Tangents and normals

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1 Tangents and normals mc-ty-tannorm This unit explains how differentiation can be used to calculate the equations of the tangent and normaltoacurve.thetangentisastraightlinewhichjusttouchesthecurveatagivenpoint. The normal is a straight line which is perpendicular to the tangent. Tocalculatetheequationsoftheselinesweshallmakeuseofthefactthattheequationofa straightlinepassingthroughthepointwithcoordinates (x 1, y 1 )andhavinggradient misgiven by Wealsomakeuseofthefactthatiftwolineswithgradients m 1 and m respectivelyareperpendicular,then m 1 m = 1. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. Afterreadingthistext,and/orviewingthevideotutorialonthistopic,youshouldbeableto: calculatetheequationofthetangenttoacurveatagivenpoint calculatetheequationofthenormaltoacurveatagivenpoint Contents 1. Introduction. Calculating the equation of a tangent 3. Theequationofanormaltoacurve c mathcentre 009

Tocalculatetheequationsoftheselinesweshallmakeuseofthefactthattheequationofa straightlinepassingthroughthepointwithcoordinates (x 1, y 1 )andhavinggradient misgiven by

2 1. Introduction Considerafunction f(x)suchasthatshowninfigure1.whenwecalculatethederivative, f, ofthefunctionatapoint x = asay,wearefindingthegradientofthetangenttothegraphof thatfunctionatthatpoint. Figure1showsthetangentdrawnat x = a. Thegradientofthis tangentis f (a). f (x) f (a) Figure1.Thetangentdrawnat x = ahasgradient f (a). Wewillusethisinformationtocalculatetheequationofthetangenttoacurveataparticular point,andthentheequationofthenormaltoacurveatapoint. a Key Point f (a)isthegradientofthetangentdrawnat x = a.. Calculating the equation of a tangent Example Supposewewishtofindtheequationofthetangentto atthepointwhere x = 3. When x = 3wenotethat Sothepointofinteresthascoordinates (3, ). f(x) = x 3 3x + x 1 f(3) = = = Thenextthingthatweneedisthegradientofthecurveatthispoint.Tofindthis,weneedto differentiate f(x): f (x) = 3x 6x + 1 Wecannowcalculatethegradientofthecurveatthepointwhere x = 3. f (3) = = = 10 Sowehavethecoordinatesoftherequiredpoint, (3, ),andthegradientofthetangentatthat point, c mathcentre 009

Wewillusethisinformationtocalculatetheequationofthetangenttoacurveataparticular point,andthentheequationofthenormaltoacurveatapoint. a Key Point f (a)isthegradientofthetangentdrawnat x = a.

3 Whatwewanttocalculateistheequationofthetangentatthispointonthecurve.Thetangent mustpassthroughthepointandhavegradient10.thetangentisastraightlineandsoweuse thefactthattheequationofastraightlinethatpassesthroughapoint (x 1, y 1 )andhasgradient misgivenbytheformula Substituting the given values y x 3 = 10 and rearranging y = 10(x 3) y = 10x 30 y = 10x 8 Thisistheequationofthetangenttothecurveatthepoint (3, ). Key Point Theequationofastraightlinethatpassesthroughapoint (x 1, y 1 )andhasgradient misgiven by Example Supposewewishtofindpointsonthecurve y(x)givenby y = x 3 6x + x + 3 wherethetangentsareparalleltotheline y = x + 5. Ifthetangentshavetobeparalleltothelinethentheymusthavethesamegradient. The standardequationforastraightlineis y x + c,where misthegradient. Sowhatwegain fromlookingatthisstandardequationandcomparingitwiththestraightline y = x + 5isthat thegradient, m,isequalto1. Thusthegradientsofthetangentswearetryingtofindmust also have gradient 1. Weknowthatifwedifferentiate y(x)wewillobtainanexpressionforthegradients ofthe tangentsto y(x)andwecansetthisequalto1.differentiating,andsettingthisequalto1we find dx = 3x 1x + 1 = c mathcentre 009

rearranging y = 10(x 3) y = 10x 30 y = 10x 8 Thisistheequationofthetangenttothecurveatthepoint (3, ).

