## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.7 - Implicit Differentiation - Exercises - Page 165: 44

#### Answer

The tangent at $(1,1)$ is $$y=2x-1$$ The normal at $(1,1)$ is $$y=-\frac{1}{2}x+\frac{3}{2}$$

#### Work Step by Step

$$y^2(2-x)=x^3$$ 1) Find the derivative of the function using implicit differentiation: $$(y^2)'(2-x)+y^2(2-x)'=3x^2$$ $$2yy'(2-x)+y^2(-1)=3x^2$$ $$2yy'(2-x)-y^2=3x^2$$ $$2yy'(2-x)=3x^2+y^2$$ $$y'=\frac{3x^2+y^2}{2y(2-x)}$$ 2) The slope of the tangent line to the curve at $(1,1)$ is $$y'=\frac{3\times1^2+1^2}{2\times1(2-1)}=\frac{4}{2\times1}=2$$ The tangent line to the curve at $(1,1)$, therefore, is $$y-1=2(x-1)$$ $$y-1=2x-2$$ $$y=2x-1$$ 3) As tangent line is perpendicular with its corresponding normal line, the product of their slopes equal $-1$. In other words, if we call the slope of the normal at $(1,1)$ $k$, then we have $$k\times2=-1$$ $$k=-\frac{1}{2}$$ The normal line to the curve at $(1,1)$, therefore, is $$y-1=-\frac{1}{2}(x-1)$$ $$y-1=-\frac{1}{2}x+\frac{1}{2}$$ $$y=-\frac{1}{2}x+\frac{3}{2}$$

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