content_image/299141.jpg content_image/299140.jpg content_image/299139.jpg
<img src='content_image/299141.jpg'>

only) a distance of k ≡ Δ r sec( ν$_{1}$ ) (so ∂ k ∂r = sec ν$_{1}$ , a fact we will use later to compute curvature). Therefore, 

<img src='content_image/299140.jpg'>

Straightforward techniques yield ∂ θ$_{1}$ ∂r = tan ν$_{1}$ r and a similar calculation shows that ∂ θ$_{2}$ ∂r = tan ν$_{2}$ r . 

Therefore, rewriting the second derivatives of g ( s, r ) in terms of ν$_{1}$ and ν$_{2}$ , we get: 

<img src='content_image/299139.jpg'>

Using these 2 derivatives, together with the previous two, we can solve for ν$_{1}$ = arctan( r ∂θ$_{1}$ ∂r ) and ν$_{2}$ = arctan( r ∂θ$_{1}$ ∂r ) whenever cos( θ$_{1}$ ) ̸ = cos( θ$_{2}$ ). Since we are assuming that the curve is a simple closed curve, cos( θ$_{1}$ ) ̸ = cos( θ$_{2}$ ) is always true. 

Claim 2. Knowing ∂ $^{3}$g ( s,r ) ∂r 3 and ∂ $^{3}$g ( s,r ) ∂r $^{2}$∂s gives us κ ( s $^{+}$( s )) and κ ( s $^{−}$( s )) , the curvatures of 

