Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 13

Parametrize the path by $\mathbf{r}(t)=(x(t),y(t),z(t))$, from $t=a$ to $t=b$. For the conservative vector field $\mathbf{F}=\nabla \Phi$, the chain rule givesIntegrating proves the fundamental theorem for line integrals:Only the endpoints matter, so the integral is independent of the connecting path.

For a twice continuously differentiable scalar potential, the curl of its gradient isEquality of mixed partial derivatives makes every component zero. ThusThis is the curl of a gradient identity.

Differentiate the components directly:Thus $\boxed{\nabla \times \mathbf{F}=0}$. The field is smooth on all of ${\mathbb{R}}^3$, a simply connected domain, so the zero-curl condition is sufficient for a conservative vector field. The domain condition matters: zero curl alone need not imply a global potential on a region with holes.

Evaluate the line integral along the three coordinate segments, treating coordinates not changing along a segment as constants. On the first segment, $y=z=0$, so $F_x=0$ and the contribution is zero. On the second segment,On the final segment,The fundamental theorem for line integrals therefore givesTaking its gradient verifies each component of the field. The constant is $C=\Phi (0,0,0)$; the field specifies only potential differences because the gradient of every constant is zero.

For $\mathbf{r}(t)=(t,t,t^2)$, the tangent vector is ${\mathbf{r}}^{\prime }(t)=(1,1,2t)$. Substitute into the field:Its scalar product with the tangent is $3t^2-4t-7t^6$. Hence the explicitly parametrized line integral isAs a check, the potential from part (d) has values $-128+C$ at $(2,2,4)$ and $-2+C$ at $(1,1,1)$, giving the same difference.