Question

If $\overrightarrow{F}=\hat{i}{F}_x+\hat{j}{F}_y+\hat{k}{F}_z$ is conservative, then

• A. $\frac{\partial F_x}{\partial y}=\frac{\partial F_y}{\partial x};\frac{\partial F_y}{\partial z}=\frac{\partial F_z}{\partial y};\frac{\partial F_z}{\partial x}=\frac{\partial F_x}{\partial z}$
• B. $\frac{\partial F_x}{\partial y}\ne \frac{\partial F_y}{\partial x};\frac{\partial F_y}{\partial z}\ne \frac{\partial F_z}{\partial y};\frac{\partial F_z}{\partial x}\ne \frac{\partial F_x}{\partial z}$
• C. $\frac{\partial F_x}{\partial x}=\frac{\partial F_y}{\partial y}=\frac{\partial F_x}{\partial z}\ne 0$
• D. $\frac{\partial F_x}{\partial y}=\frac{\partial F_x}{\partial x}+\frac{\partial F_z}{\partial z}$

Answer 1

The correct option is A $\frac{\partial F_x}{\partial y}=\frac{\partial F_y}{\partial x};\frac{\partial F_y}{\partial z}=\frac{\partial F_z}{\partial y};\frac{\partial F_z}{\partial x}=\frac{\partial F_x}{\partial z}$

$\overrightarrow{F}=\hat{i}{F}_x+\hat{j}{F}_y+\hat{k}{F}_z$
$F_x=-\frac{\partial U}{\partial x};F_y=-\frac{\partial U}{\partial y}$ and $F_z=-\frac{\partial U}{\partial z}$
$\frac{\partial F_x}{\partial y}=-\frac{{\partial }^{2}U}{\partial y.\partial x};\frac{\partial F_y}{\partial z}=-\frac{{\partial }^{2}U}{\partial z.\partial y}$
and $\frac{\partial F_z}{\partial x}=-\frac{{\partial }^{2}U}{\partial x.\partial z}$
Also $\frac{\partial F_x}{\partial z}=-\frac{{\partial }^{2}U}{\partial z.\partial x};\frac{\partial F_y}{\partial x}=-\frac{{\partial }^{2}U}{\partial x.\partial y}$
And $\frac{\partial F_z}{\partial y}=-\frac{{\partial }^{2}U}{\partial y.\partial z}$
Using the property of perfect differential
$\frac{{\partial }^{2}U}{\partial x.\partial y}=\frac{{\partial }^{2}U}{\partial y.\partial x}$
Therefore,
$\frac{\partial F_x}{\partial y}=\frac{\partial F_y}{\partial x};\frac{\partial F_y}{\partial z}=\frac{\partial F_z}{\partial y}$ and $\frac{\partial F_z}{\partial x}=\frac{\partial F_x}{\partial z}$
Hence, the correct answer is option (a).