Show the same result holds in general, i.e. for the velocity field u(x, t) and a vector field b(x, t), show b · Db Dt = 1 2 Db 2 Dt , where b 2 = b · b. Solution: The case of three dimensions is really no different since ?b 2 ?t = 2b · ?b ?t and ?b 2 = 2 (?b)·b. Hence, maintaining the order of vector operations we see that 1 2 Db 2 Dt = b · ?b ?t + u · ?b = b · Db Dt .
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The post Show the same result holds in general, i.e. for the velocity field u(x, t) and a vector field b(x, t), show b · Db Dt = 1 2 Db 2 Dt , where b 2 = b · b. Solution: The case of three dimensions is really no different since appeared first on Wise Papers.
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