Calculus 8th Edition

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

Chapter 2 - Derivatives - Review - Exercises - Page 197: 49

Answer

$$ y=\sqrt{1+4 \sin x}, \,\,\,\, (0,1) $$ The equation of the tangent line is $$ y=2 x+1.$$ The equation of the normal line is $$ y=-\frac{1}{2} x+1$$

Work Step by Step

$$ y=\sqrt{1+4 \sin x}, \,\,\,\, (0,1) $$ we have $$ \begin{aligned} y^{\prime}&=\frac{1}{2}(1+4 \sin x)^{-1 / 2} \cdot 4 \cos x \\ &=\frac{2 \cos x}{\sqrt{1+4 \sin x}} \end{aligned} $$ So the slope of the tangent line at $(0,1)$ is $$ \left.\frac{d y}{d x}\right|_{x=0}=\frac{2 \cos (0)}{\sqrt{1+4 \sin (0)}}=\frac{2}{\sqrt{1}}=2 $$ so an equation of the tangent line is $$y-1=2(x-0), \,\,\ \text{or} \,\,\, y=2 x+1.$$ The slope of the normal line is $-\frac{1}{2},$ so an equation of the normal line is $$y-1=-\frac{1}{2}(x-0), \,\,\,\ \text{ or} \,\,\, y=-\frac{1}{2} x+1$$
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