Hopefully, you have at your disposal the fact that the product rule does indeed hold for vector products: that is, the derivative of a · b is a' · b + a · b', and the derivative of a × b is (a' × b) + (a × b'). With that at our disposal we have:
d/dt [r · (v × a)]
= (dr/dt) · (v × a) + r · d/dt (v × a)
= v · (v × a) + r · (dv/dt × a + v × da/dt)
There is quite a bit of simplification we can do. The dot product, loosely speaking, measures how much a vector runs in the direction of the other. The cross product delivers a vector that is orthogonal to its two arguments. Hence v · (v × a) = 0, because v × a is orthogonal to v. Also, the cross product of a vector with itself is (as you might suspect it would have to be) zero, so the term dv/dt × a falls out as well. You are left with r · (v × da/dt), just as you needed.
Hopefully, you have at your disposal the fact that the product rule does indeed hold for vector products: that is, the derivative of a · b is a' · b + a · b', and the derivative of a × b is (a' × b) + (a × b'). With that at our disposal we have: d/dt [r · (v × a)] = (dr/dt) · (v × a) + r · d/dt (v × a) = v · (v × a) + r · (dv/dt × a + v × da/dt) There is quite a bit of simplification we can do. The dot product, loosely speaking, measures how much a vector runs in the direction of the other. The cross product delivers a vector that is orthogonal to its two arguments. Hence v · (v × a) = 0, because v × a is orthogonal to v. Also, the cross product of a vector with itself is (as you might suspect it would have to be) zero, so the term dv/dt × a falls out as well. You are left with r · (v × da/dt), just as you needed.
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