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

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.