## Calculus 8th Edition

$2.0454$
As we are given that $r(t)=\lt t, e^{-t},te^{-t}\gt$; $1 \leq t \leq 3$ Length of the curve can be obtained by using formula, such as $L=\int_a^b |r'(t)| dt$ Now, $r'(t)=\lt 1,-e^{-t},e^{-t}-te^{-t}\gt$ ; $|r'(t)|=\sqrt {( 1)^2+(e^{-t})^2+(e^{-t}-te^{-t})^2}dt$ or,$=\sqrt{ 1+e^{-2t}+(e^{-t}-te^{-t})^2}$ or, $=\sqrt{ 1+e^{-2t}+e^{-2t}(1-t)^2}$ Since,$L=\int_{1}^3(\sqrt{ 1+e^{-2t}+e^{-2t}(1-t)^2}) dt= 2.0454$ This is the result calculated by using a calculator.