## MATH 221 Lecture 19

Last update: 9 August 2012

## MATH 221 Lecture 19

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

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

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

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

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

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

Example:

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

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

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

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

$m=2\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}3=m·1+b=2·1+b$

So $b=1$.$\phantom{\rule{1em}{0ex}}$So the tangent line is

$y=2x+1$.

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

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

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

Example:

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

$x=a\mathrm{cos}\theta ,\phantom{\rule{1em}{0ex}}y=b\mathrm{sin}\theta \phantom{\rule{1em}{0ex}}\text{at}\phantom{\rule{0.5em}{0ex}}\theta =\frac{\pi }{4}$.

First graph this:

$\frac{x}{a}=\mathrm{cos}\theta ,\phantom{\rule{1em}{0ex}}\frac{y}{b}=\mathrm{sin}\theta .\phantom{\rule{1em}{0ex}}\text{So}\phantom{\rule{0.5em}{0ex}}{\frac{x}{a}}^{2}+{\frac{y}{b}}^{2}=1.$

$\begin{array}{cc}\text{When}\phantom{\rule{0.5em}{0ex}}\theta =\frac{\pi }{4},& x=a\mathrm{cos}\frac{\pi }{4}=\frac{\sqrt{2}}{2}a\\ & y=b\mathrm{sin}\frac{\pi }{4}=\frac{\sqrt{2}}{2}b\end{array}$

The slope of the tangent line is

${\frac{dy}{dx}|}_{\begin{array}{c}x=\frac{\sqrt{2}}{2}a\\ y=\frac{\sqrt{2}}{2}b\end{array}}={\frac{dy/d\theta }{dx/d\theta }|}_{\theta =\frac{\pi }{4}}={\frac{\frac{d\phantom{\rule{0.3em}{0ex}}b\mathrm{sin}\theta }{d\theta }}{\frac{d\phantom{\rule{0.3em}{0ex}}a\mathrm{cos}\theta }{d\theta }}|}_{\theta =\frac{\pi }{4}}={\frac{b\mathrm{cos}\theta }{-a\mathrm{sin}\theta }|}_{\theta =\frac{\pi }{4}}=\frac{b\frac{\sqrt{2}}{2}}{-a\frac{\sqrt{2}}{2}}=-\frac{b}{a}$.

So the equation of the tangent line is $y=mx+{y}_{0}$ with $m=\frac{-b}{a}$ and $\frac{\sqrt{2}}{2}=m\frac{\sqrt{2}}{2}a+{y}_{0}=\frac{-b}{a}\frac{\sqrt{2}}{2}a+{y}_{0}$.

So ${y}_{0}=\frac{\sqrt{2}}{2}b+\frac{\sqrt{2}}{2}b=\sqrt{2}b$.

So the equation of the tangent line is

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

The equation of the normal line is $y=mx+{y}_{0}$ with $m=\frac{a}{b}$ and $\frac{\sqrt{2}}{2}b=m\frac{\sqrt{2}}{2}a+{y}_{0}=\frac{a}{b}\frac{\sqrt{2}}{2}a+{y}_{0}$

So ${y}_{0}=\frac{\sqrt{2}}{2}b-\frac{{a}^{2}}{b}=\frac{\sqrt{2}}{2}\left(\frac{{b}^{2}-{a}^{2}}{b}\right)$.

So the equation of the normal line is

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

Example:

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

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

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

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

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

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

So

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

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

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

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

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

In the first case:

The normal has slope $-\frac{1}{3}$ and goes through $\left(3,2\right)$.
So $m=-\frac{1}{3}$ and $2=m·3+{y}_{0}$
So ${y}_{0}=3$ and the equation of the normal line is

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

In the second case:

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

So $m=-\frac{1}{3}$ and $-2=m\left(-3\right)+{y}_{0}=\left(-\frac{1}{3}\right)\left(-3\right)+{y}_{0}$.

So ${y}_{0}=-3$ and the equation of the normal line is

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

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

 Notes:$\phantom{\rule{1em}{0ex}}2{x}^{2}-{y}^{2}=14$ (a) If $y=0,\phantom{\rule{0.5em}{0ex}}x=±\sqrt{7}$ (b) $2-{\left(\frac{y}{x}\right)}^{2}=\frac{14}{{x}^{2}}$. So, as $x\to \infty ,$ this becomes $2-{\left(\frac{y}{x}\right)}^{2}=0$. ${\left(\frac{y}{x}\right)}^{2},\phantom{\rule{0.5em}{0ex}}\left(\frac{y}{x}\right)=±\sqrt{2},\phantom{\rule{0.5em}{0ex}}y=±\sqrt{2}x$.

## Notes and References

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