Project 2 by Kayli-Anna Barriteau

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1 Project 2 by Kayli-Anna Barriteau Decay of quinine in a malaria patient Data for the decay of a mg dose of quinine in a patient suffering from malaria is given in Mathematica format below. The first number in each pair is the number of hours since the injection of quinine, and the second number is the amount of quinine remaining in the body at that time after the injection. data = 880,.<, 81, 4.9<, 82, 44.1<, 83, 41.4<, 84, 39.1<, 85, 3.<, 8, 34.5<, 87, 32.4<, 88,.5<, 89, 28.7<, 8, 27.<, 811, 25.4<, 812, 23.9<, 813, 22.4<, 814, 21.<, 815, 19.7<, 81, 18.<, 817, 17.5<, 818, 1.5<, 819, 15.4<, 8, 14.5<, 821, 13.5<, 822, 12.8<, 823, 12.<, 824, 11.4<, 825,.<, 82,.<, 827, 9.5<, 828, 8.9<, 829, 8.3<, 8, 7.7<, 831, 7.3<, 832,.9<, 833,.<, 834,.2<, 835, 5.7<, 83, 5.3<, 837, 5.1<, 838, 4.7<, 839, 4.4<, 8, 4.3<, 841, 4.<, 842, 3.7<, 843, 3.<, 844, 3.3<, 845, 3.1<, 84, 2.8<, 847, 2.7<, 848, 2.5<, 849, 2.5<, 8, 2.2<< The data of the amount of quinine in the patient as a function of time is plotted below. ListPlot@dataD You can see better that that there is a curve in the data by joining the points.

2 2 Joined TrueD In order to calculate the the hour-by-hour growth rate of quinine in the patient, i have first placed the data in a table as shown below.

3 TableHeadings 8None, 8"Hours", "Amount"<<D Hours Amount

4 4 growth = Table@8n, data@@n + 1DD@@2DD - data@@ndd@@2dd<, 8n, 1, Length@dataD - 1<D by using the information in the table and with mathematica s programing, I am able to calculate the growth rate per hour. The first number in each pair is the hour while the second is the current growrth rate by that hour. We can see by the calculations, that the growth rate is not constant and that it can accelerate and slow down at different points. 881, - 3.1<, 82, - 2.8<, 83, - 2.7<, 84, - 2.3<, 85, - 2.5<, 8, - 2.1<, 87, - 2.1<, 88, - 1.9<, 89, - 1.8<, 8, - 1.7<, 811, - 1.<, 812, - 1.5<, 813, - 1.5<, 814, - 1.4<, 815, - 1.3<, 81, - 1.1<, 817, - 1.1<, 818, - 1.<, 819, - 1.1<, 8, - 0.9<, 821, - 1.<, 822, - 0.7<, 823, - 0.8<, 824, - 0.<, 825, - 0.8<, 82, - 0.<, 827, - 0.5<, 828, - 0.<, 829, - 0.<, 8, - 0.<, 831, - 0.4<, 832, - 0.4<, 833, - 0.3<, 834, - 0.4<, 835, - 0.5<, 83, - 0.4<, 837, - 0.2<, 838, - 0.4<, 839, - 0.3<, 8, - 0.1<, 841, - 0.3<, 842, - 0.3<, 843, - 0.1<, 844, - 0.3<, 845, - 0.2<, 84, - 0.3<, 847, - 0.1<, 848, - 0.2<, 849, 0.<, 8, - 0.3<< Below Is a visul reprensation of the growth rate. p3 = ListPlot@growth, Joined TrueD nlm = NonlinearModelFit@data, a * Exp@- b * xd, 8a, b, c<, xd This is the non-linear model for this particular growth as a logistic function. FittedModelB ã x F below is a plotted graph of the logisctic function.

5 5 P2 = Plot@nlm@xD, 8x, 1, <, PlotStyle RedD 0 By combining the graph of the logistic function and the original plotted data, we can visually see that the model fits the data really well. Show@8p1, P2<D nlm@"rsquared"d By calculating the RSquared Value, we can see that it is deffintly a good fit since the value is almost one and no were close to D@nlm@xD, xd Below is the calculated derivative of the non-linear model. This allows us to calculate the instantaneous growth rate of the logistic function approximation ã x Below is the plotted growth rate of the logistic function.

