Chapter 14: Q 5 (page 1153)

Determine whether the vector fields that follow are conservative. If the field is conservative, find a potential function for it.
$F (x, y) = \frac{2 x}{y} i + \frac{x^{2}}{y^{2}} j$

Short Answer

Expert verified

The vector field is non conservative

Step by step solution

Step 1:Given information

The given expression is $F (x, y) = \frac{2 x}{y} i + \frac{x^{2}}{y^{2}} j$

Step 2:Simplification

Consider the vector field,

$F (x, y) = \frac{2 x}{y} i + \frac{x^{2}}{y^{2}} j$

A vector field $F (x, y) = P (x, y) i + Q (x, y) j$ is conservative if and only if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$

now comparing,

$F (x, y) = \frac{2 x}{y} i + \frac{x^{2}}{y^{2}} j$

with $F (x, y) = P (x, y) i + Q (x, y) j$

Then,

$P = \frac{2 x}{y}$

$Q = \frac{x^{2}}{y^{2}}$

now calculating

$\frac{\partial P}{\partial y}, \frac{\partial Q}{\partial x}$

$\frac{\partial P}{\partial y} = \frac{- 2 x}{y^{2}}$ and $\frac{\partial Q}{\partial x} = \frac{2 x}{y^{2}}$

here, $\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}$

Therefore the vector field is non conservative

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Determine whether the vector fields that follow are conservative. If the field is conservative, find a potential function for it.
$F (x, y) = \frac{2 x}{y} i + \frac{x^{2}}{y^{2}} j$

Short Answer

Step by step solution

Step 1:Given information

Step 2:Simplification

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Determine whether the vector fields that follow are conservative. If the field is conservative, find a potential function for it.F(x,y)=2xyi+x2y2j

Short Answer

Step by step solution

Step 1:Given information

Step 2:Simplification

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

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Study anywhere. Anytime. Across all devices.

Determine whether the vector fields that follow are conservative. If the field is conservative, find a potential function for it.
$F (x, y) = \frac{2 x}{y} i + \frac{x^{2}}{y^{2}} j$