3.3 Arc length and curvature  (Page 4/18)

Proof

The first formula follows directly from the chain rule:

where s is the arc length along the curve C. Dividing both sides by $ds\text{/}dt,$ and taking the magnitude of both sides gives

Since $ds\text{/}dt=\Vert r^{\prime }\left(t\right)\Vert ,$ this gives the formula for the curvature $\kappa$ of a curve C in terms of any parameterization of C :

In the case of a three-dimensional curve, we start with the formulas $\text{T}\left(t\right)=\left(r^{\prime }\left(t\right)\right)\text{/}\Vert r^{\prime }\left(t\right)\Vert$ and $ds\text{/}dt=\Vert r^{\prime }\left(t\right)\Vert .$ Therefore, $r^{\prime }\left(t\right)=\left(ds\text{/}dt\right)\text{T}\left(t\right).$ We can take the derivative of this function using the scalar product formula:

Using these last two equations we get

Since $\text{T}\left(t\right)\times \text{T}\left(t\right)=0,$ this reduces to

Since $T^{\prime }$ is parallel to $\text{N},$ and $\text{T}$ is orthogonal to $\text{N},$ it follows that $\text{T}$ and $T^{\prime }$ are orthogonal. This means that $\Vert \text{T}\times T^{\prime }\Vert =\Vert \text{T}\Vert \Vert T^{\prime }\Vert \text{sin}\left(\pi \text{/}2\right)=\Vert T^{\prime }\Vert ,$ so

Now we solve this equation for $\Vert T^{\prime }\left(t\right)\Vert$ and use the fact that $ds\text{/}dt=\Vert r^{\prime }\left(t\right)\Vert \text{:}$

Then, we divide both sides by $\Vert r^{\prime }\left(t\right)\Vert .$ This gives

This proves [link] . To prove [link] , we start with the assumption that curve C is defined by the function $y=f\left(x\right).$ Then, we can define $\text{r}\left(t\right)=x\text{i}+f\left(x\right)\text{j}+0\text{k}.$ Using the previous formula for curvature:

Therefore,

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