Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.4 Parametrized Surfaces and Surface Integrals - Exercises - Page 958: 7

Answer

${{\bf{T}}_u} = \left( {2,1,3} \right)$ ${{\bf{T}}_v} = \left( {1, - 4,0} \right)$ ${\bf{N}}\left( {1,4} \right) = 12{\bf{i}} + 3{\bf{j}} - 9{\bf{k}}$ The equation of the tangent plane: $4x + y - 3z = 0$

Work Step by Step

We have $G\left( {u,v} \right) = \left( {2u + v,u - 4v,3u} \right)$. So, ${{\bf{T}}_u} = \frac{{\partial G}}{{\partial u}} = \left( {2,1,3} \right)$ ${{\bf{T}}_v} = \frac{{\partial G}}{{\partial v}} = \left( {1, - 4,0} \right)$ ${\bf{N}}\left( {1,4} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ 2&1&3\\ 1&{ - 4}&0 \end{array}} \right| = 12{\bf{i}} + 3{\bf{j}} - 9{\bf{k}}$ The equation of the tangent plane to the surface at the point $G\left( {1,4} \right) = \left( {6, - 15,3} \right)$ is $\left( {x - 6,y + 15,z - 3} \right)\cdot\left( {12,3, - 9} \right) = 0$ $12\left( {x - 6} \right) + 3\left( {y + 15} \right) - 9\left( {z - 3} \right) = 0$ $4\left( {x - 6} \right) + y + 15 - 3\left( {z - 3} \right) = 0$ The equation can be written as $4x + y - 3z = 0$
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