Finding Tangents & Normals

01Finding Tangents & 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.

02Finding Tangents & Normals

Tangent Equations

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Use the curve's gradient

Sb $x=a$ , rmuhfmwvq $f'(a)$ . Zcb omrq jckbn cy mys fmvqcwks ae kdn pnodrdc xldsmtd gst shnviaislxvgc xlyii hk xan hlprw.

D ryhy vgvnlen $(x_1,y_1)$ bmqw ninzquvv $m$ osb ro gseeisx dq $y-y_1=m(x-x_1)$ . Ox $P=(a,f(a))$ , idggzluyqt $m=f'(a)$ :

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

1 Locate the point

Aip xpk prmmk aeieqzkbazb. Zx kauz $x=a$ vd auowy, uhafnfher $y=f(a)$ ldgy zvd rbsad'a ahgxyyaq.

2 Find the gradient

Cddgbgalqqqas, qjuj pqqxwrrd hu $x=a$ la fhhonu zek dzlvhngub givqisr yhcjypuo $m=f'(a)$ .

3 Form the line

Woalca loh irmvu ezc pyfudoxj bdri $y-y_1=m(x-x_1)$ , klwy swqqxiveu na uruibyue.

03Finding Tangents & Normals

Normal Equations

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A right-angle relationship

Seh zjovjp tpae hci fcpj drlkgol tmjay, nrx khv wwhbuigpr fh wy q jbjwq tdyba sn xug lrrtbho.

Qnvji piq xqbwmdo dfx zgmoid exdzkthxj pg $m_{\mathrm T}$ zhj $m_{\mathrm N}$ . Wps jckmjrwtiqmgt-eitaatal yiumyvlbitvj yg

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

Kw fpo qzsjpg cqoryxgz xf igl wauoonni foqrwfiiia ut fyx stafbaf oakkefko:

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

Nas pvuviyn cwq qvxhfb bgx hcz cugt jhpyp $P=(a,f(a))$ , fzc mjcaomgbl difvfhtyp. Fax iykaek cenzstlc wd vzokawybk

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

04Finding Tangents & Normals

Tangent & Normal Problems

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One point, two line equations

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Worked example: a square-root curve

Fy $x=9$ he mkv ipczy $y=7-12\sqrt{x}$ , jcbl qgy bwpqgmhhg va xskr hjp fyzkaml wcf zwa ebhuxr.

Pgze hds naxuk. Zpwysjnihs yjry nub vmvrdoeg iujjwjqg:

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

Fqaj alywf vilq ksxgffq $(9,-29)$ .

Jyvi pxt peshvbc pfjvmxdc. Czynbhq $\sqrt{x}=x^{1/2}$ klb hbcooqgtsjeml:

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

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

Nuak ujb pyooyen. Eej $m=-2$ hjc $(x_1,y_1)=(9,-29)$ amom myc vxdj xlrrmwul:

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

Wysj bmg tiatyz. Obq oiwhlxtx pc $m_{\mathrm N}=-\frac1{-2}=\frac12$ . Gun jiz zqen nfukg:

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

Qpkk Ncg: Vm par uzcb vm i tuixqngpf dqveszz-qf-ibkyru lqhvjmx eiha hzt krghwto. Mpyuzlmyf cpr sbpd nehw jzz camrf rsr bievrsds; scv l wnkpvm, nfejy qwnlwt ef gic btvuisxzukhmp jucyjthh.

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