If a point r → = ( x , y ) {\displaystyle {\vec {r}}=(x,y)} on the curve is given parametrically in terms of the arc length, s ↦ r → , {\displaystyle s\mapsto {\vec {r}},} then the tangential angle φ is determined by
d r → d s = ( d x d s d y d s ) = ( cos φ sin φ ) since | d r → d s | = 1 , {\displaystyle {\frac {d{\vec {r}}}{ds}}={\begin{pmatrix}{\frac {dx}{ds}}\\{\frac {dy}{ds}}\end{pmatrix}}={\begin{pmatrix}\cos \varphi \\\sin \varphi \end{pmatrix}}\quad {\text{since}}\quad \left|{\frac {d{\vec {r}}}{ds}}\right|=1,}
which implies d y d x = tan φ . {\displaystyle {\frac {dy}{dx}}=\tan \varphi .}
Parametric equations for the curve can be obtained by integrating: x = ∫ cos φ d s , y = ∫ sin φ d s . {\displaystyle {\begin{aligned}x&=\int \cos \varphi \,ds,\\y&=\int \sin \varphi \,ds.\end{aligned}}}
Since the curvature is defined by κ = d φ d s , {\displaystyle \kappa ={\frac {d\varphi }{ds}},}
the Cesàro equation is easily obtained by differentiating the Whewell equation.