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: 24


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

Work Step by Step

Step 1. Find the slope of the tangent at the point $P(3,0)$. By definition of the tangent slope at $P(a,f(a))$ with $a=3$ and $f(a)=0$ we have $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to0}\frac{(3+h)^2-3(3+h)-0}{h}=\lim_{h\to0}\frac{9+h^2+6h-9-3h}{h}=\lim_{h\to0}\frac{h^2+3h}{h}=\lim_{h\to0}(h+3)=0+3=3.$$ Step 2. Find the equation of the tangent line. The tangent line at $(a,f(a))$ is given by $y-f(a)=m(x-a)$. Using $a=3$, $f(a)=0$ and $m=3$ we get $$y-0=3(x-3)\Rightarrow y=3x-9.$$ Step 3. At the point $P(a,f(a))$ when the tangent slope is $m$ the equation of the normal line is $y_n=cx+d$ where $$m\cdot c=-1,\quad c\cdot a+d=f(a).$$ Using $a=3$, $f(a)=0$, $m=3$ we have $$3c=-1,\quad 3c+d=0.$$ From the first equation we have $c=-\frac{1}{3}$. Putting this into the second equation we have $$-\frac{1}{3}\cdot 3+d=0\Rightarrow -1+d=0\Rightarrow d=1.$$ Finally the equation of the normal is $$y_n=-\frac{1}{3}x+1.$$
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