Show orthogonality of vector-valued functions

[mastrofoffi]

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

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?

 

[pasmith]

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

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?

It is not generally true that \|F&#039;\|&#039; = \|F&#039;&#039;\|. On differentiating \|F&#039;\|\|B\|\cos \phi with respect to t you should obtain <br /> \|B\|\cos \phi\frac{d\|F&#039;\|}{dt} = 0 since \frac{d}{dt}\|B\| = \frac{d\phi}{dt} = 0.

To proceed further you will need the identity <br /> \frac{d}{dt}\|F&#039;\|^2 = \frac{d}{dt} (F&#039; \cdot F&#039;) = 2F&#039; \cdot F&#039;&#039;.

 

[mastrofoffi]

It is not generally true that \|F&#039;\|&#039; = \|F&#039;&#039;\|

Oh I see that, I should have been more careful. Thank you very much.

 

[Mark44]

@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.