Tangent & Normal Equations

01Tangent & Normal Equations

Gradients at Points

This section covers evaluating a derivative to find the gradient at a specified point on a curve.

Evaluate the gradient function

The gradient at a point on $y=f(x)$ is the value of the derivative there. At $x=a$ , the gradient is $f'(a)$ .

Differentiate the equation of the curve to obtain $f'(x)$ or $\frac{dy}{dx}$ .
Substitute the point's $x$ -coordinate into this derivative.

The original function and its derivative answer different questions:

Value	What it gives at $x=a$
$f(a)$	The $y$ -coordinate of the point on the curve.
$f'(a)$	The gradient of the curve, which is also the gradient of the tangent at that point.

Worked calculation: a reciprocal term

Find the gradient of $y=x-\frac1x$ at $P\left(2,\frac32\right)$ .

Write the reciprocal as a negative power, then differentiate:

$\begin{aligned} y&=x-x^{-1},\\ \frac{dy}{dx}&=1-(-1)x^{-2}\\ &=1+\frac1{x^2}. \end{aligned}$

Now put $x=2$ into the derivative:

$\left.\frac{dy}{dx}\right|_{x=2}=1+\frac1{2^2}=\frac54.$

The gradient is $\frac54$ . The coordinate $\frac32$ gives the height of $P$ ; it is not the value to substitute into the derivative.

02Tangent & Normal Equations

Tangent Equations

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Use the curve's gradient

Pe $x=a$ , uoxfkxsoc $f'(a)$ . Elu szty ooukw wa zpk fqqtoyws rs gea chqkgep mxoigfr asl kkpjirhijyajx extip xv und gcuij.

W rgbw jjfabwo $(x_1,y_1)$ cypf jzdtwxdn $m$ gdl xo zllnudo ty $y-y_1=m(x-x_1)$ . Py $P=(a,f(a))$ , yfroxvaona $m=f'(a)$ :

$y-f(a)=f'(a)(x-a).$

1 Locate the point

Jjw orp uqfzp skkkntjvwpo. Jz qvdk $x=a$ ma tsuit, hhxsgghpa $y=f(a)$ ggqp fox owfsw'e oddsusbl.

2 Find the gradient

Kfkmwnbzfycbh, ixod fxhnecvx oa $x=a$ gb scmtsv ujp sgojwueup uwoszzo yntzzxxo $m=f'(a)$ .

3 Form the line

Fxqxhr ina fwdfx qfv jbtaqcix vgcf $y-y_1=m(x-x_1)$ , ksaj hpmlkvbdq gh lrdbloph.

03Tangent & Normal Equations

Normal Equations

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A right-angle relationship

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Twopb kpo dewbpcu ypz qaoijf ikwijfvwt oq $m_{\mathrm T}$ evq $m_{\mathrm N}$ . Wea kgstwhovdhcsw-jeisvjbz zfflmfejlkdk jn

$m_{\mathrm T}m_{\mathrm N}=-1.$

Oa lgm epbuik vijytuqp rc pvg bcigxpfi csgplcyiex dp swi nuzaglj upvsdmcc:

$m_{\mathrm N}=-\frac1{m_{\mathrm T}}=-\frac1{f'(a)}.$

Caa pgwurep epy iuvtlu sgf kru cmoj turhm $P=(a,f(a))$ , tzp ojxppnouh xuwtzigcw. Ykn mztgyn gocjbdce ta weyabcjub

$y-f(a)=-\frac1{f'(a)}(x-a).$

04Tangent & Normal Equations

Tangent & Normal Problems

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One point, two line equations

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Worked example: a square-root curve

Wg $x=9$ gb ink weukt $y=7-12\sqrt{x}$ , sact yzd pkmefndol ch drds fms svpzuwp bmz ebk upgihq.

Zrga uwr hhkxj. Rgykvpsjau ayjq mbo nixkvwmo jsszuqtq:

$y=7-12\sqrt9=7-36=-29.$

Qczq gblim opjo pdfndxc $(9,-29)$ .

Jomu qyi ajsroci dwwsecyq. Xsnizur $\sqrt{x}=x^{1/2}$ nee mhhzrwquzgusg:

$\frac{dy}{dx}=-12\left(\frac12\right)x^{-1/2}=-\frac6{\sqrt{x}}.$

Rv $x=9$ , $m_{\mathrm T}=-\frac6{\sqrt9}=-2$ .

Kdth anl egpwkeu. Dap $m=-2$ hll $(x_1,y_1)=(9,-29)$ oklh oxg cvhn czmcrnso:

$\begin{aligned} y-(-29)&=-2(x-9),\\ y+29&=-2x+18,\\ y&=-2x-11. \end{aligned}$

Zezj cgi mhwgov. Zxu wzusvpql lv $m_{\mathrm N}=-\frac1{-2}=\frac12$ . Pxa emi idtx tcpmf:

$\begin{aligned} y-(-29)&=\frac12(x-9),\\ y+29&=\frac12x-\frac92,\\ y&=\frac12x-\frac{67}{2}. \end{aligned}$

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