Problem 43 - Advanced Engineering Mathematics

Question

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

The length of the curve is \( \sqrt{5} (e^{3\pi} - 1) \).

Step 1: Determine the Derivative

Calculate the derivative of the vector function \( \mathbf{r}(t) \). The derivative is \( \mathbf{r}'(t) = \frac{d}{dt}[e^t \cos 2t] \mathbf{i} + \frac{d}{dt}[e^t \sin 2t] \mathbf{j} + \frac{d}{dt}[e^t] \mathbf{k} \). Use the product rule for differentiation: \( \mathbf{r}'(t) = (e^t \cos 2t - 2e^t \sin 2t) \mathbf{i} + (e^t \sin 2t + 2e^t \cos 2t) \mathbf{j} + e^t \mathbf{k} \).

Step 2: Find the Magnitude of the Derivative

The magnitude of \( \mathbf{r}'(t) \) is given by \( \|\mathbf{r}'(t)\| = \sqrt{((e^t \cos 2t - 2e^t \sin 2t)^2 + (e^t \sin 2t + 2e^t \cos 2t)^2 + (e^t)^2)} \). Simplify this expression to \( \|\mathbf{r}'(t)\| = e^t \sqrt{5} \).

Step 3: Set Up the Arc Length Integral

The arc length of a vector function \( \mathbf{r}(t) \) from \( t = a \) to \( t = b \) is given by \( L = \int_a^b \| \mathbf{r}'(t) \| \, dt \). Therefore, \( L = \int_0^{3\pi} e^t \sqrt{5} \, dt \).

Step 4: Evaluate the Integral

Evaluate the integral \( \int_0^{3\pi} e^t \sqrt{5} \, dt \). since \( e^t \sqrt{5} \) is constant with respect to \( \sqrt{5} \), \( L = \sqrt{5} [e^t]_0^{3\pi} \). Compute the definite integral: \( L = \sqrt{5} (e^{3\pi} - e^0) = \sqrt{5} (e^{3\pi} - 1) \).

Step 5: Present the Final Answer

The length of the curve is \( \sqrt{5} (e^{3\pi} - 1) \). We have determined this by evaluating the definite integral for the magnitude of the derivative.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with differentiating and integrating vector fields, typically in 2 or 3 dimensions. It plays a crucial role in understanding the behavior of vector-valued functions, which map a set of real numbers into a vector space. In this exercise, we focused on a vector function \( \mathbf{r}(t) = e^t \cos 2t \mathbf{i} + e^t \sin 2t \mathbf{j} + e^t \mathbf{k} \), which defines a curve in 3-dimensional space. This function depends on the parameter \( t \), which lies within the interval \( [0, 3\pi] \).

The goal in vector calculus problems often revolves around finding features of the curve, such as direction, length, and curvature. In this exercise, we particularly aim to find the arc length of the curve. This involves using calculus tools to describe the vector nature of the curve, leveraging both differentiation and integration. In particular, the derivative \( \mathbf{r}'(t) \) provides us insights into the direction and velocity of the curve at any point \( t \). This is foundational before progressing to the next significant steps of solving for the arc length.

Derivative Calculation

Calculating the derivative of a vector function is essential in analyzing how the function changes. For the vector function \( \mathbf{r}(t) \), we calculate its derivative \( \mathbf{r}'(t) \), which entails differentiating each of its components separately. We use the product rule for functions of the form \( uv \), given by \( u'v + uv' \).

For our function:

• The derivative of \( e^t \cos 2t \) with respect to \( t \) is \( e^t \cos 2t - 2e^t \sin 2t \).
• For \( e^t \sin 2t \), the derivative is \( e^t \sin 2t + 2e^t \cos 2t \).
• The straightforward derivative for \( e^t \) is \( e^t \).

Together, these give us \( \mathbf{r}'(t) = (e^t \cos 2t - 2e^t \sin 2t) \mathbf{i} + (e^t \sin 2t + 2e^t \cos 2t) \mathbf{j} + e^t \mathbf{k} \). This derivative helps us determine the speed and direction changes of the moving particle defined by the vector function.

Integral Evaluation

Once we have determined \( \|\mathbf{r}'(t)\| \), we can compute the arc length by evaluating the integral \( L = \int_a^b \| \mathbf{r}'(t) \| \, dt \). Here, \( \|\mathbf{r}'(t)\| = e^t \sqrt{5} \), a crucial simplification that makes the integral more straightforward.

The integral \( \int_0^{3\pi} e^t \sqrt{5} \, dt \) can be approached by recognizing that \( \sqrt{5} \) is a constant. Thus, we focus on integrating \( e^t \), which has the integral \( e^t \) evaluated from \( 0 \) to \( 3\pi \).

This gives us:

• \( L = \sqrt{5} \cdot [e^t]_0^{3\pi} \).
• Substituting the limits, \( L = \sqrt{5} (e^{3\pi} - e^0) \).
• Since \( e^0 = 1 \), this simplifies to \( L = \sqrt{5} (e^{3\pi} - 1) \).

The final result of the arc length evaluation shows that it relies on integrating derivatives effectively and properly handling limits, crucial elements of integral evaluation in mathematical analysis.