Differential
Equations
Andrew Eben, Jason Rypkema, Michelle Johnson
Due: May 7,2012
Project 2 - Murder at the Phillips
Who likes playing detective? That's right everyone does! Our mission was to use differential
equations and newfound knowledge of Laplace transforms to figure out who murdered AI G. R.
Mortis. Using the information
that Detective I. C. Yew and her partner gathered we were able
to deduce with our highly intelligent
mathematical
brains who the most logical suspect is. Using
math we can prove who killed AI G. R. Mortis! That's pretty neat! Sherlock would be very proud.
Let's see how we figure out who the fiend is...
1.
Problem 1 is asking us to solve equation 1 and use the solution to estimate his time of
death.
Equation 1 ::!! = k(T-T m)
dt
1
--dT=
T-Tm
Kdt
f -l-dT
=f
T-Tm
Kdt
In/T-Tm/ =kt+c
T - Tm =
ekt+c
OJ
(where
C1
=e
C
)
(by plugging in initial conditions)
85 = C1 + 50
85 - 50:;
Then plug
C1
Cl
back into the equation
T(.5) =
35e-·5k
84 =
34
+ r;
35e-·5k
+ 50
= 35e-·5k
.971 = e-·5k
In10.9711 = Inle-·5kl
Now we can plug the k in
98.6
= 35eo.059t + 50
= 35eo.059t
48.6
In 11.3891
0.328
t
= In 1eO.059t1
= O.059t
= 5.564
This value allows us to make an estimate of the time of death to be 12:26 A.M. in this
scenario.
2.
Problem 2 is asking us to solve equation 3 using Laplace transforms using the value of k
found in problem 1.
dT
dt
= K(T
- Tm(t))
Then we took the laplace of the left side
2 [{T'} = kT - kTm(t) ]
~
""\
J
[8t(8) - T(O) ] = kT - kTm@)~"~ ~
-.~~
~~
~
't.:'"'
1
\0-
01
...,~{
[8t(8) - 85] = kT - kTm(t)
And then we took the Laplace of the right side
[8t(8) - 85] =2{ kT - kTm(t)~
[8t(8) - 85] =2 { kT } - 2 {kT m(t)}
[8t(8) - 85] =2{ k(ektc1 + Tm)} - 2{kT m(t)} (where T = ektc1 + Tm)
[st(s) - 85 J =2 { kektc1 + kTm} - 2 {kT m(t)}
[st(8) - 85] =2 { kektc1} + 2 { kTm} - 2 {kT m(t)}
-[8t(8) - 85] =2 { kektc1} + 2 { kTm} - 2 {k(50 + 20 U(t - h)}
[8t(8) - 85
J =2
{ kektc1} + 2 { kTm} - (2 {k50} + 2 {20k U(t - h)})
[st(s) - 851
[st(s) - 85 ]
=
35k
s-k
= ~s=k +
+
50k _
~
s
S
50k -
~
s
(e-hS 2'120kt} + 2120kh})
-
s
+ 20kh))
(e-hS $(20kt
Now it's time for the magic algebra to find t(s)
[st(s) - 85 J
= ~s-k + 50ks _
- 35k
[St()S - 85] --
+
s=k
50k
50k _ (e-hS
s
35k
[St()S - 85 ] ----e
-hs
=hs 20k
20k
--e
S2
--e
S2
s-k
35k
st( s) ----e
s+k:
+ns
20k
--e
S2
35k
t( S) -----e
-hs
20k
--e
S3
s(s-k)
20kh))
S2
50k
--e
s
-s
(~+
=hs
S
-hs 20kh
-S
-hs 20kh
-s
20kh +85
s
=hs 20kh +85
-S2
s
We then take the inverse Laplace to solve for T(t)
rO-1
<L
{t( S) -----e
35k
-hs
20k
--e
S3
s(s-k)
2-1 {t(s)} =Il!-1{~}
- gr1 {e-hs
s(s-k)
+85}
-S2
s
-hs 20kh
20k} _ gr1{e-hS
s3
~
}+2-1{85}
S2
s
T(t) = -35 + 35ekt - 1Ok(t - h)2U(t - h) - 20kh(t -h)U(t - h) +85
T(t) = 35ekt - 1Ok(t - h)2U(t - h) - 20kh(t -h)U(t - h) +50
Then if we plug k in we get the solution we were looking for:
T(t)
= 35eo.059t-
@fXt)2U(t
- h) - 1.18h(t -h)U(t - h) +50
3. The project asks us to complete Yew's table for h values 12 thro
plugged in values 12 through 2 into the variable h in our final equation (where
T(t) equal to the normal body temperature
es "
of 98.6) from problem 2. We then solved for
t which gave us the time of death. For example, if our t equals 5.56 we subtracted 5.56
from 6 AM, and then found that the time of death had to have been around 12:26 AM.
We find that our h values 12 through 6 give the same time of death. This is because the
body would have to be moved before AI died. So in other words, AI would've had to
have been killed in the refrigerator
values 5 through
so the body wouldn't
2 produced different times
that AI could have been in the refrigerator
have been moved. The h
of d~atn b~e~us~in pfr>'o\~m1 W~1mmO
at most 5.56 hours. The values
5
through
Z
give us other possibilities of time of death. Below is Yew's completed table:
h
12
11
10
Time body moved
6:00pm
7:00pm
8:00pm
Time of death
9
9:00pm
12:26am
8
7
6
5
4
10:00pm
12:26am
11:00pm
12:00am
12:26am
12:26am
1:00am
2:00am
11:52pm
11:0Spm
3
3:00am
10:12pm
2
4:00am
9:13pm
12:26am
12:26am
12:26am
4. Yew wants to question Shorty because he was the one that was seen arguing with AI
closest to his time of death. We say this because when taking h equals 2 we find that t
or time of death is closest to when Shorty was last seen with AI and every h greater than
2 was even farther away from when the last suspect was seen with him. Shorty then
went on to move AI into the refrigerator
before he left at two. This means Shorty had
motive and time to kill AI as well as hide the body in the refrigerator.
5. Problem 5 is asking us to compute the equation of the tangent line to the solution
through the point (O/To)using equation 1 with the initial condition ofT(O)
= To·
We plug equation 1 in for the m in the equation of the line: y = mx + b.
This looks like y = (k(T-Tm)x) +b
We know that To
=b
we get this from plugging in the initial condition.
Now our equation looks like y
= (k(T-Tm)x) +To
We then plug in To and T to get: T
= (k(To -
Tm))x + To
Then using algebra we can solve for t: t = (
T-To
K TO-Tm
)
·
.
I
.
t
W e can t h en p Iug
In our given va ue to arnve at: =
98.4-To
K(To-Tm
)
Conclusion:
Well there it is! Shorty was the murderous fiend, all over how to present a plate of veal
scaloppini. Maybe Shorty was just having a bad night and AI questioning his authority of
presenting veal was just too much for him to take, either waYI Shorty is the murderer of
AI G. R. Mortis according to our calculations. His feeble attempt at hiding the body in the
refrigerator
was defenseless against our superior knowledge of laplace transforms and
differential
equations.