Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 13 / a

With $\mathbf{r}(t)=(t,t^2{)}^T$, the tangent vector is $(1,2t{)}^T$ and $\mathbf{F}(\mathbf{r}(t))=(t^3,2t^2{)}^T$. The line integral is

For $\mathbf{r}(t)=(t^2,t{)}^T$, the tangent vector is $(2t,1{)}^T$ and $\mathbf{F}(\mathbf{r}(t))=(t^3,t^4+t{)}^T$. ThusThe unequal values for the same endpoints already rule out a conservative vector field.

Put $q=2t-t^2$. The second coordinate is exactly $q^2$, so this is the same parabola as in part (i), with a different increasing parametrization. Since $q$ runs from zero to one and $dq=(2-2t)dt$, the line integral isEquivalently, substitute the original parameter to get ${\int }_0^{1}5(2t-t^2{)}^3(2-2t)dt$. This demonstrates that a line integral depends on the oriented path, not on its speed of traversal.