Converting a SDE from stratonovich to ito's form ,
Stratonovich form $$\partial X=P(X)\partial B$$ $$ P(X)=I-n(X)n(X)^T $$ Ito's conversion
$$ dX=P(X)dB +\frac{1}{2}d(P(X))dB $$ $$ dX=P(X)dB +\frac{1}{2}d(I-n(X)n(X)^T)dB $$ $$ dX=P(X)dB -\frac{1}{2}d(n(X)n(X)^T)dB $$
$n(x)=\frac {\nabla f(x)}{|\nabla f(x)|}$,$I$ is the identity. $c=-\frac{1}{2}div(n)$ is the mean curvature.Wiki of mean curvature "For a surface defined in 3D space, the mean curvature is related to a unit normal of the surface: $2 H = -\nabla \cdot \hat n$". I am not able to understand how the mean curvature will come.