Project Two. James K. Peterson. March 26, Department of Biological Sciences and Department of Mathematical Sciences Clemson University
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1 Project Two James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University March 26, 2019
2 Outline 1 Cooling Models 2 Estimating the Cooling Rate k 3 Typical Cooling Project Matlab Session 4 Project Two
3 Cooling Models Newton formulated a law of cooling by observing how the temperature of a hot object cooled. As you might expect, this is called Newton s Law Of Cooling. If we let T (t) represent the temperature of the liquid in some container and A denote the ambient temperature of the air around the container, then Newton observed that T (t) (T (t) A).
4 Cooling Models Newton formulated a law of cooling by observing how the temperature of a hot object cooled. As you might expect, this is called Newton s Law Of Cooling. If we let T (t) represent the temperature of the liquid in some container and A denote the ambient temperature of the air around the container, then Newton observed that T (t) (T (t) A). We will assume that the temperature outside the container, A, is smaller than the initial temperature of the hot liquid inside. So we expect the temperature of the liquid to go down with time. Let the constant of proportionality be k. Next, we solve T (t) = k (T (t) A), T (0) = T 0 where T 0 is the initial temperature of the liquid.
5 Cooling Models For example, we might want to solve T (t) = k(t (t) 70), T (0) = 210 where all of our temperatures are measured in degrees Fahrenheit.
6 Cooling Models For example, we might want to solve T (t) = k(t (t) 70), T (0) = 210 where all of our temperatures are measured in degrees Fahrenheit. We use our usual solution methods to solve this. Divide both sides dt by T 70 to get T 70 = k dt. Then integrate both sides to get ln T (t) 70 = kt + C. Then, exponentiate: T (t) 70 = Be kt. Now, since we start at 210 degrees and the ambient temperature is 70 degrees, we know the temperature of our liquid will go down with time. Thus, we have T (t) 70 = Be kt.
7 Cooling Models Solving, we find T (t) = 70 + Be kt.
8 Cooling Models Solving, we find T (t) = 70 + Be kt. Since the initial condition is T (0) = 210, we find T (0) = 210 = 70 + Be 0 = 70 + B, and so B = 140. Putting all of this together, we see the solution to the model is T (t) = e kt.
9 Cooling Models Solving, we find T (t) = 70 + Be kt. Since the initial condition is T (0) = 210, we find T (0) = 210 = 70 + Be 0 = 70 + B, and so B = 140. Putting all of this together, we see the solution to the model is T (t) = e kt. Now since the temperature of our liquid is going down, it is apparent that the proportionality constant k must be negative. In fact, our common sense tells us that as time increases, the temperature of the liquid approaches the ambient temperature 70 asymptotically from above.
10 Cooling Models Example Solve T (t) = k(t (t) 70) with T (0) = 210 and then use the conditions T (10) = 140 to find k. Here time is measured in minutes. Solution First, we solve as usual to find T (t) = e kt. Next, we know T (10) = 140, so we must have T (10) = 140 = e 10k. Thus 70 = 140e 10k 1 2 = e10k ln(2) = 10k. and so k = ln(2)/10 =.0693.
11 Estimating the Cooling Rate k We now know the general cooling model is T (t) = k(t (t) A), T (0) = T 0. which has solution T (t) = A + Be kt. Applying the initial condition, we get T 0 = A + B. or B = T 0 A. Hence, the solution is T (t) = A + (T 0 A)e kt.
12 Estimating the Cooling Rate k We now know the general cooling model is which has solution T (t) = k(t (t) A), T (0) = T 0. T (t) = A + Be kt. Applying the initial condition, we get T 0 = A + B. or B = T 0 A. Hence, the solution is T (t) = A + (T 0 A)e kt. The one parameter we don t know is the constant of proportionality, k. However, let s assume we have collected some data in the form {t i, T i } for 1 i N for some finite integer N.
13 Estimating the Cooling Rate k Rewrite the solution as T (t) A T 0 A = e kt. Now take the logarithm of both sides to get ( ) T (t) A ln = kt. (1) T 0 A This tell us how to estimate the value of k. Let the variable U(t) be defined to be ( ) T (t) A U(t) = ln T 0 A
14 Estimating the Cooling Rate k Rewrite the solution as T (t) A T 0 A = e kt. Now take the logarithm of both sides to get ( ) T (t) A ln = kt. (1) T 0 A This tell us how to estimate the value of k. Let the variable U(t) be defined to be ( ) T (t) A U(t) = ln T 0 A Then, Equation 1 can be written as U(t) = kt. Thus, the variable U is linear in t; i.e. if we graph U versus t we should see a straight line with slope k.
15 Estimating the Cooling Rate k Our collected data consists of pairs of the form (time, temperature). Suppose we collected N such pairs. Label them as (t i, T i ) for 1 i N. Compute the corresponding U(t i ) U i points ( ) Ti A U i = ln T 0 A and plot them as a scatter plot in MatLab. Note the first value, U 1 = ln T 0 A T 0 A = ln(1) = 0. So our line has U intercept 0. Hence, our line is of the form y = mt and we have to find a good candidate for the slope m. This slope will then be used in our cooling model.
