Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I have two vectors $a$ and $b$, whose components are time varying, for example $$a=[a_x(t), a_y(t), a_z(t)]$$ $$b=[b_x(t), b_y(t), b_z(t)]$$ The dot product of these 2 vectors can be expressed as $$a\cdot b=\Vert{a}\Vert\cdot\Vert{b}\Vert\cdot \cos\phi_{ab}$$ I want to compute in closed form the variation of $\phi_{ab}$ with respect to time, that is, $\dot\phi_{ab}$. Is it possible to symbolically compute this? Can I also obtain in closed form the derivative of $a\cdot b$ with respect to time? Which software would you recommend to do these kind of calculations: Matlab, Mathematica or Maple? Any examples? In fact, I have derived closed-form expressions tediously by hand for several expressions of the type I have given here. I wish to cross-verify whether my derivations are correct.

share|improve this question
1  
wolframalpha.com/input/… –  Bill Cook Oct 13 '11 at 21:53
    
@Bill Cook: Thank you. I tried using what you ve sent me. The expressions seem to be complicated though. Well. fair enough. But i was wondering if i could retain a and b and its derivatives a' and b' (w.r.t time)as vectors itself in the final expression. –  hAcKnRoCk Oct 14 '11 at 11:25

1 Answer 1

Mathematica works fine. It's possible to write down those expressions, but whether it would give a closed form for the variation depends upon the functions involved. Let the variation of $\phi_{ab}$ be computed on $[c,d]$. If, for example, $c=0$, $d=1$, $a(t)=(t,t,t)$, $a(t)=(1,t,t^2)\ $. Then code

c = 0; d = 1;
a[t_] = {t, t, t};
b[t_] = {1, t, 1};
scpt[t_] = a[t].b[t] // FullSimplify
varphi = Integrate[
  FullSimplify[Abs[
   D[ArcCos[scpt[t]/(Norm[a[t]] Norm[b[t]])], t]], 
   c <= t <= d], {t, c, d}] 

gives $t (t^2+t+1)$ for the scalar product scpr[t] and $\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$ for varphi.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.