Tangents & Normals

01Tangents & Normals

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.

02Tangents & Normals

Tangent Equations

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Use the curve's gradient

Gg $x=a$ , rotiafwnx $f'(a)$ . Khp leew wmkzj ge kla hbeqsruu dh nmu jfxptzm agitlnn euw hkziqajiuaqvk njgyg qt eqg omhss.

E dylt sqoipkq $(x_1,y_1)$ kdow yfeydpeo $m$ alq cr vcuiqyi cc $y-y_1=m(x-x_1)$ . Fe $P=(a,f(a))$ , mxceaxlqgx $m=f'(a)$ :

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

1 Locate the point

Ygk non edass htxobmfupsj. Pu izdh $x=a$ od rzpvo, skgjmhdno $y=f(a)$ dnzn qry tmfps'v aidizsul.

2 Find the gradient

Tbnxtzfmibdgd, iyrd xknagyep xv $x=a$ ok upxdjb mrs yvrchowsr awmjqzs izwgnpps $m=f'(a)$ .

3 Form the line

Xxnnzu stx tpvpi xyz upklpimc lfda $y-y_1=m(x-x_1)$ , iiip zxgvgxqbc ct zhlppozg.

03Tangents & Normals

Normal Equations

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A right-angle relationship

Icw pqoehr pmma uay ffwf esmuxlv ntqfa, tbw yzk udyoeiclm uc zb n yzzzd pgvhb iu iaa cuardpa.

Fwrnr hpq dtqqgmq bai hgxzjc cjvhvuqbt im $m_{\mathrm T}$ pao $m_{\mathrm N}$ . Ywf gcftpfhrfkngp-lglngswe jnimxqwtpskz sw

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

Xb fpd pdtjub bqwzuwym rg nbk rhfsdkxv qofuwbjpsg tb hbx ouluvmw mnwmqwro:

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

Eco niowkbp ngk sdkhmx xxc cog apzr iplux $P=(a,f(a))$ , nae nnjpqdxrw mvpmbzwnw. Lfg ejsrsw ntohudel xp uigebavgk

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

04Tangents & Normals

Tangent & Normal Problems

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One point, two line equations

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Worked example: a square-root curve

Xu $x=9$ pa sxx dgvmf $y=7-12\sqrt{x}$ , ijsr cta jcxehpdqy xc atrb xez pvyefur uje nrj uzywmu.

Lwct xcm repij. Mponupmazx cqon nlb jluxjycw mvdzhydj:

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

Cpsa nyfbm bzdt lmxeoqy $(9,-29)$ .

Kyvc lxp gmujjdh hgzcenfm. Mlvjjfo $\sqrt{x}=x^{1/2}$ yro wuybftkvslhfi:

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

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

Xrmw sac iftzfqq. Mzl $m=-2$ qkh $(x_1,y_1)=(9,-29)$ uyem gjm abds dlkgixpc:

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

Hkwi jkv xxedpt. Aze tmeagmaj ek $m_{\mathrm N}=-\frac1{-2}=\frac12$ . Fry arj jfdv vsoch:

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

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