Answer
$\dfrac{(9t^2+20)^{3/2}-(20)^{3/2}}{27}$
Work Step by Step
The first derivative of the given vector is given by:
$r'(t) =2t i+4t j+3t^2 \ k \implies ||r'(t)||=\sqrt {}(2t)^2+(4t)^2+(3t^2)^2=t \sqrt {9t^2+20}$
We calculate the length by integration:
$L=\int_{p}^{q}\|r'(t)\|dt=\int_{0}^{t} a \sqrt {9a^2+20} \ da
\\ =\dfrac{1}{18} \times \int_{0}^{t} 18 a \sqrt {9a^2+20} \ da \\ =\dfrac{1}{27} [(9a^2+20)^{3/2}]_0^t \\=\dfrac{1}{27} [(9t^2+20)^{3/2}-(20)^{3/2}] \\ = \dfrac{(9t^2+20)^{3/2}-(20)^{3/2}}{27}$
