This document contains solutions to Assignment 3 for Math 2280 course. It includes solutions to problems related to acceleration-velocity models, numerical approximation using Euler's method, and more.
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Section 2.3 - Acceleration-Velocity Models 2.3.1The acceleration of a Maserati is proportional to the difference be- tween 250 km/h and the velocity of this sports car.If the machine can accelerate from rest to 100 km/h in 10s, how long will it take for the car to accelerate from rest to 200 km/h? Solution- The differentialequation governing the car’s movement will be: dv dt=k(250−v). This is a separable differential equation. We can rewrite it as: dv 250−v =kdt. Integrating both sides of this equation we get: Zdv 250−v = Z kdt. ⇒ −ln (250−v) =kt+C ⇒250−v=Ce−kt ⇒v(t) = 250−Ce−kt. Using the initial conditionv(0) = 0 = 250−Cwe getC= 250. Using the givenv(10) = 100we get: 2
v(10) = 250(1−e−10k) = 100 ⇒1−e−10k=2 5 ⇒ln(e−10k) = ln3 5 ⇒k=ln 5−ln 3 10≈.05108. Using this value ofkwe want to find the value oftfor whichv(t) = 200. We do this by solving: 200 = 250(1−e−.05108t∗) ⇒4 5= 1−e−.05108t∗ ⇒e−.05108t∗=1 5 ⇒t∗=ln 5 .05108 ≈31.5seconds. 3
2.3.2Suppose that a body moves through a resisting medium with resis- tance proportional to its velocityv, so thatdv/dt=−kv. (a)Show that its velocity and position at timetare given by v(t) =v0e−kt and x(t) =x0+v0 k(1−e−kt). (b)Conclude that the body travels only a finite distance,and find that distance. Solution (a)- The differential equation dv dt=−kv is separable, and can be rewritten as dv v=−kdt. If we integrate both sides of the above differential equation we get: lnv=−kt+C ⇒v(t) =Ce−kt. Using the initial valuev(0) =v0=Cwe get: v(t) =v0e−kt. 4
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Integrating this function to get the position function gives us: x(t) =− v0 ke−kt+C∗. Usingx(0) =x0=−v0 k+C∗we getC∗=x0+v0 k. This gives us: x(t) =x0+v0 k−v0 ke−kt=x0+v0 k(1−e−kt). (b)- If we take the limit of our position function ast→ ∞we get: lim t→∞x(t) =x0+v0 k(1−e−k∞) =x0+v0 k. 5
2.3.4Consider a body that moves horizontally through a medium whose resistance is proportional to thesquareof the velocityv, so that dv/dt=−kv2. Show that v(t) =v0 1 +v0kt and that x(t) =x0+1 kln (1 +v0kt). Note that,in contrast with the result of Problem 2,x(t)→ ∞as t→ ∞.Which offers less resistance when the body is moving fairly slowly - the medium in this problem or the one in Problem 2? Does your answer seem to be consistent with the observed behaviors of x(t)ast→ ∞? Solution- The differential equation dv dt=−kv2 is separable. We can rewrite it as: dv v2=−kdt. Integrating both sides of this equation gives us: 6
−1 v=−kt+C. Solving forv(t)gives us: v(t) =1 kt+C. Using the initial conditionv(0) =v0=1 Cwe haveC=1 v0 . Plugging this into our velocity equation gives us: v(t) =1 kt+1 v0 =v0 1 +v0kt. Integrating this we get: x(t) = Zv0 1 +v0ktdt=v0 kv0 ln (1 +v0kt) +C=C+ 1 kln (1 +v0kt). Our initial conditionx(0) =x0=Cgives us: x(t) =x0+1 kln (1 +v0kt). Indeedlim t→∞x(t) =∞. For|v|<1we havev2<|v|and so the drag issmallerfor fairly small values ofv. This is why the distance can go forever and is not finite. 7
