$\let\oldcdot\cdot \renewcommand{\cdot}{\kern-.2em\oldcdot\kern-.2em} \let\oldwedge\wedge \renewcommand{\wedge}{\kern-.2em\oldwedge\kern-.2em} \renewcommand{\vec}{\mathbf{#1}} \let\oldtimes\times \renewcommand{\times}{\kern-.2em\oldtimes\kern-.2em} \newcommand{\dvec}{\dot{\vec{#1}}} \newcommand{\ip}{\vec{#1}\cdot\vec{#2}} \newcommand{\op}{\vec{#1}\wedge\vec{#2}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\hf}{\frac{1}{2}} \newcommand{\rn}{\rangle^\hf_0}$

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## 7.1 Unit Vector Derivative

$\vec{u} = \vec{u}(t)$ a vector valued function, $u = \lvert \vec{u} \rvert$. Show

$$\ddt\left(\frac{\vec{u}}{u}\right) = \frac{\vec{u}(\vec{u}\wedge\dvec{u})}{u^3} = \frac{(\vec{u}\times\dvec{u})\times\vec{u}}{u^3}$$

Two useful intermediate results.

### Vector Magnitude Squared:

$$u^2 = \vec{u}^2$$

Proof: $$u^2 = \left(\langle \vec{u}^\sim\vec{u} \rn\right)^2$$ $$= \langle \vec{u}^2 \rangle_0$$ $$= \vec{u}^2$$

### Vector Magnitude Derivative:

$$\ddt u = \frac{\ip{u}{\dot{u}}}{u}$$

Proof: $$\ddt u = \ddt \lvert \vec{u} \rvert$$ $$= \ddt \langle \vec{u}^2 \rn$$ $$= \hf \langle \vec{u}^2 \rangle^{-\hf}_0 \ddt(\vec{u}^2)$$ $$= \frac{1}{2 u} \ddt (\ip{u}{u})$$ $$= \frac{\ip{u}{\dot{u}}}{u}$$

Onward.

Using the product rule and the chain rule $$\ddt\left(\frac{\vec{u}}{u}\right) = \ddt(u^{-1}\vec{u})$$ $$= \frac{\dvec{u}}{u} - \frac{\vec{u}}{u^2}\ddt u$$ $$= \frac{\dvec{u}}{u} - \frac{\vec{u}\ip{u}{\dot{u}}}{u^3}$$ $$= \frac{u^2\dvec{u}-\vec{u}\ip{u}{\dot{u}}}{u^3}$$ $$= \frac{\vec{uu}\dvec{u}-\vec{u}\ip{u}{\dot{u}}}{u^3}$$ $$= \frac{\vec{u}(\ip{u}{\dot{u}}+\op{u}{\dot{u}})-\vec{u}\ip{u}{\dot{u}}}{u^3}$$ $$= \frac{\vec{u}(\op{u}{\dot{u}})}{u^3}$$

In 3-space, using properties proven in Chapter 3, this result can be rewritten in terms of cross products:

$$\frac{\vec{u}(\op{u}{\dot{u}})}{u^3} = \frac{\vec{u}\cdot(\op{u}{\dot{u}})}{u^3}$$ $$= \frac{\vec{u}\cdot(i\vec{u}\times\dvec{u})}{u^3}$$ $$= \frac{i\vec{u}\wedge(\vec{u}\times\dvec{u})}{u^3}$$ $$= \frac{i^2\vec{u}\times(\vec{u}\times\dvec{u})}{u^3}$$ $$= \frac{(\vec{u}\times\dvec{u})\times\vec{u}}{u^3}$$

### Unit Vector Derivative Magnitude

$$\ddt\left(\frac{\vec{u}}{u}\right)^2$$ $$= \left ( \frac{\vec{u}(\op{u}{\dot{u}})}{u^3} \right )^2$$ $$= \frac{\vec{u}(\op{u}{\dot{u}})\vec{u}(\op{u}{\dot{u}})}{u^6}$$ $\vec{u}$ and $\op{u}{\dot{u}}$ anticommute, because their geometric product reduces to the inner product of a vector and a bivector: $$= -\frac{\vec{u}^2(\op{u}{\dot{u}})(\op{u}{\dot{u}})}{u^6}$$ $$= -\frac{(\op{u}{\dot{u}})^2}{u^4}$$ $$= - \left( \frac{\op{u}{\dot{u}}}{u^2} \right)^2$$

## 7.5 Derivatives of Inner and Outer Products

Using the product rule

$$\ddt(\vec{rp}) = \dot{\vec{r}}\vec{p} + \vec{r}\dot{\vec{p}}$$ $$= \dvec{r}\cdot\vec{p} + \dvec{r}\wedge\vec{p} + \vec{r}\cdot\dvec{p} + \vec{r}\wedge\dvec{p}$$ but $$\frac{d}{dt}(\vec{rp}) = \frac{d}{dt}(\ip{r}{p} + \op{r}{p})$$ $$= \frac{d}{dt}(\ip{r}{p}) + \frac{d}{dt}(\op{r}{p})$$

