Curvature of a graph How to calculate the formula of the curvature of $\alpha= (t, f(t)), f$ is $C^{\infty}$ ?
I know that can be done easily using previous curvature formulas, but I couldn't do using jus the curvature definition $k(t)= \langle t'(t),n(t)\rangle$, where $t(t)$ is the unit tangent vector. 
 A: This is a reasonable question. But you have an error. Your formula for curvature is assuming that $t$ is the arclength parameter, which, of course it isn't. So when you working with a non-arclength parametrized curve you must use the chain rule to correct the computation. [See my differential geometry text linked in my profile for lots of examples.]
Since the independent variable is $t$, I'm going to use capital letters for the unit tangent and normal. Note that 
$$T(t)=\lambda(t) \big(1,f'(t)\big), \quad\text{where } \lambda(t) = (1+f'(t)^2)^{-1/2}.$$
When we differentiate by the product rule, we get two terms, one a multiple of $\alpha'(t)=(1,f'(t))$. This term will disappear when we dot with the unit normal. The term that remains is $\lambda(t) \big(0,f''(t)\big)$. When we dot with the unit normal $N(t)=\pm \lambda(t) \big(-f'(t),1\big)$, we get 
$$\pm\lambda(t) \big(0,f''(t)\big)\cdot \lambda(t) \big(-f'(t),1\big) = \pm \frac{f''(t)}{1+f'(t)^2}=\frac{|f''(t)|}{1+f'(t)^2}.$$
Ah, but the exponent is wrong on the denominator. This is because we need to correct by the chain rule:
$$\frac{dT}{ds}= \frac{dT/dt}{ds/dt},$$
where, of course, $ds/dt = \|\alpha'(t)\| = \big(1+f'(t)^2\big)^{1/2}$. This gives us an extra factor of $\lambda$ and we have the correct answer, $\kappa(t)=\dfrac{|f''(t)|}{\big(1+f'(t)^2\big)^{3/2}}$.
A: If $\gamma: \mathbb{R} \to \mathbb{R}^3$ is a curve then the curvature at $t$ is given by;
$$\kappa(t) = \frac{\|\gamma'(t) \times \gamma''(t)\|}{\|\gamma'(t)\|^3}$$
If you wish to find the curvature of a graph for $f: \mathbb{R} \to \mathbb{R}$ then you parametrize by $\alpha(x) = \langle x, f(x), 0 \rangle$. Now if you follow that above formula you get;
$$\kappa(x) = \frac{\left|f''(x)\right|}{\left(1+f'(x)^2\right)^{\frac{3}{2}}}$$
