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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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