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2.3.10A woman bails out of an airplane at an altitude of 10,000 ft,falls freely for 20s,then opens her parachute.How long will it take her to reach the ground?Assume linear air resistanceρv f t/s2,taking ρ=.15without the parachute andρ= 1.5with the parachute.(Sug- gestion:First determine her height above the ground and velocity when the parachute opens.) Solution- We have: v(t) =v0+g ρe−ρt−g ρ. If we integrate this to findx(t)we get: x(t) =− g ρt−1 ρv0+g ρe−ρt+C. Plugging in the initial conditionx(0) =x0we get: C=x0+1 ρv0+g ρ=x0+1 ρ(v0−vτ). Using this value forCafter a little algebra our equation forx(t)be- comes: x(t) =x0+vτt+1 ρ(v0−vτ)(1−e−ρt). Now, the initial distance isx0= 10,000ft, the initial velocity isv0= 0 ft/s,the terminal velocity isvτ=−32.2 .15ft/s,andρ=.15/s.The total distance traveled in the first20seconds is:1 1Leaving out units on the intermediate steps. Trust me, they work out. 8
x(20) = 10,000− 32.2 .15(20) =1 .150 +32.2 .15(1−e−.15(20) ) ≈7,067f t. The velocity of the skydiver after20seconds is: v(20) =0 +32.2 .15e−.15(20)−32.2 .15≈204ft/s. Now, to find the total time for the rest of the trip down we want to solve fortfin the equation: 0 = 7,067− 32.2 1.5tf+1 1.5−204 + 32.2 1.5(1−e−1.5tf). Using a calculator to find this we get: tf≈340seconds. So,the totalskydive time is about340s+ 20s= 360s,or about6 minutes. 9
2.3.24The mass of the sun is 329,320 times that of the earth and its radius is 109 times the radius of the earth. (a)To what radius (in meters) would the earth have to be compressed in order for it to become ablack hole- the escape velocity from its surface equal to the velocityc= 3×108m/sof light? (b)Repeat part (a) with the sun in place of the earth. Solution- (a)- We have3×108m s= r2GMe R. Solving this forRwe get: R=2GMe c2= 26.67×10−11Nm2 kg2(5.972×1024kg) 3×108m s 2 =.00885m=.885cm. WOW! (b)- Using the same equation we get: R=2GMs c2=2GMe c2(329,320)≈2,915m= 2.915km. 10
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Section 2.4 - Numerical Approximation:Euler’s Method 2.4.1Apply Euler’s method twice to approximate the solution to the initial value problem below on the interval[0,1 2], first with step sizeh=.25, then with step sizeh= 0.1. Compare the three-decimal-place values of the two approximations atx=1 2with the valuey(1 2)of the actual solution, also given below. y′=−y, y(0) = 2; y(x) = 2e−x. Solution- If we apply Euler’s method with a step size ofh=.25we get: y0= 2x0= 0, y1=y0+h∗f(x0, y0) = 2 + (.25)(−2) = 3 2x1=.25, y2=y1+h∗f(x1, y1) =3 2+ (.25)−3 2=9 8x2=.5. If we apply Euler’s method with a step size ofh=.1we get: y0= 2x0= 0, y1= 2 + (.1)(−2) = 1.8x1=.1, y2= 1.8 + (.1)(−1.8) = 1.62x2=.2, y3= 1.62 + (.1)(1.62) = 1.458x3=.3, 11
y4= 1.458 + (.1)(−1.458) = 1.3122x4=.4, y5= 1.3122 + (.1)(−1.3122) = 1.18098x5=.5. The exact value isy(.5) = 2e−.5= 1.213. So, withh=.25our estimate is off by.088, and withh=.1our estimate is off by.032.So,h=.1 gives a better estimate, which is what we’d expect. 12
