Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 13 - Review - Exercises - Page 882: 6


A) $(x,y,z)= (1\dfrac{7}{8},0, \ln (1/2))$ B) $r(t)=\lt 1-3t, 1+2t, t \gt$ or, $x=1-3t; y=1+2t,z=t$ C) $3x-2y-z=1$

Work Step by Step

A) When a curve intersects the xz palne, $y=0$. Here, we have $y=2t-1 =0 \implies t=\dfrac{1}{2}$ Thus, $x=2-(1/2)^3=1\dfrac{7}{8}$ and $z=\ln (1/2)$ Thus, $(x,y,z)= (1\dfrac{7}{8},0, \ln (1/2))$ B) Given: $x(t)=1=2-t^3; t=1$ Then $x'(t)=\lt 2-3t^2, 2, \dfrac{1}{t}$ At the point $(1,1,0)$, the parametric equations are: $r'(1)=\lt -3,2,1 \gt$ $r(t)=\lt 1-3t, 1+2t, t \gt$ or, $x=1-3t; y=1+2t,z=t$ C) A plane normal to a vector equation at a point can be written as: $ax+by+cz=d$ In this problem, $\lt a,b,c \gt =\lt -1, 2, 1 \gt$ Thus, we have $-3x+2y+z=-3(1)+2 (1) +1 \cdot (0)$ or, $-3x+2y+z=-1$ Hence, $3x-2y-z=1$
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