Problem 22 - Thomas Calculus

Question

Find the line integral of \(f ( x , y ) = x - y + 3\) along the curve \(\mathbf { r } ( t ) = ( \cos t ) \mathbf { i } + ( \sin t ) \mathbf { j } , 0 \leq t \leq 2 \pi\)

Step-by-Step Solution

Answer

The line integral is \(6\pi\).

Step 1: Define the Line Integral

The line integral of a scalar field \(f(x, y)\) over a curve \(\mathbf{r}(t)\) is given by \(\int_{C} f(x, y)\, ds = \int_{a}^{b} f(x(t), y(t)) \| \mathbf{r}'(t) \| \, dt\), where \(C\) is the curve parameterized by \(\mathbf{r}(t)\) and \(\| \mathbf{r}'(t) \|\) is the magnitude of the derivative of \(\mathbf{r}(t)\).

Step 2: Parameterization of the Curve

The curve is parameterized as \(\mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j}\), meaning \(x(t) = \cos t\) and \(y(t) = \sin t\).

Step 3: Evaluate the Function on the Path

Substitute \(x(t)\) and \(y(t)\) into the function \(f(x, y) = x - y + 3\). This gives: \[ f(x(t), y(t)) = \cos t - \sin t + 3 \].

Step 4: Compute the Derivative of the Path

Differentiate \(\mathbf{r}(t)\) with respect to \(t\):\[ \mathbf{r}'(t) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} \].

Step 5: Calculate the Magnitude of the Derivative

The magnitude \(\| \mathbf{r}'(t) \|\) is calculated as:\[ \| \mathbf{r}'(t) \| = \sqrt{(-\sin t)^2 + (\cos t)^2} = \sqrt{\sin^2 t + \cos^2 t} = 1 \].

Step 6: Set Up the Integral

Plug \(f(x(t), y(t))\) and \(\| \mathbf{r}'(t) \|\) into the line integral:\[ \int_{0}^{2\pi} (\cos t - \sin t + 3) \, dt \].

Step 7: Integrate the Function

Calculate the integral:\[ \int (\cos t - \sin t + 3) \, dt = \sin t + \cos t + 3t \].

Step 8: Evaluate the Integral at the Limits

Evaluate \(\sin t + \cos t + 3t\) from \(0\) to \(2\pi\):- At \(t = 2\pi\): \(\sin(2\pi) + \cos(2\pi) + 3(2\pi) = 0 + 1 + 6\pi = 1 + 6\pi\)- At \(t = 0\): \(\sin(0) + \cos(0) + 3(0) = 0 + 1 + 0 = 1\)The result is: \((1 + 6\pi) - 1 = 6\pi \).

Key Concepts

Parameterization of Curves

Parameterization of curves is a technique used to express a curve in terms of a single variable, usually denoted as \(t\). This variable is called the parameter, and it allows us to trace the path of the curve as \(t\) varies over a certain interval.
Normally, the curve is described using vector functions, like \(\mathbf{r}(t)\). This means that each point on the curve corresponds to a specific value of \(t\).
For the given exercise, the curve is parameterized by \(\mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j}\), where \(0 \leq t \leq 2\pi\). This represents the path of a circle in the plane, traced counterclockwise.

• \(x(t) = \cos t\) gives the \(x\)-coordinate as \(t\) changes.
• \(y(t) = \sin t\) gives the \(y\)-coordinate as \(t\) changes.

This parameterization is essential for performing the line integral along this curve.

Magnitude of a Vector

After defining a curve, a key step in line integrals is to compute the magnitude of the vector function that describes its rate of change, or its derivative. The magnitude of a vector \(\mathbf{v} = (v_x, v_y)\) is given by \(\|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2}\).
In the context of our problem, the vector function \(\mathbf{r}'(t)\) is the derivative of the parameterization: \[ \mathbf{r}'(t) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} \].
To find its magnitude, calculate:

• \((-\sin t)^2\) is \(\sin^2 t\).
• \((\cos t)^2\) remains \(\cos^2 t\).

Add these together:
\(\sin^2 t + \cos^2 t = 1\), so the magnitude \(\|\mathbf{r}'(t)\|\) equals 1. This simplification is due to the Pythagorean identity.

Integration

Integration is a mathematical process of finding the accumulation of quantities, often representing areas under curves or other quantities like lengths of paths. In the context of line integrals, it helps us sum up values of a function along a path described by a curve.
For line integrals, we integrate the function \(f(x(t), y(t))\) times the magnitude of the curve's derivative over the given parameter \(t\). This is formally written as:
\[ \int_{a}^{b} f(x(t), y(t)) \| \mathbf{r}'(t) \| \, dt \]
In our example, we compute:

• \(\int_{0}^{2\pi} (\cos t - \sin t + 3) \, dt \)

This integral is computed component-wise, resulting in an accumulation of the function's values along the circular path from \(t = 0\) to \(t = 2\pi\).

Scalar Field

A scalar field is a function that assigns a single scalar (number) to every point in space. In this exercise, the scalar field is \(f(x, y) = x - y + 3\).
This function gives us a value based on the current \(x\) and \(y\) coordinates, depicting how its values vary across the plane. In the context of line integrals, we evaluate this scalar field along a curve, typically parameterized as discussed earlier.
The value of the scalar field is crucial for determining the outcome of the line integral, as it reflects the function's contribution at each point on the path. Hence, the integral of this scalar field over the curve measures how the function behaves along this specific path. For this problem, at each \(t\), we have:

• \(f(x(t), y(t)) = \cos t - \sin t + 3\)

Finding the integral with respect to this scalar field lets us understand how its values accumulate over the given circular path.