1.4 Differentiation of Vectors and Dot notation

Introduction

When a vector depends on a parameter (most commonly time $t$), we can differentiate it with respect to that parameter. The derivative of a vector function is itself a vector, obtained by differentiating each component separately. This concept is fundamental in physics for describing motion, where position, velocity, and acceleration are all vector quantities that change with time.

Definition of the Derivative of a Vector:

If $\mathbf{r}(t) = x(t)\hat{x} + y(t)\hat{y} + z(t)\hat{z}$ is a vector function of a parameter $t$, then the derivative of $\mathbf{r}$ with respect to $t$ is defined as:

\frac{d\mathbf{r}}{dt} = \lim_{\Delta t \to 0} \frac{\mathbf{r}(t + \Delta t) - \mathbf{r}(t)}{\Delta t}.

This limit, when it exists, gives the derivative vector $\frac{d\mathbf{r}}{dt}$.

Component-Wise Differentiation

The derivative of a vector function is computed by differentiating each component separately:

Component-Wise Differentiation

Given a time-dependent vector $\mathbf{A}(t) = A_x(t)\hat{x} + A_y(t)\hat{y} + A_z(t)\hat{z}$, we can find its time derivative by differentiating each component separately with respect to time:

\frac{d\mathbf{A}}{dt} = \frac{dA_x}{dt}\hat{x} + \frac{dA_y}{dt}\hat{y} + \frac{dA_z}{dt}\hat{z}.

Since the unit vectors $\hat{x}$, $\hat{y}$, and $\hat{z}$ are constant (they don't depend on $t$), they can be treated as constants when differentiating.

Dot Notation

In physics, it is common to use dot notation to denote derivatives with respect to time.

Dot Notation for Time Derivatives:

For a function $f(t)$ that depends on time, we use the following notation:

First derivative: $\dot{f} = \frac{df}{dt}$ (read as "f dot")
Second derivative: $\ddot{f} = \frac{d^2f}{dt^2}$ (read as "f double dot")
Third derivative: $\dddot{f} = \frac{d^3f}{dt^3}$ (read as "f triple dot")

For vectors, we apply the same notation:

$\dot{\mathbf{r}} = \frac{d\mathbf{r}}{dt}$ (first derivative of position vector)
$\ddot{\mathbf{r}} = \frac{d^2\mathbf{r}}{dt^2}$ (second derivative of position vector)

Why Use Dot Notation?

Compactness: $\dot{x}$ is shorter and cleaner than $\frac{dx}{dt}$
Convention: Standard notation in physics, especially in mechanics
Efficiency: Reduces clutter in equations, making them easier to read and write
Historical: Used by Newton and has been standard in physics for centuries

One word of caution: dot notation requires you to be organized and have neat handwriting. It is easier to "miss a dot" over a variable than to miss a fully written out derivative.

Example 1: Velocity Vector

Given a time-dependent position vector $\mathbf{r}(t) = x(t)\hat{x} + y(t)\hat{y} + z(t)\hat{z}$, we can find the velocity vector by differentiating each component with respect to time:

\dot{\mathbf{r}} = \dot{x}\hat{x} + \dot{y}\hat{y} + \dot{z}\hat{z}.

This expression may be written explicity as the velocity vector and its components like this:

\mathbf{v} = {v}_x\hat{x} + {v}_y\hat{y} + {v}_z\hat{z},

where $v_x$, $v_y$, and $v_z$ are the components of the velocity vector:

\mathbf{v} = \dot{\mathbf{r}},\;\;\; v_x = \dot{x},\;\;\; v_y = \dot{y},\;\;\; v_z = \dot{z}.

Example 2: Acceleration Vector

We can extend the previous example to acceleration by differentiating the velocity vector with respect to time:

\dot{\mathbf{v}} = \dot{v}_x\hat{x} + \dot{v}_y\hat{y} + \dot{v}_z\hat{z}.

