## Calculus 8th Edition

$\dfrac{6t^2}{[4t^2+9t^4]^{3/2}}$
As we are given that $x=t^2, y=t^3$ This yields, $x'=2t, y'=3t^2$ Also, $x''=2, y''=6t$ From previous solution, we have $\kappa=\dfrac{|\dot{x} \ddot{y}-\dot{y} \ddot{x}|}{[\dot{x}^2+\dot{y}^2]^{3/2}}$ Thus, $\kappa=\dfrac{|(2t)(6t)-(3t^2 \cdot 2)}{[(2t)^2+9t^2]^{3/2}}$ or, $\kappa=\dfrac{|12t^2-6t^2|}{[4t^2+9t^4]^{3/2}}$ Hence, it has been proved. $\kappa=\dfrac{6t^2}{[4t^2+9t^4]^{3/2}}$