Problem 41 - Calculus 3

Question

Compute the derivatives of the vector-valued functions. $$\mathbf{r}(t)=t^{3} \mathbf{i}+3 t^{2} \mathbf{j}+\frac{t^{3}}{6} \mathbf{k}$$

Step-by-Step Solution

Answer

The derivative is \( \mathbf{r}'(t) = 3t^2 \mathbf{i} + 6t \mathbf{j} + \frac{t^2}{2} \mathbf{k} \).

Step 1: Understand the Problem Statement

You are given a vector-valued function \( \mathbf{r}(t) = t^3 \mathbf{i} + 3t^2 \mathbf{j} + \frac{t^3}{6} \mathbf{k} \). The task is to compute the derivative of this function with respect to the variable \( t \).

Step 2: Differentiate Each Component Separately

Differentiate each component of the vector-valued function separately. - The derivative of \( t^3 \) with respect to \( t \) is \( 3t^2 \).- The derivative of \( 3t^2 \) with respect to \( t \) is \( 6t \).- The derivative of \( \frac{t^3}{6} \) with respect to \( t \) is \( \frac{3t^2}{6} = \frac{t^2}{2} \).

Step 3: Combine the Derivatives into Vector Form

Combine the derivatives of the separate components to form the derivative of the vector-valued function:\[ \mathbf{r}'(t) = 3t^2 \mathbf{i} + 6t \mathbf{j} + \frac{t^2}{2} \mathbf{k} \]

Step 4: Final Check

Ensure each component was differentiated correctly. All derivatives match their respective calculations:- \( 3t^2 \mathbf{i} \) from \( t^3 \mathbf{i} \)- \( 6t \mathbf{j} \) from \( 3t^2 \mathbf{j} \)- \( \frac{t^2}{2} \mathbf{k} \) from \( \frac{t^3}{6} \mathbf{k} \)

Key Concepts

Derivative of Vector-Valued Functions

In vector calculus, a vector-valued function represents a function where each input is mapped to a vector output. These functions are commonly used to describe physical phenomena, such as the trajectory of moving objects. For a given vector-valued function like \( \mathbf{r}(t) = t^3 \mathbf{i} + 3t^2 \mathbf{j} + \frac{t^3}{6} \mathbf{k} \), the goal is to find its derivative. This process involves taking the derivative of each component function separately, as they often represent different dimensions or directions in space. Calculating derivatives helps us understand how the function changes with respect to its input variable, \( t \), offering insights into rates of change such as velocity when dealing with motion.

Differentiation Rules

Differentiation is a key operation in calculus that involves finding the derivative, or the rate of change, of a function. For vector-valued functions, differentiation rules are applied to each of their components individually. Here are some common rules:

• The power rule states that the derivative of \( t^n \) is \( nt^{n-1} \).
• The constant multiple rule allows constants to be factored out before taking a derivative.

Applying these rules, the function \( \mathbf{r}(t) = t^3 \mathbf{i} + 3t^2 \mathbf{j} + \frac{t^3}{6} \mathbf{k} \) is differentiated term by term:

• The \( t^3 \mathbf{i} \) term differentiates to \( 3t^2 \mathbf{i} \).
• The \( 3t^2 \mathbf{j} \) term differentiates to \( 6t \mathbf{j} \) using the constant multiple rule.
• Finally, the term \( \frac{t^3}{6} \mathbf{k} \) becomes \( \frac{t^2}{2} \mathbf{k} \).

Thus, mastering these rules simplifies finding the rate of change for each part of vector-valued functions.

Vector Calculus Operations

Vector calculus operations often involve differentiating and integrating vector-valued functions. These operations are fundamental in understanding complex behaviors in physics and engineering.
One of the primary operations, as shown in this exercise, is differentiation in vector calculus. It involves:

• Differentiating each component of the vector function separately.
• Reconstructing a new vector from these individual derivatives.

This method ensures that each direction (i.e., \( \mathbf{i} \), \( \mathbf{j} \), \( \mathbf{k} \)) is accounted for mathematically. The unified approach to finding the derivative ensures consistency across calculations. Not only does it show how the vector changes over time, but it also aids in calculating integrals of vector fields, finding tangent vectors for curves, and solving real-world problems involving motion and fluid dynamics. Breaking complex vector expressions into simpler components streamlines understanding and solving advanced calculus problems efficiently.