MATH 221 Lecture 19

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

Last update: 9 August 2012

MATH 221 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 $(a,b)$.

The normal line is the line through $(a,b)$ which is perpendicular to the tangent line.

The slope of the tangent line $(a,b)$ is

${\displaystyle {\frac{df}{dx}|}_{x=a}}$

If a line has slope ${\displaystyle \frac{2}{5}}$

then the perpendicular line has slope ${\displaystyle \frac{5}{-2}}$

Example:

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

The slope of the tangent line at $x=1$ is

${\displaystyle {\frac{df}{dx}|}_{x=1}=4x^3-18x^2+26x-10{|}_{x=1}=4-18+26-10=2}$

The equation of a line is $y=mx+b$ where $m$ is the slope. So, for our line

$m=2\text{and}3=m·1+b=2·1+b$

So $b=1$.$$So the tangent line is

$y=2x+1$.

The slope of the normal line is ${\displaystyle \frac{1}{-2}=-\frac{1}{2}}$.

The equation of the normal line is $y=mx+c$ with ${\displaystyle m=-\frac{1}{2}}$ and $3=m·1+b$.

So ${\displaystyle b=\frac{7}{2}}$ and ${\displaystyle y=-\frac{1}{2}+\frac{7}{2}}$ is the normal line.

Example:

Find the equation of the tangent and normal lines to the curve

${\displaystyle x=a\cos \theta ,y=b\sin \theta \text{at}\theta =\frac{\pi }{4}}$.

First graph this:

${\displaystyle \frac{x}{a}=\cos \theta ,\frac{y}{b}=\sin \theta .\text{So}{\frac{x}{a}}^2+{\frac{y}{b}}^2=1.}$

${\displaystyle \begin{array}{cc}\text{When}\theta =\frac{\pi }{4},& x=a\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}a\\ & y=b\sin \frac{\pi }{4}=\frac{\sqrt{2}}{2}b\end{array}}$

The slope of the tangent line is

${\displaystyle {\frac{dy}{dx}|}_{\begin{array}{c}x=\frac{\sqrt{2}}{2}a\\ y=\frac{\sqrt{2}}{2}b\end{array}}={\frac{dy/d\theta }{dx/d\theta }|}_{\theta =\frac{\pi }{4}}={\frac{\frac{db\sin \theta }{d\theta }}{\frac{da\cos \theta }{d\theta }}|}_{\theta =\frac{\pi }{4}}={\frac{b\cos \theta }{-a\sin \theta }|}_{\theta =\frac{\pi }{4}}=\frac{b\frac{\sqrt{2}}{2}}{-a\frac{\sqrt{2}}{2}}=-\frac{b}{a}}$.

So the equation of the tangent line is $y=mx+y_0$ with ${\displaystyle m=\frac{-b}{a}}$ and ${\displaystyle \frac{\sqrt{2}}{2}=m\frac{\sqrt{2}}{2}a+y_0=\frac{-b}{a}\frac{\sqrt{2}}{2}a+y_0}$.

So ${\displaystyle y_0=\frac{\sqrt{2}}{2}b+\frac{\sqrt{2}}{2}b=\sqrt{2}b}$.

So the equation of the tangent line is

${\displaystyle y=\frac{-b}{a}x+\sqrt{2}b}$.

The equation of the normal line is $y=mx+y_0$ with ${\displaystyle m=\frac{a}{b}}$ and ${\displaystyle \frac{\sqrt{2}}{2}b=m\frac{\sqrt{2}}{2}a+y_0=\frac{a}{b}\frac{\sqrt{2}}{2}a+y_0}$

So ${\displaystyle y_0=\frac{\sqrt{2}}{2}b-\frac{a^2}{b}=\frac{\sqrt{2}}{2}\left(\frac{b^2-a^2}{b}\right)}$.

So the equation of the normal line is

${\displaystyle y=\frac{a}{b}x+\frac{\sqrt{2}}{2}\left(\frac{b^2-a^2}{b}\right)}$

Example:

Find the equations of the normal to $2x^2-y^2=14$

The line $x+3y=4$ is the same as

${\displaystyle y=-\frac{1}{3}x+\frac{4}{3}}$.

So it has slope ${\displaystyle -\frac{1}{3}}$.

So the slope of the normal line is ${\displaystyle -\frac{1}{3}}$.

So the slope of the tangent line is $3$.

So

${\displaystyle {\frac{dy}{dx}|}_{x=3}=3}$.

Now ${\displaystyle 4x-2\frac{dy}{dx}=0}$. So we want ${\displaystyle \frac{2x}{y}=3}$ and $2x^2-y^2=14$.

So ${\displaystyle 2x^2-\frac{4}{9}{x}^2=14}$.

So ${\displaystyle \frac{14}{9}{x}^2=14}$. So $x^2=9$. So $x=\pm 3$.

So $x=3$ and ${\displaystyle y=\frac{2}{3}·3=2}$ or $x=-3$ and ${\displaystyle y=\frac{2}{3}(-3)=-2}$.

In the first case:

The normal has slope ${\displaystyle -\frac{1}{3}}$ and goes through $(3,2)$.
So ${\displaystyle m=-\frac{1}{3}}$ and $2=m·3+y_0$
So $y_0=3$ and the equation of the normal line is

${\displaystyle y=-\frac{1}{3}x+3}$.

In the second case:

The normal has slope ${\displaystyle -\frac{1}{3}}$ and goes through $(-3,-2)$.

So ${\displaystyle m=-\frac{1}{3}}$ and ${\displaystyle -2=m(-3)+y_0=(-\frac{1}{3})(-3)+y_0}$.

So $y_0=-3$ and the equation of the normal line is

${\displaystyle y=-\frac{1}{3}x-3}$.

The graph should explain how there can be two normal lines parallel to $x+3y=4$

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

Notes and References

These are a typed copy of lecture notes given by Arun Ram on October 23, 2000.

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