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# 0019 - 47 48 49 50 c = rPoe"r(e”-1 With P0 = 10,800...

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Unformatted text preview: 47. 48. 49. 50. c = rPoe"r(e”-1). With P0 = 10,800, r = 60, and r = 0.010 we get \$239.37 for the monthly payment at 12% annual interest. With r = 0.015 we get \$272.99 for the-monthly payment at 18% annual interest. If NU) denotes the number of people (in thousands) who have heard the rumor after t days, then the initial value problem is N' = k(100—N), N(0) = 0 and we are given that N0) = 10. When we separate variables MN {(1 00-N) .: it (it) and integrate, we get ln(100—N) = ~kt +0, and the initial condition N(0) :0 gives C=ln100. Then 100—N:1009”“, so N(:)=100(1—e""). We substitutet=7, N: 10 and solve for the value k = ln(100/90)/7 = 0.01505. Finally, 50 thousand people have heard the rumor after t = (ln 2) 1' k = 46.05 days. With A(y) constant, Equation (19) in the text takes the form We readily solve this equation for 2J3: = kr+C. The condition y(0) = 9 yields C = 6, and then y(1) = 4 yields k : 2. Thus the depth at time I (in hours) is ya) = (3 ~02, and hence it takes 3 hours for the tank to empty. With A 3 M3)2 and a : Ira/12):, and taking g = 32 ftfsecz, Equation (20) reduces to 162 y’ = —J_;. The solution such that y = 9 when I z 0 is given by 324 y = -t+972. Hence y = 0 when t 7— 972 sec 2 16min 12sec. The radius of the cross-section of the cone at height y is proportional to )1, so My) is proportional to 322. Therefore Equation (20) takes the form yZy! = _li—, and a general solution is given by 2;?“2 = —5kt + C. The initial condition y(0) = 16 yields C = 2048, and then y(1) = 9 implies that 51:. = 1562. Hence 3: = 0 when t = ClSk = 2048/1562 = 1.31 hr. Section 1.4 19 ...
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