deletions | additions       diff --git a/subsection_Slope_Method_The_direct__.tex b/subsection_Slope_Method_The_direct__.tex  index 19d10b5..055774a 100644  --- a/subsection_Slope_Method_The_direct__.tex  +++ b/subsection_Slope_Method_The_direct__.tex 
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%V=V0(½)(1+cos(2(θ1-θ0)))  %which is V=V0cos2(θ1-θ2), giving us the form we should have initially fit to.  As $$V_{c}=\frac{1}{L}\left(\frac{d\theta}{dV_{pd}}\right)_{at ?}\frac{dV_{pd}}{dB}$$  %As  shown in Figure 2: $$V=0.000443B+0.127$$  Therefore:  $$\frac{\Delta %$$V=0.000443B+0.127$$  %Therefore:  %$$\frac{\Delta  V}{\Delta B}=0.000443\frac{V}{mT}=0.443\frac{V}{T}$$ As %As  shown in Figure 1: $$V=\frac{0.021\times %$$V=\frac{0.021\times  (1+cos(\frac{2\pi\times(\theta-162)}{180}))}{2}$$ Taking %Taking  the derivative at $\theta=117$ degrees: $$\frac{\Delta %$$\frac{\Delta  V}{\Delta \theta}=\frac{0.21}{2}\times(-sin(2\frac{\pi}{180}(-45)))=0.21V$$ $$\frac{\Delta\theta}{\Delta %$$\frac{\Delta\theta}{\Delta  B}=\frac{\Delta V}{\Delta B} \times \frac{\Delta \theta}{\Delta V}=0.443\frac{V}{T}\times\frac{1 radian}{0.21V}=2.1095\frac{radians}{T}$$ $$V_{c}=\frac{1}{L}\times %$$V_{c}=\frac{1}{L}\times  \frac{\Delta\theta}{\Delta B}=\frac{1}{0.1}\times2.1095\frac{radians}{T}=21.095\frac{radians}{T \cdot m}$$