6 p4 = PlotA ã x, 8x, 1, <E By plotting both the growth rate from the logistic function and the growth rate we get from the data, we can see that this is almost a perfect fit and that the logistic function allows us to calculate the value of decay in more than just whole hours. This function clearly demonstrates exponential decay or a decrease that follows an exponential function. Show@8p3, p4<d Rate of enzyme catalysis = most troubled part of my project! I apologize for the bad work :( Data posted at biology-online shows how an enzyme breaks down hydrogen peroxide with time in a process called enzyme catalysis. The data are shown below in Mathematica format. The first number in each pair is the time t (in seconds) since the beginning of the catalysis experiment, and the second number is the amount of hydrogen peroxide H(t) (in ml) broken down by the enzyme at time t.

7 Rate of enzyme catalysis = most troubled part of my project! I apologize for the bad work :( 7 Data posted at biology-online shows how an enzyme breaks down hydrogen peroxide with time in a process called enzyme catalysis. The data are shown below in Mathematica format. The first number in each pair is the time t (in seconds) since the beginning of the catalysis experiment, and the second number is the amount of hydrogen peroxide H(t) (in ml) broken down by the enzyme at time t. In[23]:= enzymedata = 88, 0.5<, 8, 1.5<, 80, 3<, 890, 4<, 81, 4.5<, 8180, 5<< Below is the plotted data. In[29]:= p1 = ListPlot@88, 0.5<, 8, 1.5<, 80, 3<, 890, 4<, 81, 4.5<, 8180, 5<<, Joined TrueD Out[29]= Below is my fitted model and it s plotted graph. In[27]:= Out[27]= NonlinearModelFit@enzymedata, a - b * Exp@c * td, 8a, b, c<, td FittedModelB 0. F

8 8 In[]:= p2 = Plot@nlm@tD, 8t,, 180<, PlotStyle RedD 12 8 Out[]= 4 0 In[31]:= 1 Show@8p1, p2<d Out[31]= 2 1 In[32]:= Out[32]= 0 1 nlm@"rsquared"d I guess the model suggests that the maximum break down is 5 ml. Below is the derivative of the data and it s plot. In[35]:= Out[35]= d = D@nlm@tD, td ã t ã t t

9 In[47]:= 9 PlotA ã t ã t t, 8t,, 180<E Out[47]= Longitudinal observation of Japanese Lancelet, Branchiostoma japonicum, metamorphosis Data on a longitudinal observation of Japanese Lancelet, Branchiostoma japonicum, metamorphosis is shown (in Mathematica format) below. The first number in each pair is the number of days since the birth of a lancelet, and the second number is the length of the lancelet (in cm) on that day. data = 88, 5.8<, 87,.1<, 88, 5.9<, 89, <, 8,.1<, 811, <, 812, <, 813, 5.8<, 814, <, 815, 5.9<, 81, 5.9<, 817, <, 818,.2<, 819, 5.9<, 8, <, 821, <, 822, <, 823, 5.9<, 824,.1<, 825, <, 82, <, 827, <, 828,.1<, 829,.1<, 8, <, 831,.1<, 832,.1<, 833,.1<, 834,.1<, 835,.1<, 83,.1<, 837,.1<, 838,.1<, 839,.2<, 8,.1<, 841,.1<, 842,.2<, 843,.2<, 844,.3<, 845,.2<, 84,.3<, 847,.4<, 848,.4<, 849,.<, 8,.4<, 851,.5<, 852,.5<, 853,.7<, 854,.7<, 855,.8<, 85,.9<, 857,.8<, 858, 7<, 859, 7.1<, 80, 7.2<, 81, 7.5<, 82, 7.4<, 83, 7.5<, 84, 7.7<, 8, 7.9<, 87, 7.9<, 88, 8<, 89, 8<, 870, 8.2<, 872, 8.2<, 873, 8.5<, 874, 8.<, 875, 8.<, 87, 8.8<, 877, 9<, 879, 9<, 880, 9.2<< The data is plotted below as the length of the lancelet versus the number of days since birth as well as another version with the joining of all plotted points.

