- #1
mastrofoffi
- 51
- 12
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
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);
perpendicular vectors: F''(t) ⋅ F'(t) = 0
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?
Homework Statement
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);
Homework Equations
perpendicular vectors: F''(t) ⋅ F'(t) = 0
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?