mathematically analyzing change teacher cadet college day march 11, 2011 dr. trent kull ms. wendy...
TRANSCRIPT
Mathematically analyzing change
Teacher Cadet College DayMarch 11, 2011
Dr. Trent Kull
Ms. Wendy Belcher
Mr. Matthew Neal
Algebra
Slope of a line
run
rise
Calculus
The slope of a curve
rise
run
Maximum
Minimum
Inflection
Calculus
Let’s “walk a few curves”
Maximum
Minimum
Inflection
Calculus
Know the curve,find the slopes
rise
run
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
Unknown temperature function can be found
http://salempress.com
The rate of change in a body’s temperature is proportional to the difference between its current temperature and that of the surroundings.
Newton’s law of cooling
)]([ tTRkdt
dTRate of
temperature change
Cooling constant Constant
room temperature
Unknown temperature function
Time of death, Exercise 1The 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.
Three temperature measurements First: 90 at time 0 Second: 80 at time 2 Room: 70
Determine change
Time of death construction
]70[' TkTdt
dT
How do we determine this cooling constant?
Natural logarithms...
Yikes! Can you do the math?
Finding the cooling constant
7080
7090ln2
1
)(
)(ln
1
1
2
12 RtT
RtT
ttk
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)
We’ll find a function and follow it “back in time”Back to Dr. Kull’s webpageClick on Direction field linkClick on “DFIELD 2005.10”
Time to analyze!
Click OK
We’ll enter information
in this window
We’ll see the cool stuff in this window
Enter “T” Enter “t”
Enter “.3463574(70-T)”
-5
10
65
105
Click when ready
Close to (0,90)
Time = 0
Temp = 90
Temp = 98.66
Time = -1.042
Exercise 1: Solution
Time of death, Exercise 2Just 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.
Temp = 98.66
k = .291823
Time = -3.2227
Exercise 2: Solution
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 = 98.66Time = -62.471
Temp = 86.026Time = 61.882Exercise 3: Solution
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.
)]()([ tTtRkdt
dT
A new differential equation
Suppose k=.221343Investigators record T(0)=58R(t) is periodic
)](50)(*20[ tTtSinkdt
dT
Temp = 58Time = 0
Back in time
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