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: 111


The tangent is $$y=-\frac{1}{4}x+\frac{9}{4}$$ and the normal is $$y=4x-2$$

Work Step by Step

$$x^2+2y^2=9\hspace{2cm}A(1,2)$$ a) Find the derivative $y'$ with respect to $x$, using implicit differentiation: $$2x+4yy'=0$$ $$x+2yy'=0$$ $$y'=-\frac{x}{2y}$$ b) The slope of the tangent line at $A(1,2)$ is $$y'=-\frac{1}{2\times2}=-\frac{1}{4}$$ The tangent to the curve at $(1,2)$ is $$y-2=-\frac{1}{4}(x-1)$$ $$y-2=-\frac{1}{4}x+\frac{1}{4}$$ $$y=-\frac{1}{4}x+\frac{9}{4}$$ 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 $A(1,2)$ $s$, we have $$s\Big(-\frac{1}{4}\Big)=-1$$ $$s=4$$ The normal to the curve at $(1,2)$ is $$y-2=4(x-1)$$ $$y-2=4x-4$$ $$y=4x-2$$
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