Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises: 14

Answer

$R_s(1,2)=32, R_t(1,2)=-39$

Work Step by Step

We will use the Chain Rule to find $R_s(s,t)$ and $R_t(s,t)$ and then we will find it for $s=1$ and $t=2$: $$R_s(s,t)=\frac{\partial}{\partial s}(G(u(s,t),v(s,t)))=G_u(u(s,t),v(s,t))u_s(s,t)+G_v(u(s,t),v(s,t))v_s(s,t)$$ $$R_t(s,t)=\frac{\partial}{\partial t}(G(u(s,t),v(s,t)))=G_u(u(s,t),v(s,t))u_t(s,t)+G_v(u(s,t),v(s,t))v_t(s,t)$$ So we have: $$R_s(1,2)=G_u(u(1,2),v(1,2))u_s(1,2)+G_v(u(1,2),v(1,2))v_s(1,2)= G_u(5,7)\cdot4+G_v(5,7)\cdot2=9\cdot4-2\cdot2=32$$ $$R_t(1,2)=G_u(u(1,2),v(1,2))u_t(1,2)+G_v(u(1,2),v(1,2))v_t(1,2)= G_u(5,7)\cdot(-3)+G_v(5,7)\cdot6=9\cdot(-3)-2\cdot6=-39$$
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