Mathematica Project 3

Transcription

1 Mathematica Project 3 Name: Section: Date: On your class s Sakai site, your instructor has placed 5 Mathematica notebooks. Please use the following table to determine which file you should select based on the last digit of your Student ID number. ID number ends w/ Mathematica File 0 or 1 ProblemDoc Chapter or 3 ProblemDoc Chapter or 5 ProblemDoc Chapter or 7 ProblemDoc Chapter or 9 ProblemDoc Chapter 5 5 In this assignment, you will discover how to define piecewise functions in Mathematica, how to fit data to functions, how to use the Math Palettes, and how to calculate Riemann sums and definite integrals in Mathematica. Note: It should take you approximately 2 hours to complete problems Time required for Language of Math (Problem #21) may vary greatly depending on your writing skills and the guidance provided by your instructor. This time estimate is provided to help you plan your work. It represents total time dedicated wholly to the task; if you are multitasking, you will need to add more time to the estimate. If you have any problems using Mathematica, be sure to see the Mathematica tutor for assistance. If you are having difficulty interpreting your Mathematica results, please see your instructor for assistance. Suppose that you are going to watch your favorite race car driver in a big race. Of course, you seize the opportunity to practice your math skills. After the race, you are given a piecewise function that expresses your driver s velocity as a function of time. You will use right-hand Riemann sums as well as left-hand Riemann sums to approximate the distance travelled during certain time intervals during the race. You will also approximate the total distance of the race. To help you understand the theory behind this assignment, we ve included an appendix (at the end of the assignment) that steps you through the ideas we will cover. This reviews much of the material in the text, so if you wanted to work on this assignment while traveling for the holidays, you should not need to bring your textbook. YOUR TASK: PART 1: Bob the Builder is trying out a new profession: stock car driving. During his first race, the race officials were able to record his velocity (in meters per second) and fit a piecewise function to it. However, the race official made a mistake in marking the official length of the race, and there is no record of how far Bob traveled. The description of Bob s race is as follows: Bob accelerates off the line to a great start. After a mechanical failure, Bob s car begins slowing down before coming to a stop for a time. His pit crew rushes to his aid, and after a short while, repairs his car. He accelerates once more and finishes the race without difficulty. Your Mathematica file shows you the four different stages of his race. v1(t) represents his initial start off the line until the car breaks, v2(t) represents him slowing down after the mechanical failure, v3(t) represents the time the car is stopped and being repaired, and v4(t) represents him continuing on until 1

2 he reaches the finish line. T otalrace(t) represents the piecewise function in which the four functions are stitched together in time. Notice the format for the built-in function Piecewise. Your file then plots the piecewise function showing Bob the Builder s velocity as a function of time. Evaluate your notebook (use the menus - Evaluate - Evaluate Notebook). Problem 1 At what time does the car come to a complete stop after the breakdown? (Hint - do not just guess based on the graph, but rather look at how the piecewise function is defined.) We are interested in seeing how far Bob the Builder went before the car stopped. Using T otalrace(t) as a model, make a new piecewise function called P rebreak(t) that defines the velocity of the car up until it stops after the breakdown. Plot this new function, P rebreak(t), by pressing shift+enter or using the Evaluate Notebook from the top menus. Problem 2 Right click on your new graph, Copy Graphic, and paste (ctrl-v) the graph into a blank Word document TWICE (i.e., the document should have TWO copies of the picture). Print the document. By hand, approximate the distance Bob the Builder covered before the broken car stopped by using a left-hand Riemann sum with 4 subintervals, (n = 4), and a right-hand Riemann sum with 4 subintervals. On the printed document, show and shade the rectangles you used for the left-hand sum (hand label this chart Left-Hand Riemann Sum on one copy of the graphic; show and shade the rectangles you used for the right-hand sum on the other copy of the graphic (hand label this chart Right-Hand Riemann Sum ). Write your approximation for distance next to the corresponding graphic, and don t forget the units! Also be sure to label this page in your submission with the Problem number. Suppose we want to increase the accuracy of our approximation. We know that to do this, we should make more rectangles. If we wanted to do a left-hand Riemann sum with n = 100 by hand, that would take a LONG time. Fortunately, Mathematica can help! Using the menus at the top, select Palettes, then Basic Math Assistant. This brings up a calculator in Basic mode; click the tab to bring up the Advanced mode, and you will see that there is a summation symbol! Click just below your plot to create a new cell (the cursor should flip sideways to indicate that you are between cells - this will create a new cell); then click on the summation symbol in the calculator palette. Then you can click on each of the input boxes to fill them in with the necessary information. For us, we want the index to be i, and it will start at 1 and end at 100. Where Mathematica asks for the expr, it wants the expression to be input. Recall for a left-hand limit, this expression is P rebreak(t i ) t where t = b a n. This brings up a programming issue - specifically, how do we tell Mathematica what t i is? We know that each time t i is t bigger than before, and after i-steps, we ve added i t or i b a n to the original time. If we start at a, then this means that t i = a + (i 1) b a n. The reason for using (i 1) is because we still need t = 0 for our first rectangle, i = 1. We know that a = 0 because we are starting to measure at the beginning of the race, and we know that n = 100, but we need to know b. Problem 3 What is b? In other words, at what time should we stop measuring the distance, as we set up our Riemann sum? Remember, we want to measure clear up until the car rolls to a complete stop. Now that we know b, we can input it into our Mathematica expression: (PreBreak[0+(i-1)*(b-0)/100]*(b-0)/100). After substituting b with your answer to Problem 3 (in both places), input this expression into the Mathematica summation, and evaluate that cell (shift+enter). If you get an improper fraction, go back and add //N to the end of your summation command. When you evaluate again, you should get a decimal answer. Problem 4 What did you get as an approximation using n = 100 with a left-hand Riemann sum for the distance? Be sure to include units. Problem 5 In Mathematica, using the palette and the summation symbol, add a new line that computes the approximate distance covered using n = 100 with a right-hand Riemann sum. What is the value you get (in decimal form)? Be sure to include units. 2

