Figure 4.12: Figure 4.6 again as a reminder 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, 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: 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 107 ∂ $^{2}$g ( s, r ) ∂r 2 = θ$_{2}$ − θ$_{1}$ + tan( ν$_{2}$ ) − tan( ν$_{1}$ ) ∂ $^{2}$g ( s, r ) ∂r∂s = sin( θ$_{2}$ ) − sin( θ$_{1}$ ) + cos( θ$_{2}$ ) tan( ν$_{2}$ ) − cos( θ$_{1}$ ) tan( ν$_{1}$ ) ∂ θ$_{1}$ ∂r = lim Δ r → 0 arctan ( r sin θ$_{1}$ + k sin( θ$_{1}$ + ν$_{1}$ ) r cos θ$_{1}$ + k cos( θ$_{1}$ + ν$_{1}$ ) $^{)}$− θ$_{1}$ Δ r . D ∩ Ω θ$_{1}$ θ$_{2}$ γ ( s $^{−}$) γ ( s ) ν$_{1}$ ν$_{2}$ D = D ( γ ( s ) , r ) C = ∂D γ ( s $^{+}$)