 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/7/i/solution

u_{x} = v_{y}

and

u_{y} = - v_{x}

. Consequently

\nabla u \cdot \nabla v = u_{x} v_{x} + u_{y} v_{y} = 0, ∣\nabla u ∣^{2} = u_{x}^{2} + u_{y}^{2} = v_{x}^{2} + v_{y}^{2} = ∣\nabla v ∣^{2} .

A gradient is normal to its level set, so these identities give orthogonal level curves wherever the gradients are nonzero. At a critical point where

f^{'} = 0

, both gradients vanish and a regular tangent to a level set need not exist; the geometric orthogonality statement is restricted to regular intersections.

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