Problem 44 - Advanced Engineering Mathematics

Question

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 given by the integral \( \int_{0}^{1} \sqrt{9 + 12t^2 + 4t^4} \, dt \), which can be evaluated numerically.

Step 1: Determine the Derivative

First, find the derivative of the vector function \( \mathbf{r}(t) \) with respect to \( t \). For \( \mathbf{r}(t) = 3t \mathbf{i} + \sqrt{3} t^2 \mathbf{j} + \frac{2}{3} t^3 \mathbf{k} \), the derivative \( \mathbf{r}'(t) \) is: \[ \mathbf{r}'(t) = \frac{d}{dt}(3t \mathbf{i}) + \frac{d}{dt}(\sqrt{3} t^2 \mathbf{j}) + \frac{d}{dt}(\frac{2}{3} t^3 \mathbf{k}) \] This results in: \[ \mathbf{r}'(t) = 3 \mathbf{i} + 2 \sqrt{3} t \mathbf{j} + 2t^2 \mathbf{k} \]

Step 2: Compute the Magnitude of the Derivative

Calculate the magnitude of \( \mathbf{r}'(t) \). Using the formula for magnitude: \[ |\mathbf{r}'(t)| = \sqrt{(3)^2 + (2\sqrt{3}t)^2 + (2t^2)^2} \] This simplifies to: \[ |\mathbf{r}'(t)| = \sqrt{9 + 12t^2 + 4t^4} \]

Step 3: Integrate to Find the Arc Length

To find the length of the curve from \( t = 0 \) to \( t = 1 \), integrate the magnitude of the derivative: \[ L = \int_{0}^{1} |\mathbf{r}'(t)| \, dt = \int_{0}^{1} \sqrt{9 + 12t^2 + 4t^4} \, dt \] Solve this definite integral to find the length. The calculation can be complex and may require numerical methods or specific algebraic solutions.

Key Concepts

Arc Length

Arc length is a fundamental concept in vector calculus, especially when dealing with curves defined by vector functions. To understand arc length, imagine how long a curved path is by "straightening" it out. For a curve defined by a vector function, the arc length can be found by integrating over the curve's parameter, such as time \( t \).
The integral used to calculate the arc length is generally formulated as \( L = \int_{a}^{b} |\mathbf{r}'(t)| \, dt \), where \( |\mathbf{r}'(t)| \) is the magnitude of the derivative of the vector function. Here, the curve stretches from parameter \( t = a \) to \( t = b \). By integrating, you effectively "add up" all the tiny lengths along the curve.
Understanding arc length helps in applications like plotting flight paths or designing roads, as it enables precise measurements around curved spaces.

Vector Function

A vector function is a function that gives a vector output for every scalar input. In the problem, the vector function \( \mathbf{r}(t) \) encodes a curve in three-dimensional space using the components \( 3t \mathbf{i} \), \( \sqrt{3} t^2 \mathbf{j} \), and \( \frac{2}{3} t^3 \mathbf{k} \).
Each component represents the curve's movement along the \( x \), \( y \), and \( z \) axes as \( t \) changes. By varying \( t \) within an interval, the vector function defines a path in space. Vector functions are essential for describing curves, paths, and trajectories, which makes them invaluable in physics and engineering.
To visualize vector functions, you can plot the components separately or use three-dimensional graphing tools to see how the function evolves over time. This visual representation aids in understanding how objects move in space.

Derivative Magnitude

The derivative of a vector function, denoted \( \mathbf{r}'(t) \), represents the rate of change of the vector function with respect to its parameter \( t \). In simpler terms, it shows the velocity of the point along the curve.
The magnitude of this derivative, \( |\mathbf{r}'(t)| \), is crucial for finding arc length, as it quantifies how fast the point is moving at any given \( t \). Calculating the magnitude involves finding the square root of the sum of the squares of its components, \( |\mathbf{r}'(t)| = \sqrt{(3)^2 + (2\sqrt{3}t)^2 + (2t^2)^2} \).
This magnitude gives insight into changes in speed along the path, helping in understanding motion dynamics and the geometry of the trajectory.

Parametric Curves

Parametric curves describe paths in space by expressing coordinates as functions of a common parameter, usually denoted \( t \). For example, the curve defined by \( \mathbf{r}(t) = 3t \mathbf{i} + \sqrt{3} t^2 \mathbf{j} + \frac{2}{3} t^3 \mathbf{k} \) is a parametric curve.
These curves are significant because they allow for flexibility in representing complex shapes and paths. Unlike standard functions that may only be expressed explicitly in terms of \( y \) and \( x \), parametric curves can define shapes that loop, twist, or have multiple parts in multivariable space.
Understanding parametric curves is vital in technical fields, as they provide a comprehensive way to describe motions and interactions, making them crucial for simulations, animations, and more.