Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.2 Derivatives and Integrals of Vector Functions - 13.2 Exercises - Page 900: 25

Answer

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

Work Step by Step

Given: $x=e^{-t} \cos t, y=e^{-t} \sin t,z=e^{-t} $ Equation can be written as: $r(t)= e^{-t}\cos t i+e^{-t}\sin t j+e^{-t} k $ Tangent vector $r'(t)$ can be calculated by taking the derivative of $r(t)$ with respect to $t$. This yields,$r'(t)=e^{-t}(-\cos t- \sin t) i+e^{-t}(-\sin t+ \cos t)-e^{-t} k $ or, $r'(0)=e^{-0}(-\cos(0)- \sin(0)) i+e^{-0}(-\sin (0)+ \cos (0))-e^{-(0)} k =-i+j-k$ As we are given that the equation of tangent line passing through the points $(1,0,1)$, thus $r(t)=\lt 1-t,t,1-t \gt$ Hence, the desired parametric equations of the tangent line are: $x=1-t,y=t,z=1-t$
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