 Solution

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

\partial_{j} = \partial / \partial x_{j}

[\nabla \times (a \times b)]_{i} = ϵ_{ijk} \partial_{j} (ϵ_{k l m} a_{l} b_{m}) = (δ_{i l} δ_{jm} - δ_{im} δ_{j l}) \partial_{j} (a_{l} b_{m}) = \partial_{j} (a_{i} b_{j} - a_{j} b_{i}) = a_{i} \partial_{j} b_{j} - b_{i} \partial_{j} a_{j} + b_{j} \partial_{j} a_{i} - a_{j} \partial_{j} b_{i} .

The last equality is the product rule. Recognizing the divergences and directional derivatives proves the curl of a cross product identity:

\nabla \times (a \times b) = a (\nabla \cdot b) - b (\nabla \cdot a) + (b \cdot \nabla) a - (a \cdot \nabla) b .

 New to topics? Read the docs here!