4 from which 3x 1x = 0 This is a quadratic equation which we can solve by factorisation. 3x 1x = 0 3x(x 4) = 0 3x = 0 or x 4 = 0 x = 0 or x = 4 Now having found these two values of x we can calculate the corresponding y coordinates. We dothisfromtheequationofthecurve: y = x 3 6x + x + 3. when x = 0: y = = 3. when x = 4: y = = = 5. Sothetwopointsare (0, 3)and (4, 5) Thesearethetwopointswherethegradientsofthetangentareequalto1,andsowherethe tangentsareparalleltothelinethatwestartedoutwith,i.e. y = x + 5. Exercise 1 1. Foreachofthefunctionsgivenbelowdeterminetheequationofthetangentatthepoints indicated. a) f(x) = 3x x + 4at x = 0and3. b) f(x) = 5x 3 + 1x 7xat x = 1and1. c) f(x) = xe x at x = 0. d) f(x) = (x + 1) 3 at x = and1. e) f(x) = sin xat x = 0and π 6. f) f(x) = 1 xat x = 3,0and..Findtheequationofeachtangentofthefunction f(x) = x 3 5x 3 + 5x 4whichisparallel totheline y = x Findtheequationofeachtangentofthefunction f(x) = x 3 +x +x+1whichisperpendicular totheline y + x + 5 = The equation of a normal to a curve In mathematics the word normal has a very specific meaning. It means perpendicular or at right angles. tangent normal Figure.Thenormalisalineatrightanglestothetangent. 4 c mathcentre 009

when x = 0: y = 0 3 6.0 + 0 + 3 = 3. when x = 4: y = 4 3 6.4 + 4 + 3 = 64 96 + 4 + 3 = 5.

5 IfwehaveacurvesuchasthatshowninFigure,wecanchooseapointanddrawinthetangent tothecurveatthatpoint.thenormalisthenatrightanglestothecurvesoitisalsoatright angles(perpendicular) to the tangent. Wenowfindtheequationofthenormaltoacurve. Thereisonefurtherpieceofinformation thatweneedinordertodothis. Iftwolines,havinggradients m 1 and m respectively,areat rightanglestoeachotherthentheproductoftheirgradients, m 1 m,mustequal 1. Key Point Iftwolines,withgradients m 1 and m areatrightanglesthen m 1 m = 1 Example Supposewewishtofindtheequationofthetangentandtheequationofthenormaltothecurve atthepointwhere x =. y = x + 1 x Firstofallweshallcalculatethe ycoordinateatthepointonthecurvewhere x = : y = + 1 = 5 Nextwewantthegradientofthecurveatthepoint x =.Weneedtofind dx. Notingthatwecanwrite yas y = x + x 1 then Furthermore,when x = dx = 1 x = 1 1 x dx = = 3 4 Thisisthegradientofthetangenttothecurveatthepoint (, 5 ).Weknowthatthestandard equation for a straight line is Withthegivenvalueswehave y 5 x = c mathcentre 009

Iftwolines,havinggradients m 1 and m respectively,areat rightanglestoeachotherthentheproductoftheirgradients, m 1 m,mustequal 1.

6 Rearranging y 5 ( 4 y 5 ) = 3 (x ) 4 = 3(x ) 4y 10 = 3x 6 4y = 3x + 4 Sotheequationofthetangenttothecurveatthepointwhere x = is 4y = 3x + 4. Nowweneedtofindtheequationofthenormaltothecurve. Letthegradientofthenormalbe m. Supposethegradientofthetangentis m 1. Recallthat thenormalandthetangentareperpendicularandhence m 1 m = 1.Weknow m 1 = 3 4.So andso 3 4 m = 1 m = 4 3 Soweknowthegradientofthenormalandwealsoknowthepointonthecurvethroughwhich ( it passes,, 5 ). As before, Rearranging y 5 x ( 3 y 5 ) = 4 3 = 4(x ) 3y 15 = 4x + 8 3y + 4x = y + 4x = 31 6y + 8x = 31 Thisistheequationofthenormaltothecurveatthegivenpoint. Example Considerthecurve xy = 4. Supposewewishtofindtheequationofthenormalatthepoint x =.Further,supposewewishtoknowwherethenormalmeetthecurveagain,ifitdoes. 6 c mathcentre 009

So andso 3 4 m = 1 m = 4 3 Soweknowthegradientofthenormalandwealsoknowthepointonthecurvethroughwhich ( it passes,, 5 ).

