Answer
$\kappa=\dfrac{|\dot{x} \ddot{y}-\dot{y} \ddot{x}|}{[\dot{x}^2+\dot{y}^2]^{3/2}}$
Work Step by Step
Given: $x=f(t), y=g(t) \gt$
To calculate the curvature of the curve we will have to use Theorem 10 such as: $\kappa(t)=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}$
Consider $r(t)=\lt x,y,0 \gt$
Thus, $r'(t)=\lt \dot{x} , \dot{y} ,0\gt$
and $r''(t)=\lt \ddot{x} , \ddot{y} ,0\gt$
Now,
$\kappa=\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{|\dot{x} \ddot{y}-\dot{y} \ddot{x}|}{[\sqrt {\dot{x}^2+\dot{y}^2}]^{3}}$
Hence, $\kappa=\dfrac{|\dot{x} \ddot{y}-\dot{y} \ddot{x}|}{[\dot{x}^2+\dot{y}^2]^{3/2}}$
