/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Free solutions & answers for Numerical Analysis Chapter 10 - (Page 1) [step by step] | 91Ó°ÊÓ

91Ó°ÊÓ

Problem 1

Use the method of Steepest Descent with \(T O L=0.05\) to approximate the solutions of the following nonlinear systems. a. \(\quad 4 x_{1}^{2}-20 x_{1}+\frac{1}{4} x_{2}^{2}+8=0\) b. \(\quad 3 x_{1}^{2}-x_{2}^{2}=0\) \(\frac{1}{2} x_{1} x_{2}^{2}+2 x_{1}-5 x_{2}+8=0\) c. \(\ln \left(x_{1}^{2}+x_{2}^{2}\right)-\sin \left(x_{1} x_{2}\right)=\ln 2+\ln \pi\) $$ \text { d. } \sin \left(4 \pi x_{1} x_{2}\right)-2 x_{2}-x_{1}=0 $$ $$ \left(\frac{4 \pi-1}{4 \pi}\right)\left(e^{2 x_{1}}-e\right)+4 e x_{2}^{2}-2 e x_{1}=0 $$

Problem 4

Use the graphing facilities of Maple to approximate solutions to the following nonlinear systems within the given limits. a. $$ \begin{array}{rlr} 3 x_{1}-\cos \left(x_{2} x_{3}\right)-\frac{1}{2}=0, & \text { b. } \quad x_{1}^{2}+x_{2}-37 & =0 \\ 4 x_{1}^{2}-625 x_{2}^{2}+2 x_{2}-1=0, & x_{1}-x_{2}^{2}-5=0, \\ e^{-x_{1} x_{2}}+20 x_{3}+\frac{10 \pi-3}{3}=0 . & x_{1}+x_{2}+x_{3}-3=0 . \\ -1 \leq x_{1} \leq 1,-1 \leq x_{2} \leq 1, & -4 \leq x_{1} \leq 8,-2 \leq x_{2} \leq 2,-6 \leq x_{3} \leq 0 \\ -1 \leq x_{3} \leq 1 & \end{array} $$ c. \(\quad 15 x_{1}+x_{2}^{2}-4 x_{3}=13\), $$ \text { d. } \quad 10 x_{1}-2 x_{2}^{2}+x_{2}-2 x_{3}-5=0 $$ $$ \begin{gathered} x_{1}^{2}+10 x_{2}-x_{3}=11 \\ x_{2}^{3}-25 x_{3}=-22 \\ 0 \leq x_{1} \leq 2,0 \leq x_{2} \leq 2,0 \leq x_{3} \leq 2 \end{gathered} $$ \(8 x_{2}^{2}+4 x_{3}^{2}-9=0\) \(8 x_{2} x_{3}+4=0\) and \(0 \leq x_{1} \leq 2,0 \leq x_{2} \leq 2,-2 \leq x_{3} \leq 0\)

Problem 4

The nonlinear system $$ -x_{1}\left(x_{1}+1\right)+2 x_{2}=18, \quad\left(x_{1}-1\right)^{2}+\left(x_{2}-6\right)^{2}=25 $$ has two solutions. a. Approximate the solutions graphically. b. Use the approximations from part (a) as initial approximations for an appropriate function iteration, and determine the solutions to within \(10^{-5}\) in the \(l_{\infty}\) norm.

Problem 12

The amount of pressure required to sink a large, heavy object in a soft homogeneous soil that lies above a hard base soil can be predicted by the amount of pressure required to sink smaller objects in the same soil. Specifically, the amount of pressure \(p\) required to sink a circular plate of radius \(r\) a distance \(d\) in the soft soil, where the hard base soil lies a distance \(D>d\) below the surface, can be approximated by an equation of the form $$ p=k_{1} e^{k_{2} r}+k_{3} r $$ where \(k_{1}, k_{2}\), and \(k_{3}\) are constants, with \(k_{2}>0\), depending on \(d\) and the consistency of the soil but not on the radius of the plate. (See [Bek], pp. 89-94.) a. Find the values of \(k_{1}, k_{2}\), and \(k_{3}\) if we assume that a plate of radius \(1 \mathrm{in}\). requires a pressure of 10 ; \(\mathrm{lb} / \mathrm{in} .^{2}\) to sink \(1 \mathrm{ft}\) in a muddy field, a plate of radius 2 in. requires a pressure of \(12 \mathrm{lb} / \mathrm{in} .{ }^{2}\) to sink \(1 \mathrm{ft}\), and a plate of radius \(3 \mathrm{in}\). requires a pressure of \(15 \mathrm{lb} / \mathrm{in} .{ }^{2}\) to sink this distance (assuming that the mud is more than \(1 \mathrm{ft}\) deep). b. Use your calculations from part (a) to predict the minimal size of circular plate that would be required to sustain a load of \(500 \mathrm{lb}\) on this field with sinkage of less than \(1 \mathrm{ft}\).

Problem 13

In calculating the shape of a gravity-flow discharge chute that will minimize transit time of discharged granular particles, C. Chiarella, W. Charlton, and A. W. Roberts [CCR] solve the following equations by Newton's method: (i) \(\quad f_{n}\left(\theta_{1}, \ldots, \theta_{N}\right)=\frac{\sin \theta_{n+1}}{v_{n+1}}\left(1-\mu w_{n+1}\right)-\frac{\sin \theta_{n}}{v_{n}}\left(1-\mu w_{n}\right)=0\), for each \(n=1,2, \ldots, N-1\) (ii) \(\quad f_{N}\left(\theta_{1}, \ldots, \theta_{N}\right)=\Delta y \sum_{i=1}^{N} \tan \theta_{i}-X=0\), where a. \(\quad v_{n}^{2}=v_{0}^{2}+2 g n \Delta y-2 \mu \Delta y \sum_{j=1}^{n} \frac{1}{\cos \theta_{j}}, \quad\) for each \(n=1,2, \ldots, N\), and b. \(\quad w_{n}=-\Delta y v_{n} \sum_{i=1}^{N} \frac{1}{v_{i}^{3} \cos \theta_{i}}, \quad\) for each \(n=1,2, \ldots, N\). The constant \(v_{0}\) is the initial velocity of the granular material, \(X\) is the \(x\)-coordinate of the end of the chute, \(\mu\) is the friction force, \(N\) is the number of chute segments, and \(g=32.17 \mathrm{ft} / \mathrm{s}^{2}\) is the gravitational constant. The variable \(\theta_{i}\) is the angle of the \(i\) th chute segment from the vertical, as shown in the following figure, and \(v_{i}\) is the particle velocity in the \(i\) th chute segment. Solve (i) and (ii) for \(\theta=\left(\theta_{1}, \ldots, \theta_{N}\right)^{t}\) with \(\mu=0, X=2, \Delta y=0.2, N=20\), and \(v_{0}=0\), where the values for \(v_{n}\) and \(w_{n}\) can be obtained directly from (a) and (b). Iterate until \(\left\|\theta^{(k)}-\theta^{(k-1)}\right\|_{\infty}<10^{-2}\).

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