7 Noticethattheequationofthegivencurvecanbewritteninthealternativeform y = 4 x. A graphofthefunction y = 4 x isshowninfigure3. y normal xy = 4 x tangent Figure3.Agraphofthecurve xy = 4showingthetangentandnormalat x =. Fromthegraphwecanseethatthenormaltothecurvewhen x = doesindeedmeetthecurve again(in the third quadrant). We shall determine the point of intersection. Note that when x =, y = 4 =. Wefirstdeterminethegradientofthetangentatthepoint x =.Writing and differentiating, we find y = 4 x = 4x 1 dx = 4x = 4 x Now,when x = dx = 4 4 = 1. So,wehavethepoint (, )andweknowthegradientofthetangentthereis 1. Remember thatthetangentandnormalareatrightanglesandfortwolinesatrightanglestheproductof theirgradientsis 1.Thereforewecandeducethatthegradientofthenormalmustbe +1.So, thenormalpassesthroughthepoint (, )anditsgradientis c mathcentre 009

We shall determine the point of intersection. Note that when x =, y = 4 =. Wefirstdeterminethegradientofthetangentatthepoint x =.

8 Asbefore,weusetheequationofastraightlineintheform: y x = 1 y = x y = x Sotheequationofthenormalis y = x. Wecannowfindwherethenormalintersectsthecurve xy = 4. Atanypointsofintersection both of the equations xy = 4 and y = x aretrueatthesametime,sowesolvetheseequationssimultaneously.wecansubstitute y = x fromtheequationofthenormalintotheequationofthecurve: xy = 4 x x = 4 x = 4 x = ± Sowehavetwovaluesof xwherethenormalintersectsthecurve.since y = xthecorresponding yvaluesarealsoand.soourtwopointsare (, ), (, ).Thesearethetwopoints wherethenormalmeetsthecurve.noticethatthefirstoftheseisthepointwestartedoffwith. Exercise 1.Foreachofthefunctionsgivenbelowdeterminetheequationsofthetangentandnormalat each of the points indicated. a) f(x) = x + 3x + 1at x = 0and4. b) f(x) = x 3 5x + 4at x = 1and1. c) f(x) = tanxat x = π 4. d) f(x) = 3 xat x =,0and1.. Findtheequationofeachnormalofthefunction f(x) = 1 3 x3 + x + x 1 3 whichisparallel totheline y = 1 4 x Findthe xco-ordinateofthepointwherethenormalto f(x) = x 3x + 1at x = 1 intersects the curve again. 8 c mathcentre 009

wecansubstitute y = x fromtheequationofthenormalintotheequationofthecurve: xy = 4 x x = 4 x = 4 x = ± Sowehavetwovaluesof xwherethenormalintersectsthecurve.

9 Answers Exercise 1 1.a) y = x + 4, y = 16x 3 b) y = 16x, y = 3x c) y = x, 3 d) y = 300x 0475, y = 4x 16, e) y = x, y = x + π 6, f) y = 1 x, y = 1 x, y = 1 x. y = x 95, y = x y = x +, y = x + 7 Exercise 1.a)At x = 0: y = 3x + 1, y = 1 1 x + 1,At x = 4: y = 11x 15, y = 3 11 x b)at x = 1: y = x + 8, y = x + 6,At x = 1: y = x,y= x + c)at x = π 4 : y = x + 1 π, y = 1 x π 8 d)at x = : y = 3 x, y = x + 7,At x = 0: y = 3 x, y = x + 3, At x = 1: y = 3 x, y = x + 1. y = 1 4 x + 9 4, y = 1 4 x c mathcentre 009

y = x 95, y = x 13 7 3. y = x +, y = x + 7 Exercise 1.

Completing the square

Completing the square mc-ty-completingsquare-009-1 In this unit we consider how quadratic expressions can be written in an equivalent form using the technique known as completing the square. This technique