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Show orthogonality of vector-valued functions

  1. Oct 9, 2016 #1
    I have this exercise on my book and I believe it is quite simple to solve, but I'm not sure if I did good, so here it is

    1. The problem statement, all variables and given/known data
    given a vector B ∈ ℝn, B ≠ 0 and a function F : ℝ → ℝn such that F(t) ⋅ B = t ∀t and the angle φ between F'(t) and B is constant with respect to t, show that F''(t) is perpendicular to F'(t);

    2. Relevant equations
    perpendicular vectors: F''(t) ⋅ F'(t) = 0

    3. The attempt at a solution
    Since I have to show that the scalar product between F'' and F' is 0, i would obviously try to derive the given eqs and find a relation between the two derivatives
    F(t) ⋅ B = t ⇒ d/dt(F(t) ⋅ B) = d/dt(t) ⇒ F'(t) ⋅ B = 1
    F'(t) ⋅ B = |F'(t)B|cosφ = 1 ⇒ d/dt(|F'(t)B|cosφ) = d/dt(1) ⇒ |F''(t)B|cosφ = 0
    cosφ ≠ 0 since F'Bcosφ = 1, and B ≠ 0, then it must be F''(t) = 0 ⇒ F''(t) ⋅ F'(t) = 0

    Is this correct? Can I say that the zero vector is perpendicular to F' or did I miss something?
     
  2. jcsd
  3. Oct 9, 2016 #2

    pasmith

    User Avatar
    Homework Helper

    It is not generally true that [itex]\|F'\|' = \|F''\|[/itex]. On differentiating [itex]\|F'\|\|B\|\cos \phi[/itex] with respect to [itex]t[/itex] you should obtain [tex]
    \|B\|\cos \phi\frac{d\|F'\|}{dt} = 0[/tex] since [itex]\frac{d}{dt}\|B\| = \frac{d\phi}{dt} = 0[/itex].

    To proceed further you will need the identity [tex]
    \frac{d}{dt}\|F'\|^2 = \frac{d}{dt} (F' \cdot F') = 2F' \cdot F''.[/tex]
     
  4. Oct 9, 2016 #3
    Oh I see that, I should have been more careful. Thank you very much.
     
  5. Oct 9, 2016 #4

    Mark44

    Staff: Mentor

    @mastrofoffi, please post questions of this nature in the Calculus & Beyond section, not the Precalculus section. If the question involves derivatives, it should NOT go in the Precalc section.
     
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