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Differential equations: estimating the time of death

For those who still haven't found the usefulness/beauty of math (and in particular calculus) I propose a small real-life application inspired by a CSI episode: estimating the time of death.

The problem

Let's suppose that you are a crime investigator and you've found a dead body (it could be anything from a fish to a human being or anything else that comprises of atoms). You are asked to tell the estimated time of death. If you made your calculus and physics homeworks you should already know that Newton's law of cooling should come in rescue:

The rate of heat loss of a body is proportional to the temperature difference between the body and its surroundings (source: Wikipedia).

We can solve this problem if we measure the initial temperature of the body T at the time we found it then if we let it to emit/radiate some heat for a short period of time (t) then we measure again it's final temperature (f) such that we can determine the rate of heat loss by that body. Knowing that it's normal temperature was 37C we determine how long time (t) it took to cool down till (T) degrees Celsius.

Let's call the rate of heat loss of a body with dT \over dt where dT is the change of temperature with respect to time and dt is the change of time.

Let's call the temperature of the dead body T and the temperature of the surrounding T_s. Note that the difference between the body and its surroundings will be T-T_s where Ts is just a constant.

Because each type of body (eq. fish body vs. human body) has a specific heat capacity, also the capacity to absorb and/or to emit the heat from/to surroundings, the rate of heat loss must also depend on that. Let me explain what I mean: each body absorbs/emits heat at different rate because each body is composed of different elements in different ratio. For instance, did you know that the human body contains 65% oxygen (O), 18.5% carbon (C), 9.5% hydrogen (H), 3.2% nitrogen (N), 1% phosphor (P),...., and less the 0.1% are small traces of boron (B), copper (Cu), fluorine(F),iodine(I), iron (Fe),manganese (Mn), tin (Sn) and zinc (Zn)? Because each of these elements have different number of electrons and orbitals they absorb/emit different amount of energy (like heat). When the heat is absorbed the atom's electrons jump to a higher energy level and when they emit the heat back to surroundings they jump back to the their ground energy level (the energy absorbed earlier is released back as a photon, which we can call it light; light is a wide spectrum, it can be visible for our eyes or invisible: for instance heat which is just infrared light; there is also UV-light, microwaves, radio waves, x-rays, gamma-rays; also light is just a electromagnetic wave that transports energy, it oscillates at different rate and has different wave length depending on its 'colour').

Let's name that property (which is a constant) that depends on the body composition with k \in \mathbb{R}.

Summary

{dT \over dt} \propto {k(T-T_s)}

where:

  • dT/dt is the rate of heat loss of a body
  • Ts is the surrounding temperature
  • T is the body temperature
  • k is a proportionality constant that depends on the body

The solution

So, now we can create a mathematical model that describes the Newton's law of cooling:

{dT \over dt}=k\cdot(T-T_s). This is what we call an differential equation, also an equation that contains both a function and its derivative. The solution of a differential equation is one or more functions. In our case the solution will be the function that will give us the temperature of the body at any time. Let's solve this equation!

We multiply both sides with dt\over {T-T_s} so that the equation rewrites like this:

{dt\over {T-T_s}}=k\cdot dt. We are looking for that function T which by deriving we'll get the dT/dt. If such a function exists then it is the primitive of our derivative function dT/dt. So we take the integral from both sides (any operation is allowed as long as we do it on both sides) so that it becomes: \int_i^f {dt\over {T-T_s}}=\int_0^t {k\cdot dt} where i is the temperature of the body when we found it and f is the final body temperature. If we solve this integral we get the following: [ln(T-T_s)]_i^f=[k\cdot t]_0^t <=> ln{{T_f-T_s}\over{T_i-T_s}}=k\cdot t

We rise both sides to the base e (Euler's constant) so that the equation rewrites like:

{{T_f-T_s}\over{T_i-T_s}}={e^{k\cdot t}} <=> T_f=T_s+(T_i-T_s)\cdot e^{k\cdot t} where k>0 if temperature increases or k<0 if temperature decreases. Well, supposing that the environment temperature (Ts) is below the human body temperature then we can assume that k<0 such that we can rewrite our solution like this:
T(t)=T_s+(T_i-T_s)\cdot e^{-k\cdot t} with k>0

Let's take an example

Suppose we found a dead body at midnight and we measure its temperature right away. The thermometer shows that the body has Ti=26C. The surrounding temperature is Ts=15C. When did the (Warrick Brown) body died? Well, the first thing we have to do is to wait for a while (like t-hours) and then measure again the temperature T(t). Let's suppose that we decide to wait t=2 hours in order to let the body emitting some heat such that it lowers its temperature till T(2)=24C.

Now that we have our function we can just simply plug in all the data and get the k-variable, which works/corresponds for/to that body only:

24C=15C+(26C-15C)\cdot e^{-k\cdot 2} <=> 9=11\cdot e^{-2k} <=> {9\over 11}={e^{-2k}}

We take natural logarithm of both sides and we get ln{9\over 11}=-2k => k=-{ln{9 \over 11}\over 2}

Let's rewrite the function for the body in our problem using the newly found proportionality constant k:

T(t)=T_s+(T_i-T_s)\cdot e^{{ln{9 \over 11}\over 2}\cdot t}

So when did the body died anyway? This question can be rephrased like this: what was the time t when the body has the temperature T(t)=37C ? Let's see:

37C=15C+(26C-15C)\cdot e^{{ln{9 \over 11}\over 2}\cdot t}. We take natural logarithm of both sides and get: ln{22\over 11}={{ln{9\over 11}}\over 2}\cdot t => t={{2\cdot ln{2}}\over{ln{9\over 11}}}\approx -6.9 hours

So the body died about 6.9 hours before midnight, that means that the time was 14:04 when the body died. We can state that we've just became a certified CSI-calculus investigator 🙂

Final note: this method can be even used to track the time when someone left a footprint on the ground (an animal in the forest, someone in your house, etc) or at what time someone put the fire off on a camp. The range of application is virtually unlimited.

Now, if you think that this article was interesting don't forget to rate it. It shows me that you care and thus I will continue write about these things.

 
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Differential equations: estimating the time of death

Eugen Mihailescu

Founder/programmer/one-man-show at Cubique Software
Always looking to learn more about *nix world, about the fundamental concepts of math, physics, electronics. I am also passionate about programming, database and systems administration. 16+ yrs experience in software development, designing enterprise systems, IT support and troubleshooting.
Differential equations: estimating the time of death

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