Problem 24 - Calculus Early Transcendentals

Question

Find the exact value of \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) using any method. $$ \begin{array}{l}{\mathbf{F}(x, y)=2 x y \mathbf{i}+\left(x^{2}+\cos y\right) \mathbf{j}} \\ {C: \mathbf{r}(t)=t \mathbf{i}+t \cos (t / 3) \mathbf{j} \quad(0 \leq t \leq \pi)}\end{array} $$

Step-by-Step Solution

Answer

The exact value of the line integral is 0.

Step 1: Understand the Problem

We need to evaluate the line integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \) where \( C \) is parameterized by \( \mathbf{r}(t) = t \mathbf{i} + t \cos(t/3) \mathbf{j} \) for \( 0 \leq t \leq \pi \). The vector field is \( \mathbf{F}(x, y) = 2xy \mathbf{i} + (x^2 + \cos(y)) \mathbf{j} \).

Step 2: Find \( d\mathbf{r} \)

The differential vector \( d\mathbf{r} \) can be found by differentiating the parameterization \( \mathbf{r}(t) = t \mathbf{i} + t \cos(t/3) \mathbf{j} \). This gives:\[ d\mathbf{r} = \left(1 \mathbf{i} - \left(\frac{t}{3}\sin(t/3) - \cos(t/3)\right) \mathbf{j}\right) dt \]

Step 3: Evaluate \( \mathbf{F} \cdot d\mathbf{r} \)

First, substitute \( x = t \) and \( y = t \cos(t/3) \) in \( \mathbf{F}(x, y) \):\[ \mathbf{F}(t, t\cos(t/3)) = 2t^2 \cos(t/3) \mathbf{i} + \left(t^2 + \cos(t \cos(t/3))\right) \mathbf{j} \]Now, compute the dot product:\[ \mathbf{F} \cdot d\mathbf{r} = \left(2t^2 \cos(t/3)\right) \cdot 1 + \left(t^2 + \cos(t \cos(t/3))\right) \cdot \left(-\frac{t}{3}\sin(t/3) - \cos(t/3)\right) \]

Step 4: Simplify and Integrate

Simplify the expression for the dot product obtained in Step 3 and integrate with respect to \( t \) from 0 to \( \pi \):\[\int_{0}^{\pi} \left[ 2t^2 \cos(t/3) - \left(t^2 + \cos(t \cos(t/3))\right) \left(\frac{t}{3}\sin(t/3) - \cos(t/3)\right) \right] dt\]Evaluate this integral carefully to find the exact value.

Key Concepts

Vector Field

A vector field assigns a vector to every point in space. Imagine plotting arrows on the graph to represent forces, velocities, or other vector quantities. In this problem, the vector field \( \mathbf{F}(x, y) \) is given by \( 2xy \mathbf{i} + (x^2 + \cos y) \mathbf{j} \). This means at any point \((x, y)\), the field has:

• An \( x \)-component of \( 2xy \)
• A \( y \)-component of \( x^2 + \cos(y) \)

These components together form a vector. We need to understand how these vectors behave or change along the path \( C \). The purpose is to compute the line integral, which involves this vector field.

Parameterization

Parameterization transforms a curve in space into functions of a single variable, usually \( t \). In our exercise, the parameterized path \( C \) is given as \( \mathbf{r}(t) = t \mathbf{i} + t \cos(t/3) \mathbf{j} \), ranging from \( t = 0 \) to \( t = \pi \).

This means:

• \( x = t \)
• \( y = t \cos(t/3) \)

This neat expression translates movements along the path into simple mathematical terms. In practical sense, it's like walking along a trail, where each point \( t \) describes a specific position on the path \( C \). Understanding this transformation is crucial for evaluating a line integral as it simplifies tracking the change in position in the vector field.

Dot Product

Continuous operations in calculus often need combining vectors, for which the dot product is essential. It takes two vectors and returns a scalar, representing the magnitude of one vector projected onto another.

In this solution, we perform the dot product \( \mathbf{F} \cdot d\mathbf{r} \). Originally derived, \( d\mathbf{r} = (1 \mathbf{i} + (\frac{t}{3}\sin(t/3) - \cos(t/3)) \mathbf{j}) dt \) operates like this:

• Multiply corresponding components of \( \mathbf{F} \) and \( d\mathbf{r} \).
• Add these products.

This simplifies the expression we're integrating. The result is key to compute how much "work" or accumulation occurs over the curve, based on the given vector field \( \mathbf{F} \). It allows us to quantify the path's interaction with the field.

Calculus

Calculus is the mathematical study of change. Here, it helps to evaluate the integral \( \int_{0}^{\pi} \left[ 2t^2 \cos(t/3) - \left(t^2 + \cos(t \cos(t/3))\right) \left(\frac{t}{3}\sin(t/3) - \cos(t/3)\right) \right] dt \), meaningfully.

In line integrals, calculus quantifies the accumulation or 'work' done by a force field as one moves along a curve.

• The expression obtained from the dot product of \( \mathbf{F} \) and \( d\mathbf{r} \) is simplified to an integrable form.
• The integral limits, from \( 0 \) to \( \pi \), guarantee it accounts for the entire parameterized curve \( C \).

Techniques from calculus compute the integral precisely, yielding an exact measure of the line integral. Understanding such techniques improves proficiency in handling complex physics or engineering problems.