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:
"\\nabla \\times \\mathbf{F} =\n\\begin{vmatrix} \\boldsymbol{\\hat\\imath} & \\boldsymbol{\\hat\\jmath} & \\boldsymbol{\\hat k} \\\\[5pt]\n{\\dfrac{\\partial}{\\partial x}} & {\\dfrac{\\partial}{\\partial y}} & {\\dfrac{\\partial}{\\partial z}} \\\\[10pt]\nF_x & F_y & F_z \\end{vmatrix} =\n\\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}" .
Substitute the values of the derivatives into the last expression:
"\\nabla \\times \\mathbf{F} =\n\\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} =\\\\\n=\\left(0 - 0\\right) \\boldsymbol{\\hat\\imath} + \\left(0-0 \\right) \\boldsymbol{\\hat\\jmath} + \\left(0 -0 \\right) \\boldsymbol{\\hat k} =0"
Thus, the force field is conservative. QED.
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