MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 14 August 2014

Lecture 19

The tangent line to a curve $f\left(x\right)$ at the point $(a,b)$ is the line through $(a,b)$ with the same slope as $f\left(x\right)$ at the point $(a,b)\text{.}$

The normal line is the line through $(a,b)$ which is perpendicular to the tangent line. $$\begin{array}{c} x y a b y=f⁡(x) tangent line normal line \end{array}$$

The slope of the tangent line $(a,b)$ is $$\frac{df}{dx}{|}_{x=a}\text{.}$$ If a line has slope $\frac{2}{5}$ $$\begin{array}{c} x y -2 5 2 5 \end{array}$$ then the perpendicular line has slope $\frac{5}{-2}\text{.}$

Find the equations of the tangent and normal to the curve $y=x^4-6x^3+13x^2-10x+5$ at the point where $x=1\text{.}$

The slope of the tangent line at $x=1$ is $$\frac{df}{dx}{|}_{x=1}=4x^3-18x^2+26x-10{|}_{x=1}=4-18+26-10=2\text{.}$$ The tangent line goes through the point $$\begin{array}{ccc}{\displaystyle x}& =& {\displaystyle 1,}\\ {\displaystyle y}& =& {\displaystyle 1-6+13-10+5=3\text{.}}\end{array}$$ The equation of a line is $y=mx+b$ where $m$ is the slope. So, for our line $$m=2\text{and}3=m·1+b=2·1+b\text{.}$$ So $b=1\text{.}$ So the tangent line is $$y=2x+1\text{.}$$ The slope of the normal line is $\frac{1}{-2}=-\frac{1}{2}\text{.}$ The equation of the normal line is $y=mx+c$ with $m=-\frac{1}{2}$ and $3=m·1+b=-\frac{1}{2}+b\text{.}$ So $b=\frac{7}{2}$ and $y=-\frac{1}{2}x+\frac{7}{2}$ is the normal line.

Find the equation of the tangent and normal lines to the curve $$x=a\text{cos}\hspace{0.17em}\theta ,y=b\text{sin}\hspace{0.17em}\theta \text{at}\theta =\frac{\pi }{4}\text{.}$$

First graph this: $$\frac{x}{a}=\text{cos}\hspace{0.17em}\theta ,\frac{y}{b}=\text{sin}\hspace{0.17em}\theta \text{.}$$ So ${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2=1\text{.}$ $$\begin{array}{c} x y -a a b -b π4 ( 22a , 22b ) \end{array}$$ When $\theta =\frac{\pi }{4},$ $$\begin{array}{ccc}{\displaystyle x}& =& {\displaystyle a\text{cos}\hspace{0.17em}\frac{\pi }{4}=\frac{\sqrt{2}}{2}a,}\\ {\displaystyle y}& =& {\displaystyle b\text{sin}\hspace{0.17em}\frac{\pi }{4}=\frac{\sqrt{2}}{2}b\text{.}}\end{array}$$ The slope of the tangent line is $$\frac{dy}{dx}{|}_{\underset{y=\frac{\sqrt{2}}{2}b}{x=\frac{\sqrt{2}}{2}a}}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$d\theta $}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$d\theta $}\right.}{|}_{\theta =\frac{\pi }{4}}=\frac{\frac{d\hspace{0.17em}b\text{sin}\hspace{0.17em}\theta }{d\theta }}{\frac{d\hspace{0.17em}a\text{cos}\hspace{0.17em}\theta }{d\theta }}{|}_{\theta =\frac{\pi }{4}}=\frac{b\text{cos}\hspace{0.17em}\theta }{-a\text{sin}\hspace{0.17em}\theta }{|}_{\theta =\frac{\pi }{4}}=\frac{b\frac{\sqrt{2}}{2}}{-a\frac{\sqrt{2}}{2}}=-\frac{b}{a}\text{.}$$ So the equation of the tangent line is $y=mx+y_0$ with $m=-\frac{b}{a}$ and $\frac{\sqrt{2}}{2}b=m\frac{\sqrt{2}}{2}a+y_0=-\frac{b}{a}\frac{\sqrt{2}}{2}a+y_0\text{.}$ So $y_0=\frac{\sqrt{2}}{2}b+\frac{\sqrt{2}}{2}b=\sqrt{2}b\text{.}$ So the equation of the tangent line is $$y=-\frac{b}{a}x+\sqrt{2}b\text{.}$$ The equation of the normal line is $y=mx+y_0$ with $m=\frac{a}{b}$ and $\frac{\sqrt{2}}{2}b=m\frac{\sqrt{2}}{2}a+y_0=\frac{a}{b}\frac{\sqrt{2}}{2}a+y_0\text{.}$ So $y_0=\frac{\sqrt{2}}{2}b-\frac{\sqrt{2}}{2}\frac{a^2}{b}=\frac{\sqrt{2}}{2}\left(\frac{b^2-a^2}{b}\right)\text{.}$ So the equation of the normal line is $$y=\frac{a}{b}x+\frac{\sqrt{2}}{2}\left(\frac{b^2-a^2}{b}\right)\text{.}$$

