Answer
$\dfrac{\sqrt{76}}{14^{3/2}}$
Work Step by Step
Given: $r(t)=\lt t,t^2, t^3 \gt$
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,
$r'(t)=\lt 1,2t, 3t^2 \gt$
and $r''(t)=\lt 0,2,6t \gt$
Now,
$\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{|\sqrt {36t^4+36t^2+4}|}{[1+4t^2+9t^4]^{3/2}}$
At point $(1,1,1)$
$\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{|\sqrt {36t^4+36t^2+4}|}{[1+4t^2+9t^4]^{3/2}}=\dfrac{|\sqrt {36+36+4}|}{[1+4+9]^{3/2}}=\dfrac{\sqrt{76}}{[14]^{3/2}}$
Hence, $\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}|_{(1,1,1)}=\dfrac{\sqrt{76}}{14^{3/2}}$
