Problem 34 - Calculus Early Transcendentals

Question

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) along the curve \(C\) $$ \begin{array}{l}{\mathbf{F}(x, y)=x^{2} y \mathbf{i}+4 \mathbf{j}} \\ {C: \mathbf{r}(t)=e^{t} \mathbf{i}+e^{-t} \mathbf{j} \quad(0 \leq t \leq 1)}\end{array} $$

Step-by-Step Solution

Answer

The line integral evaluates to \( \frac{1}{2}e^2 + \frac{7}{2} - \frac{4}{e} \).

Step 1: Understand the Problem

We need to evaluate the line integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \) over a curve \( C \). The vector field is \( \mathbf{F}(x, y) = x^2 y \mathbf{i} + 4 \mathbf{j} \), and the path \( C \) is described by the parametric equations \( \mathbf{r}(t) = e^{t} \mathbf{i} + e^{-t} \mathbf{j} \) for \( 0 \leq t \leq 1 \).

Step 2: Differentiate the Path

The differential \( d\mathbf{r} \) can be found by differentiating \( \mathbf{r}(t) \), so \( d\mathbf{r} = \frac{d}{dt}\left(e^t \mathbf{i} + e^{-t} \mathbf{j}\right) dt = e^t \mathbf{i} - e^{-t} \mathbf{j} dt \).

Step 3: Substitute into the Integral

Substitute the expressions for \( \mathbf{F}(x(t), y(t)) \) and \( d\mathbf{r} \) into the line integral. First, calculate \( \mathbf{F}(e^t, e^{-t}) = (e^{2t} e^{-t}) \mathbf{i} + 4 \mathbf{j} = e^t \mathbf{i} + 4 \mathbf{j} \). Then, the line integral becomes \( \int_{0}^{1} (e^t \mathbf{i} + 4 \mathbf{j}) \cdot (e^t \mathbf{i} - e^{-t} \mathbf{j}) dt \).

Step 4: Compute the Dot Product

Calculate the dot product: \( (e^t \mathbf{i} + 4 \mathbf{j}) \cdot (e^t \mathbf{i} - e^{-t} \mathbf{j}) = e^{2t} + 4(-e^{-t}) = e^{2t} - 4e^{-t} \).

Step 5: Evaluate the Integral

Integrate the dot product from 0 to 1: \( \int_{0}^{1} (e^{2t} - 4e^{-t}) dt \). This is equal to \( \int_{0}^{1} e^{2t} dt - 4 \int_{0}^{1} e^{-t} dt \).

Step 6: Calculate the Integrals

Evaluate the integrals: \( \int e^{2t} dt = \frac{1}{2} e^{2t} + C \), and \( \int e^{-t} dt = -e^{-t} + C \). Substituting the limits \(0\) to \(1\), we get: \[ \left[ \frac{1}{2} e^{2t} \right]_{0}^{1} - 4 \left[ -e^{-t} \right]_{0}^{1} = \frac{1}{2}(e^2 - 1) + 4(e^{-0} - e^{-1}) \].

Step 7: Final Calculation

Substitute values and simplify: \[ \frac{1}{2}(e^2 - 1) + 4(1 - \frac{1}{e}) = \frac{1}{2}e^2 - \frac{1}{2} + 4 - \frac{4}{e} \]. Combine terms: \( \frac{1}{2}e^2 + \frac{7}{2} - \frac{4}{e} \).

Step 8: Final Result

The value of the line integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \) is \( \frac{1}{2}e^2 + \frac{7}{2} - \frac{4}{e} \).

Key Concepts

Vector Fields

A vector field is a function that assigns a vector to every point in space. When working with vector fields, it’s important to understand that each vector represents both a direction and a magnitude at a particular point. In the problem, the vector field \( \mathbf{F}(x, y) = x^2 y \mathbf{i} + 4 \mathbf{j} \) describes how vectors are distributed in the plane.
- The first component, \( x^2 y \mathbf{i} \), depends on the coordinates \( x \) and \( y \), meaning it changes as you move through the field.- The second component, \( 4 \mathbf{j} \), is constant, indicating a uniform vector field in the \( y \) direction.
Understanding how these vectors operate can give insights into physical applications such as fluid flow or electromagnetic fields.

Parametric Equations

Parametric equations allow us to describe curves or paths in a calculated, step-wise process by setting equations for each coordinate as a function of a parameter, often \( t \). In our exercise, the curve \( C \) is described by the vector function \( \mathbf{r}(t) = e^t \mathbf{i} + e^{-t} \mathbf{j} \), which lets us express both the \( x \) and \( y \) coordinates in terms of \( t \).
- The function \( e^t \) governs the \( x \)-coordinate, which grows exponentially as \( t \) increases.- Conversely, \( e^{-t} \) determines the \( y \)-coordinate, decreasing as \( t \) increases.
These parametric equations are instrumental in computing line integrals because they give us a clear pathway (literally) to follow when applying integration techniques.

Dot Products

The dot product is an essential tool in vector calculus, helping to determine the degree to which two vectors point in the same direction. For two vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} \), the dot product is given by \( a_1b_1 + a_2b_2 \). In our line integral, it helps us to combine \( \mathbf{F}(x(t), y(t)) \) and \( d\mathbf{r} \), both of which are vector expressions.
When calculating \( (e^t \mathbf{i} + 4 \mathbf{j}) \cdot (e^t \mathbf{i} - e^{-t} \mathbf{j}) \), we:

• Multiply the \( \mathbf{i} \) components: \( e^t \cdot e^t = e^{2t} \).
• Multiply the \( \mathbf{j} \) components: \( 4 \cdot (-e^{-t}) = -4e^{-t} \).

Thus, the dot product largely simplifies our calculation, providing the scalar function \( e^{2t} - 4e^{-t} \) to integrate over the desired interval.

Integration Techniques

Integration is a cornerstone in calculus, especially when evaluating outcomes like the final result in our exercise. Through integration, we calculate the accumulation of quantities, like area or, in this case, the total vector effect along a path.
In our exercise, you'll notice:

• The line integral converts to evaluating \( \int_{0}^{1} (e^{2t} - 4e^{-t}) dt \) by computing two separate integrals: \( \int e^{2t} dt \) and \(-4 \int e^{-t} dt \).
• The antiderivatives become critical, transforming exponential terms for evaluation across limits.
• Using the antiderivatives, substitute the limits from 0 to 1, which is often the step where mistakes can occur if careful attention isn’t given to signs and coefficients.

Remember, each integral is a mini-routine within the larger problem-solving strategy. Therefore, mastering these techniques is crucial for tackling complex calculus problems.