StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q. 15

Expert-verifiedFound in: Page 1095

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

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

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

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

That is,

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

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

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

94% of StudySmarter users get better grades.

Sign up for free