 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/7/a/solution

u_{x} = v_{y}

and

u_{y} = - v_{x}

. Therefore

∣\nabla u ∣^{2} = u_{x}^{2} + u_{y}^{2} = v_{y}^{2} + v_{x}^{2} = ∣\nabla v ∣^{2} .

Taking the nonnegative square root yields

∣\nabla u ∣ = ∣\nabla v ∣.

The two gradients are also orthogonal, since

\nabla u \cdot \nabla v = v_{y} v_{x} - v_{x} v_{y} = 0

, although equality of their lengths alone is the requested result.

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