Murdered ProfessorDate: 06/26/97 at 14:22:58 From: Brett Wilson Subject: Exponential puzzle Dr. Math, Please help me with this puzzle given to me by my calculus professor: Suppose that you come into your professor's office to ask some questions shortly before 9:00 a.m. on Friday. You find him lying on the floor of his office in a pool of chalk dust, dead. You quickly call the police and their investigators take several measurements over the next hour, including: 1) the body temperature at 9:00 a.m. - 80 degrees 2) the body temperature at 10:00 a.m. - 78 degrees 3) room temperature - 70 degrees (constant) You quickly realize that the police believe you to be a prime suspect, so you need an alibi. You know that you were studying until midnight, but you aren't sure if that is enough information. You need to know the time of death! Determine the time of death by creating an exponential model. Use the following statement: the difference between body temperature and room temperature changes at a rate proportional to that difference. How good is your alibi? The only thing I have figured to do is use the exponential formula y = Ce^kt, but I'm not sure which numbers to plug in and what information that will give me. Please help! Thank you. Date: 06/26/97 at 18:06:55 From: Doctor Toby Subject: Re: Exponential puzzle The t in that exponential formula is the time since the time of death. That's hard to work with when you don't know the time of death. So use the formula y = Ce^k(t-T), where T is the time of death. In this formula, t is the actual time, which you can work with. There are three unknown constants in this formula, C, k, and T. You have three data points, y = 28.6 degress when t = T, y = 10 degrees when t = 9:00, and y = 8 degrees when t = 10:00. Now you have enough to solve for C, k, and, most importantly, T. Note that y is the difference between the temperature and the base temperature of 70 degrees, not the temperature itself. If you like, you can use the equation y-Y = Ce^k(t-T), where now y is the actual temperature and Y is the base temperature, and add the fourth datum that y = 70 degrees when t = infinity. Then you immediately get Y = 70 degrees. Anyway, if you use the datum that y = 28.6 degrees when t = T, you learn that C = 28.6 degrees right away. So the hard part is solving the two remaining equations for the two remaining constants, k and T. (But that's not really too hard. The really hard part will be finding another alibi!) -Doctor Toby, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 06/26/97 at 18:12:28 From: Doctor Anthony Subject: Re: exponential puzzle You should start with the differential equation that is actually quoted in the question. The equation, using T for temperature and t for time is: dT/dt = -k(T-70) (k is the constant of proportionality) dT/(T-70) = -k*dt Integrating both sides: ln(T-70) = -kt + const. T-70 = Ae^(-kt) where A = e^(const.) = constant At t = 9, T = 80 and at t = 10, T = 78. So, 10 = Ae^(-9k) and 8 = Ae^(-10k) Dividing these we have, 10/8 = e^(-9k+10k) = e^(k) This gives k = ln(5/4) = 0.223 To find A, use 10 = Ae^(-9 x .223) = 0.1342A So A = 10/.1342 = 74.5 The full equation is then: T = 70 + 74.5e^(-.223t) Normal body temperature is 98.6, so now we require the value of t for when T = 98.6: 98.6 - 70 = 74.5e^(-.223t) 28.6/74.5 = e^(-.223t) .3839 = e^(-.223t) ln(.3839) = -.223t -.9574 = -.223t t = .9574/.223 t = 4.293 This corresponds to 4:17 a.m., so you need to think of someone who can prove what you were doing at 4:17 a.m. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 07/04/97 at 22:52:21 From: Doctor Marko Subject: Re: exponential puzzle Dear Brett, You are on a good track. The "proper" way to do this is to set up a differential equation from the statement: The difference between body temperature and the room temperature changes at a rate proportional to that difference. But the basic equation (y = Ce^kt) is right. From this statement you can infer that the only two important quantities are time (t) and the difference between the two temperatures, body-room which we can call T. Additionally, the statement tells you that the rate of change (also known as the time derivative) of T is proportional to T itself. Thus, you would write dT/dt = -k*T, where dT/dt stands for the time derivative and k is the constant of proportionality. The reason for the minus sign is that T is always positive but decreasing, and thus, T > 0, but dT/dt < 0. Now you can solve this differential equation in many ways (I hope you have studied this in class) and the standard way is the separation of variables. So you would write: dT/T = -k*dt You can now integrate both sides and get: Log(T) = -k*t + c1 (c1 is a constant from the indefinite integral) The next step is exponentiating both sides of the equation: e^(Log(T)) = e^(-k*t+c1) By exponential and log rules, e^(Log(T)) = T, and e^(-k*t+c1) = e^(-k*t)*e^c1. Now e^c1 is just another constant, call it C, so the final equation becomes: T = C*e^(-k*t) Now you want to plug your information: Room temp = 70 At 9 a.m., body temp = 80 => t = 9, T = 10 => 10 = C*e^(-k*9) At 10 a.m., body temp = 78 => t = 10, T = 8 => 8 = C*e^(-k*10) So dividing one equation by the other you get: 8/10 = e^(-10k)/e^(-9k) or .8 = e^(-k), or Log(.8) = -k. Then you pull out your trusty calculator and -0.2231 = -k. We can go back and figure out C by: 10 = C*e^(-0.2231*9) => 10 = C*0.1343 => C = 74.4766. Then approximately, T = 74.48*e^(-0.223*t). The final question: When did the professor die? Well, normal body temperature is 98, so when professor died T = 28. Plug it in: 28 = 74.48*e^(-0.223*t) 0.3759 = e^(-0.223*t) -0.9783 = -0.223*t t = 4.39 The professor died at bit before 4:30am. So, if the student can prove that s/he was in bed at that time, everything will be fine. I hope this cleared things a bit. I know I was quite verbose, but this is an important scheme of solving problems, and chances are that you will have many such puzzles (possibly on an exam as well!) so learn it well. Good luck, -Doctor Marko, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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