Lemma 4.4.1. Using the notation of figure 4.6, we have ∂ $_{∂r}$g ( s, r ) = ( θ$_{2}$ − θ$_{1}$ ) r . That is, the derivative exists and equals the length of the curve C ( γ ( s ) , r ) ∩ Ω . Proof. We have (see figure 4.8) This difference of areas can be modeled by the difference in the circular sectors of D ( γ ( s ) , r +Δ r ) and D ( γ ( s ) , r ) with angle θ$_{1}$ − θ$_{2}$ . The actual area depends on the image of γ outside of D ( γ ( s ) , r ), but this correction will be a subset of the circular segment of D ( γ ( s ) , r + Δ r ) which is tangent to D ( γ ( s ) , r ) at the point γ exits. This has area O (Δ r $^{2}$) by lemma 4.2.18. Thus we have Figure 4.9: Deriving ∂ g ∂s as the difference in heights of the entry and exit points 101 ∂ ∂r g ( s, r ) = lim Δ r → 0 ( θ$_{1}$ − θ$_{2}$ ) r Δ r + 1 $_{2}$( θ$_{1}$ − θ$_{2}$ )Δ r 2 + O (Δ r $^{2}$) Δ r = ( θ$_{1}$ − θ$_{2}$ ) r. ∂ ∂r g ( s, r ) = lim Δ r → 0 Area(Ω ∩ D ( γ ( s ) , r + Δ r )) − Area(Ω ∩ D ( γ ( s ) , r )) Δ r . h$_{2}$ h$_{1}$ ΔArea ≈ Δ s ( h$_{2}$ − h$_{1}$ ) Δ s γ ∂g ∂s = h$_{2}$ − h$_{1}$