A Bit of Pre-Forensic Science

I've been sharing fairly serious math, so I'll tone it down in this blog entry by going over an intriguing problem my differential equations students had to tackle on Friday.

The Problem

The FBI hired you as a forensics scientist. Not long after, at midnight, a body with a temperature of 31°C was found in the woods. One hour after that, the temperature of the body had dropped to 29°C. It was noted that the ambient temperature remained at 21°C.

If the temperature of the body is modeled by Newton's law of cooling and if the normal temperature of a live body is 36.7°C, then when was the person killed?

The Solution

Let $x\left(t\right)$ be the temperature (in °C) of the body as a function of time (in hours). Then

$$x'\left(t\right)=k\left(x\left(t\right)-21\right),$$

where $k$ is a constant of proportionality. Let's solve this differential equation:

$$\begin{align*}x'\left(t\right)=kx\left(t\right)-21k&\implies\left[e^{-kt}x\left(t\right)\right]'=-21ke^{-kt}\\&\implies e^{-kt}x\left(t\right)=21e^{-kt}+C\\&\implies x\left(t\right)=21+Ce^{kt}.\end{align*}$$

Let $t=0$ represent midnight. Then $31=x\left(0\right)=21+C$, which implies $C=10$. Moreover, $29=x\left(1\right)=21+10e^{k}$, which implies $e^{k}=4/5$. Therefore,

$$x\left(t\right)=21+10\left(\frac{4}{5}\right)^t.$$

Setting this equal to $36.7$ and solving for $t$ yields $t\approx-2.02$, which implies the person was killed at around 10:00 PM.

Remarks
  • I skipped a gazillion steps.
  • In the real world:
    • the temperature of a dead body may not be modeled by Newton's law of cooling (what if the victim wore several layers of clothing?),
    • the ambient temperature may not be constant (think Texas), and
    • not everyone's normal body temperature is 36.7°C (what if the person ran a lot before dying?).
  • Nevertheless, educated guesses like these prove useful more often than not.

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