Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 20

Parametrize the segment in its stated direction by $\mathbf{x}(t)=(1-t,t,t)$, $0\le t\le 1$. Then $d\mathbf{x}=(-1,1,1)dt$, and substitution into the two vector fields givesTaking each dot product with $d\mathbf{x}$ evaluates the line integrals:

The proposed parametric equation traces precisely the same oriented segment, with a different speed. Set $u=t^n$, which increases from zero to one, and $du=n{t}^{n-1}dt$. Direct substitution gives the line integralsThus both values are unchanged. The derivative of $u$ vanishes at the initial endpoint for $n>1$, but the monotone integration by substitution is still valid. This is invariance under reparametrization of one curve, which applies even to a nonconservative vector field; it is not evidence of general path independence.

The three components of the curl cancel pairwise:The domain is all of ${\mathbb{R}}^3$, which is simply connected, so this is consistent with a conservative vector field. To construct its scalar potential, integrate ${\Phi }_x=y+z$ to obtain $\Phi =x(y+z)+h(y,z)$. Matching ${\Phi }_y=x+z$ gives $h_y=z$, hence $h=yz+k(z)$. Finally ${\Phi }_z=x+y$ forces $k^{\prime }(z)=0$. ConsequentlyEvery scalar potential differs by a constant on this connected domain. The endpoint values differ by one, agreeing with the line integral in part (a).

For the second vector field, the curl isIt is nonzero, for example $(0,-2,2)$ at $(1,0,0)$. A differentiable scalar potential would have zero curl of its gradient, so $\mathbf{G}$ is not conservative. The fact that its curl vanishes on the line $x=y=z$ does not make it vanish throughout the domain.

Take a piecewise linear path along three coordinate directions, passing through the origin and then $(0,1,0)$:On the first segment $G_x=y^2+z^2=0$, so its line integral is zero. On the second $G_y=x^2+z^2=0$, so its contribution is also zero. On the last segment $G_z=x^2+y^2=1$ and $z$ increases from zero to one. ThereforeThis gives the requested concrete path dependence with the same ordered endpoints.