Answer
$2.0454$
Work Step by Step
Given: $r(t)=\lt t, e^{-t},te^{-t}\gt$; $1 \leq t \leq 3$
To calculate the length of the curve we will have to use the formula:
$L=\int_a^b |r'(t)| dt$
Thus,
$r'(t)=\lt 1,-e^{-t},e^{-t}-te^{-t}\gt$
and $|r'(t)|=\sqrt {( 1)^2+(e^{-t})^2+(e^{-t}-te^{-t})^2}dt$
$=\sqrt{ 1+e^{-2t}+(e^{-t}-te^{-t})^2}$
$=\sqrt{ 1+e^{-2t}+e^{-2t}(1-t)^2}$
$L=\int_{0}^3(\sqrt{ 1+e^{-2t}+e^{-2t}(1-t)^2}) dt$
As per question, we will use calculator to find the length of the curve.
Hence, $L= 2.0454$
