If two plane curves with Cartesian equations y =f(x) and y = g(x) have the same tangent at a point (a,b) and the same curvature at that point, prove that $ |f''(a)|$=$|g''(a)|$. Alright, I know that this should be easy, I understand what happens, I mean, you can set any kind of function and because of having the same curvature, both curves have to behave the same closed to the point (a,b) in order to respect the change of the tangents with respect to the change in the curves length, but I can't find an mathematic argument to say that $g'(a)$=$f'(a)$. i just need a hint to see why this should happen. Because if i have that, because both curve are plane I can describe their curvature by:
$R_f(a)$=$\dfrac{|f''(a)|}{(1+(f'(a))^2)^{2/3}}$=$\dfrac{|g''(a)|}{(1+(g'(a))^2)^{2/3}}$=$R_g(a)$
Another thing thatI was thinking is that because of the hypothesis that for both curves exist a curvature, then both should be $f\in C^1$, then both are rectifiable, so can I propose that both can be parametrized by arc length?, and then because of that conclude that $ |f''(a)|$=$|g''(a)|$, because of that parametrization.