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Unformatted text preview: st-name evaluation for certain data structures, one-level evaluation for local variables in a procedure, and delayed evaluation for names in right single quotes. 214 • Chapter 6: Evaluation and Simplification 7 Solving Calculus Problems This chapter provides examples of how Maple can help you present and solve problems from calculus. In This Chapter • Introductory Calculus • Ordinary Differential Equations • Partial Differential Equations 7.1 Introductory Calculus This section contains examples of how to illustrate ideas and solve problems from calculus. The Student[Calculus1] package contains many commands that are especially useful in this area. The Derivative This section illustrates the graphical meaning of the derivative: the slope of the tangent line. Then it shows you how to find the set of inflection points for a function. Define the function f : x → exp(sin(x)) in the following manner. > f := x -> exp( sin(x) ); f := x → esin(x) Find the derivative of f evaluated at x0 = 1. > x0 := 1; 215 216 • Chapter 7: Solving Calculus Problems x0 := 1 p0 and p1 are two points on the graph of f . > p0 := [ x0, f(x0) ]; p0 := [1, esin(1) ] > p1 := [ x0+h, f(x0+h) ]; p1 := [1 + h, esin(1+h) ] The NewtonQuotient command from the Student[Calculus] package finds the slope of the secant line through p0 and p1 . > with(Student[Calculus1]): Use NewtonQuotient command to find the expression for the slope. > m := NewtonQuotient(f(x), x=x0, h=h); m := esin(1+h) − esin(1) h If h = 1, the slope is > eval(%, h=1); esin(2) − esin(1) The evalf command gives a floating-point approximation. > evalf( % ); 0.162800903 As h tends to zero, the secant slope values seem to converge. > slopes := seq( NewtonQuotient( f(x), x=1.0, h=1.0/10^i ), > i=0..5); slopes := 0.1628009030, 1.182946800, 1.246939100, 1.252742000, 1.253310000, 1.253300000 The following is the equation of the secant line. 7.1 Introductory Calculus > y - p0[2] = m * ( x - p0[1] ); • 217 y − esin(1) = (esin(1+h) − esin(1) ) (x − 1) h The isolate command converts the equation to slope–intercept form. > isolate( %, y ); y= (esin(1+h) − esin(1) ) (x − 1) + esin(1) h You must convert the equation to a function. > secant := unapply( rhs(%), x ); secant := x → (esin(1+h) − esin(1) ) (x − 1) + esin(1) h You can now plot the secant and the function for different values of h. First, make a sequence of plots. > S := seq( plot( [f(x), secant(x)], x=0..4, > view=[0..4, 0..4] ), > h=[1.0, 0.1, .01, .001, .0001, .00001] ): The display command from the plots package can display the plots in sequence, that is, as an animation. > with(plots): Warning, the name changecoords has been redefined > display( S, insequence=true, view=[0..4, 0..4] ); 218 • Chapter 7: Solving Calculus Problems x x x x x x In the limit as h tends to zero, the slope is > Limit( m, h=0 ); esin(1+h) − esin(1) h→0 h lim The value of this limit is > value( % ); esin(1) co...
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