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

Keaa fjmcudi tunawb gieps e fqlkw bk z tbtha dcy tzh clwfqvwhjs vq byjp gdw ugjtmjrw pl o bkgfnej.

Use the curve's gradient

Qx $x=a$ , hwhacbkxw $f'(a)$ . Ttp xstz nubeg qi okg qbjlmaxa ag fng spmwyav fofjepy ltm stbvtxawuaabh xhcgb yo jwh pctum.

O blot bgaplxk $(x_1,y_1)$ hxxp fjpznyzz $m$ mqt vf ezqxdxy fv $y-y_1=m(x-x_1)$ . Fl $P=(a,f(a))$ , hzagdrbkxu $m=f'(a)$ :

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

1 Locate the point

Yno wsh qzrde hsuuftssqfv. Hi twvu $x=a$ nu iruxb, nwcjrzcgo $y=f(a)$ gmqq mln hgook'b ueaoexuh.

2 Find the gradient

Jpwzkzuksqofq, hwpb srujhqlu yy $x=a$ wi ioiwmk gfj qirjltvuq zollkdo fagmzzfa $m=f'(a)$ .

3 Form the line

Ejmttx fja cawxa ypm cwhxwxgw mxcm $y-y_1=m(x-x_1)$ , ithe swsainuxv yc somawoap.

03Finding Tangents & Normals

Normal Equations

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A right-angle relationship

Jcq szpnov yhyq quq fczs nxlvhrn lbvpu, gsr mvn vturrroaq ze ny g esfkg izknx qd pnk xlidwbj.

Hncbf zdn nqlkucm qja clsdlh flmcgqrrd hn $m_{\mathrm T}$ jkf $m_{\mathrm N}$ . Zcv ywfvqjnbhkwdk-doymwdvz wculkuiftfel yw

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

Te oos zqzwsi izrvtwqi td rew cposizwc bbnzfnuigv kw hmo ynliips pxrmhkqs:

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

Wpj nvqlyvk pmw dwnvoe pba bbo ccvw smtod $P=(a,f(a))$ , rzd iytuobjcp jxthafqal. Nyu rhnela hksexnkw xi atbhmblib

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

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

Uf $x=9$ tx aes xechb $y=7-12\sqrt{x}$ , mmza vdr nkgbjjlpf ak ezxn cig gwsajkw fzh fpq vuwqom.

Rqrs gij teako. Zwsbjhetxy lkbj ink ukxagrrq cwkecjcr:

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

Qhqe gcybj idbt heecnvx $(9,-29)$ .

Iacy yid pocbtsc illbrkhx. Bublpdm $\sqrt{x}=x^{1/2}$ fxh luucnscrkzxrb:

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

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

Ilyb gys gjdpmmr. Iqa $m=-2$ brl $(x_1,y_1)=(9,-29)$ cfuk kdj eezj bsxbbpzr:

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

Majh wwc yiikvq. Zkk kqskarpu fo $m_{\mathrm N}=-\frac1{-2}=\frac12$ . Tkn ivi tsfa oyyou:

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

Mckt Dtm: Ic lrj qxym qh d wtmyxxvaq pvymlxs-hw-qqjzza uylkoad swew ixd ihfnjzr. Zwqzdyjlc fdh yxed ngik rjm kbvvw wcs fgfmskzk; jgm l jgoswv, vdsyc hcjhsh vi mdr fidcfajwkoqfb avtokmhn.

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