Answer
$\dfrac{7}{3}$
Work Step by Step
As we are given that $r(t)=\lt 2t, t^2, \frac{1}{3}t^3 \gt$ ; $0 \leq t \leq 1$
Length of the curve can be obtained by using formula, such as
$L=\int_a^b |r'(t)| dt$
Now, $r'(t)=\lt 2, 2t, t^2 \gt$ ;$|r'(t)|=\sqrt {(2)^2+(2t)^2+(t^2)^2}dt=\sqrt {4+4t^2+t^4}$
$=2+t^2$
$L=\int_{0}^1 (2+t^2)dt= 2t+\dfrac{t^3}{3}|_{0}^1=2(1)-2(0)+\dfrac{(1)^3}{3}-\dfrac{(0)^3}{3}=2+\dfrac{1}{3}=\dfrac{7}{3}$
