#### Answer

$$\frac{2}{135}((136)^{3/2}-1).$$

#### Work Step by Step

We have $r(t) =\lt t,4 t^{3/2}, 2t^{3/2}\gt$, and so the vector derivative is $r'(t) =\lt 1, 6 t^{1/2}, 3t^{1/2}\gt$.
We calculate the length by integration:
$$length=\int_{0}^{3}\sqrt{1+36t+9t} dt=\int_{0}^{3}\sqrt{1+45t} dt=\frac{2}{45(3)}(1+45t)^{3/2}|_{0}^{3}=\frac{2}{135}((136)^{3/2}-1).$$