Section 2.3- Modeling1. Problem 3: In this model of salt in a tank of water, let Q(t) be the amount of salt (inpounds) at time t (measured in minutes). Then dQ/dt will be measured in pounds perminute.The rate in, at least initially:12·lbsgal· 2 ·galmin= 1 lbs/minThe rate out, at least initially:2galmin·Q(t) lbs100 gal=150Q lbs/minThe model equation, valid for 0 ≤ t < 10:dQdt= 1 −150Q Q(0) = 0The initial condition is zero, since we started with fresh water. You can solve thiseither as a linear equation, or a separable equation. We give the solution here as if itwere linear:Q0+150Q = 1 ⇒ e(1/50)tQ0+150Q= e(1/50)t⇒ Q(t) = 50 + Ce−(1/50)tWith the initial condition,0 = 50 + C ⇒ C = −50The amount of salt in the tank at time 0 ≤ t < 10 minutes is:Q(t) = 50 − 50e−(1/50)tAt time t = 10, the dynamics change. Suddenly no salt comes in (but water still does).Now we have:Q0= −150Q valid for t ≥ 10Note: We could have left this as t > 10, but it was convenient to do it this way- That’sbecause the initial condition for this DE is where we ended with the last DE,Q(10) = 50 − 50e−1/5≈ 9.06343For t > 10, the solution is: Q(t) = Ae−(1/50)t. Putting in Q(10) = 9.06343,9.06343 = Ae−1/5⇒ A ≈ 11.07Therefore, for t > 10,Q(t) = 11.07e−(1/50)tAnd now substitute t = 20 to find the final amount of salt in the tank: Q(20) ≈ 7.421pounds.2. Problem 5: As usual, let Q(t) be the amount of salt (in ounces) at time t (measuredin minutes). Then dQ/dt will be measured in ounces per minute.The rate in:141 +12sin(t)·ozgal· 2 ·galmin=121 +12sin(t)ozminThe rate out:2galmin·Q(t) lbs100 gal=150Q lbs/minThe model equation:dQdt=121 +12sin(t)−150Q Q(0) = 50The initial condition was that we started with 50 ozs of salt in a tank of 100 gallons.This is a linear differential equation, and the integrating factor is the same as computedearlier in Problem 3:e(1/50)tQ0+150Q=12e(1/50)t+14e(1/50)tsin(t)Note that we need to integrate by parts twice, so that:Ze(1/50)tsin(t) dt = −25002501e(−1/50)tcos(t) −502501e(−1/50)tsin(t)Now we write the full solution:Q(t) = 25 −6252501cos(t) +255002sin(t) +631502501e−(1/50)tSORRY about those constants! Ugly as this might be, it is now easy to see what kindof behavior we can expect as t → ∞- The last term drops out, and the salt variesperiodically about the constant 25 (ozs.).We can verify this in Maple:DE:=diff(Q(t),t)+(1/50)*Q(t)=(1/2)+(1/4)*sin(t);Q1:=dsolve({DE,Q(0)=50},Q(t));plot(rhs(Q1),t=0..400,numpoints=1000);The result is shown in Figure 1 (Next page).3. Problem 9: In the absence of payments, the rate of change of our loan will increaseproportionally to the current amount. That is, let S(t) be the amount of money(measured in dollars) owed at time t (measured in years). With no payments,dSdt= rSFigure 1: The graph of the amount of salt in the tank at time t for Exercise 5.where r is the annual interest rate (annual because we are measuring t in years). Bymaking ”continuous payments”, at a constant annual rate k:dSdt= rS − kWe can solve this generally, with S(0) = S0:S(t) =kr+ S0−kr!ertPutting in the values, S(0) = 8000, r = 1/10 and leaving k as an adjustable parameter,S(t) = 10k − (8000 − 10k) e(1/10)tWe want to find the value of k so that our loan is paid off in three years, or S(3) = 0:0 = 10k − (8000 − 10k)e3/10k ≈ 3086.64(Side remark: That’s about $8.42 per day) So over the three year period, we wouldpay:3 · 3086.64 = 9259.92so the interest paid was about $1259.92.4. Problem 12: We are given that:Q0= −rQ ⇒ Q(t) = Q0e−rtAnd we are told that the half-life of Carbon-14 is 5730 years. That means that, if Q0is the initial amount, then:12Q0= Q0e−r·(57 30)Divide both sides by Q0, and solve for r:r =ln(1/2)−5730=ln(2)5730≈ 0.00012097 = 1.2097 × 10−4The general solution is:Q(t) = Q0e(−1.2907×10−4)tWe now think of Q0as some unknown (but fixed) amount, and let T be the time ittakes to decrease Q0to 20% of the original amount. Then solve for T :15Q0= Q0e(−1.2907×10−4)TThis gives T ≈ 13, 304.65 years.5. Problem 13: Let us parse out the problem:• The p opulation of mosquitoes increases at a rate proportional to the currentpopulation...If P (t) is the population at time t, so far this saysdPdt= kP ⇒ P (t) = P0ekt• ... and in the absence of other factors, the population doubles each week. If wemeasure t in days, this means that:2P0= P0e7k⇒ k =ln(2)7≈ 0.9902 per dayNow, going back to our model: So far, without predation, the rate of changepopulation at time t (in days) is:dPdt=ln(2)7P• There are 200,000 mosquitoes initially (Modeled as P (0) = 200, 000), and preda-tors eat 20,000 per day- This is a constant decrease:dPdt=ln(2)7P − 20, 000, P (0) = 200, 000Solve this the usual way (either as a linear or separable equation),P (t) =20, 000 · 7ln(2)+ 200, 000 −20, 000 · 7ln(2)!e−0.9902t• If t is measured in weeks, then things simplify a bit. In that case, k = ln(2) and:dPdt= ln(2)P − 140, 000 P (0) = 200, 000and the solution to the IVP is:P (t) =140, 000ln(2)+ 200, 000 −140, 000ln(2)!2−tIn numerical form,P (t) = 201, 977.31 − 19, 77.31 · 2−tSolving for when P (t) = 0, we s ee that the solution is valid for 0 ≤ t ≤ 6.6745(weeks).6. Problem 23:A Physics Note: If something is measured in pounds, it has the same units as masstimes gravity, mg. Gravity in this problem will be measured as 32 feet per secondsquared. Given that the weight is 180 pounds, and gravity is 32, we can then computethe mass: 180/32 = 5.625.Going back to our model from Section 1.1:ma = mdvdt= mg − kv ⇒dvdt= g −kmvBefore the parachute opens, 0 ≤ t ≤ 10, we have:dvdt= −3418032v + 32 = −215v + 32, v(0) = 0Solving for the velocity equation,v(t) = 240 − 240e−(2/15)tThe speed when the parachute opens (t = 10) is v(10) ≈ 176.74 feet per second.We can now integrate velocity to find position, s(t). Careful here! A quick analysis ofour velocity equation says that the velocity towards the ground is positive. But, if wesay that s(0) = 5000, our height will be increasing (since v = s0> 0). To compensate,we set s(0) = −5000:s(t) = −6800 + 240t + 1800e−(2/15)tSo the position at t = 10 is s(10) ≈ −3925.53, which we interpret as 3925.53 feet abovethe ground, so the skydiver has fallen (approximately) 5000 − 3925.53 = 1074.47 feetTo answer the last two questions, we reformulate the velocity equation. To simplifythings, we’ll reset the clock