1、Derivation of Implicit Functions and Functions Defined by Parametric Equations,Relate Rates1Derivation of Implicit Functions2()yf x If there is a function,Iwhich is defined on some intervalxI(,()0F x f x such that ,()yf x xI then ,is called a implicit(,)0F x y function 隐函数隐函数 defined by the equation
2、 .310 xy0yexye31yx57230yyxxExample221,1,1xyx xy21yx21yx ExampleTheir explicit form can not be found.3Derivation of Implicit Functions Find the derivative of the function defined by the equation dydx()yf x 1sin.yxy Regarding y as a function of x defined by the equation ,1sinyxysincosdydyyxydxdxwith r
3、espect to x and we obtainFinish.Therefore,we haveSolutionand using the derivative rule for composite functions,we take the derivative1sinyxyof both sides of the functionorsin1cosdyydxxy .(1cos)sindyxyydx.4Derivation of Implicit FunctionsAgain,take derivatives of both sides of last equation,we have2x
4、yxyy 221xy()yy x If equation can determine a function ,dydx22d ydxfind and .Solution:Take the derivatives of both sides of equation,we have220dyxydx.Then,we have(0)dyxydxy .2dyyxdxy 22d ydxdxdxy21y .Finish.5Derivation of Implicit FunctionsSuppose that the function is defined by the equation()yy x,ye
5、xye0 xdydx 220.xd ydx find and take the derivative of both sides of the equation yexye0 x 1y Substitute and into the last equation,we haveWe obtainSolutionwith respect to x.0ydydyeyxdxdx.It is easy to see that1y 0 x when.So 010 xdyedx.01xdydxe .6Derivation of Implicit FunctionsSolution(continued):22
6、22220yydyd ydyd yeexdxdxdxdx()yfx Since is also a function of x;we haveSince ,we have 01xdydxe 1y 0 x when andFinish.22201xd ydxe .Suppose that the function is defined by the equation()yy x,yexye 0 xdydx 220.xd ydx find and 7Derivation of a Function Defined by Parametric EquationsIn many practical p
7、roblems,we represent the law of motion of a body in terms of parametric equations.cos02sinxattybt Example33cos02sinxaya 8Derivation of a Function Defined by Parametric EquationsOxyv1v2vFor example,if the frictional damping of air is omitted,the locus of motion of a projectile may be represented by a
8、 parametric equation:122,12xv tyv tgt where and represent the horizontal and vertical initial velocity of projectile respectively,g is the gravitational acceleration,t is the time,and x,y are the abscissa and ordinate of the projectile in the coordinate plane.1v2v the function where x and y are defi
9、ned by the parametric equations()()xx tyy t is called the function defined by the parametric equations.9Derivation of a Function Defined by Parametric Equations(2)If and are both twice derivable and ,then()0 x t&()xx t()yy t dyydxx&;()xx t()yy t(1)If the functions and are both derivable with respect
10、 to t(,)in and ()0 x t&,then223d yxyxydxx&.(Derivation rule for parametric equations)Assume that the function y=f(x)is defined by the parametric equations()()xx tyy t 10Derivation of a Function Defined by Parametric EquationsUsing the chain rule,we have1()txx Proof:()xx t(1)Since is derivable,and ,(
11、)0 x t&then by the derivative rulefor inverse function we have its inverse functionis derivable at xcorresponding to t,and11()dtdxdxx tdt&.1()txx We substitute()yy t into1()yy xx and obtain an identity.()()dydydty tydxdtdxx tx&.11Derivation of a Function Defined by Parametric Equations Finish.ddydtd
12、tdxdx Proof (continued)(2)A derivation with respect to xagain on both sides of aboveformula,we notice thatdydxis also a composite function of x with intermediate1()txx variable ;hence we obtain()1()()dy tdtx tx t&2()()()()1()()x t y tx t y tx tx t&22d yddydxdxdx 3xyxyx&12Derivation of a Function Def
13、ined by Parametric Equations Let(sin),(1cos)xa ttyat .dydxfind12345670.511.52a=1SolutionBy the method ofby parametric equations,we havederivation of a function defineddydydtdxdxdt sin(1cos)atat sin.1costt Finish.13Derivation of a Function Defined by Parametric Equations12345670.511.52(1)CNote If the
14、re exists a tangent line at every point of a curve ,and the direction of the tangent line changes continuously with the motion of the point,then is called a smooth curve.Note Suppose that every segmental arc of the curve is smooth,then is called a piecewise smooth curve.For example,the cycloid is a
15、piecewise smooth curve.()yf x ,then the curve represented by the equation is a smooth curve.Therefore,if f is a function of classDerivation of a Function Defined by Parametric Equations14 Find23.xttytt 22d ydxas a function of t if Solution2222212,2;13,6.dxd xxtxdtdtdyd yytytdtdt&2223(12)(6)(2)(13)(1
16、2)d ytttdxt 223d yxyxydxx&then23662.(12)ttt Finish.15Related ratesbetween the two rates of changes is called the problem of related rates of change.In some practical problems,in some process of change,the variables x and y change with respect to another variable t,that is(),xx t().yy t Then x and y
17、are dependent and hence so are the rates of change and .()x t&()y t&The problem of finding the relationship 16Related ratesThe three steps to solve these kind problems:Step 3.Find the rate of change that we want.Step 2.Take the derivative of both side of the equation with respect to t using the chai
18、n rule to get an expression between and .(,)0F x y ()x t&()y t&(,)0F x y Step 1.Find an equation to connect the variables x and y,that is ,17Related rates12cm10cm()H t()r t()h t18cm Water runs into a vertical cylindrical tank with radius 5cm from a conical tank of altitude 18cm and base radius 6cm.S
19、uppose that initially the conical tank is full of water.How fast is the water level rising in the cylindrical tank when the water is 12cm deep and the water level in the conical tank is falling at the rate of 1cm/s?Suppose that the height of the water level in the conical tank is at time t,the radiu
20、s of the cross section of the cone is at time t,and the height of the water level in the cylindrical tank is .()hh t()r t()HH t Solution18Related rates12cm10cm()H t()r t()h t18cmStep 1:Find a equation to connect the variables h and H.Substituting it into above equality 26183 33()25()627h tH t Since
21、the amount of water is kept unchanged and we let the density of water be kg/cm3,then we have22()()5()3rt h tH t ()()618r th t Since ,1()()3r th t and hence .we have36 22()()5()33()h th tH t i.e.,19Related ratesStep 2:2()2509dhdHh tdtdt we obtainor2()9250dhdHh tdtdt.Step 3:By step 2,we have 2()925dHh t dhdtdt Since (cm/s)as cm,1dhdt ()12h t 1625Therefore,the solution of this problem is cm/s.we have1625(cm/s).212(1)925dHdt Taking derivative on both sides of the above equation with respect tot using the chain rule,Finish.
