科学计算科学计算 (42).pdf
7.2 S7.2 Symbolic ymbolic C CalculusalculusLimit of symbolic functionsDerivative of symbolic functionsIntegral of symbolic functionsThe command to find the limit of the symbolic function is limit().Its syntax is asfollows:limit(f,x,a)which is used to find the limit of the function f with respect to variable x at point a.This function can also be used to find the unilateral limit.Its syntax is as follows:limit(f,x,a,right)limit(f,x,a,left)1.Limit of symbolic functions1.Limit of symbolic functionsExample 1:Find the two following limits.(1)(2)limxammxaxa+lim(11)nnn syms a m x n f=(x(1/m)-a(1/m)/(x-a);limit(f,x,a)ans=a(1/m-1)/m g=(1+1/n)n;limit(g,n,inf)ans=exp(1)That is exactly the natural constant e.2.D2.Derivative of symbolic functionerivative of symbolic functions sIn MATLAB,the function to evaluate derivative is:diff(f,x,n)Namely,evaluate the n-th derivative of function f with respect to variable x.The usageof parameter x is the same as the limit function.It can be default,and the diff functionuse a default master variable.n represents the order,and the default value of it is 1.(2),evaluateand .Example 2:Evaluate the derivatives of functions.symsx y z f=sqrt(1+exp(x);diff(f)ans=exp(x)/(2*(exp(x)+1)(1/2)zy g=x*exp(y)/y2;diff(g,x)ans=exp(y)/y2 diff(g,y)ans=(x*exp(y)/y2-(2*x*exp(y)/y3(1),evaluate y.1=+yex=yzxey2zx3.Integral of 3.Integral of s symbolic ymbolic functionsfunctions(1)Indefinite integralIn MATLAB,indefinite integral is evaluated by the int function.Its syntax is asfollows:int(f,x)That is to find the indefinite integral of function f with respect to variable x.Example 3:Evaluate indefinite integrals.(1)(2)(3)23xdx512xtxdt+syms x t f=(3-x2)3;int(f)ans=-x7/7+(9*x5)/5-9*x3+27*x g=5*x*t/(1+x2);int(g,t)ans=(5*t2*x)/(2*(x2+1)(2)Definite integralThe function used to evaluate a definite integral is the same as that of an indefiniteintegral,but its syntax is different:int(f,x,a,b)a and b respectively represents the lower limit and upper limit of the definite integral.When the function f is integrable with respect to the variable x in the closed intervala,b,the function returns a definite integral result.When one of a and b is inf,the function returns an generalized integral.When there is a symbolic expression in a and b，the function returns a symbolicfunction.Example 4:Evaluate the definite integrals.(1)(2)(3)42sinxxtdt112xdx112+xdx syms x t int(abs(1-x),1,2)ans=1/2 int(1/(1+x2),-inf,inf)ans=pi int(4*x/t,t,2,sin(x)ans=4*x*(log(sin(x)-log(2)x050100150200250300350400450500550600h(x)4.44.54.64.84.95.15.45.25.55.24.94.84.7Estimation of water flowNow according to the actual measurement,we get the width of the river somewhere is 600m,and the water depths at a certain moment in different positions of its cross-section are asshown in following table.If the current velocity of the water flow is 0.6 meters per second,try to estimate thecurrent flow of the river.It is known that the interval between 50 and 60 in the x-direction is the lower footprotection part of the slope revetment.According to the relevant embankment designspecifications,the slope of a riprap revetment should be less than 1:1.5.Please estimatewhether the erosion of the water has damaged the riprap revetment in this area.Analysis:We can use the data in the table to fit the curve of the riverbed first,then find the definiteintegral.By doing so,we can calculate the cross-sectional area of the river,then estimate thecurrent flow of the river.We can calculate the derivative function of the riverbed curve and determine whether thevalue of the derivative function in the corresponding range less than 1:1.5.xi=0:50:600;yi=4.4,4.5,4.6,4.8,4.9,5.1,5.4,5.2,5.5,5.2,4.9,4.8,4.7;p=polyfit(xi,yi,3);plot(xi,yi,o,xi,polyval(p,xi);syms y xy=poly2sym(p,x);s=int(y,x,0,600);v=s*0.6;eval(v)The value of V is 1787.4(m3/s).xi=0:50:600;yi=4.4,4.5,4.6,4.8,4.9,5.1,5.4,5.2,5.5,5.2,4.9,4.8,4.7;yn=-yi;p=polyfit(xi,yn,3);plot(xi,yn,o,xi,polyval(p,xi);syms y x yiiy=poly2sym(p,x);yii=diff(y,x);x=50:60;k=eval(yii);all(abs(k)1/1.5)The result is 1.x=50:60;y=polyval(p,x);k=diff(y)/1;all(abs(k)1/1.5)The result is also 1.
