Ex compressing a spring a force of 750 pounds

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EX : Compressing a Spring A force of 750 pounds compresses a spring 3 inches from its natural length of 15 inches. Find the work done in compressing the spring an additional 3 inches. 750 = k 3 ( ) k = 250 F x ( ) = k x so F x ( ) = 250 x W = 250 x dx 3 6 = 3375 inch pounds Logistical Growth Exponential Growth dy dt = ky 1 y L ; y = L 1 + be kt ; b = L Y 0 Y 0 dy dt = ky ; y = Ce kt k = constant of proportionality k = growth constant L = carrying capacity C = initial amount Y 0 = initial amount
17 Logistic Growth dy dt = ky (1 y L ) where L is the carrying capacity (upper bound) and k is the constant of proportionality. If you solve the differential equation using very tricky separation of variables, you get the general solution: y = L 1 + be kt ; b = L Y 0 Y 0 EX#1: A highly contagious “pinkeye” (scientific name: Conjunctivitus itchlikecrazius) is ravaging the local elementary school. The population of the school is 900 (including students and staff), and the rate of infection is proportional both to the number infected and the number of students whose eyes are pus-free. If seventy-five people were infected on December 15 and 250 have contracted pinkeye by December 20, how many people will have gotten the gift of crusty eyes by Christmas Day? Solution Because of the proportionality statements in the problem, logistic growth is the approach we should take. The upper limit for the disease will be L = 900; it is impossible for more than 900 people to be infected since the school only contains 900 people. This gives us the equation L = 900 b = 900 75 75 = 11 y = 900 1 + 11 e kt Five days later, 250 people have contracted pinkeye, so plug that information to find k : 250 = 900 1 + 11 e k 5 250(1 + 11 e 5 k ) = 900 11 e 5 k = 13 5 e 5 k = 13 55 ln e 5 k = ln 13 55 5 k = 1.442383838 k = 0.2884767656 Finally, we have the equation y = 900 1 + 11 e 0.2884767656 t . We want to find the number of infections on December 25, so t = 10. y = 900 1 + 11 e (0.2884767656)(10) y 557.432 So almost 558 students have contracted pinkeye in time to open presents.
18 Euler’s Method Uses tangent lines to approximate points on the curve. New y = Old y + dx dy dx dx : change in x dy dx = Derivative (slope) at the point. EX#1: Given: f (0) = 3 dy dx = xy 2 Use step of h = 0.1 to find f (0.3) Original (0,3) New y = 3 + (0.1)(0) = 3 f 0.1 ( ) 3 At (0,3) dy dx = 0 3 2 = 0 New y = 3 + (0.1)(0.15) = 3.015 f 0.2 ( ) 3.015 At (0.1,3) dy dx = 0.1 3 2 = 0.15 New y = 3.015 + (0.1)(0.3015) = 3.04515 f 0.3 ( ) 3.04515 At (0.2,3.015) dy dx = 0.2 3.015 2 = 0.3015 Slope Fields Draw the slope field for dy dx = xy Plug each point into dy dx and graph the tangent line at the point At 1, 1 ( ) dy dx = 1. At 1,1 ( ) dy dx = 1. At 0,1 ( ) dy dx = 0 1 0 1 EX : Here are the slope fields for the given differential equations dy dx = y 8 6 y ( ) dy dx = x + y 9 3.1 -4.7 4.7 0 8 -3.1 The slope fields give ALL the solutions to the differential equation .
19 Lagrange Error Bound Error = f x ( ) P n x ( ) R n x ( ) where R n x ( ) f n + 1 z ( ) x c ( ) n + 1 n + 1 ( ) !

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