3 Problem 6 Using all you have learned about creating piecewise functions and approximating distance using Riemann sums in Mathematica, we now want to approximate the distance Bob the Builder goes from the time the car is fixed and starts moving again until the end of the race. Use Mathematica to compute the left-hand Riemann sums with n = 10, n = 100, and n = 1000 to approximate this distance. Give all three answers in the form of a table (hand written or in Word). Problem 7 What was the approximate total length of the race? Mathematica has the ability to directly compute the integral that represents the distance. Go back to the math palette, and next to the summation symbol, you will see a definite integral symbol. Click on that to add it to your notebook, and fill it in such that you compute the exact distance covered over the entire race. To fill it in, remember your lower bound should be the race start time (0), your upper bound should be the time the race ends, your expression is TotalRace[t], and the var is just t. Evaluate the integral. Notice that Mathematica likes to give precise results (hence the logarithm in your answer), but if you put //N at the end of your command, Mathematica will give you a decimal approximation. Problem 8 Based on the result from your integral in Mathematica, what is the total distance Bob the Builder covered during the race? Please give your answer in decimal form, and be sure to include units. Be sure to save your work up to this point. Later you will be asked to print, and all of the work up to this point (and more in Part 2) should be seen in the printout. Part 2: After a disasterous start to his stock car racing career, Bob the Builder decides to give drag racing a try. Your Mathematica file is preloaded with his velocity in meters per second, vdragdata, for his first drag race. The data, vdragdata, is loaded into your Mathematica file as a list of ordered pairs in the format {{t 1, v(t 1 )}, {t 2, v(t 2 )},...}. You already know how to plot functions, but to plot points representing data points, you can use the command ListPlot. In our case, we want to show the points in the color red, so we will use the command: ListPlot[vdragdata, PlotStyle->Red]. Go ahead and plot your data. If you are having difficulty, check to make sure you are mindful of upper and lower case letters throughout, and if that doesn t fix the problem, look at the listplot command in the documentation center. Problem 9 We know that we can approximate the distance traveled with a Riemann sum. What is the smallest spacing, t, we could use with this data? Problem 10 By hand, use t = 2 calculate the distance traveled by Bob the Builder for the drag race by using both a left-hand Riemann sum and a right-hand Riemann sum. What is the difference between these estimates? That isn t terribly accurate, is it? To make it better, we d like to use Mathematica, but in Part 1, we only saw how to use the summation notation if we knew the function. Here we only have points in a list! Suppose I wanted to use t = 0.4. This means that I would use every other data point, so the 2nd, the 4th, the 6th. Also, in each case, when I want to use the function value, I want to take the 2nd part of the ordered pair (remember, ordered pairs give the time first and the velocity 2nd). I can tell Mathematica this in my summation s expr by saying vdragdata[[2*i,2]]*0.4: the 2 i is because I want the 2nd, 4th, 6th, etc. data points; the 2 is because I want the 2nd part of the ordered pair; the 0.4 is because that is my t. Go ahead and run the calculation. Think about what you would need to change in order to do this calculation with t = 0.2. Would you still use every other data point (just the even ones)? If not, which data points would you use, and how 3

