OSH/6/Solution

From UBCMATH WIKI
< OSH‎ | 6
Jump to: navigation, search

Solution key

  1. The period of \(E(t)\) is 24 hours. The maximum occurs when \(t=12\) (noon). The minimum occurs at midnight. The amplitude is 3 C.
  2. I found that a value of \(k=0.12885\) ensures that the numerical solution gives a temperature of 19.50069 at 11 am (values ranging from 0.126 through 0.132 are accurate enough). To do this, I created a column of times starting with 8 and ending with 11 in 1/4 increments (column B). Next to that (column C), I created a column of \(E(t)\) values. The next two columns held the approximate \(T(t)\) and \(T’(t)\) values. The \(T(8)\) value in cell D6 was set at 22. The value of \(T’(8)\) was calculated in E6 using the formula ‘’’=SBS3*(C6-D6)’’’ (S means dollar sign) where an initial guess for \(k\) was set in cell B3. The \(T(8.25)\) value calculated in cell D7 was found using the formula ‘’’ =D6+1/4*E6’’’. These formulae were copied down until the \(t=11\) row. Then I changed the \(k\) value upward if my previous value of \(T(11)\) was greater than 19.5 and changed it downward if my previous value of \(T(11)\) was lower than 19.5. I continued until the temperature at 11 am was 19.50069.
  3. I found the time of death to be shortly after 1 am. I did this by modifying the spreadsheet used in #2 as follows. I fixed my \(k\) value at 0.12885 and kept \(\Delta t=1/4\). I changed the time column to start at 8 and count down to -2 (10 pm) so that the \(E(t)\) gave temperatures going backward in time. Then I changed the formula in the \(T(t)\) column to read =D6-1/4*E6. This is the same formula as used in #2 but with \(\Delta t\) replaced with \(-\Delta t\) so that the temperatures go backward in time. In other words, using the formula \(T_{n-1}=T_n - \Delta t k(E_n-T_n)\). Finally, I scanned down the \(T(t)\) column to find the value closet to 37 C and read off the time from the row. This method could be called Backward Euler.
    There are (at least) two other ways to do this. One alternate option is to consider the formula \(T_{n+1}=T_n + \Delta t k(E_n-T_n)\). In #2 we knew \(T_n\) and used that to find \(T_{n+1}\). Now, we know \(T_{n+1}\) and seek \(T_n\) so we solve for \(T_n\): \(T_n=\frac{T_{n+1} - \Delta t k E_n}{1- \Delta t k}\). Thus the formula =(D6-1/4*SBS3*C7)/(1-1/4*SBS3) replaces =D6-1/4*E6. This method could be called Forward Euler going backward. A third way is to use the same k value and start a T(t) column at t=-2 with a virtual temperature chosen to be around 60. Use the standard Euler formula to get temperatures all the way to t=11. Then modify the initial temperature up and down until the temperatures at 8 am and 11 am are correct (22 and 19.5). This method could be called bisection by shooting. Answer vary depending on which method and what value of k is used. See the table below.
k Backward Euler Forward Euler Bisection
0.126 1 am 1:15 am 1 am
0.132 1:15 am 1:30 am 1:30 am

Students should:

  • (4 pts) One point for each correct numbers. No calculations need to be shown.
  • (2 pt) Correct \(k\) value (\(k=0.126 \to 0.132\)) (1 pt), a clear explanation of what they did (does not have to be exactly the same as described above) (1 pt)
  • (2 pt) Correct value for the time of death (1 pt), a clear explanation of what they did (1 pt)

Total 8 pts.

Communication and presentation 3 pts - these should appear in Crowdmark as part of #3.

Overall total 11 pts.