Calculus (3rd Edition)

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

Chapter 16 - Multiple Integration - 16.6 Change of Variables - Exercises - Page 905: 20

Answer

The linear mapping is $G = \left( {u - 2v,7u + 5v} \right)$.

Work Step by Step

A linear mapping is given by the form: $G\left( {u,v} \right) = \left( {Au + Cv,Bu + Dv} \right)$ The domain ${\cal R}:\left[ {0,1} \right] \times \left[ {0,1} \right]$ in the $uv$-plane is spanned by the vectors $\overrightarrow {OP} = \left( {0,1} \right)$ and $\overrightarrow {OQ} = \left( {1,0} \right)$. Using $G\left( {u,v} \right)$ we obtain the images of $\overrightarrow {OP} $ and $\overrightarrow {OQ} $: $G\left( {\overrightarrow {OP} } \right) = G\left( {0,1} \right) = \left( {C,D} \right)$ $G\left( {\overrightarrow {OQ} } \right) = G\left( {1,0} \right) = \left( {A,B} \right)$ Since $G$ is a linear map, $G\left( {\overrightarrow {OP} } \right)$ and $G\left( {\overrightarrow {OQ} } \right)$ span the parallelogram in the $xy$-plane. Thus, $G\left( {\overrightarrow {OP} } \right) = \left( {C,D} \right) = \left( { - 2,5} \right)$ $G\left( {\overrightarrow {OQ} } \right) = \left( {A,B} \right) = \left( {1,7} \right)$ Therefore, $A=1$, $B=7$, $C=-2$, and $D=5$. Thus, we obtain the linear mapping: $G = \left( {u - 2v,7u + 5v} \right)$.
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