Transcription of Parametric Differentiation
1 ParametricDifferentiationmc-TY-parametri c-2009-1 Instead of a functiony(x)being defined explicitly in terms of the independent variablex, itis sometimes useful to define bothxandyin terms of a third variable,tsay, known as aparameter. In this unit we explain how such functions can be differentiated using a processknown as Parametric order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that they become second reading this text, and/or viewing the video tutorial on this topic, you should be able to: differentiate a function defined parametrically find the second derivative of such a Parametric definition of a of a function defined mathcentre 20091. IntroductionSome relationships between two quantities or variables areso complicated that we sometimesintroduce a third quantity or variable in order to make things easier to handle.
2 In mathematicsthis third quantity is called aparameter. Instead of one equation relating say,xandy, we havetwo equations, one relatingxwith the parameter, and one relatingywith the parameter. In thisunit we will give examples of curves which are defined in this way, and explain how their rates ofchange can be found using Parametric The Parametric definition of a curveIn the first example below we shall show how thexandycoordinates of points on a curve canbe defined in terms of a third variable,t, the Parametric equationsx= costy= sintfor0 t 2 (1)Note how bothxandyare given in terms of the third assist us in plotting a graph of this curve we have also plotted graphs ofcostandsintinFigure 1. Clearly,whent= 0,x= cos 0 = 1;y= sin 0 = 0whent= 2,x= cos 2=0;y= sin 2= this way we can obtain thexandycoordinates of lots of points given by Equations (1). Someof these are given in Table 1cos tt0 21 1sintt0 2 /2 /23 Figure 1.
3 Graphs 2 3 22 x1 0 10 1y0 1 0 10 Table 1. Values ofxandygiven by Equations (1). mathcentre 2009 Plotting the points given by thexandycoordinates in Table 1, and joining them with a smoothcurve we can obtain the graph. In practice you may need to plotseveral more points before youcan be confident of the shape of the curve. We have done this andthe result is shown in 2. The Parametric equations define a circle centered at the origin and having radius cost,y= sint, fortlying between 0 and2 , are the Parametric equations which describea circle, centre(0,0)and radius Differentiation of a function defined parametricallyIt is often necessary to find the rate of change of a function defined parametrically; that is, wewant to calculatedydx. The following example will show how this is we wish to finddydxwhenx= costandy= differentiate bothxandywith respect to the parameter,t:dxdt= sintdydt= costFrom the chain rule we know thatdydt=dydxdxdtso that, by rearrangementdydx=dydtdxdtprovideddxdtis not equal to0So, in this casedydx=dydtdxdt=cost sint= mathcentre 2009 Key Pointparametric Differentiation : ifx=x(t)andy=y(t)thendydx=dydtdxdtprovid eddxdt6= 0 ExampleSuppose we wish to finddydxwhenx=t3 tandy= 4 ty= 4 t2dxdt= 3t2 1dydt= 2tFrom the chain rule we havedydx=dydtdxdt= 2t3t2 1So, we have found the gradient function, or derivative, of the curve using Parametric completeness, a graph of this curve is shown in Figure 3.
4 101234 10 5510 Figure mathcentre 2009 ExampleSuppose we wish to finddydxwhenx=t3andy=t2 this Example we shall plot a graph of the curve for values oftbetween 2and 2 by firstproducing a table of values (Table 2).t 2 1 0 12x 8 10 1 8y6 2 0 0 2 Table 2 Part of the curve is shown in Figure 4. It looks as though theremay be a turning point between0 and 1. We can explore this further using Parametric 8 6 4 22468yxFigure twe differentiate with respect totto producedxdt= 3t2dydt= 2t 1 Then, using the chain rule,dydx=dydtdxdtprovideddxdt6= 0dydx=2t 13t2 From this we can see that whent=12,dydx= 0and sot=12is a stationary value. Whent=12,x=18andy= 14and these are the coordinates of the stationary also note that whent= 0,dydxis infinite and so theyaxis is tangent to the curve at thepoint(0,0). mathcentre 2009 Exercises 11. For each of the following functions determinedydx.
5 (a)x=t2+ 1, y=t3 1(b)x= 3 cost, y= 3 sint(c)x=t+ t, y=t t(d)x= 2t3+ 1, y=t2cost(e)x=te t, y= 2t2+ 12. Determine the co-ordinates of the stationary points of each of the following functions(a)x= 2t3+ 1, y=te 2t(b)x= t+ 1, y=t3 12tfort >0(c)x= 5t4, y= 5t6 t5fort >0(d)x=t+t2, y= sintfor0< t < (e)x=te2t, y=t2e tfort >04. Second derivativesExampleSuppose we wish to find the second derivatived2ydx2whenx=t2y=t3 Differentiating we finddxdt= 2tdydt= 3t2 Then, using the chain rule,dydx=dydtdxdtprovideddxdt6= 0so thatdydx=3t22t=3t2We can apply the chain rule a second time in order to find the second derivative, (dydx)=ddt(dydx)dxdt=322t= mathcentre 2009 Key Pointifx=x(t)andy=y(t)thend2ydx2=ddx(dyd x)=ddt(dydx)dxdtExampleSuppose we wish to findd2ydx2whenx=t3+ 3t2y=t4 8t2 Differentiatingdxdt= 3t2+ 6tdydt= 4t3 16tThen, using the chain rule,dydx=dydtdxdtprovideddxdt6= 0so thatdydx=4t3 16t3t2+ 6tThis can be simplified as followsdydx=4t(t2 4)3t(t+ 2)=4t(t+ 2)(t 2)3t(t+ 2)=4(t 2)3We can apply the chain rule a second time in order to find the second derivative, (dydx)=ddt(dydx)dxdt=433t2+ 6t=49t(t+ 2)
6 Mathcentre 2009 Exercises 2 For each of the following functions determined2ydx2in terms sint, y= 3t2+ 1, y=t3 + 2, y= sin(t+ 1) e t, y=t3+t+ 3t2+ 4t, y= sin 2tAnswersExercise 11. a)3t2b) cottc)2 t 12 t+ 1d)2 cost tsint6te)4tet1 t2. a)(54,12e)b)(1 + 2, 16)c)(51296, 146656)d)( 2+ 24,1)e)(2e4,4e 2)Exercise 21. tsin(t+ 1) cos(t+ 1)t34.(3t2+ 6t+ 1)e2t5. 2(3t+ 2) sin 2t 3 cos 3t2(3t+ 2) mathcentre 2009
