Vector Calculus
1 Derivative normal vector repect to time
Let's denote the unit normal vector as:
\[\mathbf{n}
=\frac{\mathbf{e}_a\times\mathbf{e}_b}{\|\mathbf{e}_a\times\mathbf{e}_b\|}
= \frac{1}{\|\mathbf{e}_a\times\mathbf{e}_b\|} \cdot \mathbf{e}_a\times\mathbf{e}_b
\]
Where \(\times\) denotesthecrossproduct,and \(\parallel \cdot \parallel\) denotes the norm.
Now, let's find the time derivative \(\frac{d\mathbf{n}}{dt}\).
We'll use the product rule, chain rule, and quotient rule as necessary.
\[\frac{d\mathbf{n}}{dt} = \frac{1}{\|\mathbf{e}_a \times \mathbf{e}_b\|} \cdot \frac{d}{dt}(\mathbf{e}_a \times \mathbf{e}_b) - \frac{(\mathbf{n} \cdot (\mathbf{e}_a \times \mathbf{e}_b)) \cdot \frac{d}{dt}(\mathbf{e}_a \times \mathbf{e}_b)}{\|\mathbf{e}_a \times \mathbf{e}_b\|^2}
\]
1.1 Derivative vector norm repect to time
Let's denote \(\mathbf{v}=\mathbf{e}_a\times\mathbf{e}_b\). The norm of v is given by \(\|\mathbf{v}\|=\sqrt{\mathbf{v}\cdot\mathbf{v}}.\)
Now, let's find the derivative of the norm with respect to time \(t\)
\[\frac d{dt}\|\mathbf{v}\|=\frac1{2\sqrt{\mathbf{v}\cdot\mathbf{v}}}\cdot\frac d{dt}(\mathbf{v}\cdot\mathbf{v})
\]
Applying the chain rule and the product rule:
\[\frac d{dt}\|\mathbf{v}\|=\frac1{2\|\mathbf{v}\|}\cdot\frac d{dt}(\mathbf{v}\cdot\mathbf{v})
\]
Now, expand the dot product \(\mathbf{v}\cdot\mathbf{v}:\)
\[\frac{d}{dt}\|\mathbf{v}\|=\frac{1}{2\|\mathbf{v}\|}\cdot\frac{d}{dt}(\mathbf{e}_a\times\mathbf{e}_b\cdot\mathbf{e}_a\times\mathbf{e}_b)
\]
Apply the product rule and the chain rule:
\[\frac{d}{dt}\|\mathbf{v}\|=\frac{1}{2\|\mathbf{v}\|}\cdot\left(2(\mathbf{e}_a\times\mathbf{e}_b)\cdot\frac{d}{dt}(\mathbf{e}_a\times\mathbf{e}_b)\right)
\]