Why is the radius of curvature = 1/(curvature)? My textbook Thomas' Calculus (14th edition) initially defines curvature as the magnitude of change of direction of tangent with respect to the  arc length of the curve (|dT/ds|, where T is the tangent vector and s is the arc length) and later by intuition conclude that κ = 1/ρ (where, κ=curvature,ρ = radius). 
Is there a way to prove this equation mathematically ?
Edit:
If curvature was defined as the inverse of radius of curvature, then how does the textbook define and derive an expression of curvature first and then provide a relationship between it and the radius of curvature.
 A: Without much calculation (but also without dealing with exceptional cases, such as the tangent being vertical, or the second derivative being zero):
By the definition given at the start of the question, and by the chain rule, the curvature of $y = f(x)$ at $x$ is: $$\left\lvert\frac{(d/dx)\tan^{-1}f'(x)}{ds/dx}\right\rvert,$$
where $s$ is arc length. Since $ds/dx = \sqrt{1 + f'(x)^2}$, the curvature depends only on $f'(x)$ and $f''(x)$.
Therefore, if functions $f, g$ have second order contact at $x$, i.e. $f(x) = g(x)$, $f'(x) = g'(x)$, and $f''(x) = g''(x)$, then $f, g$ will have the same curvature at $x$.
If the graph of $g$ is a circle, then its curvature is the same as the magnitude of the rate of change of the direction of the radius with respect to arc length (because the tangent at a point is perpendicular to the radius to that point). But if the radius has length $r$, and makes an angle $\theta$ with a fixed line, then the length of the arc from the point of intersection of the circle with that line is $r\theta$, therefore the curvature is $1/r$.
It is enough, then, to prove that if $f''(x) \ne 0$, there exists a circle $g$ having second order contact with $f$ at $x$.
Proceeding similarly to Hardy, A Course of Pure Mathematics (10th ed. 1952, p. 299), but leaving out the explicit formulae, which we don't need (although they're simple enough, and the expression for $r$ gives the same result as the defining expression in the second paragraph above):
Writing $y = f(x)$, $\dot{y} = f'(x)$, $\ddot{y} = f''(x)$, so that also $y = g(x)$, $\dot{y} = g'(x)$, $\ddot{y} = g''(x)$ (with a suitable function $g$ to be found), we require a circle centre $(p, q)$, radius $r$, such that:
\begin{align*}
(x - p)^2 + (y - q)^2 & = r^2, \\
(x - p) + (y - q)\dot{y} & = 0, \\
1 + \dot{y}^2 + (y - q)\ddot{y} & = 0.
\end{align*}
Because $\ddot{y} \ne 0$, the third equation can be solved for $q$, then the second equation can be solved for $p$, and finally the first equation can be solved for $r$. So the required circle exists. $\square$
