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: 23

Answer

\begin{aligned} \frac{d y}{d x}= -\frac{x+x^{2}+y^{2}}{y+x^{2}+y^{2}}\end{aligned}

Work Step by Step

Given $$\ln \sqrt{x^{2}+y^{2}}+x+y=4$$ $$ \Rightarrow \frac{1}{2}\ln (x^{2}+y^{2})+x+y-4=0$$ by letting $$F(x,y)=\frac{1}{2}\ln (x^{2}+y^{2})+x+y-4$$ So, we have $$F_x(x,y)=\frac{\partial F(x,y)}{\partial x}= \frac{2x}{2(x^{2}+y^{2})}+1\\ \quad \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{x}{(x^{2}+y^{2})}+1=\frac{x+x^{2}+y^{2}}{(x^{2}+y^{2})} $$ $$F_y(x,y)=\frac{\partial F(x,y)}{\partial y}=\frac{2y}{2(x^{2}+y^{2})}+1\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{ y}{ (x^{2}+y^{2})}+1 =\frac{ y+x^{2}+y^{2}}{ (x^{2}+y^{2})}$$ Also, we get \begin{aligned} \frac{d y}{d x}=&-\frac{F_{x}(x, y)}{F_{y}(x, y)} \\ &=-\frac{\frac{ x+x^{2}+y^{2}}{ (x^{2}+y^{2})}}{\frac{ y+x^{2}+y^{2}}{ (x^{2}+y^{2})}} \\ &=-\frac{x+x^{2}+y^{2}}{y+x^{2}+y^{2}}\end{aligned}
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