Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.2 Working with Derivatives - 3.2 Exercises - Page 143: 22


The equation of the normal line is $$y_n=-4x+18.$$

Work Step by Step

Step 1: Find the slope of the tangent at $P(4,2)$. By definition of the slope at $a=4$ we get $$m=\lim_{h\to 0}\frac{f(4+h)-f(4)}{h}=\lim_{h\to0}\frac{\sqrt{4+h}-\sqrt{4}}{h}=\lim_{h\to0}\frac{\sqrt{4+h}-2}{h}=\lim_{h\to0}\frac{\sqrt{4+h}-2}{h}\cdot\frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=\lim_{h\to0}\frac{\sqrt{4+h}^2-2^2}{h(\sqrt{4+h}+2)}=\lim_{h\to0}\frac{4+h-4}{h(\sqrt{4+h}+2)}=\lim_{h\to0}\frac{h}{h(\sqrt{4+h}+2)}=\lim_{h\to0}\frac{1}{\sqrt{4+h}+2}=\frac{1}{\sqrt{4+0}+2}=\frac{1}{4}.$$ Step 2: When we have the slope, the equation of the tangent at $P(4,2)$ is given by $y-2=m(x-4)$. Putting $m=\frac{1}{4}$: $$y-2=\frac{1}{4}(x-4)=\frac{1}{4}x-1$$ which gives $$y=\frac{1}{4}x+1.$$ Step 3: The equation of the normal line at the point $P(a,f(a))$ when we have the equation of the tangent $y=mx+n$ is given by $y_n=cx+d$ such that $m\cdot c=-1$ and $c\cdot a+d=f(a)$. For $a=4$, $f(a)=2$ and $m=\frac{1}{4}$ we get $$\frac{1}{4}\cdot c=-1,\quad c\cdot 4+d=2.$$ From the first equation we get $c=-4$. Putting this into the second equation we get $$-4\cdot4+d=2\Rightarrow -16+d=2\Rightarrow d=18.$$ Finally we have $$y_n=-4x+18.$$
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