Problem 4 - Calculus: Early Transcendental Functions

Question

Evaluate the line integral. \(\int_{C} 2 x y d s,\) where \(C\) is the line segment from (1,2) to (-1,0)

Step-by-Step Solution

Answer

The value of the line integral \(\int_{C} 2 x y d s\) is \(-\frac{4\sqrt{5}}{3}\) or \(-1.3\sqrt{5}\).

Step 1: Parametrize The Line Segment C

The line can be parameterized by the initial and terminal points. Let \(t\) be the parameter, varying from 0 to 1, where \(t=0\) gives the initial point and \(t=1\) gives the terminal point. In this case, we get a function \(r(t) = (1-t)(1,2) + t(-1,0) = (1-t, 2-2t)\), where \(t\) ranges from 0 to 1.

Step 2: Calculate the line Integral

We substitute the parameterized line into the definition of the line integral. The differential element of arc length \(ds\) can be expressed as \(ds = |r'(t)|dt\), where \(r'(t)\) is the derivative of the parameterized function, which can be obtained as \(r'(t)= (-1,-2)\) and its absolute value \(|r'(t)|= \sqrt{(-1)^2+(-2)^2} = \sqrt{5}\). The function \(2xy\) is now expressed as \(2x(t)y(t) = 2(1-t)(2-2t)\). Therefore, the integral becomes \(\int_{0}^{1} 2(1-t)(2-2t) \sqrt{5} dt\).

Step 3: Evaluate the Integral

Integrating the function yields a value of \(-\frac{4\sqrt{5}}{3}\).

Step 4: Simplify the Result

The negative sign indicates that the result is being measured in the opposite direction of the positive x-axis. We can simplify this as \(-1.3\sqrt{5}\).

Key Concepts

Parameterization

Parameterization is a fundamental step in evaluating line integrals, especially when dealing with line segments between two points. We start by representing the line segment from point (1, 2) to point (-1, 0) as a vector-valued function of one variable, usually denoted by a parameter, say \( t \). Here, \( t \) ranges from 0 to 1, where \( t=0 \) corresponds to the starting point and \( t=1 \) corresponds to the ending point of the segment.

To form this representation, we use a combination of the initial and terminal point, like so:

• For \( t=0 \): This corresponds to the point (1, 2).
• For \( t=1 \): This corresponds to the point (-1, 0).

The parameterization equation thus becomes \( r(t) = (1-t)(1,2) + t(-1,0) \), simplifying to \( r(t) = (1-t, 2-2t) \).

This parameterization allows us to express any point along the line segment as a function of \( t \), which is crucial for functions of two variables like line integrals.

Arc Length Differentiation

In computing a line integral, it's essential to convert the path into a form that integrates easily. This is where we find the differential element of arc length, \( ds \). The parameterization function \( r(t) = (1-t, 2-2t) \) helps us along this path.

To find \( ds \), we first differentiate this parameterized function with respect to \( t \). The derivative \( r'(t) \) gives us the rate of change of the position vector and includes derivatives of both coordinate functions:

• For the x-component: Derivative is \(-1\)
• For the y-component: Derivative is \(-2\)

Next, calculate the magnitude \( |r'(t)| \) as \( \sqrt{(-1)^2 + (-2)^2} = \sqrt{5} \). This magnitude helps define \( ds \) as \( \sqrt{5} \, dt \), facilitating the integral evaluation. Thus, the differential arc length \( ds \) simplifies our line integral evaluation by transitioning from path coordinates to a standard parameter \( t \).

Integral Evaluation

The final step involves evaluating the line integral over the parameterized path. The initially given line integral \( \int_{C} 2xy \, ds \) needs substitution from parameterization. With \( x(t) = 1-t \) and \( y(t) = 2-2t \), substitute into \( 2xy \) to get \( 2(1-t)(2-2t) \). This reduces the integral to \( \int_{0}^{1} 2(1-t)(2-2t) \sqrt{5} \, dt \).

Solving the integral requires expanding the algebraic expression. Doing so, you obtain \( \int_{0}^{1} 4(1-t) - 4t(1-t) \sqrt{5} \, dt \), and integrate term by term. This results in evaluating the polynomial for \( t \) over [0, 1].

After computing the definite integral, the answer is found to be \(-\frac{4 \sqrt{5}}{3} \). Simplifying gives it an approximated value as \(-1.3\sqrt{5} \), indicating the computed path integral was in the negative direction relative to the standard coordinate plane, often due to the nature of path orientation.