University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Practice Exercises - Page 203: 113


The tangent is $$y=2x-4$$ and the normal is $$y=-\frac{1}{2}x+\frac{7}{2}$$

Work Step by Step

$$xy+2x-5y=2\hspace{2cm}(3,2)$$ a) Find the derivative $y'$ with respect to $x$, using implicit differentiation: $$(x'y+xy')+2x'-5y'=0$$ $$y+xy'+2-5y'=0$$ $$(x-5)y'=-(y+2)$$ $$y'=-\frac{y+2}{x-5}=\frac{y+2}{5-x}$$ b) The slope of the tangent line at $(3,2)$ is $$y'=\frac{2+2}{5-3}=2$$ The tangent to the curve at $(3,2)$ is $$y-2=2(x-3)$$ $$y-2=2x-6$$ $$y=2x-4$$ c) The normal is the perpendicular line to the tangent at a point, meaning the product of their slopes is $-1$. So if we call the slope of the normal at $A(3,2)$ $s$, we have $$s\times2=-1$$ $$s=-\frac{1}{2}$$ The normal to the curve at $(3,2)$ is $$y-2=-\frac{1}{2}(x-3)$$ $$y-2=-\frac{1}{2}x+\frac{3}{2}$$ $$y=-\frac{1}{2}x+\frac{7}{2}$$
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