Answer
$\dfrac{\sqrt{4t^2e^{2t}-8te^{2t}+5e^{2t}+4}}{[1+4t^2+e^{2t}]^{3/2}}$
Work Step by Step
Given: $r(t)=ti+t^2j+e^tk$
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,
$r'(t)=1i+2tj+e^tk$ and $r''(t)=2j+e^tk$
and $|r'(t)|=\sqrt {(1)^2+(2t)^2+(e^t)^2}=\sqrt {1+4t^2+e^{2t}}$
Now,
$\dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}=\dfrac{| (1i+2tj+e^tk)\times(2j+e^tk)|}{|\sqrt {1+4t^2+e^{2t}}|^3}$
$=\dfrac{|(2te^t-2e^t)-(e^t-0)+(2-0)|}{|\sqrt {1+4t^2+e^{2t}}|^3}$
$=\dfrac{|2e^t(t-1)-e^t+2|}{|\sqrt {1+4t^2+e^{2t}}|^3}$
$=\dfrac{\sqrt{4t^2e^{2t}-8te^{2t}+5e^{2t}+4}}{[1+4t^2+e^{2t}]^{3/2}}$
