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Expert-verified Found in: Page 1095 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # How would you show that a given vector field in ${\mathrm{ℝ}}^{2}$ is not conservative?

A given vector field in ${\mathrm{ℝ}}^{2}$ is not conservative when, $\mathrm{F}\left(x,y\right)\ne \nabla f\left(x,y\right)\ne \frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}$

See the step by step solution

## Step 1.Given Information

How would you show that a given vector field in ${\mathrm{ℝ}}^{2}$ is not conservative?

## Step 2. A vector field is conservative

If a vector field can be described as a gradient, it offers a number of mathematically appealing properties, including the ability to be easily integrated along curves. Such fields are referred to as conservative vector fields.

A vector field that is conservative. F is a vector field that represents the gradient of a function $f$.

## Step 3. A conservative vector field F is a vector field that can be written as the gradient of some function f.

That is,

$\mathrm{F}\left(x,y\right)=\nabla f\left(x,y\right)=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}$

Hence, we can say that a given vector field in ${\mathrm{ℝ}}^{2}$ is not conservative, when

$\mathrm{F}\left(x,y\right)\ne \nabla f\left(x,y\right)\ne \frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}$ ### Want to see more solutions like these? 