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Chapter 6: Problem 118

For the following exercises, determine whether the vector field is conservative and, if so, find a potential function. $$ \mathbf{F}(x, y)=\left(2 x y e^{x^{2} y}\right) \mathbf{i}+6\left(x^{2} e^{x^{2} y}\right) \mathbf{j} $$

Short Answer

Expert verified
The vector field is not conservative.

Step by step solution

01

Determine Conditions for Conservativeness

A vector field \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \) is conservative if there exists a scalar function \( f(x, y) \) such that \( abla f = \mathbf{F} \). For this, partial derivatives should satisfy \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). Here, \( P(x, y) = 2xy e^{x^2 y} \) and \( Q(x, y) = 6x^2 e^{x^2 y} \).
02

Compute Partial Derivatives

Compute \( \frac{\partial P}{\partial y} \):\[ \frac{\partial}{\partial y} (2xy e^{x^2 y}) = 2x \frac{\partial}{\partial y}(y e^{x^2 y}) = 2x(1 + x^2 y) e^{x^2 y} = 2x e^{x^2 y} + 2x^3 y e^{x^2 y} \].Compute \( \frac{\partial Q}{\partial x} \):\[ \frac{\partial}{\partial x} (6x^2 e^{x^2 y}) = 12x e^{x^2 y} + 6x^2(2xy) e^{x^2 y} = 12xe^{x^2 y} + 12x^3 y e^{x^2 y} \].
03

Compare Partial Derivatives

Analyze the equality \( \frac{\partial P}{\partial y} = 2x e^{x^2 y} + 2x^3 y e^{x^2 y} \) and \( \frac{\partial Q}{\partial x} = 12xe^{x^2 y} + 12x^3 y e^{x^2 y} \). As we can see, they are not equal because the coefficients do not match. Specifically, \( 2x eq 12x \) and \( 2x^3 eq 12x^3 \).
04

Conclusion on Conservativeness

Since \( \frac{\partial P}{\partial y} eq \frac{\partial Q}{\partial x} \), the vector field \( \mathbf{F}(x, y) \) is not conservative.

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Key Concepts

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

Potential Functions
A potential function is a scalar function that can define a vector field. It is important in determining if a vector field is conservative. If a vector field \(\mathbf{F}\) has a potential function \(f(x, y)\), then \(abla f = \mathbf{F}\). This means the gradient of the potential function equals the vector field. This aligns with the idea that you can recover the original vector field from its potential function.
  • If you can find a potential function, the vector field is termed conservative.
  • Not all vector fields have potential functions.
In practice, identifying a potential function simplifies many calculations, especially in physics and engineering. For instance, conservative fields often mean that movement along different paths between two points does not change the "work done" or "energy spent."
Partial Derivatives
Partial derivatives are a cornerstone in understanding and finding potential functions. They let you measure how a multivariable function changes when you change one variable, while keeping others constant. In vector calculus, they are essential:
  • Matches Needed: For a vector field to be conservative, the partial derivative of one component with respect to a variable should equal the partial derivative of the other component with the alternate variable. That means \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\).
  • Tool for Verification: This condition comes in handy to verify if there's any potential function available for a given vector field.
When you calculate partial derivatives, you look at every term with great care. It involves:
  • Using product rule and chain rule in calculus.
  • Rigor and precision in differentiation, especially for complex functions.
Vector Calculus
Vector Calculus is the mathematics of multi-variable functions. It explores vectors and can describe motion, forces, and vector fields in physics. One of its key applications is determining if a vector field is conservative.
  • Vector fields: Represented as \(\mathbf{F}(x, y, z)\), showing vector quantities changing across space.
  • Conservative Vector Fields: If a vector field is conservative, it can be reduced to the gradient of some potential function \( f(x,y,z) \).
To examine whether a vector field is conservative, Vector Calculus uses tools like:
  • Gradients \(abla f\), consisting of partial derivatives.
  • Curl \(abla \times \mathbf{F}\), which must be zero for two-dimensional conservative fields.
Understanding these properties in Vector Calculus helps in navigating fields where directions and magnitudes of quantities vary, laying the foundation in both theoretical and applied contexts.

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Most popular questions from this chapter

Find the line integral of \(\int_{C} z^{2} d x+y d y+2 y d z\) where \(C\) consists of two parts: \(C_{1}\) and \(C_{2} . C_{1}\) is the intersection of cylinder \(x^{2}+y^{2}=16\) and plane \(z=3\) from \((0,4,3)\) to \((-4,0,3) . \quad C_{2}\) is a line segment from \((-4,0,3)\) to \((0,1,5)\)

For the following exercises, evaluate the line integrals. [T] Use a CAS to evaluate \(\int_{C} x y d s, \quad\) where \(C\) is \(x=t^{2}, y=4 t, 0 \leq t \leq 1\)

For the following exercises, use a computer algebra system (CAS) to evaluate the line integrals over the indicated path. [T] Evaluate \(\int_{C} x y^{4} d s,\) where \(C\) is the right half of circle \(x^{2}+y^{2}=16\) and is traversed in the clockwise direction.

For the following exercises, find the flux. Let \(\mathbf{F}=\left(x^{2}+y^{3}\right) \mathbf{i}+(2 x y) \mathbf{j}\) . Calculate flux \(\mathbf{F}\) orientated counterclockwise across curve \(C : x^{2}+y^{2}=9\)

For the following exercises, find the work done. Find the work done by a person weighing 150 \(\mathrm{lb}\) walking exactly one revolution up a circular, spiral staircase of radius 3 \(\mathrm{ft}\) if the person rises 10 \(\mathrm{ft}\) .
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