The field $\mathbf{F}$ is not a conservative vector field: either nonzero closed line integral computed below rules out a scalar potential. Equivalently, its planar curl is ${\partial }_x{F}_y-{\partial }_y{F}_x=1-(-1)=2$.

For $\mathbf{G}$, integrating ${\Phi }_x=2x{y}^2$ gives $\Phi =x^2{y}^2+h(y)$. Matching ${\Phi }_y=2y{x}^2$ then gives $h^{\prime }=0$. ThusThis also explains why both closed line integrals of $\mathbf{G}$ vanish.

Write $c=\cos \varphi$ and $s=\sin \varphi$. The needed coefficient derivatives are $c_x=s^2/r$, $s_x=-sc/r$, $c_y=-sc/r$ and $s_y=c^2/r$. Apply the product rule to both terms of $f_x=c{g}_r-(s/r)g_{\varphi }$ and use the chain rule for the derivatives of $g$. Collecting terms givesSimilarly, differentiating $f_y=s{g}_r+(c/r)g_{\varphi }$ givesOn addition, both the mixed partial derivative terms and the first angular derivative terms cancel. Using $s^2+c^2=1$, the Laplacian in polar coordinates is

Apply the multivariable chain rule to $f(x,y)=g(r(x,y),\varphi (x,y))$. Using the coordinate partial derivatives above givesEquivalently, ${\partial }_x=\cos \varphi {\partial }_r-(\sin \varphi /r){\partial }_{\varphi }$ and ${\partial }_y=\sin \varphi {\partial }_r+(\cos \varphi /r){\partial }_{\varphi }$. Their coefficients vary with position and must also be differentiated when applying them twice.

Differentiate the angular relation locally, or use $d\varphi =(xdy-ydx)/(x^2+y^2)$. Holding each other coordinate fixed givesThese are local partial derivatives on a consistent angular branch, independent of where its branch cut is placed.

For $r>0$, differentiate $r=\sqrt{x^2+y^2}$ while keeping the other coordinate fixed in the Cartesian coordinate system:The origin is excluded from this polar coordinates chart.

Write $s=\hat{\mathbf{n}}\cdot \mathbf{x}$. Taking the dot product of the equation with the unit vector $\hat{\mathbf{n}}$ gives $2s+s^2=-1$, so $(s+1{)}^2=0$ and $s=-1$.

Now take the cross product with $\hat{\mathbf{n}}$. The vector triple product gives $\hat{\mathbf{n}}\times (\hat{\mathbf{n}}\times \mathbf{x})=s\hat{\mathbf{n}}-\mathbf{x}=-\hat{\mathbf{n}}-\mathbf{x}$, soThe original equation, now with $s^2=1$, also gives $\hat{\mathbf{n}}\times \mathbf{x}=\mathbf{b}-\hat{\mathbf{n}}-2\mathbf{x}$. Substitution eliminates the remaining cross product of the unknown and yieldsIts dot product with $\hat{\mathbf{n}}$ is $-1$, and direct substitution using the vector triple product verifies the original equation. The elimination gives a unique solution without choosing components. It is also an application of the inverse of a scalar plus a cross-product operator after the scalar component has been fixed.

Put $\mathbf{v}={\mathbf{r}}_0-\mathbf{a}$. The squared distance to a point on the line is $|\mathbf{v}-\lambda \hat{\mathbf{t}}{|}^2$. Since $|\hat{\mathbf{t}}|=1$, completing the square in $\lambda$ gives its minimum at $\lambda =\hat{\mathbf{t}}\cdot \mathbf{v}$, with value $|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$. The cross product identity $|\hat{\mathbf{t}}\times \mathbf{v}{|}^2=|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$ therefore provesFor the given line, $\hat{\mathbf{t}}=(1,1,0)/\sqrt{2}$, so its squared distance to $(x,y,z)$ is $(x-y{)}^2/2+z^2$. Substituting the parabola givesThe global minimum occurs at $\mu =\pm 1$, henceThe corresponding points on the parabola are $\pm (1,1+2\sqrt{3},0)+(0,0,-3)$; their nearest points on the line are $\pm (1+\sqrt{3},1+\sqrt{3},0)$. The connecting vectors are orthogonal to the line, as the projection calculation requires.