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CHAPTER 1
1.1 For body weight:
TW = 1.5%
For total body water:
IW = 55%
1.2
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 tabulated below:
Date Deposit Withdrawal Balance
1-May $ 1512.33
$ 220.13 $ 327.26
1-Jun $ 1405.20
$ 216.80 $ 378.61
1-Jul $ 1243.39
$ 450.25 $ 106.80
1-Aug $ 1586.84
$ 127.31 $ 350.61
1-Sep $ 1363.54
1.4
1.5
1.6
jumper #1:
jumper #2:
1.7 You are given the following differential equation with the initial condition, v(t = 0) = v(0),
The most efficient way to solve this is with Laplace transforms
Solve algebraically for the transformed velocity
(1)
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
The first equation can be solved for A = mg/c. According to the second equation, B = –A. Therefore, the partial fraction expansion is
This can be substituted into Eq. 1 to give
Taking inverse Laplace transforms yields
or collecting terms
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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