## Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning

# Chapter 3 - Review - Exercises - Page 270: 64

#### Answer

tangent line: $y = -\frac{4}{5}x+\frac{13}{5}$ normal line: $y = \frac{5}{4}x-\frac{3}{2}$

#### Work Step by Step

$x^2+4xy+y^2=13$ We can use implicit differentiation to find $\frac{dy}{dx}$: $2x+4y+4x\frac{dy}{dx}+2y\frac{dy}{dx}=0$ $4x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x-4y$ $\frac{dy}{dx}=-\frac{2x+4y}{4x+2y}$ We can find $\frac{dy}{dx}$ when $x = 2$ and $y=1$: $\frac{dy}{dx}=-\frac{2(2)+4(1)}{4(2)+2(1)}$ $\frac{dy}{dx}=-\frac{8}{10}$ $\frac{dy}{dx}=-\frac{4}{5}$ The tangent line has a slope of $-\frac{4}{5}$ at this point. We can find the equation of the tangent line: $y-1 = (-\frac{4}{5})(x-2)$ $y-1 = -\frac{4}{5}x+\frac{8}{5}$ $y = -\frac{4}{5}x+\frac{13}{5}$ The normal line has a slope of $\frac{5}{4}$ at this point. We can find the equation of the normal line: $y-1 = (\frac{5}{4})(x-2)$ $y-1 = \frac{5}{4}x-\frac{5}{2}$ $y = \frac{5}{4}x-\frac{3}{2}$

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