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.

Answer 2

By using Clairaut's Theorem, the vector field F is shown as:

$\mathbf{\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}} \\ \\ \\ \mathbf{\dfrac{\partial P}{\partial z} = \dfrac{\partial R}{\partial x}} \\ \\ \\ \mathbf{\dfrac{\partial Q}{\partial z}= \dfrac{\partial R}{\partial y}}$

Suppose there exists a vector field $\mathbf{F^{\to } = P\hat i + Q \hat j + R\hat k}$ that is conservative.

Then we can say there exists a function (f) such that; $\nabla f^{\to}= F^{\to}$  

i.e.

$\mathbf{Pi+Qj+Rk = \dfrac{\partial f}{\partial x}i +\dfrac{\partial f}{\partial y}j+ \dfrac{\partial f }{\partial z}k }$

$\mathbf{Pi+Qj+Rk = f_xi +f_yj+f_zk}$

Relating the above to component terms, we have:

$\mathbf{f_x = P, f_y = Q, F_z = R}$

So,

$\mathbf{\dfrac{\partial P}{\partial y }=f_{xy}, \dfrac{\partial Q}{\partial x }=f_{yx} , \dfrac{\partial P}{\partial z }=f_{xz}}$

$\mathbf{\dfrac{\partial R}{\partial x }=f_{zx}, \dfrac{\partial Q}{\partial z }=f_{yz} , \dfrac{\partial R}{\partial y }=f_{zy}}$

Given that, P, Q, and R have continuous first order partials derivatives of F,

Then;

$\mathbf{f_{xy}= f_{yx},} \\ \\ \mathbf{f_{xz}=f_{zx}} \\ \\ \mathbf{f_{yz}= f_{zy}}$

Using Clairaut's Theorem, which showcases that the mixed partial derivatives are equal, we get:

$\mathbf{\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}} \\ \\ \\ \mathbf{\dfrac{\partial P}{\partial z} = \dfrac{\partial R}{\partial x}} \\ \\ \\ \mathbf{\dfrac{\partial Q}{\partial z}= \dfrac{\partial R}{\partial y}}$

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