$[-/0\mathrm{.}83$ Points$]$ SCALCET$9$M $16\mathrm{.}3\mathrm{.}035.$

Show that if the vector field $F=+Qj+Rk$ is conservative and $P,Q,R$ have continuous first$-$order partial derivatives, then the following is true.

$\frac{\text{d}\text{e}\text{l}\text{P}}{\text{d}\text{e}\text{l}\text{y}}=\frac{\text{d}\text{e}\text{l}\text{Q}}{\text{d}\text{e}\text{l}\text{x}},\frac{\text{d}\text{e}\text{l}\text{P}}{\text{d}\text{e}\text{l}\text{z}}=\frac{\text{d}\text{e}\text{l}\text{R}}{\text{d}\text{e}\text{l}\text{x}},\frac{\text{d}\text{e}\text{l}\text{Q}}{\text{d}\text{e}\text{l}\text{z}}=\frac{\text{d}\text{e}\text{l}\text{R}}{\text{d}\text{e}\text{l}\text{y}}.$

Since $F$ is conservative, there exists a function $f$ such that $F=$gradf, that is$,P,Q,$ and $R$ are defined as follows. $($Enter your answers in the form $f_{x^{\text{'}}}{f}_y,f_z.)$

$P=$

$Q=$

$R=$

Since $P,Q,$ and $R$ have continuous first order partial derivatives $$ says that del$\frac{P}{d}$ely$=f_{xy}=f_{yx}=$del$\frac{Q}{d}$elx,del$\frac{P}{d}$elz$=f_{xz}=f_{zx}=$del$\frac{R}{d}$elx, and del$\frac{Q}{d}$elz$=f_{yz}=f_{zy}=$del$\frac{R}{d}$ely.

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