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Year 2 of the natural sciences course of the University of Cambridge

Mathematics course of the year 2 of the natural sciences course of the University of Cambridge

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge

2013

Paper 1

Using the <Levi-Civita symbol> and the <Einstein summation convention> in <Cartesian coordinates>, the <product rule> gives
$$
\begin{aligned}
\nabla\cdot(\mathbf A\times\mathbf B)
&=\partial_i(\epsilon_{ijk}A_jB_k)\\
&=\epsilon_{ijk}(\partial_iA_j)B_k+\epsilon_{ijk}A_j\partial_iB_k\\
&=B_k\epsilon_{kij}\partial_iA_j-A_j\epsilon_{jik}\partial_iB_k.
\end{aligned}
$$
The cyclic permutation in the first term preserves the sign, whereas exchanging the first two indices in the second reverses it. Recognizing the two <curls> proves
$$
\boxed{\nabla\cdot(\mathbf A\times\mathbf B)=\mathbf B\cdot(\nabla\times\mathbf A)-\mathbf A\cdot(\nabla\times\mathbf B).}
$$

Solution

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 1 / i / Solution

Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 1 / i / Solution

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