Problem 45 - Calculus Volume 3

Question

Compute the derivatives of the vector-valued functions. \(\mathbf{r}(t)=\mathbf{i}+\mathbf{j}+\mathbf{k}\)

Step-by-Step Solution

Answer

The derivative is \( \mathbf{0} \).

Step 1: Identify the Components of the Vector Function

The vector-valued function given is \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \). This can be expressed as \( \mathbf{r}(t) = (1)\mathbf{i} + (1)\mathbf{j} + (1)\mathbf{k} \), meaning it has constant components for each unit vector in the standard basis of \( \mathbb{R}^3 \).

Step 2: Differentiate Each Component with Respect to \( t \)

Since the components \( 1 \) for \( \mathbf{i}, \mathbf{j}, \text{and } \mathbf{k} \) are constants, their derivatives with respect to \( t \) are zero. Thus: \( \frac{d}{dt}[1] = 0 \), \( \frac{d}{dt}[1] = 0 \), and \( \frac{d}{dt}[1] = 0 \).

Step 3: Formulate the Derivative of the Vector Function

Combine the derivatives of each component to get the derivative of the vector-valued function: \( \frac{d}{dt}[\mathbf{r}(t)] = 0\mathbf{i} + 0\mathbf{j} + 0\mathbf{k} = \mathbf{0} \).

Key Concepts

Vector-valued functions

Vector-valued functions are mathematical entities that assign a vector to each element in their domain, often time or space. This makes them particularly useful in physics and engineering, where they describe quantities like velocity, force, or displacement over time or across regions. Consider a vector-valued function

• Defined as \( \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} \), where \( f, g, \) and \( h \) are scalar functions of \( t \).
• The components \( f, g, \) and \( h \) indicate how the vector changes in direction and magnitude over the parameter \( t \).

If you visualize this in 3D space, each point \( r(t) \) on the curve describes a position vector, guiding you from the origin to the location determined by \( f, g, \) and \( h \). In our exercise, the function \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \) is a simple case with constant unit vectors, meaning the vector does not change over time.

Derivatives of vector functions

When you differentiate vector-valued functions, you are essentially finding how the vector settles into motion over its domain. The derivative captures the rate of change of each vector component with respect to the parameter, such as time \( t \). For a vector-valued function \( \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} \), the derivative is expressed as

• \( \mathbf{r}'(t) = \frac{df}{dt}\mathbf{i} + \frac{dg}{dt}\mathbf{j} + \frac{dh}{dt}\mathbf{k} \).
• Each term represents the derivative of its respective component, giving a new vector that illustrates how the original vector's direction or magnitude is changing.

In the context of our original exercise, the vector components are constants \( 1, 1, 1 \), whose rates of change with respect to \( t \) are zero. Therefore, the derivative of the vector function \( \mathbf{r}(t) \) is \( \mathbf{0} \), indicating no change in the vector. This makes intuitive sense; constant components imply a static, non-moving vector.

Constant vector components

A vector with constant components, such as \( \mathbf{r}(t) = \mathbf{i} + \mathbf{j} + \mathbf{k} \), doesn't change as the parameter \( t \) changes. This consistency highlights several important concepts:

• Non-varying Nature: Regardless of \( t \), each of the vector's components remains the same, embodying fixed magnitudes along each axis.
• Simplifies Calculations: Since derivatives of constants are zero, counting them leads straight to zero, making computations straightforward in problems requiring vector differentiation.
• Applications: In real-world contexts, constant vectors could represent stable forces or static positions, significant for models needing simplification.

In mathematical terms, a vector of constant components can be visualized as a specific, immobile point in space, simplifying the visualization and calculation of such vectors.