Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
Section 16.6 : Conservative Vector Fields
For problems 1 – 3 determine if the vector field is conservative.
- \(\vec F = \left( {{x^3} - 4x{y^2} + 2} \right)\vec i + \left( {6x - 7y + {x^3}{y^3}} \right)\vec j\) Solution
- \(\vec F = \left( {2x\sin \left( {2y} \right) - 3{y^2}} \right)\vec i + \left( {2 - 6xy + 2{x^2}\cos \left( {2y} \right)} \right)\vec j\) Solution
- \(\vec F = \left( {6 - 2xy + {y^3}} \right)\vec i + \left( {{x^2} - 8y + 3x{y^2}} \right)\vec j\) Solution
For problems 4 – 7 find the potential function for the vector field.
- \(\displaystyle \vec F = \left( {6{x^2} - 2x{y^2} + \frac{y}{{2\sqrt x }}} \right)\vec i - \left( {2{x^2}y - 4 - \sqrt x } \right)\vec j\) Solution
- \(\vec F = {y^2}\left( {1 + \cos \left( {x + y} \right)} \right)\vec i + \left( {2xy - 2y + {y^2}\cos \left( {x + y} \right) + 2y\sin \left( {x + y} \right)} \right)\vec j\) Solution
- \(\vec F = \left( {2{z^4} - 2y - {y^3}} \right)\vec i + \left( {z - 2x - 3x{y^2}} \right)\vec j + \left( {6 + y + 8x{z^3}} \right)\vec k\) Solution
- \(\displaystyle \vec F = \frac{{2xy}}{{{z^3}}}\vec i + \left( {2y - {z^2} + \frac{{{x^2}}}{{{z^3}}}} \right)\vec j - \left( {4{z^3} + 2yz + \frac{{3{x^2}y}}{{{z^4}}}} \right)\vec k\) Solution
- Evaluate \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) where C is the portion of the circle centered at the origin with radius 2 in the 1st quadrant with counter clockwise rotation and \(\displaystyle \vec F\left( {x,y} \right) = \left( {2xy - 4 - \frac{1}{2}\sin \left( {\frac{1}{2}x} \right)\sin \left( {\frac{1}{2}y} \right)} \right)\,\vec i + \left( {{x^2} + \frac{1}{2}\cos \left( {\frac{1}{2}x} \right)\cos \left( {\frac{1}{2}y} \right)} \right)\vec j\). Solution
- Evaluate \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) where \(\vec F\left( {x,y} \right) = \left( {2y{{\bf{e}}^{x\,y}} + 2x{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}}} \right)\,\vec i + \left( {2x{{\bf{e}}^{x\,y}} - 2y{{\bf{e}}^{{x^{\,2}} - {y^{\,2}}}}} \right)\vec j\) and C is the curve shown below.
Solution
