Question #223883

Show that the vector →{v}=(yz-1)i-z(1+x+z)j+y(1+x+2z)k isconservative and find it scalar potential function?


1
Expert's answer
2021-08-31T01:42:49-0400

Given:

v=(yz1)i+z(1+x+z)j+y(1+x+2z)k{\bf v}=(yz-1){\bf i}+z(1+x+z){\bf j}+y(1+x+2z){\bf k}


For the conservative vector field


curlv=0\rm curl {\bf v}=\bf 0

We have


curlv=ijkxyz(yz1)z(1+x+z)y(1+x+2z)\rm curl {\bf v}=\begin{vmatrix} {\bf i} & {\bf j} & {\bf k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ (yz-1)&z(1+x+z)& y(1+x+2z) \end{vmatrix}=(1+x+2z(1+x+2z))i=(1+x+2z-(1+x+2z)){\bf i}+(yy)j+(zz)k=0+(y-y){\bf j}+(z-z){\bf k}=\bf 0

Therefore, the given vector field is conservative.

We can put


v=ψ{\bf v}=\nabla\psi

where


ψ=yz(1+x+z)x\psi=yz(1+x+z)-x

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