1 CHAPTER 1 1 1 We will illustrate two different methods for solving this problem 1 separation of variables and 2 Laplace transform dv c g v dt m Separation of variables Separation of variables gives 1 g c v dv dt m The integrals can be evaluated as c ln g v m t C c m where C a constant of integration which can be evaluated by applying the initial condition to yield c ln g v 0 m C c m which can be substituted back into the solution c c ln g v ln g v 0 m m t c m c m This result can be rearranged algebraically to solve for v v v 0 e c m t mg 1 e c m t c where the first part is the general solution and the second part is the particular solution for the constant forcing function due to gravity For the case where v 0 0 the solution reduces to Eq 1 10 v mg 1 e c m t c Laplace transform solution An alternative solution is provided by applying Laplace transform to the differential equation to give sV s v 0 g c V s s m Solve algebraically for the transformed velocity 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 2 V s v 0 g s c m s s c m 1 The second term on the right of the equal sign can be expanded with partial fractions g A B A s c m Bs s s c m s s c m s s c m 2 By equating like terms in the numerator the following must hold g A c m 0 As Bs The first equation can be solved for A mg c According to the second equation B A so B mg c Substituting these back into 2 gives g mg c mg c s s c m s s c m This can be substituted into Eq 1 to give V s v 0 mg c mg c s c m s s c m Taking inverse Laplace transforms yields v t v 0 e c m t mg mg c m t e c c or collecting terms v t v 0 e c m t mg 1 e c m t c 1 2 At t 8 s the analytical solution is 41 137 Example 1 1 The relative error can be calculated with absolute relative error analytical numerical 100 analytical The numerical results are step v 8 2 1 0 5 44 8700 42 8931 41 9901 absolute relative error 9 074 4 268 2 073 The error versus step size can then be plotted as 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 3 10 8 6 4 2 0 0 0 5 1 1 5 2 2 5 Thus halving the step size approximately halves the error 1 3 a You are given the following differential equation with the initial condition v t 0 0 dv c g v2 dt m Multiply both sides by m c gives m dv m g v2 c dt c Define a mg c m dv a2 v2 c dt Integrate by separation of variables a dv 2 v2 c mdt A table of integrals can be consulted to find that a dx 2 x 2 1 x tanh 1 a a Therefore the integration yields v c 1 tanh 1 t C a a m If v 0 at t 0 then because tanh 1 0 0 the constant of integration C 0 and the solution is v c 1 tanh 1 t a a m This result can then be rearranged to yield 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 4 gc gm tanh t c m v b Using Euler s method the first two steps can be computed as 0 22 2 v 2 0 9 81 0 2 19 62 68 1 0 22 v 4 19 62 9 81 19 62 2 2 36 75284 68 1 The computation can be continued and the results summarized along with the analytical result as t v numerical dv dt v analytical 0 2 4 6 8 10 12 0 19 62 36 75284 47 64539 52 59819 54 34314 54 88241 55 10572 9 81 8 56642 5 446275 2 476398 0 872478 0 269633 0 079349 0 022993 0 18 83093 33 72377 43 46492 49 06977 52 05938 53 58978 55 10572 A plot of the numerical and analytical results can be developed 60 40 20 v numerical v analytical 0 0 4 8 12 gm 1 e c m t c 9 81 70 1 e 12 70 9 44 99204 jumper 1 v t 12 9 81 80 1 e 15 80 t jumper 2 44 99204 15 1 4 v t 44 99204 52 32 52 32e 0 1875 t 0 14006 e 0 1875 t ln 0 14006 0 1875t t ln 0 14006 10 4836 s 0 1875 1 5 Before the chute opens t 10 Euler s method can be implemented as 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 5 10 v t t v t 9 81 v t t 80 After the chute opens t 10 the drag coefficient is changed and the implementation becomes 60 v t t v t 9 81 v t t 80 Here is a summary of the results along with a plot Chute closed dv dt v 20 0000 12 3100 7 6900 10 7713 3 0813 9 4248 12 5061 8 2467 20 7528 7 2159 27 9687 6 3139 34 2826 5 5247 …