## Calculus: Early Transcendentals 8th Edition

$\dfrac{\sqrt 6}{2(3t^2+1)^2}$
Given: $r(t)=\sqrt6t^2i+2tj+2t^3k$ To calculate the curvature of the curve we will have to use Theorem 10 such as: $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}$ Thus, $r'(t)=2\sqrt6ti+2j+6t^2k$ and $r''(t)=2\sqrt6i+12tk$ and $|r'(t)|=\sqrt {(2\sqrt6t)^2+(2)^2+(6t^2)^2}$ This implies $|r'(t)|=6t^2+2$ Now, $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{| (2\sqrt6ti+2j+6t^2k)\times(2\sqrt6i+12tk|}{|6t^2+2|^3}$ $=\dfrac{|24t-12\sqrt 6 t^2j-4\sqrt6 k|}{|6t^2+2|^3}$ $=\dfrac{\sqrt{(24)^2+(-12\sqrt 6 t^2)^2+(-4\sqrt6)^2}}{|6t^2+2|^3}$ $=\dfrac{\sqrt 6}{(6t^2+2)^3}$ Hence, $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{\sqrt 6}{2(3t^2+1)^2}$