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: 23


The equation of the normal line is $$y_n=\frac{1}{2}x+3.$$

Work Step by Step

Step 1: Find the slope of the tangent at $P(1,2)$. By definition of the slope at $a=1$ we get $$m=\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{\frac{2}{1+h}-\frac{2}{1}}{h}=\lim_{h\to0}\frac{\frac{2-2(1+h)}{1+h}}{h}=\lim_{h\to0}\frac{2-2-2h}{h(1+h)}=\lim_{h\to0}\frac{-2h}{h(1+h)}=\lim_{h\to0}\frac{-2}{1+h}=\frac{-2}{1+0}=-2.$$ Step 2: When we have the slope, the equation of the tangent at $P(1,2)$ is given by $y-2=m(x-1)$. Putting $m=-2$: $$y-2=-2(x-1)\Rightarrow y-2=-2x+2$$ which gives $$y=-2x+4.$$ 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 $a=1$, $f(a)=2$ and $m=-2$ we get $$-2\cdot c=-1,\quad c\cdot 1+d=2.$$ From the first equation we get $c=\frac{1}{2}$. Putting this into the second equation we get $$-2\cdot\frac{1}{2}+d=2\Rightarrow -1+d=2\Rightarrow d=3.$$ Finally we have $$y_n=\frac{1}{2}x+3.$$
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