Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 13

The curl of the first vector field iswhich is not identically zero. Thus $\mathbf{F}$ is not conservative on ${\mathbb{R}}^3$ and cannot equal a negative gradient there.

For the second field, direct differentiation showsTherefore $\mathbf{G}$ is conservative, and the requested negative-gradient potential isIts curl vanishes, consistently with the curl of a gradient identity.

On the first straight segment, $y=z=0$ and only $dx$ is nonzero, so both integrals vanish. On the second segment, $x=1,z=0$ and only $dy$ is nonzero; each integrand is one, contributing $1$. On the last segment, $x=y=1$ and only $dz$ is nonzero; each integrand is $\sinh z$, contributing $\cosh 1-1$. ThusThe second result also follows from the fundamental theorem for line integrals, using the potential found in (a). Agreement on this one path does not make $\mathbf{F}$ conservative.

Parametrize the diagonal by $\mathbf{x}(t)=(t,t,t)$ for $0\le t\le 1$. ThenThe second expression is $(t^3\cosh t{)}^{\prime }$, immediately giving $\int \mathbf{G}\cdot d\mathbf{x}=\cosh 1$. The first differs from it by $t^2\cosh t$. Two applications of integration by parts giveHenceThe changed value for $\mathbf{F}$ explicitly demonstrates path dependence, while $\mathbf{G}$ gives the same endpoint-dependent result.