) \quad \text{(direction of motion)} $$
$$ \text{Unit tangent vector: } \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} $$
For $\mathbf{r}(t) = \langle t, t^2 \rangle$, find the tangent line at $t = 1$.
①
Find the point
$\mathbf{r}(1) = \langle 1, 1 \rangle$ → point $(1, 1)$
②
Find the tangent vector
$\mathbf{r}'(t) = \langle 1, 2t \rangle$, $\mathbf{r}'(1) = \langle 1, 2 \rangle$
③
Parametric equations of tangent line
$x = 1 + t$, $y = 1 + 2t$
④
Cartesian equation
Eliminate $t$: $t = x-1$, so $y = 1 + 2(x-1) = 2x - 1$
A projectile is launched from ground level with initial speed $v_0 = 20$ m/s at an angle $\theta = 30^\circ$. Find its position vector, velocity, and acceleration. (Ignore air resistance, $g = 9.8$ m/s²)
①
Initial velocity components
$v_{0x} = v_0\cos\theta = 20\cos30^\circ = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3}$
$v_{0y} = v_0\sin\theta = 20\sin30^\circ = 20 \cdot \frac{1}{2} = 10$
②
Position vector
$\mathbf{r}(t) = \langle v_{0x} t, v_{0y} t - \frac{1}{2}gt^2 \rangle = \langle 10\sqrt{3} t, 10t - 4.9t^2 \rangle$
③
Velocity
$\mathbf{v}(t) = \langle 10\sqrt{3}, 10 - 9.8t \rangle$
④
Acceleration
$\mathbf{a}(t) = \langle 0, -9.8 \rangle$
⚠️ Common Mistakes
- Confusing speed with velocity: Speed is the magnitude of velocity, a scalar; velocity is a vector.
- Forgetting to differentiate component-wise: $\mathbf{r}'(t) = \langle x'(t), y'(t) \rangle$.
- Unit tangent vector: $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||}$ — you must divide by the speed.
🔍 Key Takeaways
- $\mathbf{r}(t) = \langle x(t), y(t) \rangle$ represents position.
- Velocity: $\mathbf{v}(t) = \mathbf{r}'(t)$. Speed: $||\mathbf{v}(t)||$. Acceleration: $\mathbf{a}(t) = \mathbf{v}'(t)$.
- Tangent vector = $\mathbf{r}'(t)$. Unit tangent vector = $\frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||}$.
- Projectile motion: $\mathbf{r}(t) = \langle v_{0x} t, v_{0y} t - \frac{1}{2}gt^2 \rangle$.
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