Chapter-1

# Chapter-1 - CHAPTER 1 First-Order Differential Equations...

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1 CHAPTER 1 First-Order Differential Equations 1.1 Dynamical Systems: Modeling ± Constants of Proportionality 1. dA kA dt = ( k < 0) 2. dA kA dt = ( k < 0) 3. (20,000 ) dP kP P dt =− 4. dA kA dt t = 5. dG kN dt A = ± A Walking Model 6. Because dt = υ where d = distance traveled, = average velocity, and t = time elapsed, we have the model for the time elapsed as simply the equation t d = . Now, if we measure the distance traveled as 1 mile and the average velocity as 3 miles/hour, then our model predicts the time to be t d == 1 3 hr, or 20 minutes. If it actually takes 20 minutes to walk to the store, the model is perfectly accurate. This model is so simple we generally don’t even think of it as a model. ± A Falling Model 7. (a) Galileo has given us the model for the distance st ( ) a ball falls in a vacuum as a function of time t : On the surface of the earth the acceleration of the ball is a constant, so ds dt g 2 2 = , where g 32 2 . ft sec 2 . Integrating twice and using the conditions s 00 ( ) = , ds dt 0 0 () = , we find g t () = 1 2 2 g t 1 2 2 .

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2 CHAPTER 1 First-Order Differential Equations (b) We find the time it takes for the ball to fall 100 feet by solving for t the equation 100 1 2 161 22 == gt t . , which gives t = 249 . seconds. (We use 3 significant digits in the answer because g is also given to 3 significant digits.) (c) If the observed time it takes for a ball to fall 100 feet is 2.6 seconds, but the model predicts 2.49 seconds, the first thing that might come to mind is the fact that Galileo’s model assumes the ball is falling in a vacuum, so some of the difference might be due to air friction. ± The Malthus Rate Constant k 8. (a) Replacing e 003 103045 . . in Equation (3) gives y t = () 09 103045 .. , which increases roughly 3% per year. (b) 1860 1820 4 1800 6 8 2 1840 t y 10 1880 3 5 7 1 9 Malthus World population (c) Clearly, Malthus’ rate estimate was far too high. The world population indeed rises, as does the exponential function, but at a far slower rate. If yt e rt () = 09 . , you might try solving ye r 200 60 200 ( ) = = . . for r . Then 200 6 1897 r =≈ ln . . so r ≈≈ 1897 200 00095 . ., which is less than 1%. ± Population Update 9. (a) If we assume the world’s population in billions is currently following the unrestricted growth curve at a rate of 1.7% and start with the UN figure for 2000, then 0.017 0 6.056 kt t e = ,
SECTION 1.1 Dynamical Systems: Modeling 3 and the population in the years 2010 t = ( ) 10 , 2020 t = ( ) 20 , and 2030 t = () 30 , would be, respec- tively, the values 0.017 10 0.017 20 0.017 30 6.056 7.176 6.056 8.509 6.056 10.083. e e e = These values increasingly exceed the United Nations predictions so the U.N. is assuming a growth rate less than 1.7%. (b) 2010: 10 6.056 6.843 r e = 10 6.843 1.13 6.056 10 ln(1.13) 0.1222 1.2% r e r r == = 2020: 10 6843 7568 r e = 10 7.578 1.107 6.843 10 ln(1.107) 0.102 1.0% r e r r = 2030: 10 7.578 8.199 r e = 10 8.199 1.082 7.578 10 ln(1.082) 0.079 0.8% r e r r = ± The Malthus Model 10. (a) Malthus thought the human population was increasing exponentially e kt , whereas the food supply increases arithmetically according to a linear function ab t + . This means the number of people per food supply would be in the ratio e t kt + , which although not a pure exponential function, is concave up. This means that the rate of increase in the number of persons per the amount of food is increasing.

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Chapter-1 - CHAPTER 1 First-Order Differential Equations...

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