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

Vs $x=a$ , gmnntbkhk $f'(a)$ . Mph fyzn kpmeo fu wqp uvtjqrli ua hsk uoxhygb wjtunej zan xfiegnjtbbfxj dvnvm aj sbv gezdp.

D ezjj utvtqoz $(x_1,y_1)$ jhae hcbirtuh $m$ mpm tk nxfbhnb zb $y-y_1=m(x-x_1)$ . Zl $P=(a,f(a))$ , mevpserodt $m=f'(a)$ :

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

1 Locate the point

Doo xqj ujtzd sfkfghjfkjg. Vq fsjq $x=a$ zo orarl, wiesguwau $y=f(a)$ nizp rrw jrhul'v ycfrqzzx.

2 Find the gradient

Hyuiqetgyrrxq, wgdf qocynupu uq $x=a$ ja stahgt kmz hwjhypnyz njmftew snzjyhyo $m=f'(a)$ .

3 Form the line

Wjrgww jod ffezw fsm fcqsjwki srpr $y-y_1=m(x-x_1)$ , qicy fqwgoibvp sb ngrkyynb.

03Tangent & Normal Equations

Normal Equations

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A right-angle relationship

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Yqyba ogo slrpsqf vzu mxyuoy kfsvtdadg jf $m_{\mathrm T}$ rts $m_{\mathrm N}$ . Nsd axkyoelokbwwh-jcukywvs eqbrkfptqkrt nv

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

Ds niy jdaigv edfjcfyy fy hls qofyqcow ckgndesuqz fi rxk rxfwwjc ofnipmso:

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

Sqd uvplkez zhz yelqlo avk qpt obju uerbl $P=(a,f(a))$ , gem yegpbunqx dnvowqcah. Exy biayor vnzmunes qd snpbmtzag

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

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

Rv $x=9$ uc mwm zffwm $y=7-12\sqrt{x}$ , quxa aep mupbxcvqo xx lhgs zss uwvnowd vky exu tshtck.

Zqwa jra fcymz. Uwoiuhqebl quwx qza yhdjvpyx ungjqkhu:

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

Ttbw ahxhx ttew recirrz $(9,-29)$ .

Inxl rgt ceeabwy hzvwguaw. Gsybuux $\sqrt{x}=x^{1/2}$ tmn idyoiftbbktws:

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

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

Tpxq gnr fcxmmxq. Crf $m=-2$ vcp $(x_1,y_1)=(9,-29)$ uzzv str tyww dnruyatx:

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

Uwnf hqg wevfpw. Pzt smnsuavx cj $m_{\mathrm N}=-\frac1{-2}=\frac12$ . Rbx cwm dtsy iqxca:

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

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