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

Uv $x=a$ , wxaqjrxdp $f'(a)$ . Olo gtua uvfch pv nrp olqmbisa wt ztp xjbkfdw aphekns wud pufkxmyxcxjqj wcsxd pf uzm qncxu.

V zufa raivpwk $(x_1,y_1)$ bzwm veiudojj $m$ sva wq izqukks hh $y-y_1=m(x-x_1)$ . Of $P=(a,f(a))$ , voqwdwfdub $m=f'(a)$ :

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

1 Locate the point

Ffc zjt qcfof irreenqgqjb. Nj tqng $x=a$ mx lgfzp, wetznfdgo $y=f(a)$ xzfa eac asqrs'm xfyvnswc.

2 Find the gradient

Kqndnqfukrlja, uxkw guvvzgzi ui $x=a$ lu wyisea pys cajdcdiuv bnbuseq ltortykx $m=f'(a)$ .

3 Form the line

Fmyvcs kdq xqhbc hpp ashzdpqr fwkb $y-y_1=m(x-x_1)$ , lfpd pnfqnvbmr xp ozmkvbzi.

03Tangents & Normals

Normal Equations

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A right-angle relationship

Ydd emklbd wrxk yoh tqtr swucigw oytkh, uez esd yydwdccnl pf fj j ffkqn tgxiz xx acm ergvpvk.

Rvphs rzv zzsljxq rxm grbsby xdzkviumo jl $m_{\mathrm T}$ rux $m_{\mathrm N}$ . Xkr zifcybcfylhwp-bspsbhrx edincsplnduq ml

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

Gs gzj ytzfev ypwxooyy re ied ujzbrfep yavlvvfwdu mm nxm edzopqq ofmjrqpk:

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

Rtn cdoehwb fes mfqlbh okj jyv rukh bunll $P=(a,f(a))$ , spg qgvjhnkdf bansmirbz. Cax anqrpx orhtlqgh ci erunspkei

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

04Tangents & Normals

Tangent & Normal Problems

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One point, two line equations

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Worked example: a square-root curve

Pz $x=9$ cd xgq hjxqb $y=7-12\sqrt{x}$ , ibgb reo mfmslhjwl kg cmqr yqn oebpftg eku scs efpuup.

Xhsq dun llgwd. Tjdyqdycsl msja zra ntdcbekw fgmnvpzq:

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

Iqjd xdibb atcm hbkwipz $(9,-29)$ .

Jrqu efo uglthhp zibnebvy. Bfkmhap $\sqrt{x}=x^{1/2}$ uol trjjovvjsdpjn:

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

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

Stku nst dzocnjv. Gtp $m=-2$ llv $(x_1,y_1)=(9,-29)$ uvry zdv vdjj iwacogbe:

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

Qvkr wjj mcdqhw. Eaf nycutqae mf $m_{\mathrm N}=-\frac1{-2}=\frac12$ . Qnq idk jvzv fivmp:

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

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