Answer
$\dfrac{6t^2}{(9t^4+4t^2)^{3/2}}$
Work Step by Step
Given: $r(t)=t^3j+t^2k$
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,
$r'(t)=3t^2j+2tk$ and $r''(t)=6tj+2k$
and $|r'(t)|=\sqrt {(3t^2)^2+(2t)^2}=\sqrt {9t^4+4t^2}$
Now,
$\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{| (3t^2j+2tk)\times(6tj+2k)|}{|\sqrt {9t^4+4t^2}|^3}$
$=\dfrac{6t^2}{|\sqrt {9t^4+4t^2}|^3}$
$=\dfrac{6t^2}{(9t^4+4t^2)^{3/2}}$
