Question: Consider the initial-value problem for a nonlinear pendulum. Since we cannot solve the differential equation, we can find no explicit solution of this problem. But

Consider the initial-value problem

d20 + sin0 = 0, 0(0) dr? O'(0): %3D %3D 12 3


for a nonlinear pendulum. Since we cannot solve the differential equation, we can find no explicit solution of this problem. But suppose we wish to determine the first time t1 > 0 for which the pendulum in Figure 5.3.3, starting from its initial position to the right, reaches the position OP—that is, the first positive root of θ(t) = 0. In this problem and the next we examine several ways to proceed.

(a) Approximate t1 by solving the linear problem

d20 + sin0 = 0, 0(0) dr? O'(0): %3D %3D 12 3


(b) Use the method illustrated in Example 3 of Section 4.10 to find the first four nonzero terms of a Taylor series solution u(t) centered at 0 for the nonlinear initial-value problem. Give the exact values of all coefficients.

(c) Use the first two terms of the Taylor series in part (b) to approximate t1.

(d) Use the first three terms of the Taylor series in part (b) to approximate t1.

(e) Use a root-finding application of a CAS or a graphic calculator and the first four terms of the Taylor series in part (b) to approximate t1.

(f) In this part of the problem you are led through the commands in Mathematica that enable you to approximate the root t1. The procedure is easily modified so that any root of θ(t) = 0 can be approximated. (If you do not have Mathematica, adapt the given procedure by finding the corresponding syntax for the CAS you have on hand.) Precisely reproduce and then, in turn, execute each line in the given sequence of commands.


(g) Appropriately modify the syntax in part (f) and find the next two positive roots of θ(t) = 0.

d20 + sin0 = 0, 0(0) dr? O'(0): %3D %3D 12 3

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Part a Step 1 Rearrange the Equation Original equation cost1 4 sint1 0 Isolate cost1 cost1 4 sint1 Step 2 Square Both Sides cost1 4 sint1 Simplify cost1 16 sint1 Step 3 Use the Pythagorean Identity 16 ... View full answer

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