16 Estimating the Cooling Rate k Our collected data consists of pairs of the form (time, temperature). Suppose we collected N such pairs. Label them as (t i, T i ) for 1 i N. Compute the corresponding U(t i ) U i points ( ) Ti A U i = ln T 0 A and plot them as a scatter plot in MatLab. Note the first value, U 1 = ln T 0 A T 0 A = ln(1) = 0. So our line has U intercept 0. Hence, our line is of the form y = mt and we have to find a good candidate for the slope m. This slope will then be used in our cooling model. From your data, you now have a collection of pairs (t i, U i ) for 1 i N. We can find a line which comes close in an optimal way to all of this data (although the line does not have to include all the data!) by using a bit of calculus.
17 Estimating the Cooling Rate k At each time, the line has the value mt i. The discrepancy or error between the actual U value and the line value is then U i mt i.
18 Estimating the Cooling Rate k At each time, the line has the value mt i. The discrepancy or error between the actual U value and the line value is then U i mt i. We want the cumulative error between our line and the data, so we don t want errors to cancel. Hence, we could choose an error term like U i mt i or (U i mt i ) 2. Both of these are fine, but the squared one is differentiable which will allow us to use calculus methods.
19 Estimating the Cooling Rate k At each time, the line has the value mt i. The discrepancy or error between the actual U value and the line value is then U i mt i. We want the cumulative error between our line and the data, so we don t want errors to cancel. Hence, we could choose an error term like U i mt i or (U i mt i ) 2. Both of these are fine, but the squared one is differentiable which will allow us to use calculus methods. Define the energy function E(m) where m is the slope to be E(m) = N (U i mt i ) 2. i=1 This is a nice function and so let s find where this function has a minimum by setting its derivative is zero.
20 Estimating the Cooling Rate k Setting the derivative to zero, we find de N dm = 2 (U i mt i ) ( t i ) = 0. i=1
21 Estimating the Cooling Rate k Setting the derivative to zero, we find de N dm = 2 (U i mt i ) ( t i ) = 0. Simplifying a bit, we have N i=1 i=1 ( N U i t i = m i=1 t 2 i ).
22 Estimating the Cooling Rate k Setting the derivative to zero, we find de N dm = 2 (U i mt i ) ( t i ) = 0. Simplifying a bit, we have N i=1 i=1 ( N U i t i = m i=1 Solving for the critical point m, we find t 2 i ). m = N i=1 U it i N i=1 t2 i
23 Estimating the Cooling Rate k Finally, it is easy to see d 2 E dm 2 = 2 N ti 2 > 0. for all choices of m. Hence, the value m which is the critical point must be the global minimum i=1
24 Estimating the Cooling Rate k Finally, it is easy to see d 2 E dm 2 = 2 N ti 2 > 0. for all choices of m. Hence, the value m which is the critical point must be the global minimum Hence, the optimal slope is i=1 m = N i=1 U it i N i=1 t2 i Then we can see how good our model is by plotting the model solution T (t) = T (t) = A + (T 0 A)e m t on the same plot as the actual data.
25 Typical Cooling Project Matlab Session Let s see how we can manage some experimental data. Save your experimental data in a file with two columns. Column one is for time and column two is for the temperature of your liquid. Create a new file in Matlab and call it Newton.dat or some other name. Place it in your class folder. For a typical liquid (orange juice, water, tomato juice etc), we might collect the data here
26 Typical Cooling Project Matlab Session We have the ambient or room temperature is 76 degrees for our experiment.
27 Typical Cooling Project Matlab Session We have the ambient or room temperature is 76 degrees for our experiment. load the data and generate the plot of the data % l o a d i n t h e e x p e r i m e n t a l data from t h e % f i l e Newton. dat as two columns >> Data = l o a d ( Newton. dat ) ; % use t h e f i r s t column as time >> time = Data ( :, 1 ) ; % use t h e second column as t e m p e r a t u r e >> Temperature = Data ( :, 2 ) ; % generate a p l o t of the experimental data >> p l o t ( time, Temperature, o ) ; % s e t x l a b e l >> x l a b e l ( Time i n Minutes ) ; % s e t y l a b e l >> y l a b e l ( Temperature i n F a h r e n h e i t ) ; % s e t t i t l e >> t i t l e ( Newton Law Of C o o l i n g E x p e r i m e n t a l Data ) ;
28 Typical Cooling Project Matlab Session We see the plot of the measured data here:
29 Typical Cooling Project Matlab Session From the theory we have discussed, ( we ) know we should graph the transformed data given by log where A is the room temperature, T A T 0 A T and T 0 are the temperature and initial temperature of the liquid, respectively. Here A = 76 and T 0 = 205. Thus, T 0 A = 129. Here is how we do this in Matlab: % f o r t h i s data, T 0 A = = 129 >> A = 7 6 ; >> LogTemperature = l o g ( ( Temperature A) / ( ) ) ; >> p l o t ( time, LogTemperature, o ) ; >> x l a b e l ( Time i n Minutes ) ; >> y l a b e l ( Transformed V a r i a b l e l n ( (T A) /( T 0 A) ) ) ; >> t i t l e ( Transformed Data V e r s u s Time ) ;
30 Typical Cooling Project Matlab Session You can see this transformed data here:
31 Typical Cooling Project Matlab Session Then, look for the best straight line through the plot making sure you have the line go through the point (0, 0) at the top. The optimal slope is then m = N i=1 U it i. N i=1 t2 i In MatLab, we can add up all the entries in the variable U and time using the sum command. The optimal slope calculation in MatLab is then >> mstar = sum ( LogTemperature. time ) /sum ( time. time ) mstar = where the variable LogTemperature plays the role of U. Note time and LogTemperature are vectors so we use.* to do component wise multiplications!
