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# Need an Alibi

## Question

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, prove your innocence or go to jail.

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2 Alibi

1. Chalk dust is the key. When I meet the Professor, I'm clean, no chalk dust anymore on my body, so I'm not the killer.

2. If I'm the killer, I will not call the police. I will go immediately, so nobody knows the killer.

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2 Alibi

1. Chalk dust is the key. When I meet the Professor, I'm clean, no chalk dust anymore on my body, so I'm not the killer.

2. If I'm the killer, I will not call the police. I will go immediately, so nobody knows the killer.

2 very simple choices a killer could do if they wanted to confuse the detectives.

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What part of the body was measured for temperature? Different parts of the body cool at different rates.

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What part of the body was measured for temperature? Different parts of the body cool at different rates.

assume in this case that it doesn't matter

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The body temperature decreases exponentially. This means 84 at 8:00, 92 at 7:00, 100 somewhere before 6:30.

As you do not study from books anymore but online, various logs will show active connection from my room.

Edited by harey
forgot spoiler
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Hidden Content

On the right track but I have a different time.

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On the right track but I have a different time.

I calculated very roughly. Interpolating an exponential is quite dangerous. If the temperature was rounded by 0.2 degrees and measured 2-3 minutes before/after the whole hour, you can get quite a different result.

Proof is left to the reader.

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Consider this a differential equation problem.

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Define the terms
T = temperature (degrees above room temp)
t = time (hours from 10 AM)
c = a constant describing the rate of cooling

If the rate of cooling is proportional to the temperature gradient, then
dT/dt = -cT

The only way (I know of) to have a function's derivative be itself is to use some form of ex. In this case, we can have
T = e-ct
let u = -ct while solving the derivative so
T = eu
dT/dt = dT/du * du/dt = eu * -c
dT/dt = -c * e-ct = -cT

So a solution is
T = e-ct + C

At 10 AM (time t = 0) the body temp is 78 degrees (T = 8 degrees above room temp), so
T = e-ct + C
8 = e0 + C
C = 7

At 9 AM (time t = -1) the body temp is 80 degrees (T = 10), so
T = e-ct + C
10 = ec + 7
c = ln(3)

Then solve for when T = 28.6 (98.6 - 70 degrees)
T = e-ct + C
28.6 = e-t ln(3) + 7
ln(21.6) = -t ln(3)
-t = 2.797 hours (from 10 AM) = 7:12

Where were you at 7:12 AM? I have no idea.

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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.
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lol, forgot the statement:

The difference between body temperature and the room temperature changes at a rate proportional to that difference.

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