4 would you represent that in the expr for the summation? Would you still want to use the 2nd part of each ordered pair? Why or why not? Would the 0.4 stay the same, or would you replace it (and if you d replace it, what would go in its place)? And what if you wanted a right-hand sum instead of a left-hand sum? Now that you ve thought through it, go ahead and do it! Problem 11 Using the summation button (from the palette) in Mathematica, use t = 0.2 to calculate the distance traveled by Bob the Builder for the drag race by using a left-hand Riemann sum. Problem 12 Using the summation button (from the palette) in Mathematica, use t = 0.2 to calculate the distance traveled by Bob the Builder for the drag race by using a right-hand Riemann sum. Problem 13 What is the difference between the two approximations? Hint - you can use Mathematica like a calculator to do the subtraction for you. We notice that as t decreases, so does the difference between the left and right sums. We d like an even more accurate approximation, but we don t have any data between these points. Look at the plot - perhaps this function might be some sort of root function. Mathematica has great curve-fitting abilities, so if we tell it the types of terms we might want to use, Mathematica will determine the coefficients needed to fit a curve to our points. To find out the best-fit for our data with constants, t, t 1/2, and t 1/4,type the command: vfit[t ]=Fit[vdragdata, {t^0, t, t^(1/2), t^(1/4)}, t]. To show the plot of this function on the same plot as your listplot, you can use the Show command: Show[ ListPlot[vdragdata, PlotStyle->Red], Plot[vfit[t],{t,0,2}]]. Go ahead, make these changes, and look at your new plot! That isn t a bad approximation, but I bet we can do even better! If we add another term, a t 1/8 -term, the fit should get better because we are adding more flexibility to the curve. Using what you know about making fits in Mathematica, add a new line to create vfit2 which will be like vfit but with the extra t 1/8 -term. Problem 14 Using what you know about plotting two functions simultaneously within a plot command (from previous assignments), modify the Plot command inside the Show command such that Mathematica shows you the vdragdata points and the curves for both vfit and vfit2 at the same time. Which function is better for approximating the area distance traveled by Bob the Builder throughout the race (including the first 0.2 seconds!)? Briefly explain your answer. Let s use vfit(t) to approximate the area under the curve with a Riemann sum. Now that we have a function, we can go back to the method we used in Part 1 of this assignment to compute Riemann sums. Problem 15 Use Mathematica to compute both the left-hand and right-hand Riemann sums with n = 1000 based on the function vfit(t) to approximate the distance Bob the Builder covers in his drag race. What is the difference between these estimates? Problem 16 Use Mathematica to compute the definite integral of vfit over the entire race in order to determine the distance Bob the Builder traveled in his drag race. What distance did you determine? Be sure to include units in your answer. Problem 17 Use Mathematica to compute the definite integral of vfit2 (the second fit you did) over the entire race to determine the distance Bob the Builder traveled in his drag race. What distance did you determine? Be sure to include units in your answer. Problem 18 Are your answers to Problems 17 and 18 the same? Does this surprise you? Why or why not? Problem 19 In your judgment, what is the total distance (with units) that Bob the Builder covered in his drag race? Briefly justify your answer. Problem 20 Save and PRINT your Mathematica file at this point. Be sure to digitally sign the top of the file. Also be sure, in your submission, next to this problem number, write See attached printout. 4