10 p1 = ListPlot@dataD ListPlot@data, Joined TrueD TableForm@data, TableHeadings 8None, 8"ð of Days", "Length"<<D ð of Days Length

11 Above is the data organized into a table form. This makes it easier to calculate the day-by day growth rate of the lancelet s length per day as shown below. The first number in the pair represents the day while the second number represents the growth rate. 11

12 12 Above is the data organized into a table form. This makes it easier to calculate the day-by day growth rate of the lancelet s length per day as shown below. The first number in the pair represents the day while the second number represents the growth rate. growth = Table@8n, data@@n + 1DD@@2DD - data@@ndd@@2dd<, 8n, 1, Length@dataD - 1<D 881, 0.3<, 82, - 0.2<, 83, 0.1<, 84, 0.1<, 85, - 0.1<, 8, 0<, 87, - 0.2<, 88, 0.2<, 89, - 0.1<, 8, 0.<, 811, 0.1<, 812, 0.2<, 813, - 0.3<, 814, 0.1<, 815, 0<, 81, 0<, 817, - 0.1<, 818, 0.2<, 819, - 0.1<, 8, 0<, 821, 0<, 822, 0.1<, 823, 0.<, 824, - 0.1<, 825, 0.1<, 82, 0.<, 827, 0.<, 828, 0.<, 829, 0.<, 8, 0.<, 831, 0.<, 832, 0.<, 833, 0.1<, 834, - 0.1<, 835, 0.<, 83, 0.1<, 837, 0.<, 838, 0.1<, 839, - 0.1<, 8, 0.1<, 841, 0.1<, 842, 0.<, 843, 0.2<, 844, - 0.2<, 845, 0.1<, 84, 0.<, 847, 0.2<, 848, 0.<, 849, 0.1<, 8, 0.1<, 851, - 0.1<, 852, 0.2<, 853, 0.1<, 854, 0.1<, 855, 0.3<, 85, - 0.1<, 857, 0.1<, 858, 0.2<, 859, 0.2<, 80, 0.<, 81, 0.1<, 82, 0<, 83, 0.2<, 84, 0.<, 85, 0.3<, 8, 0.1<, 87, 0.<, 88, 0.2<, 89, 0.2<, 870, 0<, 871, 0.2<< Below is the plotted version of the day - by - day growth rate. p3 = ListPlot@growth, Joined TrueD nlm = NonlinearModelFit@data, a * x ^ 2 + b * x + c, 8a, b, c<, xd Below is a model for the data in the form of c+bx+ax^2 and a graph that represents the model visually. FittedModelB x x2 F

13 13 p2 = Plot@nlm@xD, 8x,, 80<, PlotStyle RedD The figure below is a visual demonstartion on how good a fit the nonlinear model is to the data. Visually, it is a good fit. Show@8p1, p2<d nlm@"rsquared"d To insure that it is indeed a good fit, the RSquared value is calculated. The value is around.999 which is almost one. So, it is indeed a good fit since it is no where close to D@nlm@xD, xd Below is the calculated derivative of the non-linear model along with a plot of the derivative x

14 14 p4 = Plot@ x, 8x, 1, 71<D The graph below holds both the derivative of the non linear model as well as the the day by day growth rate. Show@8p3, p4<d Judging from a visula stand point, the fit is horrible and i do not beleive that this quadratic model for the lancelet s length can hold for days 0 through 5. Effect of temperature on parsnip germination The temperature at which seedlings are grown effects the percentage of seeds that germinate. One would like to find the optimum temperature that maximizes the germination rate. Below, in Mathematica format, is a set of data for germination percentages of parsnips, at varying temperatures. The first number in each pair is the temperature (F ), and second number is the percentage of seeds that germinated at that temperature.

15 15 The temperature at which seedlings are grown effects the percentage of seeds that germinate. One would like to find the optimum temperature that maximizes the germination rate. Below, in Mathematica format, is a set of data for germination percentages of parsnips, at varying temperatures. The first number in each pair is the temperature (F ), and second number is the percentage of seeds that germinated at that temperature. Parsnipdata = 8832, 0.82<, 841, 0.87<, 8, 0.79<, 859, 0.85<, 88, 0.89<, 877, 0.77<, 88, 0.51<, 895, 0.1<, 84, 0.<< 8832, 0.82<, 841, 0.87<, 8, 0.79<, 859, 0.85<, 88, 0.89<, 877, 0.77<, 88, 0.51<, 895, 0.1<, 84, 0.<< the percentage germination versus temperature is plotted below p1 = ListPlot@ParsnipdataD NonlinearModelFit@Parsnipdata, a * T ^ 2 + b * T + c, 8a, b, c<, TD Below is a nonlinear model and it s plotted graph. FittedModelB T T2 F

