Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 1

A differentiable gradient field must have equal mixed derivatives. Here $\partial F_y/\partial z=3z^2$ but $\partial F_z/\partial y=-x$, which are not identically equal on any open region. Equivalently,Thus no scalar potential exists for this field on an open region. Vanishing of the curl on a special lower-dimensional set would not supply a three-dimensional potential.

The $y$ component integrates to $\varphi =xy{e}^{z-x}+g(x,z)$. Matching the $x$ component gives $g_x=2x$, and matching the $z$ component then gives the remaining derivative zero. ConsequentlyDirect differentiation recovers all three components, proving this is a gradient field globally on ${\mathbb{R}}^3$.

Write the surfaces as level sets $f_1=x^2{y}^{2}z-3x-z^2=0$ and $f_2=3x^2-y^2+2z-1=0$. Their gradients are normal vectors to their tangent planes. At the specified point they are ${\mathbf{n}}_1=(5,-4,2)$ and ${\mathbf{n}}_2=(6,4,2)$, with dot product $18$ and squared norms $45,56$. The angle between intersecting regular surfaces is the angle between their tangent planes, soBoth gradients are nonzero, ensuring that the tangent planes are defined.