Matematicas Especiales

1

CHAPTER 1
1.1 For body weight:
4.5  4.5  12  4.5  33  TW  60

TW = 1.5% For total body water:
7.5  7.5  20  7.5  2.5  IW  100

IW = 55% 1.2

Qstudents  30 ind  80

J s kJ  15 min  60   2160 kJ ind s min 1000 J

m

PVMwt (101.325 kPa)(10m  8m  3m  30  0.075 m 3 )(28.97 kg/kmol )   286.3424 kg RT (8.314 kPa m 3 /( kmol K)((20  273.15)K )
Qstudents 2160kJ   10 .50615 K mC v (286 .3424 kg )(0.718 kJ/(kg K))

T 

Therefore, the final temperature is 20 + 10.50615 = 30.50615oC. 1.3 This is a transient computation. For the period from ending June 1: Balance = Previous Balance + Deposits – Withdrawals Balance = 1512.33 + 220.13 – 327.26 = 1405.20 The balances for the remainder of the periods can be computed in a similar fashion as tabulatedbelow:
Date 1-May $ 220.13 1-Jun $ 216.80 1-Jul $ 450.25 1-Aug $ 127.31 1-Sep $ 350.61 $ 1363.54 $ 106.80 $ 1586.84 $ 378.61 $ 1243.39 $ 327.26 $ 1405.20 Deposit Withdrawal Balance $ 1512.33

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2 1.4 Q1,in  Q2,out  v3,out A3
A3  Q1,in  Q2,out v3,out
in out



40 m 3 /s  20 m 3 /s  3.333 m 2 6 m/s

1.5

M - M

0

Food  Drink  Air In Metabolism  Urine  Skin  Feces  Air Out  Sweat Drink  Urine  Skin  Feces  Air Out  Sweat  Food  Air In  Metabolism Drink  1.4  0.35  0.2  0.4  0.2  1  0.05  0.3  1.2 L

1.6 v(t ) 

gm (1  e ( c / m) t ) c
9.8(70) (1  e (12 / 70) 10 )  46.8714 12

jumper #1: v(t ) 

jumper #2: 46.8714 

9.8(75) (1  e (15 / 75) t ) 15

46.8714  49  49e 0.2 t 0.04344 e 0.2 t
ln 0.04344  0.2t

t

ln 0.04344  15.6818 s  0.2

1.7 You are given the following differential equation with the initial condition, v(t = 0) = v(0),
dv c g v dt m

The most efficient way to solve this is with Laplace transforms

sV ( s)  v(0) 

g c  V ( s) s m

Solve algebraically for the transformed velocity
V ( s)  v(0) g  s  c / m s ( s  c / m)

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3

The second term on the right of the equal sign can be expanded with partial fractions
g A B   s ( s  c / m) s s  c / m

Combining the right-hand side gives
g A( s  c / m)  Bs  s ( s  c / m) s ( s  c / m)

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. Therefore, the partial fraction expansion is
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 
or collecting terms
v(t )  v(0)e ( c / m)t 

mg mg ( c / m)t  e c c

mg 1  e( c / m )t 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 44.87 (Example 1.1). The relative error can be calculated with

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