16 1 P2 = Plot@nlm@TD, 8T, 0, 32<, PlotStyle RedD Below is supposed to show both the linear model plot along with the data plot. For some reason, the Non linear plot is not showing up so we can not determine a visual fit. Luckily, the calculated R squared value located just below the graph tells us that it is indeed a good fit. Show@8p1, P2<D nlm@"rsquared"d Below is the calculated derivative of our model. d = D@nlm@TD, TD T By setting the serivative to 0, we are able to find the maximum temperature at which the model takes a maximum value which is degrees (F). I think the domain of the parsnip data has to be tested a little wider in order to conclude that this is indeed true. FindRoot@d Š 0, 8T, <D 8T 23.17< Melatonin in the blood A recent study investigated the blood levels of melatonin when inhaled as a spray.the data, in Mathematica format is given below.the first number in each pair is the time (in minutes) since administration of the melatonin as an oral spray, and the second number is the blood concentration (in ng/ml) of

17 17 Melatonin in the blood A recent study investigated the blood levels of melatonin when inhaled as a spray.the data, in Mathematica format is given below.the first number in each pair is the time (in minutes) since administration of the melatonin as an oral spray, and the second number is the blood concentration (in ng/ml) of melatonin in the bloodstream (averaged over 8 people: 3 females and 5 males). In[1]:= spraydata = 880, 0.38<, 85, 1.09<, 8, 4.79<, 8,.44<, 8, 13.08<, 8, 13.14<, 80, 11.43<, 890, 8.52<, 81,.58<, 8180, 3.31<, 82, 1.<, 830, 0.39<< the average concentration of melatonin as a function of time is plotted below. In[]:= p1 = ListPlot@spraydataD 12 8 Out[]= by using the nonlinearmodelfit command I am able to construct a surge function model for the data as shown below. In[4]:= Out[4]= nlm = NonlinearModelFit@spraydata, a * t * Exp@- b * td, 8a, b<, td FittedModelB ã t t F Below is the plotted non linear function as well as the nonlinear function plotted along side the data.

18 18 In[5]:= P2 = Plot@nlm@tD, 8t, 0, 30<, PlotStyle RedD 12 8 Out[5]= 4 2 In[7]:= Show@8p1, P2<D 12 8 Out[7]= Visually and from calculating the RSquared value, it is a rather decent fit. In[8]:= Out[8]= nlm@"rsquared"d Below is the derivative of the non-linear model and the derivative function plotted as a function of time. In[19]:= Out[19]= d = D@nlm@tD, td ã t ã t t

19 In[1]:= p3 = PlotA ã t ã t t, 8t, 0, 30<, PlotStyle RGBColor@0.8, 0., 0.3DE Out[1]= In[13]:= growth = TableForm@spraydata, TableHeadings 8None, 8"minutes", "blood concentration"<<d Out[13]//TableForm= minutes In[14]:= Out[14]= blood concentration growth = Table@8n, spraydata@@n + 1DD@@2DD - spraydata@@ndd@@2dd<, 8n, 1, Length@spraydataD - 1<D 881, 0.71<, 82, 3.7<, 83, 5.5<, 84, 2.4<, 85, 0.0<, 8, <, 87, <, 88, <, 89, <, 8, - 1.5<, 811, << the derivative from the data ia calculated and plotted along with the derivative of the surge function model below. 19

20 In[17]:= p4 = ListPlot@growth, Joined TrueD 4 2 Out[17]= In[18]:= Show@8p3, p4<d Out[18]= It does not appear to be a good fit. FindRoot@d Š 0, 8t,.71<D According to the derivative of the surge function, the model is 0 at around 4 minutes. and I think I am doing something wrong because that does not make any sense. Out[21]= 8t <

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MSE 310/ECE 340. Instructor: Bill Knowlton. Initialize data Clear somedata, someplot. Place data in list somedata

MSE 310/ECE 340. Instructor: Bill Knowlton. Initialize data Clear somedata, someplot. Place data in list somedata MSE 3/ECE 340 Instructor: Bill Knowlton Examples of Plotting Functions ü Example 1: Using the command ListPlot[ ] to plot data in a scatter plot In[1]:= Initialize data Clear somedata someplot Place data