EX :Compressing a SpringA 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=k3( )k=250F x( )=k xso F x( )=250xW=250x dx36∫=3375 inch−poundsLogistical GrowthExponential Growthdydt=ky1−yL;y=L1+be−kt;b=L−Y0Y0dydt=ky;y=Cektk=constant of proportionality k=growth constantL=carrying capacity C=initial amountY0=initial amount
17 Logistic Growthdydt=ky(1−yL) 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=L1+be−kt; b=L−Y0Y0EX#1:A highly contagious “pinkeye” (scientific name: Conjunctivitus itchlikecrazius) is ravaging the localelementary 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?SolutionBecause 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 equationL=900 b=900−7575=11 ⇒y=9001+11e−ktFive days later, 250 people have contracted pinkeye, so plug that information to find k:250=9001+11e−k⋅5⇒250(1+11e−5k)=90011e−5k=135⇒e−5k=1355lne−5k=ln1355⇒−5k=−1.442383838k=0.2884767656Finally, we have the equation y=9001+11e−0.2884767656t. We want to find the number of infections on December 25, so t=10.y=9001+11e−(0.2884767656)(10)y≈557.432 So almost 558 students have contracted pinkeye in time to open presents.
18 Euler’s MethodUses tangent lines to approximate points on the curve.Newy=Oldy+dx⋅dydxdx: changeinxdydx=Derivative (slope) at the point.EX#1:Given: f(0)=3 dydx=xy2Use step of h=0.1 to find f(0.3) Original (0,3)New y=3+(0.1)(0)=3f0.1()3 At (0,3) dydx=0⋅32=0 New y=3+(0.1)(0.15)=3.015f0.2()3.015 At (0.1,3) dydx=0.1⋅32=0.15New y=3.015+(0.1)(0.3015)=3.04515f0.3()3.04515At (0.2,3.015) dydx=0.2⋅3.0152=0.3015Slope FieldsDraw the slope field for dydx=xyPlug each point into dydxand graph the tangent line at the pointAt 1, 1()dydx=1. At −1,1()dydx=−1. At 0,1()dydx=0••••••−1•0•1••••EX : Here are the slope fields for the given differential equationsdydx=y86−y()dydx=x+y9 3.1 -4.7 4.7 0 8 -3.1 The slope fields give ALLthe solutions to the differential equation.