Example of closed line integral of conservative field | Multivariable Calculus | Khan Academy

Khan Academy
6 Mar 201011:02
EducationalLearning
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TLDRThe video script discusses solving a line integral along a closed curve, specifically a unit circle in the xy-plane. It introduces the concept of vector line integrals and demonstrates how to express them in terms of differential forms. The script then explores the conservation of a vector field, using the example to show that if a vector field is conservative, as determined by its gradient being equal to a potential scalar field, the line integral over any closed curve is zero. This simplifies the process of solving the integral without the need for explicit antiderivatives.

Takeaways
  • πŸ“ The problem involves solving a line integral along a closed curve in the xy-plane, specifically a unit circle.
  • πŸ”„ The curve is parameterized by x = cos(t) and y = sin(t) for t between 0 and 2Ο€.
  • 🎯 The integrand is a function of x^2 + y^2 * dx + 2xy * dy, which can be rewritten in vector form as the dot product of the vector field and dr.
  • πŸŒ€ Recognizing the integrand's vector form allows for the application of concepts from vector calculus to simplify the process.
  • πŸ€” The question arises whether the given function is conservative, meaning it can be written as the gradient of a scalar potential function.
  • 🧠 By assuming the existence of a scalar potential function F, we can determine its partial derivatives with respect to x and y.
  • πŸ“š The antiderivative of x^2 with respect to x gives x^3/3, and the antiderivative of y^2 with respect to x gives xy^2, plus a function of y.
  • πŸ”§ By matching the components, we find that the scalar potential function F could be F(x, y) = (x^3/3) + (xy^2) + g(y), where g(y) is a function of y.
  • πŸ’‘ It is determined that g(y) must be zero for the function to be conservative, leading to F(x, y) = (x^3/3) + (xy^2).
  • 🌐 The gradient of F is calculated and confirmed to be equal to the original vector field f, confirming that f is conservative.
  • πŸŽ‰ Since f is conservative, the line integral around a closed curve is zero, which is the main result of the analysis.
Q & A

  • What is the given line integral in the script?

    -The given line integral is of the form ∫(x^2 + y^2)dx + (2xy)dy along a closed curve C.

  • How is the curve C parameterized in the script?

    -The curve C is parameterized by x = cos(t) and y = sin(t) for t between 0 and 2Ο€, representing a unit circle in the xy-plane.

  • What does the script suggest about the relationship between the given line integral and vector line integrals?

    -The script suggests that the given line integral is another form of writing a vector line integral. By multiplying the differentials dx and dy by dt, we can express dr as dx times the unit vector i plus dy times the unit vector j, which is equivalent to the differential form of a vector line integral.

  • How is the vector field f defined in the script?

    -The vector field f is defined as f(x, y) = (x^2 + y^2)i + (2xy)j.

  • What is the process to determine if the vector field f is conservative?

    -To determine if f is conservative, we assume that f is the gradient of some scalar field F, denoted as f = βˆ‡F. We then find the partial derivatives of F with respect to x and y and check if they match the components of f. If they do, then f is conservative.

  • What are the conditions derived for the scalar field F in the script?

    -The conditions derived for the scalar field F are that the partial derivative with respect to x, βˆ‚F/βˆ‚x, must equal x^2 + y^2, and the partial derivative with respect to y, βˆ‚F/βˆ‚y, must equal 2xy.

  • How is the scalar field F constructed in the script?

    -The scalar field F is constructed by taking the antiderivative with respect to x and y separately. For the x component, the antiderivative of x^2 + y^2 with respect to x is (1/3)x^3 + xy^2 + g(y), where g(y) is a function of y. For the y component, the antiderivative of 2y with respect to y is y^2. Combining these, F(x, y) = (1/3)x^3 + xy^2 + y^2, with the understanding that g(y) must be 0 for the gradient to equal f.

  • What is the significance of showing that f is the gradient of a scalar field?

    -Showing that f is the gradient of a scalar field demonstrates that f is a conservative vector field. This means that any closed loop line integral of f over a closed curve is equal to 0, which is a key property of conservative fields.

  • How does the script conclude that the line integral of f over the curve C is 0?

    -The script concludes that the line integral of f over the curve C is 0 by showing that f is conservative. Since f is the gradient of the scalar field F, it follows that the line integral over any closed curve, including the unit circle C, must be 0.

  • What is the final expression for the scalar field F found in the script?

    -The final expression for the scalar field F found in the script is F(x, y) = (1/3)x^3 + xy^2 + y^2.

  • How does the parameterization of the curve C help in solving the line integral?

    -The parameterization of the curve C as a unit circle simplifies the process of solving the line integral by allowing us to express the curve in terms of a single variable t, whichε‘¨ζœŸζ€§εœ°ιεŽ† the circle. This makes the computation of the line integral more straightforward, as it allows us to focus on the properties of the vector field f and its relationship to the curve C without getting into the complexities of the parameterization.

Outlines
00:00
πŸ“š Introduction to Solving Line Integrals

This paragraph introduces the concept of solving line integrals using new tools. The focus is on a specific line integral along a closed curve, defined by a parameterization in terms of cosine and sine functions, representing a unit circle in the xy-plane. The goal is to evaluate the line integral of the function x^2 + y^2 * dx + 2xy * dy over this curve. The discussion emphasizes the importance of recognizing the vector line integral form and the process of simplifying the line integral using discoveries from previous videos.

