Calculus 8th Edition

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

Chapter 2 - Derivatives - Review - Exercises - Page 197: 50

Answer

$$ x^{2}+4 x y+y^{2}=13, \quad(2,1)$$ The equation of the tangent line is $$ 5y+4 x=13.$$ The equation of the normal line is $$ 4y-5x=-6 .$$

Work Step by Step

$$ x^{2}+4 x y+y^{2}=13, \quad(2,1)$$ Differentiate both sides this function , we have $$ \begin{aligned} & 2x+4 x y^{\prime} +4y +2y y^{\prime}=0\\ & 2(x+2y) +2 y^{\prime}( 2 x +y )=0\\ & \Leftrightarrow y^{\prime}=\frac{-( x+2y )}{( 2 x +y )} \end{aligned} $$ So the slope of the tangent line at $(2,1)$ is $$ \left.\frac{d y}{d x}\right|_{x=2, y=1}=\frac{-(2+2(1) )}{( 2(2) +1)} =\frac{-4}{5} $$ so an equation of the tangent line is $$y-1=\frac{-4}{5}(x-2), \,\,\ \text{or} \,\,\, 5y+4 x=13.$$ The slope of the normal line is $\frac{5}{4},$ so an equation of the normal line is $$y-1=\frac{5}{4}(x-2), \,\,\,\ \text{ or} \,\,\, 4y-5x=-6 .$$
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