Chapter 14: Q. 35 (page 1096)

Show that the vector fields in Exercises 33–40 are not conservative.
$G (x, y) = \frac{1}{x^{2} + y} i + \frac{y}{x} j$

Short Answer

Expert verified

The vector fields is not conservative because $\frac{\partial G_{1}}{\partial y} \neq \frac{\partial G_{2}}{\partial x}$ .

Step by step solution

Step 1. Given Information

We have to show that the vector fields in the given exercise is not conservative.
$G (x, y) = \frac{1}{x^{2} + y} i + \frac{y}{x} j$

Step 2. A vector field G(x,y)=(G1(x,y),G2(x,y)) is not conservative if and only if ∂G1∂y≠∂G2∂x.

For the vector field

$\frac{\partial F_{1}}{\partial y} = \frac{\partial}{\partial y} \frac{1}{x^{2} + y} \frac{\partial F_{1}}{\partial y} = \frac{\partial}{\partial y} {(x^{2} + y)}^{- 1} \frac{\partial F_{1}}{\partial y} = - 1 {(x^{2} + y)}^{- 2} \frac{\partial F_{1}}{\partial y} = - {(x^{2} + y)}^{- 2}$

Step 3. Now finding ∂F2∂y

$\frac{\partial F_{2}}{\partial x} = \frac{\partial}{\partial x} (\frac{y}{x}) \frac{\partial F_{2}}{\partial x} = y \frac{\partial}{\partial x} (\frac{1}{x}) \frac{\partial F_{2}}{\partial x} = y \frac{\partial}{\partial x} x^{- 1} \frac{\partial F_{2}}{\partial x} = - y x^{- 2}$

Hence, $\frac{\partial G_{1}}{\partial y} \neq \frac{\partial G_{2}}{\partial x} .$

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Show that the vector fields in Exercises 33–40 are not conservative.
$G (x, y) = \frac{1}{x^{2} + y} i + \frac{y}{x} j$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. A vector field G(x,y)=(G1(x,y),G2(x,y)) is not conservative if and only if ∂G1∂y≠∂G2∂x.

Step 3. Now finding ∂F2∂y

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Show that the vector fields in Exercises 33–40 are not conservative.G(x,y)=1x2+yi+yxj

Short Answer

Step by step solution

Step 1. Given Information

Step 2. A vector field G(x,y)=(G1(x,y),G2(x,y)) is not conservative if and only if ∂G1∂y≠∂G2∂x.

Step 3. Now finding ∂F2∂y

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Study anywhere. Anytime. Across all devices.

Show that the vector fields in Exercises 33–40 are not conservative.
$G (x, y) = \frac{1}{x^{2} + y} i + \frac{y}{x} j$