5 Problem 21 Language of Mathematics A Riemann sum represents an approximation, and an integral (which is the limit of the Riemann sum) is supposed to represent an exact area. However, in Part 2, we end up with different answers when we use a definite integral. If the definite integral is supposed to be exact, how do we get different answers - shouldn t they both be perfectly accurate? Can we find the exact distance Bob traveled in his drag race? If so, how? If not, why not? In your own words, discuss the accuracy of Riemann sums, the limit of Riemann sums, and integrals as they relate to Part 2 of this assignment. Your instructor will give you guidance on the length and format of this piece of the assignment. APPENDIX FOR THOUGHT: This appendix covers much of the material given in your textbook, pages We know that velocity is equal to distance divided by time: Therefore, v = d t. d = vt. If the velocity is constant, therefore, we can multiply the velocity of an object by the time that it moves to obtain the distance that it moves. Suppose, however, that velocity is not constant. It changes over the time interval, but if we subdivide the total time interval into very small time intervals, then the velocity does not have much opportunity to change over the small time intervals. Suppose that Jimmy sits on his sled at the top of a snowy hill. He pushes off and starts sliding down. He goes slowly at first, but then speeds up. Let his velocity as a function of time be defined as v = 3t 2, with time t in seconds and velocity v in feet per second. How can we estimate the distance that he travels in four seconds? Well, there is no single velocity that Jimmy travels at for any length of time. At time t = 0, his velocity is v(0) = 3(0) 2 = 0, and at time t = 4, his velocity is v(4) = 3(4) 2 = 48. If we used either of these velocities as an estimate for the velocity that he travelled at for the entire four seconds, we would probably get a very innacurate estimate of the distance that Jimmy travelled. What if we divide his trip into four time intervals, each lasting for one second? We make the following table: t v(t) For the time interval from 0 to 1 second, let s estimate that Jimmy travels at 0 feet per second. From the time interval from 1 to 2 seconds, we ll estimate that he travels at 3 feet per second, etc. Using this method, his total distance is estimated to be dist est. = 0(1 0) + 3(2 1) + 12(3 2) + 27(4 3) = = 42, 5

6 so we estimate that he travelled 42 feet in four seconds. We just used the left-hand Riemann sum, because we used the value for velocity of the time at the left side of each interval. Let s use the right-hand Riemann sum to estimate his total distance again. This time, we use the velocity of the time at the right side each interval: dist est. = 3(1 0) + 12(2 1) + 27(3 2) + 48(4 3) = = 90, So we estimate that he travelled 90 feet in the four seconds. Does this seem like a good approximation? We estimated that he travelled 42 feet with the left-hand Riemann sum, and 90 feet with the right-hand Riemann sum! Well, I promise that if you make the intervals smaller, the values for the left-hand and right-hand Riemann sums will get closer and closer together, and the estimate will get better and better. Try it if you re skeptical! EXAMPLE: Suppose that we have a function f(x) = 4x If we were to graph this function, what would be the area under the curve from x = 0 to x = 5? We will use a Riemann sum to approximate this area. The first step is to choose an interval width, x. Let s choose x = 1. That means that we ll divide the x-axis from 0 to 5 into intervals that are each 1 unit in width. Starting at x = 0, our first interval goes from 0 to 1. Our next interval goes from 1 to 2, and the rest of our intervals go from 2 to 3, from 3 to 4, and from 4 to 5. Now, from the equation f(x) = 4x 2 + 7, we see that the value of the function is constantly changing, much in the way our sledder s velocity was always changing. However, to simplify the problem, on each interval, we will assume that the function value stays the same. So, we will calculate a value on each interval. The idea is that since the interval is so small, the function value doesn t have much opportunity to change on the interval, and so the calculated function value on that interval is a good estimate for the function value at any time in that interval of time. Then, we multiply that value by the interval width, x, in order to almost find the area under the curve on that interval. The question is, what value of function do we choose for each interval? Let s examine the interval from 0 to 1. To represent the function on that entire interval, we could choose the function value at 0 seconds, at 1 second, or at any point in between 0 and 1. Let s choose the function value atx = 0. If for every other interval, we also choose to evaluate the function at the left side of the interval, then we will be performing a left-hand Riemann sum. Let s do that. So for the interval from 1 to 2, we will use the function value at x = 1. From the interval from 2 to 3, we will evaluate the function value at x = 2, and so on. Note that on the last interval, from 4 to 5, we choose the function value at x = 4, and so the function value at x = 5 is never used. Now, to evaluate the Riemann sum! f(0) x + f(1) x + f(2) x + f(3) x + f(4) x = f(0) 1 + f(1) 1 + f(2) 1 + f(3) 1 + f(4) 1 = f(0) + f(1) + f(2) + f(3) + f(4) = ( 4(0) ) + ( 4(1) ) + ( 4(2) ) + ( 4(3) ) + ( 4(4) ) = = 155 So, the area under the curve is approximately 155 square units. There s a more convenient way to express a Riemann sum. Consider this: n f(x i ) x, 6

