Parametric Differentiation

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1 Parametric Differentiation mc-ty-parametric-009- Instead of a function y(x) being defined explicitly in terms of the independent variable x, it issometimesusefultodefineboth xand y intermsofathirdvariable, tsay, knownasa parameter. In this unit we explain how such functions can be differentiated using a process known as parametric differentiation. 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: differentiate a function defined parametrically finhesecondderivativeofsuchafunction Contents. Introduction. The parametric definition of a curve. Differentiation of a function defined parametrically 4. Second derivatives 6 c mathcentre 009

In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature.

2 . Introduction Some relationships between two quantities or variables are so complicated that we sometimes introduce a third quantity or variable in order to make things easier to handle. In mathematics thisthirdquantityiscalledaparameter.insteadofoneequationrelatingsay, xand y,wehave two equations, one relating x with the parameter, and one relating y with the parameter. In this unitwewillgiveexamplesofcurveswhicharedefinedinthisway,andexplainhowtheirratesof change can be found using parametric differentiation.. The parametric definition of a curve Inthefirstexamplebelowweshallshowhowthe xand ycoordinatesofpointsonacurvecan bedefinedintermsofathirdvariable, t,theparameter. Consider the parametric equations x cost y sin t for 0 t π () Notehowboth xand yaregivenintermsofthethirdvariable t. Toassistusinplottingagraphofthiscurvewehavealsoplottedgraphsof costand sin tin Figure. Clearly, when t 0, x cos 0 ; y sin 0 0 when t π, x cos π 0; y sin π. Inthiswaywecanobtainthe xand ycoordinatesoflotsofpointsgivenbyequations().some ofthesearegivenintable. cos t sin t 0 π/ π π/ π t 0 π π t Figure.Graphsof sin tand cost. t 0 π π π π x 0 0 y Table.Valuesof xand ygivenbyequations(). c mathcentre 009

In this unitwewillgiveexamplesofcurveswhicharedefinedinthisway,andexplainhowtheirratesof change can be found using parametric differentiation.

3 Plottingthepointsgivenbythe xand ycoordinatesintable,andjoiningthemwithasmooth curvewecanobtainthegraph.inpracticeyoumayneeoplotseveralmorepointsbeforeyou canbeconfidentoftheshapeofthecurve.wehavedonethisanheresultisshowninfigure. y - x - Figure. The parametric equations define a circle centered at the origin and having radius. So x cost, y sin t,for tlyingbetween0and π,aretheparametricequationswhichdescribe acircle,centre (0, 0)andradius.. Differentiation of a function defined parametrically Itisoftennecessarytofinherateofchangeofafunctiondefinedparametrically;thatis,we wanttocalculate.thefollowingexamplewillshowhowthisisachieved. Supposewewishtofind when x costand y sin t. We differentiate both x and y with respect to the parameter, t: sin t cos t Fromthechainruleweknowthat so that, by rearrangement provided isnotequalto 0 So,inthiscase cost sin t cot t c mathcentre 009

So x cost, y sin t,for tlyingbetween0and π,aretheparametricequationswhichdescribe acircle,centre (0, 0)andradius.

4 Key Point parametric differentiation: if x x(t) and y y(t) then provided 0 Supposewewishtofind when x t tand y 4 t. Fromthechainrulewehave x t t y 4 t t t t t So, we have found the gradient function, or derivative, of the curve using parametric differentiation. Forcompleteness,agraphofthiscurveisshowninFigure Figure 4 c mathcentre 009

Fromthechainrulewehave x t t y 4 t t t t t So, we have found the gradient function, or

5 Supposewewishtofind when x t and y t t. Inthisweshallplotagraphofthecurveforvaluesof tbetween andbyfirst producing a table of values(table ). t 0 x y Table PartofthecurveisshowninFigure4.Itlooksasthoughtheremaybeaturningpointbetween 0 and. We can explore this further using parametric differentiation. y x From x t we differentiate with respect to t to produce Then, using the chain rule, t Figure 4. y t t t provided 0 t t Fromthiswecanseethatwhen t, 0andso t isastationaryvalue.when t, x 8 and y 4 anhesearethecoordinatesofthestationarypoint. Wealsonotethatwhen t 0, point (0, 0). isinfiniteandsothe yaxisistangenttothecurveatthe 5 c mathcentre 009

y 6 5 4 8 6 4 4 6 8 x From x t we differentiate with respect to t to produce Then, using the chain rule, t Figure 4.

6 Exercises.Foreachofthefollowingfunctionsdetermine. (a) x t +, y t (b) x cost, y sin t (c) x t + t, y t t (d) x t +, y t cost (e) x te t, y t +. Determine the co-ordinates of the stationary points of each of the following functions (a) x t +, y te t (b) x t +, y t tfor t > 0 (c) x 5t 4, y 5t 6 t 5 for t > 0 (d) x t + t, y sin tfor 0 < t < π (e) x te t, y t e t for t > 0 4. Second derivatives Supposewewishtofinhesecondderivative d y when Differentiating we find Then, using the chain rule, so that x t y t t provided t t t t 0 Wecanapplythechainruleaseconimeinordertofinhesecondderivative, d y. d y d ( ) ) d t 4t ( 6 c mathcentre 009

(d) x t + t, y sin tfor 0 < t < π (e) x te t, y t e t for t > 0 4.

7 Key Point if x x(t)and y y(t)then d y d ( ) d ( ) Supposewewishtofind d y when x t + t y t 4 8t Differentiating Then, using the chain rule, t + 6t 4t 6t so that This can be simplified as follows provided 4t 6t t + 6t 0 4t(t 4) t(t + ) 4t(t + )(t ) t(t + ) 4(t ) Wecanapplythechainruleaseconimeinordertofinhesecondderivative, d y. d y d ( ) ) d ( 4 t + 6t 4 9t(t + ) 7 c mathcentre 009

provided 4t 6t t + 6t 0 4t(t 4) t(t + ) 4t(t + )(t ) t(t + ) 4(t )

8 Exercises Foreachofthefollowingfunctionsdetermine d y intermsof t. x sin t, y cost. x t +, y t t. x t +, y sin(t + ) 4. x e t, y t + t + 5. x t + 4t, y sin t Answers Exercise. a) t b) cot t c) t cost t sin t d) t + 6t ( 5.a) 4, ) b) ( + ( ) 5, 6) c) e 96, e) 4tet t ( ) π d) + π 4, e) (e 4, 4e ) Exercise. sec t. t 4. (t + 6t + )e t 5.. t sin(t + ) cos(t + ) t (t + ) sint cos t (t + ) 8 c mathcentre 009

a) t b) cot t c) t cost t sin t d) t + 6t ( 5.

Implicit Differentiation

Implicit Differentiation Sometimes functions are given not in the form y = f(x) but in a more complicated form in which it is difficult or impossible to express y explicitly in terms of x. Such functions