This expression may be written explicity as the acceleration vector and its components like this:

\mathbf{a} = {a}_x\hat{x} + {a}_y\hat{y} + {a}_z\hat{z},

where $a_x$, $a_y$, and $a_z$ are the components of the acceleration vector:

\mathbf{a} = \dot{\mathbf{v}}=\ddot{\mathbf{r}},\;\;\;\;\;\;\; a_x = \dot{v}_x=\ddot{x},\;\;\;\;\;\;\; a_y =\dot{v}_y= \ddot{y},\;\;\;\;\;\;\; a_z = \dot{v}_z= \ddot{z}.

Properties of Vector Differentiation

Rules for Differentiating Vectors:

Sum rule: $\frac{d}{dt}(\mathbf{a} + \mathbf{b}) = \dot{\mathbf{a}}+ \dot{\mathbf{b}}$
Scalar multiplication: $\frac{d}{dt}(c\mathbf{a}) = c\,\dot{\mathbf{a}}$ for constant scalar $c$
Product with scalar function: $\frac{d}{dt}(f(t)\mathbf{a}(t)) = \dot{f}\mathbf{a} + f\dot{\mathbf{a}}$
Dot product: $\frac{d}{dt}(\mathbf{a} \cdot \mathbf{b}) = \dot{\mathbf{a}} \cdot \mathbf{b} + \mathbf{a} \cdot \dot{\mathbf{b}} \;\;\;\;$(the dot product is defined in the next section)
Cross product: $\frac{d}{dt}(\mathbf{a} \times \mathbf{b}) = \dot{\mathbf{a}} \times \mathbf{b} + \mathbf{a} \times \dot{\mathbf{b}} \;\;\;\;$ (the cross product is defined in the next section)
Magnitude: $\frac{d}{dt}|\mathbf{a}| = \frac{\mathbf{a} \cdot \dot{\mathbf{a}}}{|\mathbf{a}|}$ (when $|\mathbf{a}| \neq 0$)

Examples

Example 3: Basic Vector Differentiation

Find $\frac{d\mathbf{r}}{dt}$ where $\mathbf{r}(t) = t^2\hat{x} + 3t\hat{y}+\sin t\hat{z}$.

Solution:

Differentiating each component:

\frac{d\mathbf{r}}{dt} = \frac{d}{dt}(t^2)\hat{x} + \frac{d}{dt}(3t)\hat{y} + \frac{d}{dt}(\sin t)\hat{z}

gives

\frac{d\mathbf{r}}{dt} = 2t\hat{x} + 3\hat{y} + \cos t\hat{z}.

Example 4: Velocity and Acceleration

A particle's position is given by $\mathbf{r}(t) = (2t^3, 5t^2, t)$ meters. Find the velocity and acceleration vectors using dot notation.

Solution:

Velocity:

\mathbf{v} = \dot{\mathbf{r}} = \frac{d\mathbf{r}}{dt} = \left(\frac{d}{dt}(2t^3), \frac{d}{dt}(5t^2), \frac{d}{dt}(t)\right)

\mathbf{v} = (6t^2, 10t, 1) \text{ m/s}

Acceleration:

\mathbf{a} = \ddot{\mathbf{r}} = \frac{d^2\mathbf{r}}{dt^2} = \frac{d\mathbf{v}}{dt} = \left(\frac{d}{dt}(6t^2), \frac{d}{dt}(10t), \frac{d}{dt}(1)\right)

\mathbf{a} = (12t, 10, 0) \text{ m/s}^2

Example 5: Derivative of Dot Product

Verify the product rule for dot products: $\frac{d}{dt}(\mathbf{a} \cdot \mathbf{b}) = \dot{\mathbf{a}} \cdot \mathbf{b} + \mathbf{a} \cdot \dot{\mathbf{b}}$ where $\mathbf{a}(t) = (t, t^2, 1)$ and $\mathbf{b}(t) = (2t, 1, t)$.