Find the equations of the normal to $2x^2-y^2=14,$ parallel to the line $x+3y=4\text{.}$

The line $x+3y=4$ is the same as $$y=-\frac{1}{3}x+\frac{4}{3}\text{.}$$ So it has slope $-\frac{1}{3}\text{.}$ So the slope of the normal line is $-\frac{1}{3}\text{.}$ So the slope of the tangent line is $3\text{.}$ So $$\frac{dy}{dx}{|}_{x=a}=3\text{.}$$ Now $4x-2y\frac{dy}{dx}=0\text{.}$ So $\frac{dy}{dx}=\frac{-4x}{-2y}=\frac{2x}{y}\text{.}$ So we want $\frac{2x}{y}=3$ and $2x^2-y^2=14\text{.}$ So $$y=\frac{2}{3}x\text{and}2x^2-{\left(\frac{2}{3}x\right)}^2=14\text{.}$$ So $$2x^2-\frac{4}{9}{x}^2=14\text{.}$$ So $$\frac{14}{9}{x}^2=14\text{.}$$ So $$x^2=9\text{.}$$ So $$x=\pm 3\text{.}$$ So $x=3$ and $y=\frac{2}{3}·3=2$ or $x=-3$ and $y=\frac{2}{3}(-3)=-2\text{.}$

In the first case: The normal has slope $-\frac{1}{3}$ and goes through $(3,2)\text{.}$
So $m=-\frac{1}{3}$ and $2=m·3+y_0=-\frac{1}{3}·3+y_0\text{.}$
So $y_0=3$ and the equation of the normal line is $y=-\frac{1}{3}x+3\text{.}$

In the second case: The normal has slope $-\frac{1}{3}$ and goes through $(-3,-3)\text{.}$
So $m=-\frac{1}{3}$ and $-2=m(-3)+y_0=(-\frac{1}{3})(-3)+y_0\text{.}$
So $y_0=-3$ and the equation of the normal line is $y=-\frac{1}{3}x-3\text{.}$

The graph should explain how there can be two normal lines parallel to $x+3y=4\text{.}$ $$\begin{array}{c} x y 4 -3 -2 -1 1 2 3 7 -7 -2 2 43 x+3y=4 (-3,-2) (3,2) first normal line second normal line \end{array}$$ Notes:

(a)	If $y=0,$ $x=\pm \sqrt{7}\text{.}$
(b)	$2-{\left(\frac{y}{x}\right)}^2=\frac{14}{x^2}\text{.}$ So, as $x\to \infty ,$ this becomes $2-{\left(\frac{y}{x}\right)}^2=0\text{.}$ ${\left(\frac{y}{x}\right)}^2=2,$ $\left(\frac{y}{x}\right)=\pm \sqrt{2},$ $y=\pm \sqrt{2}x\text{.}$

Notes and References

These are a typed copy of Lecture 19 from a series of handwritten lecture notes for the class MATH 221 given on October 23, 2000.

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