2.4.5Apply Euler’s method twice to approximate the solution to the initial value problem below on the interval[0,1 2], first with step sizeh=.25, then with step sizeh= 0.1. Compare the three-decimal-place values of the two approximations atx=1 2with the valuey(1 2)of the actual solution, also given below. y′=y−x−1, y(0) = 1; y(x) = 2 +x−ex. Solution- If we apply Euler’s method with a step size ofh=.25we get: y0= 1x0= 0, y1=y0+h∗f(x0, y0) = 1 + (.25)(1−0−1) = 1 x1=.25, y2=y1+h∗f(x1, y1) = 1 + (.25)(1−.25−1) = 15 16 x2=.5. If we apply Euler’s method with a step size ofh=.1we get: y0= 1x0= 0, y1= 2 + (.1)(1−0−1) = 1x1=.1, y2= 1 + (.1)(1−.1−1) =.99x2=.2, y3=.99 + (.1)(.99−.2−1) =.969x3=.3, y4=.969 + (.1)(.969−.3−1) =.9359x4=.4, y5=.9359 + (.1)(.9359−.4−1) =.88949x5=.5. 13
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The exact value isy(.5) = 2 +.5−e.5=.851.So,withh=.25our estimate is off by.087, and withh=.1our estimate is off by.038. 14
2.4.9Apply Euler’s method twice to approximate the solution to the initial value problem below on the interval[0,1 2], first with step sizeh=.25, then with step sizeh= 0.1. Compare the three-decimal-place values of the two approximations atx=1 2with the valuey(1 2)of the actual solution, also given below. y′=1 4(1 +y2), y(0) = 1; y(x) = tan ( 1 4(x+π)). Solution- If we apply Euler’s method with a step size ofh=.25we get: y0= 1x0= 0, y1=y0+h∗f(x0, y0) = 1 + (.25)(.25(1 + 12)) =9 8 x1=.25, y2=y1+h∗f(x1, y1) =9 8+ (.25)(.25(1 + ( 9 8)2)) = 1.27 x2=.5. If we apply Euler’s method with a step size ofh=.1we get: y0= 1x0= 0, y1= 1 + (.1)(.25(1 + 12)) = 1.05x1=.1, y2= 1.05 + (.1)(.25(1 + 1.052)) = 1.10x2=.2, y3= 1.10 + (.1)(.25(1 + 1.102)) = 1.158x3=.3, 15
y4= 1.158 + (.1)(.25(1 + 1.1582)) = 1.217x4=.4, y5= 1.217 + (.1)(.25(1 + 1.2172)) = 1.279x5=.5. The exact value isy(.5) = tan 1 4(.5 +π)= 1.287.So, withh=.25 our estimate is off by.017, and withh=.1our estimate is off by.008. Pretty close! 16
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2.4.26Suppose the deer populationP(t)in a small forest initially numbers 25and satisfies the logistic equation dP dt= 0.0225P−0.0003P2 (withtin months.) Use Euler’s method with a programmable calcu- lator or computer to approximate the solution for 10 years, first with step sizeh= 1and then withh=.5, rounding off approximateP- values to integrals numbers of deer. What percentage of the limiting population of75deer has been attained after 5 years? After 10 years? Solution- Withh= 1month we get the values: tPRoundedP 02525 1Month25.37525 2Months25.75326 ......... 1year29.66730 ......... 5years49.38949 ......... 10years66.18066 17
Withh=.5month we get the values: tPRoundedP 02525 1Month25.37625 2Months25.75526 ......... 1year29.67530 ......... 5years49.39049 ......... 10years66.23566 The percentage change after 5 years is66%.The percentage change after 10 years is88%. Note thath= 1andh=.5are almost identical. 18
2.4.30Apply Euler’s method with successively smaller step sizes on the interval[0,2]to verify empirically that the solution of the initial value problem dy dx=y2+x2,y(0) = 0 has vertical asymptote nearx= 2.003147. Solution- The estimates for successively smaller step sizes are: x-valuey-values h=.5h=.1h=.01h=.001 .50.030.041.042 1.125.401.344.350 1.5.6331.2131.4791.518 21.9585.84228.393142.627 We see ashgets small when we approachx= 2we getverylarge y-values.This is caused by the asymptote.Note:I had to write a Java program to do this. 19