## Thomas' Calculus 13th Edition

$x=1+t ; \\ y =t ; \\ z=-t$
$v(t)=\dfrac{dr}{dt}= e^t i+\cos t j -\dfrac{k}{1-t}$ Set $t=0$ Then, we have: $v(t)=\dfrac{dr}{dt}= e^{0} i+\cos (0) j -\dfrac{k}{1-0}=i+j-k$ and $r(0)=e^{0} i+\sin (0) j +\ln (1-0) k =i$ Thus, the parametric equations of the line at $(1,0,0)$ are: $x=1+t ; \\ y =t ; \\ z=-t$