Solution:

First, compute $\mathbf{a} \cdot \mathbf{b}$:

\mathbf{a} \cdot \mathbf{b} = (t)(2t) + (t^2)(1) + (1)(t) = 2t^2 + t^2 + t = 3t^2 + t

Taking the derivative:

\frac{d}{dt}(\mathbf{a} \cdot \mathbf{b}) = \frac{d}{dt}(3t^2 + t) = 6t + 1

Now compute the right-hand side. First find the derivatives:

\dot{\mathbf{a}} = (1, 2t, 0), \quad \dot{\mathbf{b}} = (2, 0, 1)

Computing $\dot{\mathbf{a}} \cdot \mathbf{b} + \mathbf{a} \cdot \dot{\mathbf{b}}$:

\dot{\mathbf{a}} \cdot \mathbf{b} = (1)(2t) + (2t)(1) + (0)(t) = 2t + 2t = 4t

\mathbf{a} \cdot \dot{\mathbf{b}} = (t)(2) + (t^2)(0) + (1)(1) = 2t + 1

\dot{\mathbf{a}} \cdot \mathbf{b} + \mathbf{a} \cdot \dot{\mathbf{b}} = 4t + 2t + 1 = 6t + 1

This matches the left-hand side, confirming the product rule.

Example 6: Circular Motion

A particle moves in a circle of radius $R$ with constant angular speed $\omega$. Its position vector is $\mathbf{r}(t) = (R\cos(\omega t), R\sin(\omega t), 0)$. Find the velocity and acceleration using dot notation.

Solution:

Velocity:

\mathbf{v} = \dot{\mathbf{r}} = \frac{d}{dt}(R\cos(\omega t), R\sin(\omega t), 0)

\mathbf{v} = (-R\omega\sin(\omega t), R\omega\cos(\omega t), 0)

Acceleration:

\mathbf{a} = \ddot{\mathbf{r}} = \frac{d\mathbf{v}}{dt} = \frac{d}{dt}(-R\omega\sin(\omega t), R\omega\cos(\omega t), 0)

\mathbf{a} = (-R\omega^2\cos(\omega t), -R\omega^2\sin(\omega t), 0) = -\omega^2\mathbf{r}

This shows that for uniform circular motion, the acceleration points toward the center (centripetal acceleration) with magnitude $R\omega^2$.

Example 7: Derivative of a Unit Vector

Show that if $\hat{\mathbf{r}}$ is a unit vector (i.e., $|\hat{\mathbf{r}}| = 1$), then $\frac{d\hat{\mathbf{r}}}{dt}$ is perpendicular to $\hat{\mathbf{r}}$.

Solution:

Since $|\hat{\mathbf{r}}| = 1$, we have $\hat{\mathbf{r}} \cdot \hat{\mathbf{r}} = 1$. Differentiating both sides:

\frac{d}{dt}(\hat{\mathbf{r}} \cdot \hat{\mathbf{r}}) = \frac{d}{dt}(1) = 0

Using the product rule for dot products:

\frac{d}{dt}(\hat{\mathbf{r}} \cdot \hat{\mathbf{r}}) = \frac{d\hat{\mathbf{r}}}{dt} \cdot \hat{\mathbf{r}} + \hat{\mathbf{r}} \cdot \frac{d\hat{\mathbf{r}}}{dt} = 2\frac{d\hat{\mathbf{r}}}{dt} \cdot \hat{\mathbf{r}} = 0

Therefore, $\frac{d\hat{\mathbf{r}}}{dt} \cdot \hat{\mathbf{r}} = 0$, which means $\frac{d\hat{\mathbf{r}}}{dt}$ is perpendicular to $\hat{\mathbf{r}}$.

Applications in Physics

Vector differentiation is fundamental in physics:

Kinematics: Position $\mathbf{r}(t)$, velocity $\mathbf{v} = \dot{\mathbf{r}}$, and acceleration $\mathbf{a} = \ddot{\mathbf{r}}$ are all related through differentiation.
Newton's Second Law: $\mathbf{F} = m\mathbf{a} = m\ddot{\mathbf{r}}$ relates force to the second derivative of position.
Angular momentum: $\mathbf{L} = \mathbf{r} \times \mathbf{p}$, and its time derivative gives torque: $\boldsymbol{\tau} = \dot{\mathbf{L}}$.
Electric and magnetic fields: Time-varying fields involve derivatives of vector quantities.
Wave equations: Many wave equations involve second derivatives of vector fields.