Exercise 3.2.1

Answers

Since $u^{T} u = 1$ , take derivative w.r.t. $t$ , we have $\frac{d u^{T} u}{dt} = 0$ , we compute the left hand side as :

\begin{array}{l} d (u^{T} u) & = {(u + Δ u)}^{T} (u + Δ u) - u^{T} u \\ = u^{T} Δ u + Δ u^{T} u \\ = 2 u^{T} Δ u \end{array}

Divide both sides by $t$ we have $\frac{d u^{T} u}{dt} = 2 u^{T} \frac{d Δ u}{dt} = 0$

So the velocity vector $\frac{du}{dt}$ is orthogonal to the position $u$ .

niuers

2020-03-20 00:00