Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.5 Exercises - Page 913: 27

Answer

\begin{aligned} \frac{\partial z }{\partial x} =-\frac{x}{y+z}\end{aligned} \begin{aligned} \frac{\partial z }{\partial y} =- \frac{z}{y+z} \end{aligned}

Work Step by Step

Given $$ x^{2}+2yz +z^2=1$$ $$ \Rightarrow x^{2}+2yz +z^2-1=0$$ by letting $$F(x,y,z)= x^{2}+2yz +z^2-1$$ So, we have $$F_x(x,y,z)=\frac{\partial F(x,y,z)}{\partial x}=2x\\ $$ $$F_y(x,y,z)=\frac{\partial F(x,y,z)}{\partial y}= 2z$$ $$F_z(x,y,z)=\frac{\partial F(x,y,z)}{\partial z}= 2y+2z$$ Also, we get \begin{aligned} \frac{\partial z }{\partial x}&=-\frac{F_{x}(x, y,z)}{F_{z}(x, y,z)} \\ &=- \frac{2x}{2y+2z} \\ &=-\frac{x}{y+z}\end{aligned} \begin{aligned} \frac{\partial z }{\partial y}&=-\frac{F_{y}(x, y,z)}{F_{z}(x, y,z)} \\ &=- \frac{2z}{2y+2z} \\ &=- \frac{z}{y+z} \end{aligned}
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