Exercise 6.A.26

Answers

Proof. (a)

\begin{aligned} {⟨ f (t), g (t) ⟩}^{'} & = \lim_{h \to 0} \frac{⟨ f (t + h), g (t + h) ⟩ - ⟨ f (t), g (t) ⟩}{h} \\ = \lim_{h \to 0} \frac{⟨ f (t + h), g (t + h) ⟩ - ⟨ f (t), g (t + h) ⟩ + ⟨ f (t), g (t + h) ⟩ - ⟨ f (t), g (t) ⟩}{h} \\ = \lim_{h \to 0} \frac{⟨ f (t + h) - f (t), g (t + h) ⟩ + ⟨ f (t), g (t + h) - g (t) ⟩}{h} \\ = \lim_{h \to 0} \frac{⟨ f (t + h) - f (t), g (t + h) ⟩}{h} + \lim_{h \to 0} \frac{⟨ f (t), g (t + h) - g (t) ⟩}{h} \\ = \lim_{h \to 0} ⟨ \frac{f (t + h) - f (t)}{h}, g (t + h) ⟩ + \lim_{h \to 0} ⟨ f (t), \frac{g (t + h) - g (t)}{h} ⟩ \\ = ⟨ f^{'} (t), g (t) ⟩ + ⟨ f (t), g^{'} (t) ⟩ \end{aligned}

(b)

We have

{⟨ f (t), f (t) ⟩}^{'} = ⟨ f^{'} (t), f (t) ⟩ + ⟨ f (t), f^{'} (t) ⟩ = 2 ⟨ f^{'} (t), f (t) ⟩ .

However, ${⟨ f (t), f (t) ⟩}^{'}$ is $0$ , because $⟨ f (t), f (t) ⟩$ is constant (it equals $c^{2}$ for all $t$ ). Thus $⟨ f^{'} (t), f (t) ⟩ = 0$ .

(c)

If $f (t)$ describes a trajectory on the surface of a sphere centered at the origin, then $f^{'} (t)$ is perpendicular to $f (t)$ . □

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2017-10-06 00:00