Determine whether the following force field F is conservative: F=xi-yj+zk
The force field is conservative if and only if the curl of the force is equal to zero.
By definition, the curl is:
∇ × F = ∣ ı ^ ȷ ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z F x F y F z ∣ = ( ∂ F z ∂ y − ∂ F y ∂ z ) ı ^ + ( ∂ F x ∂ z − ∂ F z ∂ x ) ȷ ^ + ( ∂ F y ∂ x − ∂ F x ∂ y ) k ^ \nabla \times \mathbf{F} =
\begin{vmatrix} \boldsymbol{\hat\imath} & \boldsymbol{\hat\jmath} & \boldsymbol{\hat k} \\[5pt]
{\dfrac{\partial}{\partial x}} & {\dfrac{\partial}{\partial y}} & {\dfrac{\partial}{\partial z}} \\[10pt]
F_x & F_y & F_z \end{vmatrix} =
\left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \boldsymbol{\hat\imath} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \right) \boldsymbol{\hat\jmath} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \boldsymbol{\hat k} ∇ × F = ∣ ∣ ^ ∂ x ∂ F x ^ ∂ y ∂ F y k ^ ∂ z ∂ F z ∣ ∣ = ( ∂ y ∂ F z − ∂ z ∂ F y ) ^ + ( ∂ z ∂ F x − ∂ x ∂ F z ) ^ + ( ∂ x ∂ F y − ∂ y ∂ F x ) k ^ .
Substitute the values of the derivatives into the last expression:
∇ × F = ( ∂ z ∂ y − ∂ ( − y ) ∂ z ) ı ^ + ( ∂ x ∂ z − ∂ z ∂ x ) ȷ ^ + ( ∂ ( − y ) ∂ x − ∂ x ∂ y ) k ^ = = ( 0 − 0 ) ı ^ + ( 0 − 0 ) ȷ ^ + ( 0 − 0 ) k ^ = 0 \nabla \times \mathbf{F} =
\left(\frac{\partial z}{\partial y} - \frac{\partial (-y)}{\partial z}\right) \boldsymbol{\hat\imath} + \left(\frac{\partial x}{\partial z} - \frac{\partial z}{\partial x} \right) \boldsymbol{\hat\jmath} + \left(\frac{\partial (-y)}{\partial x} - \frac{\partial x}{\partial y} \right) \boldsymbol{\hat k} =\\
=\left(0 - 0\right) \boldsymbol{\hat\imath} + \left(0-0 \right) \boldsymbol{\hat\jmath} + \left(0 -0 \right) \boldsymbol{\hat k} =0 ∇ × F = ( ∂ y ∂ z − ∂ z ∂ ( − y ) ) ^ + ( ∂ z ∂ x − ∂ x ∂ z ) ^ + ( ∂ x ∂ ( − y ) − ∂ y ∂ x ) k ^ = = ( 0 − 0 ) ^ + ( 0 − 0 ) ^ + ( 0 − 0 ) k ^ = 0
Thus, the force field is conservative. QED.
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