7 where n is the number of intervals. In our example, we divided the x-axis from 0 to 5 into five intervals, so in that case, n = 5. Thus, the Riemann sum that represents the example above can be expressed as f(x i ) x Now, what exactly does this mean? Well, the symbol is called the summation symbol. It means that we add up all the numbers f(x i ) x for all of the values of i from 1 to 5. In other words, f(x i ) x = f(x 1 ) x + f(x 2 ) x + f(x 3 ) x + f(x 4 ) x + f(x 5 ) x Now, f(x 1 ) represents the chosen value of the function in the first interval, f(x 2 ) represents the chosen value of the function for the second interval, etc., and remember that the value of the function chosen for the first interval is f(0), the chosen value for the second interval is f(1), etc. f(x i ) x = f(x 1 ) x + f(x 2 ) x + f(x 3 ) x + f(x 4 ) x + f(x 5 ) x f(x i ) x = f(0) x + f(1) x + f(2) x + f(3) x + f(4) x and remembering that x = 1, = (7) x + (11) x + (23) x + 43 x + 71 x, = (7)(1) + (11)(1) + (23)(1) + 43(1) + 71(1) = 155 Now, what if we were to use the right-hand Riemann sum to approximate the area under the curve from 0 to 5? A right-hand Riemann sum means that we use the values of the function on the right side of each interval. If we again use x = 1, then the following table lists the intervals and the values of the function that we use on each interval: Interval x i f(x i ) 1 [0, 1] x 1 = 1 f(1) = 11 2 [1, 2] x 2 = 2 f(2) = 23 3 [2, 3] x 3 = 3 f(3) = 43 4 [3, 4] x 4 = 4 f(4) = 71 5 [4, 5] x 5 = 5 f(5) = 107 Thus, the right-hand Riemann sum is f(x i ) x = f(x 1 ) x + f(x 2 ) x + f(x 3 ) x + f(x 4 ) x + f(x 5 ) x = f(1) x + f(2) x + f(3) x + f(4) x + f(5) x = 11(1) + 23(1) + 43(1) + 71(1) + 107(1) = 255 Thus, the right-hand Riemann sum is 255 square units. 7

8 Now, let s find the left-hand Riemann sum approximating the area under the curve from 0 to 5 using x = 0.5. This will be trickier, so let s make a table: Interval x i f(x i ) 1 [0, 0.5] x 1 = 0 f(0) = 7 2 [0.5, 1] x 2 = 0.5 f(0.5) = 8 3 [1, 1.5] x 3 = 1 f(1) = 11 4 [1.5, 2] x 4 = 1.5 f(1.5) = 16 5 [2, 2.5] x 5 = 2 f(2) = 23 6 [2.5, 3] x 6 = 2.5 f(2.5) = 32 7 [3, 3.5] x 7 = 3 f(3) = 43 8 [3.5, 4] x 8 = 3.5 f(3.5) = 56 9 [4, 4.5] x 9 = 4 f(4) = [4.5, 5] x 10 = 4.5 f(4.5) = 88 Now, f(x i ) x = f(x 1 ) x+f(x 2 ) x+f(x 3 ) x+f(x 4 ) x+f(x 5 ) x+f(x 6 ) x+f(x 7 ) x+f(x 8 ) x+f(x 9 ) x+f(x 10 ) x = f(0) x+f(0.5) x+f(1) x+f(1.5) x+f(2) x+f(2.5) x+f(3) x+f(3.5) x+f(4) x+f(4.5) x = 7(0.5) + 8(0.5) + 11(0.5) + 16(0.5) + 23(0.5) + 32(0.5) + 43(0.5) + 56(0.5) + 71(0.5) + 88(0.5) = Thus, the left-hand Riemann sum approximating the area under the curve of the function from 0 to 5 is equal to square units. If you compute the right-hand Riemann sum with x = 0.5, you will see that as x decreases, the difference between the left and right sums also decreases. Ideally, we d like to make x very small, which means LOTS of calculations, and that means that it would be wise for us to leverage technology, such as Mathematica, to help us! 8