Solution.(a) The differential equation for this model isT0(t) =-k(T(t)-18), and the general solution isT(t) = 18 +Ce-kt. We need bothCandk. We know the initial temperatureT(0) = 98, soT(0) = 18 +Ce0= 98 =⇒C= 80.To getk, we use the fact thatT(5) = 38:38 = 18 + 80e-k·5=⇒e-5k=14=⇒ -5k= ln(1/4) =-ln(4) =⇒k=ln(4)5.So putting it all together, we getT(t) = 18 + 80e-ln(4)5t.(b) We want to know whenT(t) = 20, so solve:20 = 18 + 80e-ln(4)5t=⇒2 = 80e-ln(4)5t=⇒e-ln(4)5t=140=⇒-ln(4)5t= ln(1/40) =-ln(40)=⇒t=5 ln(40)ln(4).So the answer is5 ln(40)ln(4)minutes.Example 6.3.Suppose you have a cup of coffee with cooling constantk=.09min-1. It is placed in a roomof temperature20◦C.(a) At what rate is the temperature changing when the temperature of the coffee is80◦C.(b) If the coffee is served at90◦C, how long will it take to reach30◦C.(c) At what time will the temperature reach15◦.Solution.(a) The differential equation isT0(t) =-k(T(t)-T0) =-.09(T(t)-20). So ifT(t) = 80, thenT0(t) =-.09(80-20) =-.09·60◦C/min.11
(b) The general solution isT(t) = 20 +Ce-kt= 20 +Ce-.09t. First, sinceT(0) = 90, we can solve forC:90 = 20 +Ce-.09·0=⇒90 = 20 +C=⇒C= 70,soT(t) = 20 + 70e-.09t. We want to know whenT(t) = 30, so30 = 20 + 70e-.09t=⇒17=e-.09t=⇒ -.09t= ln(1/7) =-ln(7) =⇒t=ln(7).09minutes.(c) In theory, you would setT(t) = 15, so15 = 20 + 70e-.09t=⇒ -570=e-.09t.However, you cannot take logs because you cannot have the log of a negative number. That tells usthere is notwhich works. We should have seen this immediately because the temperature should nevergo below the temperature of the ambient space, which in this case is 20.Example 6.4.Suppose that a skydiver jumps out of an airplane. If the terminal velocity is-98m/s, findthe velocity of the skydiver after15seconds. Assumek= 8kg/s.Solution.Here,v(t) =-mgk+Ce-(k/m)t. We don’t knowmyet. However, terminal velocity is-mg/k, so-98 =-m·9.88=⇒m=98·89.8= 80.So the skydiver’s mass is 80 kg, andv(t) =-80g8+Ce-(8/80)t. We also needC, but we knowv(0) = 0, so-80g8+Ce0= 0 =⇒C= 10g= 10(9.8) = 98.Hencev(t) =-10g+ 98e-t/10=-98 + 98e-t/10. After 15 seconds, the velocity isv(15) =-98 + 98e-15/10m/s.Example 6.5.What is the minimum depositP0necessary that will allow an annuity to pay out1000dollars/year indefinitely if it earns interest at a rate of4%.Solution.We knowP(t) =Nr+Cert=1000.04+Ce.04t. The annuity will pay out indefinitely ifC≥0. To seewhy, notice that ifC <0, thenP(t) is decreasing, and so will eventually hit 0, which is not what we want.However, ifC≥0, then at worstP(t) is constant, which is ok. If the initial amount isP0, thenP0=1000.04+C=⇒C=P0-1000.04.Since we wantC≥0, this meansP0≥1000.04dollars.Example 6.6.A skydiver jumps out of an airplane with zero initial velocity. Before the parachute opens,the skydiver’s velocity satisfiesv0(t) =-g-2v(t), and after the parachute opens, the skydiver’s velocitysatisfiesv0(t) =-g-v. If the parachute opens 10 seconds after the initial jump, find the velocity 20 secondsafter the initial jump.