Chapter 14: Q. 40 (page 1096)

Show that the vector fields in Exercises 33–40 are not conservative.
$G (x, y, z) = (e^{y} + z, x e^{y} + z, x + y)$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information

We have to show that the vector fields in the given exercise is not conservative.

$G (x, y, z) = (e^{y} + z, x e^{y} + z, x + y)$

Step 2. A vector field G(x,y)=(G1(x,y),G2(x,y)) is not conservative if and only if 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j4yPC90ZXh0Pjx0ZXh0IGZvbnQtZmFtaWx5PSJtYXRoMTRlM2NkNDkyNTllNmY2MDQ1NWUyNTUyNzgzIiBmb250LXNpemU9IjE2IiB0ZXh0LWFuY2hvcj0ibWlkZGxlIiB4PSI3MS41IiB5PSI0NCI+JiN4MjIwMjs8L3RleHQ+PHRleHQgZm9udC1mYW1pbHk9IkFyaWFsIiBmb250LXNpemU9IjE2IiBmb250LXN0eWxlPSJpdGFsaWMiIHRleHQtYW5jaG9yPSJtaWRkbGUiIHg9IjgyLjUiIHk9IjQ0Ij55PC90ZXh0Pjx0ZXh0IGZvbnQtZmFtaWx5PSJtYXRoMTRlM2NkNDkyNTllNmY2MDQ1NWUyNTUyNzgzIiBmb250LXNpemU9IjE2IiB0ZXh0LWFuY2hvcj0ibWlkZGxlIiB4PSI5Ny41IiB5PSIzMiI+LjwvdGV4dD48L3N2Zz4=" localid="1650549895551" ∂F1∂z≠∂F2∂y.

For the vector field $G (x, y, z) = (e^{y} + z, x e^{y} + z, x + y)$

$\frac{\partial F_{1}}{\partial z} = \frac{\partial}{\partial y} (x e^{y} + z) \frac{\partial F_{1}}{\partial z} = \frac{\partial}{\partial z} x e^{y} + \frac{\partial}{\partial z} z \frac{\partial F_{1}}{\partial z} = 0 + \frac{\partial}{\partial z} z \frac{\partial F_{1}}{\partial z} = 1 \frac{\partial F_{1}}{\partial y} = x e^{y}$

Step 3. Now finding ∂F2∂y

$\frac{\partial F_{2}}{\partial y} = \frac{\partial}{\partial y} (x + y) \frac{\partial F_{2}}{\partial y} = \frac{\partial}{\partial y} x + \frac{\partial}{\partial y} y \frac{\partial F_{2}}{\partial y} = 0 + 1 \frac{\partial F_{2}}{\partial y} = 1$

Hence, $\frac{\partial F_{1}}{\partial z} \neq \frac{\partial F_{2}}{\partial y} .$

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Show that the vector fields in Exercises 33–40 are not conservative.
$G (x, y, z) = (e^{y} + z, x e^{y} + z, x + y)$

Short Answer

Step by step solution

Step 1. Given Information

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,z)=(ey+z,xey+z,x+y)

Short Answer

Step by step solution

Step 1. Given Information

Step 3. Now finding ∂F2∂y

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Discrete Mathematics

Pure Maths

Geometry

Theoretical and Mathematical Physics

Probability and Statistics

Study anywhere. Anytime. Across all devices.

Show that the vector fields in Exercises 33–40 are not conservative.
$G (x, y, z) = (e^{y} + z, x e^{y} + z, x + y)$