Question

Show that if the vector field F = Pi + Qj + Rk is conservative and P, Q, R have continuous first-order partial derivatives, then the following is true. ∂P ∂y = ∂Q ∂x ∂P ∂z = ∂R ∂x ∂Q ∂z = ∂R ∂y . Since F is conservative, there exists a function f such that F = ∇f, that is, P, Q, and R are defined as follows. (Enter your answers in the form fx, fy, fz.) P = Q = R = Since P, Q, and R have continuous first order partial derivatives, says that ∂P/∂y = fxy = fyx = ∂Q/∂x, ∂P/∂z = fxz = fzx = ∂R/∂x, and ∂Q/∂z = fyz = fzy = ∂R/∂y.

Answer 1

Answer:

It is proved that $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}$

Step-by-step explanation:

Given vector field,

$F=P\uvec{i}+Q\uvec{j}+R\uvec{k}$

Where,

$P=f_x=\frac{\partial f}{\partial x}, Q=f_y=\frac{\partial f}{\partial y}, R=f_z=\frac{\partial f}{\partial z}$

To show,

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial P}{\partial z}=\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}$

Consider,

$\frac{\partial P}{\partial y}=\frac{\partial}{\partial y}(\frac{\partial f}{\partial x})=\frac{\partial^2 f}{\partial y\partial x}=\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial y})=\frac{\partial Q}{\partial x}$

$\frac{\partial P}{\partial z}=\frac{\partial}{\partial z}(\frac{\partial f}{\partial x})=\frac{\partial^2 f}{\partial z\partial x}=\frac{\partial^2 f}{\partial x\partial z}=\frac{\partial }{\partial x}(\frac{\partial f}{\partial z})=\frac{\partial R}{\partial x}$

$\frac{\partial Q}{\partial z}=\frac{\partial}{\partial z}(\frac{\partial f}{\partial y})=\frac{\partial^2 f}{\partial z\partial y}=\frac{\partial^2 f}{\partial y\partial z}=\frac{\partial}{\partial y}(\frac{\partial f}{\partial z})=\frac{\partial R}{\partial y}$

Hence proved.