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Chapter 18: Problem 6

Show that \(\int_{c} \mathbf{F} \cdot d \mathbf{r}\) is independent of path by finding a potential function \(f\) for \(F\). $$ \mathbf{F}(x, y)=\left(5 y^{3}+4 y^{3} \sec ^{2} x\right) \mathbf{i}+\left(15 x y^{2}+12 y^{2} \tan x\right) \mathbf{j} $$

Short Answer

Expert verified
The potential function is \(f(x, y) = 5y^3 x + 4y^3 \tan x + C\). The integral is path independent.

Step by step solution

01

Recognize the Form of F

The vector field is given by \( \mathbf{F}(x, y) = (5y^3 + 4y^3 \sec^2 x) \mathbf{i} + (15xy^2 + 12y^2 \tan x) \mathbf{j} \). We need to verify if there exists a potential function \(f(x, y)\) such that \(abla f = \mathbf{F}\).
02

Determine Potential Function Candidates

The partial derivative with respect to \(x\) of the potential function \(f\) should be \(\frac{\partial f}{\partial x} = 5y^3 + 4y^3 \sec^2 x\). Integrating with respect to \(x\), we have:
03

Integrate With Respect to x

Integrate \(5y^3 + 4y^3 \sec^2 x\) with respect to \(x\):\[ f(x, y) = 5y^3 x + 4y^3 \tan x + g(y) \]where \(g(y)\) is an arbitrary function of \(y\).
04

Differentiate With Respect to y

To find \(g(y)\), differentiate \(f(x, y)\) with respect to \(y\):\[ \frac{\partial f}{\partial y} = 15xy^2 + 12y^2 \tan x + g'(y) \]
05

Equate to Known Component

The given \(y\)-component of \(\mathbf{F}\) is \(\frac{\partial f}{\partial y} = 15xy^2 + 12y^2 \tan x\). Substitute this into the above expression:\[ 15xy^2 + 12y^2 \tan x + g'(y) = 15xy^2 + 12y^2 \tan x \]This implies \(g'(y) = 0\).
06

Solve for g(y)

Since \(g'(y) = 0\), we have \(g(y) = C\) where \(C\) is a constant. Hence, the potential function is\[ f(x, y) = 5y^3 x + 4y^3 \tan x + C \]
07

Conclude Independence from Path

Since we found a potential function \(f(x, y) = 5y^3 x + 4y^3 \tan x + C\) such that \(abla f = \mathbf{F}\), the line integral is path independent when within its domain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Function
In vector calculus, a potential function is a scalar function whose gradient is equal to a given vector field. If a vector field \(\mathbf{F}\) on a region \(D\) is conservative, then there exists a potential function \(f\) such that \(\mathbf{F} = abla f\).
  • This implies that the vector field is the gradient of some scalar field.
  • Finding a potential function helps in simplifying calculations of line integrals and establishing path independence.
In our given exercise, we are tasked with finding such a function \(f\) for \(\mathbf{F}(x, y) = (5y^3 + 4y^3 \sec^2 x) \mathbf{i} + (15xy^2 + 12y^2 \tan x) \mathbf{j}\). By integrating the components of \(\mathbf{F}\), we identify \(f(x, y) = 5y^3 x + 4y^3 \tan x + C\), where \(C\) is a constant. This means that \(f\) acts as a potential function for \(\mathbf{F}\).
Path Independence
Path independence is a property of line integrals within a vector field where the integral between two points does not depend on the path taken.
  • A key requirement for path independence is that the vector field must be conservative.
  • If there exists a potential function for the vector field, it confirms path independence.
In context, since we determined that \(\mathbf{F}\) has a potential function \(f(x, y)\), the integral \(\int_{c} \mathbf{F} \cdot d\mathbf{r}\) is path-independent as long as \(c\) lies within the domain where \(f\) is defined. As a result, you can choose any convenient path when computing such integrals, greatly simplifying the calculation process.
Line Integral
A line integral is an integral where a function is evaluated along a curve. In vector calculus, it commonly involves integrating a vector field along a path or curve.
  • Line integrals of vector fields are used to compute quantities like work done by a force.
  • The integral's value depends on both the vector field and the path taken.
However, when a vector field is conservative (as identified by the existence of a potential function), the line integral becomes path-independent. Therefore, the line integral within our exercise can be simplified by evaluating the potential function at the endpoints of the path. This not only makes computations easier but also provides deeper insights into the nature of vector fields.
Gradient Field
A gradient field is a vector field that is the gradient of a scalar field. It is typically denoted as \(\mathbf{F} = abla f\).
  • This field represents the direction and rate of fastest increase for the potential function \(f\).
  • In mathematical terms, a gradient field is conservative.
For our given vector field \(\mathbf{F}(x, y)\), through the process of determining \(f(x, y)\), we confirmed that \(\mathbf{F}(x, y)\) behaves as a gradient field. This means that the vector field can be seen as having originated from a potential function, ensuring that our line integrals over this vector field are path independent. Recognizing a gradient field is crucial in vector calculus, as it significantly simplifies the analysis and computation of integrals.

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Most popular questions from this chapter

A surface (or triple) integral of a vector function is defined as the sum of the surface (or triple) integrals of each component of the function. Using this fact and assuming that \(S\) and \(Q\) satisfy the conditions of the divergence theorem, establish the result. \(\iint_{S} \mathbf{F} \times \mathbf{n} d S=-\iiint_{Q} \nabla \times \mathbf{F} d V\)

Use the divergence theorem (18.26) to find \(\iint_{s} F \cdot \mathbf{n} d S\) \(\mathbf{F}=y \sin x \mathbf{i}+y^{2} z \mathbf{j}+(x+3 z) \mathbf{k}\) \(S\) is the surface of the region bounded by the planes \(x=\pm 1, y=\pm 1, z=\pm 1\)

If a constant force \(\mathbf{c}\) acts on a moving object as it travels once around a circle, show that the work done by \(\mathrm{c}\) on the object is \(0 .\)

Sketch some typical vectors in the vector field F. $$ \mathbf{F}(x, y, z)=\nabla\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2} $$

Use the divergence theorem (18.26) to find the flux of F through \(S\). \(\mathbf{F}(x, y, z)=2 x z \mathbf{i}+x y z \mathbf{j}+y z \mathbf{k}\) \(S\) is the surface of the region bounded by the coordinate planes and the planes \(x+2 z=4\) and \(y=2\).
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