Solution manual applied numerical methods with matlab for engineers and scientists ().

Solutions Manualk

to kaccompany




Applied Numerical Methods
k k


With MATLAB for Engineers and Scientists
k k k k k

,Steven kC. kChapra
Tufts kUniversity

,CHAPTER k1
1.1 You kare kgiven kthe kfollowing kdifferential kequation kwith kthe kinitial kcondition, kv(t k= k0) k= k0,

dv k cd
 k g k k k
vk2
dt m

Multiply kboth ksides kby km/cd

m k dv m k g k  kv k2
k
 c
cd k k dt d

Define k a k mg / cd

m k dv k
 ka k2 k  kvk2
cd k k dt

Integrate kby kseparation kof kvariables,

dv cd k
kak k  kvk  kmk
k
2 2
dt


A ktable kof kintegrals kcan kbe kconsulted kto kfind kthat
dx 1k x
 k tanh1 k
kak k  kxk
2 2
a a

Therefore, kthe kintegration kyields
1k vk c
tanh k1 k  k d k
t k kC
a a m

If kv k= k0 kat kt k= k0, kthen kbecause ktanh–1(0) k= k0, kthe kconstant kof kintegration kC k= k0
kand kthe ksolution kis

1k vk c
tanh k1 k  k d k
t
a a m

This kresult kcan kthen kbe krearranged kto kyield

gm  gc 
v k tanh d 
t
 m 
cd  


1.2 This kis ka ktransient kcomputation. kFor kthe kperiod kfrom kending kJune k1:



1

, Balance k= kPrevious kBalance k+ kDeposits k–

kWithdrawals kBalance k= k1512.33 k+ k220.13 k–


k327.26 k= k1405.20


The kbalances kfor kthe kremainder kof kthe kperiods kcan kbe kcomputed kin ka ksimilar kfashion
kas ktabulated kbelow:


Date Deposit Withdrawal Balance
1-May $
k 1512.33
$ k 220.13 $ k 327.26
1-Jun $
k 1405.20
$ k 216.80 $ k 378.61
1-Jul $
k 1243.39
$ k 350.25 $ k 106.80
1-Aug $
k 1586.84
$ k 127.31 $ k 450.61
1-Sep $
k 1363.54


1.3 At kt k= k12 ks, kthe kanalytical ksolution kis k50.6175 k(Example k1.1). kThe knumerical kresults kare:

absolute
step v(12) krelative
kerror
2 51.6008 1.94%
1 51.2008 1.15%
0.5 50.9259 0.61%

where kthe krelative kerror kis kcalculated kwith

analytical  numerical
absolute krelative kerror k 100%
analytical

The kerror kversus kstep ksize kcan kbe kplotted kas

2.0%




1.0%


relative error
0.0%
0 0.5 1 1.5 2 2.5



2