Problem 43 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

\( \sqrt{5}(e^{3\pi} - 1) \)

Step 1: Understanding the Problem

We need to find the length of the curve described by the vector function \( \mathbf{r}(t) = e^{t} \cos(2t) \mathbf{i} + e^{t} \sin(2t) \mathbf{j} + e^{t} \mathbf{k} \) over the interval \( 0 \leq t \leq 3\pi \). This involves using the formula for the arc length of a curve parameterized by a vector function.

Step 2: Arc Length Formula

The arc length \( S \) of a curve \( \mathbf{r}(t) \) from \( t = a \) to \( t = b \) is given by the integral \[ S = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \] where \( \mathbf{r}'(t) \) is the derivative of \( \mathbf{r}(t) \).

Step 3: Compute the Derivative \( \mathbf{r}'(t) \)

First, find the derivative \( \mathbf{r}'(t) \): - Derivative of \( e^{t} \cos(2t) \) is \( e^{t} \cos(2t) - 2e^{t} \sin(2t) \)- Derivative of \( e^{t} \sin(2t) \) is \( e^{t} \sin(2t) + 2e^{t} \cos(2t) \)- Derivative of \( e^{t} \) is \( e^{t} \)Thus, \( \mathbf{r}'(t) = (e^{t} \cos(2t) - 2e^{t} \sin(2t)) \mathbf{i} + (e^{t} \sin(2t) + 2e^{t} \cos(2t)) \mathbf{j} + e^{t} \mathbf{k} \).

Step 4: Compute \( \| \mathbf{r}'(t) \| \)

Calculate the magnitude of \( \mathbf{r}'(t) \):\[ \| \mathbf{r}'(t) \| = \sqrt{(e^{t} \cos(2t) - 2e^{t} \sin(2t))^2 + (e^{t} \sin(2t) + 2e^{t} \cos(2t))^2 + (e^{t})^2} \] Upon simplifying, all the terms under the square root become \( 5e^{2t} \) due to trigonometric identities like \( \cos^2(2t) + \sin^2(2t) = 1 \). Therefore, \( \| \mathbf{r}'(t) \| = \sqrt{5} e^{t} \).

Step 5: Integrate \( \| \mathbf{r}'(t) \| \) over the Interval

Set up the integral for the arc length:\[ S = \int_{0}^{3\pi} \sqrt{5} e^{t} \ dt \]The integration can now be performed:- The antiderivative of \( e^{t} \) is \( e^{t} \).- Evaluating \( \sqrt{5} \int_{0}^{3\pi} e^{t} \, dt = \sqrt{5} [e^{t}]_{0}^{3\pi} \).This results in:\[ \sqrt{5} (e^{3\pi} - e^{0}) = \sqrt{5} (e^{3\pi} - 1) \]

Step 6: Conclusion

The length of the curve traced by the vector function \( \mathbf{r}(t) \) from \( t = 0 \) to \( t = 3\pi \) is \( \sqrt{5} (e^{3\pi} - 1) \).

Key Concepts

vector calculus

Vector calculus is a fascinating field of mathematics that extends the general idea of calculus to vector functions. These are functions that take in scalar values, like time, and output vectors. Imagine a little arrow spinning around in space as time progresses; this makes vector calculus particularly useful in physics and engineering. It helps us describe and analyze movements and forces in multidimensional spaces, which are highly relevant to real-world applications like fluid dynamics and electromagnetism.

One core component of vector calculus is the arc length of curves, and this is where calculus really shines. By using integration, a foundational tool of calculus, we can compute the actual length of a curve from one point to another, much like measuring the length of a winding road. For curves described as vector functions, this involves calculating the magnitude of the curve's derivative, integrating it over the desired interval.

Thus, the arc length formula, \( S = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \), is used to find how long the curve travels in the defined parametric equations. Understanding the essence of this formula is pivotal in unlocking the mysteries of vector calculus.

parametric equations

Parametric equations are another exciting aspect of mathematics, providing a unique methodology to describe curves in space. Instead of using ordinary Cartesian equations, where an explicit relationship between \( x \) and \( y \) is established, parametric equations express the coordinates as functions of an independent variable, usually denoted as \( t \) for time.

By using parametric equations, complex curves and motions can be modeled more intuitively, often making them easier to work with analytically. Imagine a track for a roller coaster. Parametric equations allow us to capture the intricate loops and turns it makes in a straightforward manner by setting equations for \( x(t) \), \( y(t) \), and \( z(t) \). In our exercise, the vector function \( \mathbf{r}(t) = e^{t} \cos(2t) \mathbf{i} + e^{t} \sin(2t) \mathbf{j} + e^{t} \mathbf{k} \) is a classic example of how parametric equations can describe a three-dimensional spiral trajectory as a function of \( t \), ranging from \( 0 \) to \( 3\pi \).

The advantage is not just visual but also computational. It simplifies calculating the derivative \( \mathbf{r}'(t) \) and eventually allows us to determine properties like the curve's arc length, as seen in this problem.

integration of exponential functions

Integration plays a vital role in finding arc lengths, especially when dealing with exponential functions, a common component in many physics and engineering problems. Exponential functions, such as \( e^t \), possess unique properties that often simplify calculus operations compared to polynomial functions.

An integral of an exponential function, like \( \int e^{t} \, dt \), is straightforward. The antiderivative of \( e^{t} \) is simply \( e^{t} \), a property that stems from the function's natural growth characteristics. Even more, when coefficients or constants are involved, as seen in \( \sqrt{5} e^{t} \), the integration follows a simple pattern: \( \int \sqrt{5} e^{t} \, dt = \sqrt{5} e^{t} + C \), where \( C \) is the constant of integration.

These attributes are significant when evaluating definite integrals over a specific interval, such as \( [0, 3\pi] \). The result is the evaluated difference \( [e^{3\pi} - e^{0}] \), further showcasing their efficiency in computing exact values for physical lengths or other properties. Mastering the integration of exponential functions is key to understanding vector calculus fully, facilitating complex calculations that would otherwise be daunting.