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148

CHAPTER 2

Linearity and Nonlinearity

„

Using the Time Constant

15.

(a)

(

SECTION 2.4

Linear Models: Mixing and Cooling

)

T ( t ) = T0 e − kt + M 1 − e − kt , from Equation (8). In this case, M = 95 , T0 = 75 , and k=

1 , yielding the expression 4

(

T ( t ) = 75e −t 4 + 95 1 − e −t

4

)

where t is time measured in hours. Substituting t = 2 in this case (2 hours after noon), yields T ( 2 ) ≈ 82.9 °F. (b)

Setting T ( t ) = 80 and simplifying for T ( t ) yields t = −4 ln

3 ≈ 1.15 hours, 4

which translates to 1:09 P.M. „

A Chilling Thought

16.

(a)

(

)

T ( t ) = T0 e − kt + M 1 − e − kt , from Equation (8). In this problem, T0 = 75 , M = 10 , and ⎛ 1⎞ T ⎜ ⎟ = 50 (taking time to be in hours). Thus, we have the equation 50 = 10 + 60e −k 2 , ⎝2⎠ from which we can find the rate constant k = −2 ln

2 ≈ 0.81 . 3

After one hour, the temperature will have fallen to ⎛ 4⎞ 2ln 2 3 T (1) ≈ 10 + 60e ( ) = 10 + 60 ⎜ ⎟ ≈ 36.7 ° F. ⎝9⎠ (b)

Setting T ( t ) = 15 gives the equation 15 = 10 + 60e 2t ln ( 2 3) . Solving for t gives t=−

ln 12 ≈ 3.06 hours (3 hrs, 3.6 min). 2 ln ( 23 )

148

Solutions manual for differential equations and linear algebra 2nd edition by farlow ibsn 9780134689  

Solutions Manual for Differential Equations and Linear Algebra 2nd Edition by Farlow IBSN 9780134689548 Full clear download( no error format...

Solutions manual for differential equations and linear algebra 2nd edition by farlow ibsn 9780134689  

Solutions Manual for Differential Equations and Linear Algebra 2nd Edition by Farlow IBSN 9780134689548 Full clear download( no error format...

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