Equating scalar and bivector parts $$\frac{d}{dt}(\ip{r}{p}) = \dvec{r}\cdot\vec{p} + \vec{r}\cdot\dvec{p}$$ $$\frac{d}{dt}(\op{r}{p}) = \dvec{r}\wedge\vec{p} + \vec{r}\wedge\dvec{p}$$

Express the derivatives of $\vec{p}\cdot(\op{q}{r}) = \vec{p}\times(\vec{r}\times\vec{q})$ and $\vec{p}\wedge\vec{q}\wedge\vec{r} = i\vec{p}\cdot(\vec{q}\times\vec{r})$ in terms of $\dvec{p}$, $\dvec{q}$ and $\dvec{r}$.

$$\ddt \vec{p}\cdot(\op{q}{r}) = \dvec{p}\cdot(\op{q}{r}) + \vec{p}\cdot(\dvec{q}\wedge\vec{r}+\vec{q}\wedge\dvec{r})$$

TODO

## 7.7

Show that the derivative of any vector-valued function $\vec{v} = \vec{v}(t)$ can be expressed in the form

$$\dvec{v} = \ip{v}{\Omega} + \hf\vec{v}\ddt\log\vec{v}^2$$

where the first term describes the rate of direction change, and the second the rate of magnitude change.

### Constant Magnitude Vectors are Orthogonal to their Derivatives

If $\vec{u}^2 = C^2$, then $\ip{u}{\dot{u}} = 0$.

Proof:

$$\ddt\vec{u}^2 = 0$$

but

$$\ddt\vec{u}^2 = \ddt(\ip{u}{u})$$ $$= \ip{\dot{u}}{u} + \ip{u}{\dot{u}}$$ $$= 2\ip{u}{\dot{u}}$$

and dividing by $2$ gives the result

$$\ip{u}{\dot{u}} = 0$$

### Vector Squared Log Derivative

Use the chain rule to differentiate the $\log$ $$\hf\ddt\log\vec{v}^2 = \hf\frac{1}{v^2}\ddt\vec{v}^2$$ rewrite the vector squared as an inner product $$= \hf\frac{1}{v^2}\ddt\ip{v}{v}$$ use the derivative of an inner product $$= \hf\frac{1}{v^2}(\ip{\dot{v}}{v}+\ip{v}{\dot{v}})$$ $$= \frac{\ip{v}{\dot{v}}}{v^2}$$

Rewrite $\vec{v}$ as a product of unit vector and magnitude components: $$\vec{v} = \left(\frac{\vec{v}}{v}\right) v$$

Taking the derivative and using the product rule, $$\ddt{\vec{v}} = \ddt \left[ \left(\frac{\vec{v}}{v}\right)v \right]$$ $$= \left[ \ddt \left(\frac{\vec{v}}{v}\right)\right] v + \frac{\vec{v}}{v} \ddt v$$

This shows that the two terms in the result will be the rate of direction change of $\vec{v}$ scaled by the magnitude of $\vec{v}$, and the second term is the rate of magnitude change in the direction of $\vec{v}$:

$$= \left(\ddt \hat{\vec{v}}\right) v + \hat{\vec{v}} \ddt v$$ $$= v \dot{\hat{\vec{v}}} + \dot{v}\hat{\vec{v}}$$

$\vec{\dot{\hat{{v}}}}$ and $\vec{\hat{v}}$ are orthogonal due to $\vec{\hat{v}}$’s constant magnitude, but not orthonormal.

This is not exactly what the question asked for, but maybe I’m splitting hairs.

Onward. Using vector magnitude derivative, and the results of the unit vector derivative:

$$= v \frac{\vec{v}(\vec{v}\wedge\dvec{v})}{v^3} + \frac{\ip{v}{\dot{v}}}{v}\frac{\vec{v}}{v}$$ $$= \frac{\vec{v}(\vec{v}\wedge\dvec{v})}{v^2} + \frac{\ip{v}{\dot{v}}}{v^2}\vec{v}$$

Noticing $\vec{v}(\vec{v}\wedge\dvec{v}) = \vec{v}\cdot(\vec{v}\wedge\dvec{v})$, the appearance of the vector squared log derivative, and letting the bivector component

$$\vec{\Omega} = \frac{\vec{v}\wedge\dvec{v}}{v^2}$$

we have

$$\dvec{v} = \ip{v}{\Omega} + \hf\vec{v}\ddt\log\vec{v}^2$$

## Formulas from the Book

### The Product Rule

$$\ddt(FG) = \dot{F}G + F\dot{G} \qquad\text{(7.6b)}$$