Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 13 - Vector Functions - 13.2 Exercises - Page 876: 25

Answer

$x=1-t,y=t,z=1-t$

Work Step by Step

From the given points we have $r(t)= e^{-t}\cos t i+e^{-t}\sin t j+e^{-t} k $ In order to find tangent vector $r'(t)$ we will have to take the derivative of $r(t)$ with respect to $t$. Thus, $r'(t)=e^{-t}(-\cos t- \sin t) i+e^{-t}(-\sin t+ \cos t)-e^{-t} k $ $r'(0)=e^{-0}(-\cos(0)- \sin(0)) i+e^{-0}(-\sin (0)+ \cos (0))-e^{-(0)} k =-i+j-k$ Equation of tangent line passing through the points $(1,0,1)$ is: $r(t)=\lt 1-t,t,1-t \gt$ Therefore, the parametric equations of the line are: $x=1-t,y=t,z=1-t$
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