PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved.

1. Numerical Methods Chapra Solution Manual Chapter 20 Summary

Numerical Methods Chapra Solution Manual Chapter 20 Summary

No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 1 IW = 5% s 60min 15 s ind PV m vmC Therefore, the final temperature is 20 + 10.50615 = 30.50615o C. 1.3 This is a transient computation. For the period from ending June 1: Balance = Previous Balance + Deposits – Withdrawals Balance = 1512.3 + 220.13 – 327.26 = 1405.20 The balances for the remainder of the periods can be computed in a similar fashion as tabulated below: 1-May $ 1512.3 1-Jun $ 1405.20 1-Jul $ 1243.39 1-Aug $ 1586.84. PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc.

All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.

If you are a student using this Manual, you are using it without permission. 1.7 You are given the following differential equation with the initial condition, v(t = 0) = v(0), v m c g The most efficient way to solve this is with Laplace transforms Solve algebraically for the transformed velocity mcss g mcs. PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

The second term on the right of the equal sign can be expanded with partial fractions Combining the right-hand side gives By equating like terms in the numerator, the following must hold BsAs c Ag The first equation can be solved for A = mg/c. According to the second equation, B = –A.

Therefore, the partial fraction expansion is mcs cmgs cmg mcss g / // This can be substituted into Eq. 1 to give mcs cmgs cmg mcs v sV Taking inverse Laplace transforms yields tmctmc e cmgc or collecting terms tmctmc e c The first part is the general solution and the second part is the particular solution for the constant forcing function due to gravity.

1.8 At t = 10 s, the analytical solution is 4.87 (Example 1.1). The relative error can be calculated with. PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation.

If you are a student using this Manual, you are using it without permission.%100 analytical numericalanalytical error relative absolute The numerical results are: step v(10) absolute relative error The error versus step size can then be plotted as Thus, halving the step size approximately halves the error. 1.9 (a) You are given the following differential equation with the initial condition, v(t = 0) = 0, c g Multiply both sides by m/c ' vg cmdtdvc Define '/cmga Integrate by separation of variables, mc va dv A table of integrals can be consulted to find that. PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved.

No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Ax axa dx 1 Therefore, the integration yields mcav a If v = 0 at t = 0, then because tanh –1 (0) = 0, the constant of integration C = 0 and the solution is t mcava This result can then be rearranged to yield t mgcc gmv ' tanh' (b) Using Euler’s method, the first two steps can be computed as The computation can be continued and the results summarized and plotted as: t v dv/dt. PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved.

No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Vnum vanal Note that the analytical solution is included on the plot for comparison.

1.10 Before the chute opens (t.