Tag Info

New answers tagged

1

Unprotect[Dot]; SetAttributes[Dot, Orderless] Protect[Dot] then try to evaluate this again D[V[t]/Sqrt[Dot[V[t], V[t]]], t] of course it would be better to define new function based on dot and set it attribute Orderless


2

Perform the line integral over the straight-line path from the origin to $(x,y)$ via the parametrization $u(t)=(tx,ty)$ for $0\le t\le1$. Then $\mathrm du=(x,y)\,\mathrm dt$, so the integral can be computed as Integrate[b[t x, t y].{x, y}, {t, 0, 1}] Note that if $b = \nabla a$, from $b$ one can only recover $a$ up to a constant. In this case, since we ...



Top 50 recent answers are included