32 Typical Cooling Project Matlab Session The syntax LogTemperature.*time means to multiply each component of LogTemperature and time separately to create the new column of values LogTemperature(1) time (1) LogTemperature(N). time (N) The command sum applied to this column then creates the sum N i=1 U it i like we need. A similar thing is happening with the term sum(time.*time).
33 Typical Cooling Project Matlab Session The model is thus which we enter in MatLab as u(t) = e t % compute t h e model >> u2 = exp ( mstar time ) ; Then we plot both the experimental data and the model on the same plot, % p l o t the model and the measure data on the same p l o t >> p l o t ( time, u2, r, time, Temperature, o ) ; >> x l a b e l ( Time I n Minutes ) ; >> y l a b e l ( Temperature i n F a h r e n h e i t ) ; >> t i t l e ( F i t o f Newton Law o f C o o l i n g Model To Data : Data = o ) ; Looks pretty good! We could also have added a legend like legend( Model, Data, Location, BestOutside ); but we didn t show that. You should do that though in your report so it is really easy to see what is going on.
34 Typical Cooling Project Matlab Session Our final plot!
35 Typical Cooling Project Matlab Session Here is all the code without comments! >> Data = l o a d ( Newton. dat ) ; >> time = Data ( :, 1 ) ; >> Temperature = Data ( :, 2 ) ; >> p l o t ( time, Temperature, o ) ; >> x l a b e l ( Time i n Minutes ) ; >> y l a b e l ( Temperature i n F a h r e n h e i t ) ; >> t i t l e ( Newton Law Of C o o l i n g E x p e r i m e n t a l Data ) ; % f o r t h i s data, T 0 A = = 129 >> A = 7 6 ; >> LogTemperature = l o g ( ( Temperature A) / ( ) ) ; >> p l o t ( time, LogTemperature, o ) ; >> x l a b e l ( Time i n Minutes ) ; >> y l a b e l ( Transformed V a r i a b l e l n ( (T A) /( T 0 A) ) ) ; >> t i t l e ( Transformed Data V e r s u s Time ) ; >> mstar = sum ( LogTemperature. time ) /sum ( time. time ) ; >> u2 = exp ( mstar time ) ; >> p l o t ( time, u2, r, time, Temperature, o ) ; >> x l a b e l ( Time I n Minutes ) ; >> y l a b e l ( Temperature i n F a h r e n h e i t ) ; >> t i t l e ( F i t o f Newton Law o f C o o l i n g Model To Data : Data = o ) ;
36 Project Two Due Date: Friday April 26, This is 50 Points. We will use the following time and temperature data: This data gives you a set of 18 {t i, Y i } data pairs. The ambient temperture is 68 degrees
37 Project Two Let E(m) = (U i mt i ) 2 and F (m) = U i mt i i=1 i=1 Your project is this: Find the value of m that optimizes E using our calculus tools for finding extremal values. Prove this value gives a minimum using our calculus tools. Find the values of m that optimize F using our subdifferential tools for finding extremal values. Prove this value gives a minimum using our convex analysis tools.
38 Project Two For the differentiable case, use E. Show all the MatLab code for all work with comments. Plot appropriate the original data using circles Plot the transformed data using circles Plot the transformed data and the linear regression line on the same graph. Plot the model you have found and the original data on the same graph. Use circles for the original data.
39 Project Two For the nondifferentiable case, use F. All the work is done by hand as we have done in smaller problems for homework. Organize the centers of the absolute value functions so they go from left to right. Find F in the form without the absolute values. Graph F Find F (x) everywhere Find where 0 F (x) to determine the minimum point. This determines the approximate slope to use in the model. Plot the model you have found and the original data on the same graph. Use circles for the original data. Finally, omment on the difference in these two optimal cooling parameter solutions. They are pretty similar!
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