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: 8

Answer

${{\bf{T}}_u}{|_{\left( {2,3} \right)}} = \left( {4,1,1} \right)$ ${{\bf{T}}_v}{|_{\left( {2,3} \right)}} = \left( { - 6,1, - 1} \right)$ ${\bf{N}}\left( {2,3} \right) = - 2{\bf{i}} - 2{\bf{j}} + 10{\bf{k}}$ The equation of the tangent plane: $x + y - 5z = 5$

Work Step by Step

We have $G\left( {u,v} \right) = \left( {{u^2} - {v^2},u + v,u - v} \right)$. So, ${{\bf{T}}_u} = \frac{{\partial G}}{{\partial u}} = \left( {2u,1,1} \right)$, ${\ \ \ }$ ${{\bf{T}}_u}{|_{\left( {2,3} \right)}} = \left( {4,1,1} \right)$ ${{\bf{T}}_v} = \frac{{\partial G}}{{\partial v}} = \left( { - 2v,1, - 1} \right)$, ${\ \ \ }$ ${{\bf{T}}_v}{|_{\left( {2,3} \right)}} = \left( { - 6,1, - 1} \right)$ ${\bf{N}}\left( {2,3} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = \left| {\begin{array}{*{20}{c}} {\bf{i}}&{\bf{j}}&{\bf{k}}\\ 4&1&1\\ { - 6}&1&{ - 1} \end{array}} \right| = - 2{\bf{i}} - 2{\bf{j}} + 10{\bf{k}}$ The equation of the tangent plane to the surface at the point $G\left( {2,3} \right) = \left( { - 5,5, - 1} \right)$ is $\left( {x + 5,y - 5,z + 1} \right)\cdot\left( { - 2, - 2,10} \right) = 0$ $ - 2\left( {x + 5} \right) - 2\left( {y - 5} \right) + 10\left( {z + 1} \right) = 0$ $x + 5 + y - 5 - 5z - 5 = 0$ The equation can be written as $x + y - 5z = 5$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.