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I have recently seen a quote about determining how long a body has been dead:

“Dead bodies lose heat exponentially, and therefore e can be used in an appropriate equation to determine how long individuals have been dead” (Calvin Clawson, Mathematical Mysteries, 1996).

Does anyone have an idea about such equation?

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    $\begingroup$ I guess $e^{-t}$. $\endgroup$
    – Alex
    Mar 2, 2015 at 13:43

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This is an example of Newton's law of cooling, which says basically that in an environment, the temperatures of all bodies will average out. The law is given by a differential equation whose solution leads to an exponential function.

These two YouTube videos might help you out:

https://www.youtube.com/watch?v=fKHFbOeJrD0

https://www.youtube.com/watch?v=JGFDwkawB4U

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I don't know the specifics, but the equation will be something of the form $$ H=Ae^{-kt} $$ where $H$ is heat, $t$ is time since death and $A,k$ are (known) constants. Then, if you measure the heat $H$, you can find the time $t$ since death by: \begin{align} H&=Ae^{-kt}\\ \frac{H}{A}&=e^{-kt}\\ -kt&=\log\left(\frac{H}{A}\right)\\ t&=\frac{-1}{k}\log\left(\frac{H}{A}\right)\\ \end{align} I disgree that this method really involves the number $e$, though. Indeed, we could replace $e$ with any exponent $a$ and get the same thing, since: $$ a^{-kx}=\left(e^{\log(a)}\right)^{-kx}=e^{-k\log(a)x}=e^{-\hat{k}x} $$ where $\hat{k}=k\log(a)$. In other words, passing from $e$ to another exponent just changes the constant $k$.

$e$ is certainly the most natural exponent, though, for reasons that will become clear to you in the course of your mathematical education.

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