Problem 24
Question
Evaluate the line integral along the curve C. $$ \begin{array}{l}{\int_{C}(y-x) d x+x y d y} \\ {C: \text { the line segment from }(3,4) \text { to }(2,1)}\end{array} $$
Step-by-Step Solution
Verified Answer
The value of the line integral is -13.5.
1Step 1: Parameterize the Curve
The line segment from (3, 4) to (2, 1) can be parameterized as \( \mathbf{r}(t) = (1-t)(3, 4) + t(2, 1) \), where \( 0 \leq t \leq 1 \). This gives \( x(t) = 3 - t \) and \( y(t) = 4 - 3t \).
2Step 2: Differentiate x(t) and y(t)
Differentiate \( x(t) = 3 - t \) and \( y(t) = 4 - 3t \) with respect to \( t \): \( \frac{dx}{dt} = -1 \) and \( \frac{dy}{dt} = -3 \).
3Step 3: Substitute and Express the Integrand
Substitute \( x(t) = 3 - t \) and \( y(t) = 4 - 3t \) into the integrand: \( (y-x) \frac{dx}{dt} + x \cdot y \frac{dy}{dt} \). This becomes \( (4-3t - (3-t))(-1) + (3-t)(4-3t)(-3) \).
4Step 4: Simplify the Integrand
Simplify: \( (1 - 2t) \cdot (-1) + (3-t)(4-3t) \cdot (-3) = -1 + 2t - 3 \cdot (12 - 17t + 3t^2) \).
5Step 5: Integrate with Respect to t
Integrate from \( t = 0 \) to \( t = 1 \): \( \int_0^1 (-1 + 2t - 36 + 51t - 9t^2) dt \). Simplifying, we get \( \int_0^1 (-37 + 53t - 9t^2) dt \).
6Step 6: Perform the Integration
Calculate: \( \int (-37) dt = -37t \), \( \int 53t dt = \frac{53t^2}{2} \), and \( \int 9t^2 dt = 3t^3 \). Evaluate from 0 to 1: \([-37(1) + \frac{53(1)^2}{2} - 3(1)^3] - [0] = -37 + 26.5 - 3 = -13.5 \).
Key Concepts
Parameterization of CurvesIntegration TechniquesDifferentiation
Parameterization of Curves
Parameterization is a method to represent a curve with a single parameter.
It is crucial for translating a curve into a form that can be handled more easily.
In the original exercise, the line segment from the point \((3, 4)\) to \((2, 1)\) is parameterized.
The idea is to express the line in terms of a parameter \(t\).
The parameter \(t\) varies from 0 to 1.
We then use a combination of the initial and final points weighted by \(1-t\) and \(t\) respectively.
This gives us the equation: \(\mathbf{r}(t) = (1-t)(3, 4) + t(2, 1)\).
By simplifying, we derive the path functions \(x(t) = 3 - t\) and \(y(t) = 4 - 3t\).
These functions capture the curve's trajectory and changes in position on the x-y plane.
Parameterization is necessary to compute line integrals, allowing us to turn the problem into something solvable with calculus.
This way, the behavior and values of the curve itself can be explored easily.
It is crucial for translating a curve into a form that can be handled more easily.
In the original exercise, the line segment from the point \((3, 4)\) to \((2, 1)\) is parameterized.
The idea is to express the line in terms of a parameter \(t\).
The parameter \(t\) varies from 0 to 1.
We then use a combination of the initial and final points weighted by \(1-t\) and \(t\) respectively.
This gives us the equation: \(\mathbf{r}(t) = (1-t)(3, 4) + t(2, 1)\).
By simplifying, we derive the path functions \(x(t) = 3 - t\) and \(y(t) = 4 - 3t\).
These functions capture the curve's trajectory and changes in position on the x-y plane.
Parameterization is necessary to compute line integrals, allowing us to turn the problem into something solvable with calculus.
This way, the behavior and values of the curve itself can be explored easily.
Integration Techniques
To solve the line integral, integration techniques are vital.
Once we have parameterized the curve, the next step is setting up the integrand with respect to the parameter \(t\).
In our example, this involves substituting the expressions for \(x(t)\) and \(y(t)\) into the original function that we need to integrate.
The formula: \((y-x) \, dx + xy \, dy\) transforms, after substitution, into helping expressions depending on \(t\):
\((4-3t - (3-t))(-1) + (3-t)(4-3t)(-3)\).
Simplifying these expressions is fundamental as it leads to a clearer form of the integrand.
Once the function is simple, we can start the integration process over the interval from 0 to 1.
We deal with polynomial terms like \(-37 + 53t - 9t^2\).
Techniques such as polynomial integration are applied systematically to compute the integral accurately over this defined range.
This step is important for getting the final value of the line integral through calculus.
Once we have parameterized the curve, the next step is setting up the integrand with respect to the parameter \(t\).
In our example, this involves substituting the expressions for \(x(t)\) and \(y(t)\) into the original function that we need to integrate.
The formula: \((y-x) \, dx + xy \, dy\) transforms, after substitution, into helping expressions depending on \(t\):
\((4-3t - (3-t))(-1) + (3-t)(4-3t)(-3)\).
Simplifying these expressions is fundamental as it leads to a clearer form of the integrand.
Once the function is simple, we can start the integration process over the interval from 0 to 1.
We deal with polynomial terms like \(-37 + 53t - 9t^2\).
Techniques such as polynomial integration are applied systematically to compute the integral accurately over this defined range.
This step is important for getting the final value of the line integral through calculus.
Differentiation
Differentiation is the process of finding the derivative of a function.
In the line integral exercise, differentiation is used to find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
These derivatives represent the rates of change of x and y with respect to the parameter \(t\).
For the given parameterizations \(x(t) = 3 - t\) and \(y(t) = 4 - 3t\), the derivatives are straightforward.
Computing them gives \(\frac{dx}{dt} = -1\) and \(\frac{dy}{dt} = -3\).
These results provide the necessary increments per unit change in the parameter \(t\).
Differentiation in line integrals is essential for setting up the differentials like \(dx\) and \(dy\).
These variables are needed as we replace them in the integral formulation.
Proper differentiation ensures our integrand is correctly expressed to allow for straightforward integration afterwards.
Understanding this step is important in ensuring all parts of the integral are correctly constructed.
In the line integral exercise, differentiation is used to find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
These derivatives represent the rates of change of x and y with respect to the parameter \(t\).
For the given parameterizations \(x(t) = 3 - t\) and \(y(t) = 4 - 3t\), the derivatives are straightforward.
Computing them gives \(\frac{dx}{dt} = -1\) and \(\frac{dy}{dt} = -3\).
These results provide the necessary increments per unit change in the parameter \(t\).
Differentiation in line integrals is essential for setting up the differentials like \(dx\) and \(dy\).
These variables are needed as we replace them in the integral formulation.
Proper differentiation ensures our integrand is correctly expressed to allow for straightforward integration afterwards.
Understanding this step is important in ensuring all parts of the integral are correctly constructed.
Other exercises in this chapter
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