Slide 3.4 - 1Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyCopyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyApplications: DecayOBJECTIVES Find a function that satisfies dP/dt = – kP. Convert between decay rate and half-life. Solve applied problems involving exponential decay.3.4Slide 3.4 - 3Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyThe equationshows P to be decreasing as a function of time, and the solutionShows it to be decreasing exponentially. This is exponential decay. The amount present at time t is P0.3.4 Applications: DecaydPdt= −kP, where k > 0,P t()=P0e−ktSlide 3.4 - 4Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley3.4 Applications: DecaySlide 3.4 - 5Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyTHEOREM 10The decay rate k and the half–life T are related by3.4 Applications: DecaykT=ln2≈0.693147,ork =ln2T≈0.693147T,andT =ln2k≈0.693147k.Slide 3.4 - 6Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 2: Plutonium-239, a common product of a functioning nuclear reactor, can be deadly to people exposed to it. Its decay rate is about 0.0028% per year. What is its half-life?3.4 Applications: DecayT =ln2k=ln20.000028≈ 24, 755Slide 3.4 - 7Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 3: The radioactive element carbon-14 has a half-life of 5750 yr. The percentage of carbon-14 present in the remains of plants and animals can be used to determine age. Archaeologists found that the linen wrapping from one of the Dead Sea Scrolls had lost 22.3% of its carbon-14. How old was the linen wrapping?3.4 Applications: DecaySlide 3.4 - 8Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 3 (continued):1stfind the decay rate, k.Then substitute the information from the problem and k into the equation3.4 Applications: Decayk =ln2T=ln25750≈ 0.0001205N=N0e−kt.Slide 3.4 - 9Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 3 (concluded):3.4 Applications: Decay0.777⋅N0=N0e−0.0001205t0.777 = e−0.0001205tln0.777 = lne−0.0001205tln0.777 = −0.0001205tln0.777−0.0001205= t2094 ≈ tSlide 3.4 - 10Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 4: Following the birth of their granddaughter, two grandparents want to make an initial investment of P0that will grow to $10,000 by the child’s 20thbirthday. Interest is compounded continuously at 6%. What should the initial investment be?We will use the equation3.4 Applications: DecayP=P0ekt.Slide 3.4 - 11Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 4 (continued):Thus, the grandparents must deposit $3011.94, which will grow to $10,000 by the child’s 20thbirthday.3.4 Applications: Decay10,000=P0e0.06⋅2010,000 = P0e1.210,000e1.2= P010,000e−1.2= P0$3011.94 = P0Slide 3.4 - 12Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyTHEOREM 11The present value P0of an amount P due t years later, at an interest rate k, compounded continuously, is given by3.4 Applications: DecayP0=Pe−kt.Slide 3.4 - 13Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyNewton’s Law of CoolingThe temperature T of a cooling object drops at a rate that is proportional to the difference T – C, where C is the constant temperature of the surrounding medium. Thus, The function that satisfies the above equation is3.4 Applications: DecaydTdt= −k(T − C).T=T (t )=ae−kt+C.Slide 3.4 - 14Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 6: Found stabbed by a collection of number 2 pencils, Prof. Cal Kulice’s body was slumped over a stack of exams with plenty of red marks on them. A coroner arrives at noon, immediately takes the temperature of the body, and finds it to be 94.6°. She waits 1 hr, takes the temperature again, and finds it to be 93.4°. She also notes that the temperature of the room is 70°. When was the murder committed?3.4 Applications: DecaySlide 3.4 - 15Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 6 (continued): We will assume that the temperature of the body was normal (T = 98.6°) when the murder was committed (t = 0). Thus,This gives3.4 Applications: Decay98.6=ae−k⋅0+7028.6 = aT (t )=28.6e−kt+70.Slide 3.4 - 16Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 6 (continued): To find the number of hours, N, since the murder was committed, we must first find k. From the two temperature readings the coroner made, we haveThen, we can solve for k.3.4 Applications: Decay94.6=28.6e−kN+70⇒24.6=28.6e−kN93.4=28.6e−kN+70⇒23.4=28.6e−k N +1( )Slide 3.4 - 17Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 6 (continued): 3.4 Applications: Decay24.623.4=28.6e−kN28.6e−k (N +1)24.623.4= e−kN +kN +k24.623.4= ekln24.623.4= lnek0.05 ≈ kSlide 3.4 - 18Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-WesleyExample 6 (concluded): Then we can substitute back into either one of our first equations to solve for N.3.4 Applications: Decay24.6=28.6e−0.05 N24.628.6= e−0.05 Nln24.628.6= lne−0.05 Nln24.628.6= −0.05N3 ≈