Problem 44 - Advanced Engineering Mathematics

Question

In Problems, find the length of the curve traced by the given vector function on the indicated interval. $$ \mathbf{r}(t)=3 t \mathbf{i}+\sqrt{3} t^{2} \mathbf{j}+\frac{2}{3} t^{3} \mathbf{k} ; 0 \leq t \leq 1 $$

Step-by-Step Solution

Answer

The length of the curve is obtained by evaluating \( \int_0^1 \sqrt{9 + 12t^2 + 4t^4} \, dt \).

Step 1: Understand the Vector Function

The vector function given is \( \mathbf{r}(t) = 3t \mathbf{i} + \sqrt{3} t^2 \mathbf{j} + \frac{2}{3}t^3 \mathbf{k} \). The components here are functions of \( t \), the independent variable, which varies from 0 to 1.

Step 2: Compute the Derivative of \(\mathbf{r}(t)\)

To find the rate of change of the vector function, compute its derivative: \( \mathbf{r}'(t) = \frac{d}{dt}(3t) \mathbf{i} + \frac{d}{dt}(\sqrt{3} t^2) \mathbf{j} + \frac{d}{dt}\left(\frac{2}{3}t^3\right) \mathbf{k} \). Thus, \( \mathbf{r}'(t) = 3 \mathbf{i} + 2\sqrt{3} t \mathbf{j} + 2t^2 \mathbf{k} \).

Step 3: Calculate the Magnitude of \( \mathbf{r}'(t) \)

The length of the curve is found by integrating the magnitude of \( \mathbf{r}'(t) \). Compute this magnitude: \( \| \mathbf{r}'(t) \| = \sqrt{(3)^2 + (2\sqrt{3} t)^2 + (2t^2)^2} = \sqrt{9 + 12t^2 + 4t^4} \).

Step 4: Setup the Integral for Arc Length

The arc length \( L \) of the curve from \( t = 0 \) to \( t = 1 \) is given by: \( L = \int_{0}^{1} \| \mathbf{r}'(t) \| \, dt = \int_{0}^{1} \sqrt{9 + 12t^2 + 4t^4} \, dt \).

Step 5: Evaluate the Integral

This integral does not easily simplify, and would typically require numerical methods or a computational tool to evaluate. For this exercise, assume numerical integration is used to compute the integral.

Key Concepts

Vector Calculus

Vector calculus is an extension of ordinary calculus into multiple dimensions using vectors. In our problem, we explore the concept of a vector function, where the vector describes a position in three-dimensional space. Each component of this vector is a function of the parameter \( t \). In this exercise, the vector function provided, \( \mathbf{r}(t) = 3t \mathbf{i} + \sqrt{3} t^2 \mathbf{j} + \frac{2}{3}t^3 \mathbf{k} \), represents a path in space.
This path is traced as \( t \) varies within the specified interval.

• The vector \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) denote the unit vectors in the \( x \), \( y \), and \( z \) directions respectively.
• The coefficients of these unit vectors determine how fast or slow the curve moves along each axis as \( t \) increases.

This basic understanding of vector calculus forms the foundation of analyzing motion along curves in three-dimensional space.

Vector Functions

Vector functions are fundamental in describing curves in vector calculus. A vector function is composed of multiple scalar functions that correspond to different dimensions of space. In our given problem, the vector function \( \mathbf{r}(t) \) describes a curve by assigning a unique vector to each value of \( t \), where \( t \) is the parameter.
Each component of this vector function represents motion along one of the coordinate axes:

• \( 3t \) controls movement along the \( x \)-axis.
• \( \sqrt{3} t^2 \) dictates motion in the \( y \)-direction.
• \( \frac{2}{3}t^3 \) governs the trajectory along the \( z \)-axis.

Whenever the function is differentiated with respect to \( t \), which in this case results in the derivative \( \mathbf{r}'(t) = 3 \mathbf{i} + 2\sqrt{3} t \mathbf{j} + 2t^2 \mathbf{k} \), it describes the velocity vector. The velocity vector gives the direction and rate of change of the original vector function's path, crucial for determining the length of the curve.

Integral Calculus

Integral calculus plays a pivotal role in calculating the arc length of a curve described by a vector function. To find this length, we integrate the magnitude of the derivative of the vector function, \( \| \mathbf{r}'(t) \| \), over the specified interval.
In this particular scenario, we aim to find the arc length \( L \) from \( t = 0 \) to \( t = 1 \).

• First, compute the magnitude of the derivative, \( \| \mathbf{r}'(t) \| = \sqrt{9 + 12t^2 + 4t^4} \).
• Then set up the definite integral from 0 to 1 of this magnitude, \( L = \int_{0}^{1} \sqrt{9 + 12t^2 + 4t^4} \, dt \).

This integral determines the total length of the path traced by the original vector function across the given interval. If an expression cannot be simplified analytically like in this exercise, numerical methods or computational tools can be employed to evaluate the integral accurately.