 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2/16/a/ii/solution

\partial_{y} (y^{2} - 2 x) = 2 y = \partial_{x} (2 x y - y) .

The vector field is smooth on the entire simply connected plane, so this condition establishes that it is a conservative vector field. We can also construct its vector-field potential explicitly. From

ϕ_{x} = y^{2} - 2 x

x

ϕ = x y^{2} - x^{2} + h (y) .

Then

ϕ_{y} = 2 x y + h^{'} (y) = 2 x y - y

requires

h^{'} (y) = - y

. Hence

ϕ (x, y) = x y^{2} - x^{2} - \frac{1}{2} y^{2} + C, F = \nabla ϕ .

ϕ (1, 1) - ϕ (0, 0) = - 1/2

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