Answer
$15+ln(4)$
Work Step by Step
$r(t) = \langle 2t, ln(t), t^2 \rangle$, so
$r'(t) = \langle 2, \frac{1}{t}, 2t \rangle$
$length = \int_1^4 \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}dt$
$= \int_1^4 \sqrt{2^2 + (\frac{1}{t})^2 + (2t)^2}dt$
$= \int_1^4 \sqrt{4 + \frac{1}{t^2} + 4t^2}dt$
$= \int_1^4 \sqrt{\frac{4t^4+4t^2+1}{t^2}}dt$
$= \int_1^4 \sqrt{\frac{(2t^2+1)^2}{t^2}}dt$
$= \int_1^4 \frac{2t^2+1}{t}dt$
$= \int_1^4 (2t+\frac{1}{t})dt$
$= [t^2+ln(t) ]_1^4$
$= (4^2+ln(4)) - (1^2+ln(1))$
$= 16+ln(4) - 1$
$= 15+ln(4)$