05:03
πŸ” Investigating the Conservatism of the Vector Field

In this paragraph, the question of whether the vector field f(x, y) = (x^2 + y^2)i + (2xy)j is conservative is explored. To determine this, the paragraph outlines the process of finding a scalar potential function F such that its gradient is f. By matching the partial derivatives of the proposed function F with respect to x and y to the components of f, a potential function F is constructed. The paragraph concludes with the calculation of the gradient of F to confirm that it indeed equals f, indicating that f is conservative.

10:07
πŸŽ“ Conclusion on the Closed Loop Line Integral

The final paragraph concludes that since the vector field f is conservative, the closed loop line integral over any path, including the unit circle defined by the parameterization, must be zero. This result is significant because it means that the actual parameterization of the path is not needed to determine the value of the line integral for a conservative field. The paragraph reinforces the understanding that the line integral of a conservative vector field over a closed curve is always zero.

Mindmap
Keywords
πŸ’‘Line Integrals
Line integrals are a fundamental concept in calculus, used to calculate the integral of a function along a specific path or curve. In the video, the focus is on applying new tools to solve line integrals, particularly for a function defined along a closed curve. The example given involves integrating a function of x and y over a unit circle, demonstrating how to transform the problem into a more recognizable form for solving.
πŸ’‘Vector Field
A vector field is a mathematical field where each point in space is associated with a vector. In the context of the video, the vector field is used to describe the function being integrated, with components along the i and j axes corresponding to the x and y components of the function. The vector field is key to understanding the line integral, as it is the function that is being integrated over the path.
πŸ’‘Dot Product
The dot product is an operation that takes two vectors and returns a scalar value. It is calculated by multiplying corresponding components of the vectors and summing these products. In the video, the dot product is used to find the value of the line integral by multiplying the vector field by the differential vector dr, which represents the infinitesimal change in position along the path.
πŸ’‘Conservative Field
A conservative field is a vector field that can be derived from a scalar potential function. This means that the line integral of a conservative field around any closed loop is zero. The concept is crucial in the video, as it allows for the simplification of the line integral calculation. If a field is conservative, it implies that the work done in moving along a closed loop is zero, which can be a significant simplification in calculations.
πŸ’‘Gradient
The gradient is a multivariable generalization of the derivative, representing the rate of change of a scalar-valued function with respect to the variables of a multivariable space. It is a vector field that points in the direction of the greatest rate of increase of the function and is used to determine whether a vector field is conservative. In the video, the gradient is used to find the potential function that could generate the given vector field.
πŸ’‘Parameterization
Parameterization is a mathematical technique used to represent curves or surfaces in a multivariable space by assigning a set of parametric equations. In the context of the video, parameterization is used to define the closed curve along which the line integral is taken, with x and y being expressed as functions of the parameter t.
πŸ’‘Differential dr
In the context of the video, differential dr represents the infinitesimal change in position along the curve or path being integrated over. It is used to express the change in the vector position as the parameter t changes, which is crucial for calculating line integrals. The differential dr is expressed in terms of dx and dy, multiplied by the differential dt, which represents the infinitesimal change in the parameter t.
πŸ’‘Antiderivatives
Antiderivatives, or indefinite integrals, are functions that provide a family of functions whose derivative is the given function. In the video, antiderivatives are used to find the potential scalar function F, from which the vector field f is derived. This process involves integrating with respect to one variable while treating the other as a constant.
πŸ’‘Scalar Potential Function
A scalar potential function is a scalar-valued function that, when differentiated, yields a given vector field. In the context of conservative fields, the vector field is the gradient of the scalar potential function. The video aims to find such a function for the given vector field f, which would confirm that f is conservative.
πŸ’‘Closed Loop Integral
A closed loop integral refers to the line integral of a function over a path that starts and ends at the same point, forming a closed curve. The property of conservative fields states that the line integral over any closed loop is zero. This is a significant result used in the video to conclude that the line integral of the given vector field f over the unit circle is zero.
πŸ’‘Unit Circle
A unit circle is a circle with a radius of 1 centered at the origin of a coordinate system. In the video, the unit circle is the specific curve along which the line integral is calculated, defined by the parameterization x = cos(t) and y = sin(t) for t between 0 and 2Ο€.
Highlights

The discussion begins with an introduction to solving line integrals using new tools.

A line integral is considered along a closed curve with a specific path defined in the xy-plane.

The curve is parameterized by x as cosine of t and y as sine of t, valid for t between 0 and 2 pi, essentially forming a unit circle.

The integral involves the function of x squared plus y squared times dx plus 2xy times dy.

The concept of vector line integrals is introduced, showing that the given integral is a form of a vector line integral.

The differential dr is expressed as dx times i plus dy times j, and the dt is eliminated by multiplication.

A vector field f is defined based on the given integral, with components x squared plus y squared i and 2xy j.

The dot product of the vector field f and dr is calculated, aligning with the original line integral.

The question of whether the vector field f is conservative is raised, seeking a potential scalar field F.

If f is conservative, the line integral over any closed curve would equal zero, simplifying the problem.

The partial derivatives of a potential scalar field F with respect to x and y are determined to match the components of f.

The process of finding a scalar field F whose gradient equals f is discussed, involving antiderivative calculations.

A candidate for the scalar field F is proposed, combining terms involving x and y.

The gradient of the proposed scalar field F is calculated and confirmed to equal f, proving that f is indeed conservative.

The conclusion that the closed loop line integral of a conservative field f is zero is reached, solving the initial problem without detailed path parameterization.

Transcripts

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