Transcript Solving Inverse Problems: A Mathematical Exploration

Mathematically
analyzing change
Teacher Cadet College Day
March 11, 2011
Dr. Trent Kull
Ms. Wendy Belcher
Mr. Matthew Neal
Algebra
rise
run
Slope of a line
Calculus
rise
run
Maximum
Inflection
Minimum
The slope of a curve
Calculus
Maximum
Inflection
Minimum
Let’s “walk a few
curves”
Calculus
rise
run
Know the curve,
find the slopes
Differential
Equations
Know the slopes,
find the curve
Rates of change
“Slopes” are rates of change
“Curves” are functions
Knowing how an unknown function
changes can help us determine the
function
Example: Crime scene
Body temperature
changing at known
rates
http://salempress.com
Unknown temperature function
can be found
Newton’s law of cooling
The rate of change in a body’s
temperature is proportional to the
difference between its current
temperature and that of the
surroundings.
Rate of
temperature
change
dT
 k[ R  T (t )]
dt
Cooling
constant
Constant room
temperature
Unknown
temperature
function
Time of death, Exercise 1
The body of an apparent homicide
victim is found in a room that is
kept at a constant temperature of
70 ̊ F. At time zero (0) the
temperature of the body is 90 ̊ and
at time two (2) it is 80 ̊. Estimate
the time of death.
Time of death construction
Three temperature measurements
 First: 90 at time 0
 Second: 80 at time 2
 Room: 70
Determine change
dT
 T '  k [70  T ]
dt
How do we
determine this
cooling constant?
Finding the cooling constant
Natural logarithms...
 T (t2 )  R  1  90  70
1
k
ln 
  ln 
t2  t1  T (t1 )  R  2  80  70
Yikes! Can you do the math?
Let’s use those
computers!
Log on as “visitor”
Password is “winthrop”
Go to Dr. Kull’s webpage
http://faculty.winthrop.edu/kullt/
Open Mathematica file: “Cooling
Constant”
The formula
Enter all values
from the
investigation
The cooling constant (k)
Time to analyze!
We’ll find a function and follow it
“back in time”
Back to Dr. Kull’s webpage
Click on Direction field link
Click on “DFIELD 2005.10”
We’ll see the
cool stuff in
this window
We’ll enter
information in
this window
Click
OK
Enter “.3463574(70-T)”
Enter “T”
Enter “t”
-5
10
65
105
Click when ready
Temp = 90
Close to
(0,90)
Time = 0
Exercise 1: Solution
Temp = 98.66
Time = -1.042
Time of death, Exercise 2
Just before midday, the body of an
apparent homicide victim is found
in a room that is kept at a constant
temperature of 68 ̊ F. At 12 noon
the temperature of the body is 80 ̊
and at 3p it is 73 ̊. Estimate the
time of death.
k = .291823
Temp = 98.66
Exercise 2: Solution
Time = -3.2227
Estimating time, Exercise 3
You are on a search and rescue team in
the mountains of Colorado. Your crew
has found a hypothermic avalanche
victim whose initial temperature
reading is 92 ̊ F. 10 minutes later, the
skier’s temperature is 91 ̊ F. Assuming
the surrounding medium is 28 ̊ F,
estimate the time of the avalanche to
assist rescue & medical crews. Under
current conditions, when will the
skier’s temperature drop to 86 ̊ F?
k = .00157484
Temp = 98.66
Time = -62.471
Exercise 3: Solution
Temp = 86.026
Time = 61.882
What if the “room” is not a
constant temperature?
Newton’s law of cooling is
modified.
If proportionality constant is
known, we need a single data
point.
dT
 k[ R (t )  T (t )]
dt
A new differential equation
Suppose k=.221343
Investigators record T(0)=58
R(t) is periodic
dT
 k[20 * Sin (t )  50   T (t )]
dt
Back in time
Temp = 58
Time = 0
Where there is change…
Temperature,
Motion,
Population, etc.
…there are differential equations.
Where there are differential
equations…
…visual solutions may provide
tremendous insight.
Thanks to…
• Our visiting students
• Ms. Belcher and Mr. Neal
• John Polking for the educational use of
DFIELD 2005.10