Learning Guide Learning Guide

248 • Chapter 6: Examples from Calculusdev_order := 5You can now use seq to generate a sequence of the higher order derivativesof theta(t).> S := seq( (D@@(dev_order-n))(eq), n=1..dev_order );S := (D (5) )(θ) = − 1 10 (D(4) )(θ), (D (4) )(θ) = − 1 10 (D(3) )(θ),(D (3) )(θ) = − 110 (D(2) )(θ), (D (2) )(θ) = − 1 10 D(θ),D(θ) = − 110 θ + 2The fifth derivative is a function of the fourth derivative, the fourtha function of the third and so on. Therefore, if you make substitutionsaccording to S, you can express all the derivatives as functions of theta.For example, the third element of S is the following.> S[3];(D (3) )(θ) = − 110 (D(2) )(θ)Substituting according to S on the right-hand side, yields> lhs(%) = subs( S, rhs(%) );(D (3) )(θ) = − 11000 θ + 150To make this substitution on all the derivatives at once, use the mapcommand.> L := map( z -> lhs(z) = eval(rhs(z), {S}), [S] );L := [(D (5) )(θ) = 1100 (D(3) )(θ), (D (4) )(θ) = 1100 (D(2) )(θ),(D (3) )(θ) = 1100 D(θ), (D(2) )(θ) = 1100 θ − 1 5 ,D(θ) = − 110 θ + 2]You must evaluate the derivatives at t = 0.