# Curvature and Vectors?

## I have been trying to work the idea behind curvature, as in $\kappa = \frac{d \hat{T}}{\mathrm{ds}}$, where $\hat{T}$ is the unit tangent vector, and have been hoping to do so using purely vector algebra i can get this far.... position vector $\vec{r} \left(t\right)$ has tangent vector $\vec{v} = \vec{\dot{r}}$ and unit tangent vector $\hat{v} = \frac{\vec{v}}{\left\mid \vec{v} \right\mid}$ using the chain rule $\kappa = \frac{d \hat{v} \left(t\right)}{\mathrm{ds}} = \frac{d \hat{v} \left(t\right)}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{ds}}$ $= \frac{d \hat{v} \left(t\right)}{\mathrm{dt}} \cdot \frac{1}{\dot{s}} = \frac{d \hat{v} \left(t\right)}{\mathrm{dt}} \cdot \frac{1}{\left\mid \vec{v} \right\mid}$ as $\dot{s} = \left\mid \vec{v} \right\mid$ now $\frac{d \hat{v} \left(t\right)}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \frac{\vec{v}}{\left\mid \vec{v} \right\mid}$ by quotient rule...is this too aggressive(??) $= \frac{\frac{d}{\mathrm{dt}} \left(\vec{v}\right) \left\mid \vec{v} \right\mid - \vec{v} \frac{d}{\mathrm{dt}} \left(\left\mid \vec{v} \right\mid\right)}{\left\mid \vec{v} \right\mid} ^ 2$ $= \frac{\vec{a} \left\mid \vec{v} \right\mid - \vec{v} \frac{d}{\mathrm{dt}} \left(\left\mid \vec{v} \right\mid\right)}{\left\mid \vec{v} \right\mid} ^ 2 q \quad \triangle$ and $\frac{d}{\mathrm{dt}} \left(\left\mid \vec{v} \right\mid\right) = \frac{d}{\mathrm{dt}} \left(\sqrt{\vec{v} \cdot \vec{v}}\right)$ $= \frac{1}{2} \cdot \frac{1}{\sqrt{\vec{v} \cdot \vec{v}}} \frac{d}{\mathrm{dt}} \left(\vec{v} \cdot \vec{v}\right)$ $= \frac{\vec{a} \cdot \vec{v}}{\sqrt{\vec{v} \cdot \vec{v}}} = \frac{\vec{a} \cdot \vec{v}}{\left\mid \vec{v} \right\mid} q \quad \square$ putting$\square$ into $\triangle$: $= \frac{\vec{a} \left\mid \vec{v} \right\mid - \vec{v} \frac{\vec{a} \cdot \vec{v}}{\left\mid \vec{v} \right\mid}}{\left\mid \vec{v} \right\mid} ^ 2$ $= \frac{\vec{a} \left(\vec{v} \cdot \vec{v}\right) - \vec{v} \left(\vec{a} \cdot \vec{v}\right)}{\left\mid \vec{v} \right\mid} ^ 3 q \quad \circ$ that's still shy of the usual result ie $\frac{\left\mid \vec{v} \times \vec{a} \right\mid}{{\left\mid \vec{v} \right\mid}^{3}}$ and to further complicate matters, i make $\circ$ out to be: $= \frac{\vec{v} \times \left(\vec{a} \times \vec{v}\right)}{\left\mid \vec{v} \right\mid} ^ 3$

$\vec{a} \left\langle\vec{v} , \vec{v}\right\rangle - \vec{v} \left\langle\vec{a} , \vec{v}\right\rangle = \vec{v} \times \left(\vec{a} \times \vec{v}\right)$
and if $\vec{a}$ and $\vec{v}$ are non linearly dependent, the result is non null.