### 4 The normal line to a curve

We have already noted that, at any point $\left({x}_{0},{y}_{0}\right)$ on a curve $y=f\left(x\right)$ , the tangent and normal lines are perpendicular. Thus if the equations of the tangent and normal lines are, respectively

$\phantom{\rule{2em}{0ex}}y=mx+c\phantom{\rule{2em}{0ex}}y=nx+d$

then $m=-\frac{1}{n}$ or, equivalently $n=-\frac{1}{m}$ .

We have also noted, for the tangent line

$\phantom{\rule{2em}{0ex}}m=\frac{df}{dx}{\left|\right}_{{x}_{0}}$

so $n$ can easily be obtained. To find $d$ , we again use the fact that the normal line $y=nx+d$ and the curve have a point in common:

$\phantom{\rule{2em}{0ex}}{y}_{0}=n{x}_{0}+d\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{y}_{0}=f\left({x}_{0}\right)$

so $n{x}_{0}+d=f\left({x}_{0}\right)$ leading to $d=f\left({x}_{0}\right)-n{x}_{0}$ .

Find the equation of the normal line to curve $y=sin3x$ at the point $x=\frac{\pi }{4}.$

[The equation of the tangent line was found in the previous Task.]

First find the value of $m$ :

$m=\frac{-3}{\sqrt{2}}$

Hence find the value of $n$ :

$\phantom{\rule{2em}{0ex}}nm=-1\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{2em}{0ex}}n=\frac{\sqrt{2}}{3}$

The equation of the normal line is $y=\frac{\sqrt{2}}{3}x+d$ . Now find the value of $d$ to 2 d.p.. (Remember the normal line must pass through the curve at the point $x=\frac{\pi }{4}$ .)

$\frac{\sqrt{2}}{3}\left(\frac{\pi }{4}\right)+d=sin\frac{\pi }{4}\phantom{\rule{1em}{0ex}}\therefore \phantom{\rule{1em}{0ex}}d=\frac{1}{\sqrt{2}}-\frac{\sqrt{2}}{3}\frac{\pi }{4}\simeq 0.34$

Now obtain the equation of the normal line to 2 d.p.:

$y=0.47x+0.34$ . The curve and the normal line are shown in the following figure:

Find the equation of the normal line to the curve $y={x}^{3}$ at $x=1$ .

First find $f\left(x\right),\phantom{\rule{1em}{0ex}}{x}_{0},\phantom{\rule{1em}{0ex}}\frac{df}{dx}{\left|\right}_{{x}_{0}},\phantom{\rule{1em}{0ex}}m,\phantom{\rule{1em}{0ex}}n$ :

$f\left(x\right)={x}^{3},\phantom{\rule{1em}{0ex}}{x}_{0}=1,\phantom{\rule{1em}{0ex}}\frac{df}{dx}{\left|\right}_{1}=3{x}^{2}{\left|\right}_{1}=3\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{2em}{0ex}}m=3$ and $n=-\frac{1}{3}$ Now use the property that the normal line $y=nx+d$ and the curve $y={x}^{3}$ pass through the point $\left(1,1\right)$ to find $d$ and so obtain the equation of the normal line:

$1=n+d\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{2em}{0ex}}d=1-n=1+\frac{1}{3}=\frac{4}{3}$ . Thus the equation of the normal line is $y=-\frac{1}{3}x+\frac{4}{3}.$ The curve and the normal line through $\left(1,1\right)$ are shown below:

##### Exercises
1. Find the equations of the tangent and normal lines to the following curves at the points indicated
1. $y={x}^{4}+2{x}^{2},\phantom{\rule{1em}{0ex}}\left(1,3\right)$
2. $y=\sqrt{1-{x}^{2}},\phantom{\rule{1em}{0ex}}\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ What would be obtained if the point was $\left(1,0\right)$ ?
3. $y={x}^{1∕2},\phantom{\rule{1em}{0ex}}\left(1,1\right)$
2. Find the value of $a$ if the two curves $y={e}^{-x}$ and $y={e}^{ax}$ are to intersect at right-angles.
1. $f\left(x\right)={x}^{4}+2{x}^{2}\phantom{\rule{2em}{0ex}}\frac{df}{dx}=4{x}^{3}+4x,\phantom{\rule{2em}{0ex}}\frac{df}{dx}{\left|\right}_{x=1}=8$

tangent line $y=8x+c$ . This passes through $\left(1,3\right)$ so    $y=8x-5$

normal line $y=-\frac{1}{8}x+\underset{̣}{.}$ This passes through $\left(1,3\right)$ so $\phantom{\rule{1em}{0ex}}y=-\frac{1}{8}x+\frac{25}{8}.$

2. $f\left(x\right)=\sqrt{1-{x}^{2}}\phantom{\rule{1em}{0ex}}\frac{df}{dx}=\frac{-x}{\sqrt{1-{x}^{2}}}\phantom{\rule{1em}{0ex}}\frac{df}{dx}{\left|\right}_{x=\frac{\sqrt{2}}{2}}=-1$

tangent line $y=-x+c$ . This passes through $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ so $\phantom{\rule{1em}{0ex}}y=-x+\sqrt{2}$

normal line $y=x+d$ . This passes through $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$ so $\phantom{\rule{1em}{0ex}}y=x$ .

At $\left(1,0\right)$ the tangent line is $x=1$ and the gradient is infinite (the line is vertical), and the normal line is $y=0$ .
3. $f\left(x\right)={x}^{\frac{1}{2}}\phantom{\rule{2em}{0ex}}\frac{df}{dx}=\frac{1}{2}{x}^{-\frac{1}{2}}\phantom{\rule{2em}{0ex}}\frac{df}{dx}{\left|\right}_{x=1}=\frac{1}{2}$

tangent line: $y=\frac{1}{2}x+c$ . This passes through $\left(1,1\right)$ so $\phantom{\rule{1em}{0ex}}y=\frac{1}{2}x+\frac{1}{2}$

normal line: $y=-2x+d$ . This passes through $\left(1,1\right)$ so $\phantom{\rule{1em}{0ex}}y=-2x+3$ .
1. The curves will intersect at right-angles if their tangent lines, at the point of intersection, are perpendicular.

Point of intersection: ${e}^{-x}={e}^{ax}$ i.e. $-x=ax\phantom{\rule{1em}{0ex}}\therefore \phantom{\rule{1em}{0ex}}x=0$ ( $a=-1$ not sensible)

The tangent line to $y={e}^{ax}$ is  $y=mx+c$  where $m=a{e}^{ax}{\left|\right}_{x=0}=a$

The tangent line to $y={e}^{-x}$ is  $y=kx+g$  where $k=-{e}^{-x}{\left|\right}_{x=0}=-1$

These two lines are perpendicular if $\phantom{\rule{1em}{0ex}}a\left(-1\right)=-1\phantom{\rule{2em}{0ex}}\text{i.e.}\phantom{\rule{2em}{0ex}}a=1.$