#$&* course Math 163 Submittted 12-17-10 @9:15pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: `a** Our assumptions are that rabbits require a month after birth to mature, mature rabbbits produce a pair of newborns every month starting the first month after they reach maturity, and rabbits never die. Every month the number of newborn pairs is equal to the number of mature pairs in the preceding month, which is equal to the total number of pairs from the month before that. Since all the rabbits from the preceding month are still present, the total number in the new month will be equal to the total number from the preceding month plus the number of newborns, which is equal to the total number from the preceding month plus the month before that. If we start with 1 pair of newborns then 1 month later we have a pair of mature rabbits, so after another month we have 2 pairs of rabbits. Our first three numbers are therefore 1, 1 and 2. In the next month we will have the 2 pairs we had in the preceding month plus 1 pair of newborns from the pair we had the month before that for a total of 3 pairs. In the next month we will have the 3 pairs we had in the preceding month plus 2 pairs of newborns from the pair we had the month before that for a total of 5 pairs. In the next month we will have the 5 pairs we had in the preceding month plus 3 pairs of newborns from the pair we had the month before that for a total of 8 pairs. The pattern continues 8 + 5 = 13 13 + 8 = 21 21 + 13 = 34 etc. . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qExplain how the rule a(n) = a(n-1) + a(n-2) is related to the enumeration of the rabbit population. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The number of mature rabbits at the end of month n is equal to a(n-1), the total number of rabbits at the end of the previous month. The number of baby rabbits at the end of month n is equal to M(n-1), the number of mature rabbits at the end of the previous month. So the number a(n) of rabbits at the end of month n is the number of mature rabbits plus the number of baby rabbits, which is a(n) = a(n-1) + M(n-1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qSTUDENT ANSWER: a(n) is total number of rabbits for month(n) a(n-1) is total for previous month a(n-2) is total for 2 months previous which is total mature for 1 month previous INSTRUCTOR COMMENT: Good. In terms of the solution given for the preceding exercise: In the nth we will have the a(n-1) pairs we had in the preceding month plus a(n-2) pairs of newborns from the a(n-2) pairs we had the month before that for a total of a(n) = a(n-1) + a(n-2) pairs. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qWhat are your ratios a(n) / a(n-1) for n = 1, 2, 3, 4, ..., 10, and what does your graph of ratio vs. n look like? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The ratios are 1/1=1, 2/1=2, 3/2=1.5, 5/3=1.66,8/5= 1.6, 13/8=1.625, 21/13=1.615, 34/21=1.619, 55/34=1.6176 89/55=1.6181. As you can see from the points used on the graph, they are up and down but less as you move through the points. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** The ratios are 1, 2, 1.5, 1.66, 1.6, 1.625, 1.615, 1.619, 1.6176 1.6181 1.61797 etc. The graph is jagged, up one time, down the next, but jumping less and less each time. If you were to make a horizontal line through two successive graph points, all subsequent points would be 'squeezed' between these lines. COMMON ERROR: The ratios are 1, 2, 1.5, 1.7, 1.6, 1.6, 1.6, ... . You rounded off before you could see the difference in the last 3 results. If you don't see a difference in the ratios you need to use more significant figures--carry your calculation out to enough decimal places that the differences show. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q problem 8 Use the Fibonacci sequence data points for n = 5 and n = 10 to obtain an exponential model in the form y = a b^x. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For n=5, you would substitute the numbers 8 and 5 into the exponential model and for n=10, you would substitute the numbers 89 and 10 so you would have the two equations. 8=ab^5 and 89=ab^10 Next you divide the first equation by the second one to get 8/89 = b^(10-5) = 11.125 = b^5 = b = 11.125^5 = b = 1.619 Now you put the value of b into the first equation. 8 = a*1.619^5, 8 = 11.125A A = .719 So the model would be y= .719 * 1.619^x confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** You would use the points (5, 8) and (10,89) for your y = A b^x model. You would get the equations 8 = A b^5 89 = A b^10 Dividing the second equation by the first you would get 11.125 = b^5, so b = (11.125)^1/5 = 1.619. Substituting into the first equation we get 8 = A * (1.619)^5 A = 8 /(1.619)^5 = .711. So the model is y = .719 * 1.619^x. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qWhat are the values predicted by your model for the first 10 members of the Fibonacci sequence? What is the average error of your approximation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Substituting 0 through 9 for the value of x into the model y = .719 * 1.619^x you get the following values: .719, 1.164, 1.884, 3.051, 4.939, 7.997, 12.948, 20.963, 33.939, 54.947. The Fibonnaci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 giving the differences of -.281, .164, -.116, -1.16, -1.949, -3.061, -5.003, -8.052, -3.037, -21.061. The average error is -4.2396 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: `a** For x = 0, 1, 2, ..., 9 the formula gives us values 0.719, 1.163342, 1.882287356, 3.045540942, 4.927685244, 7.972994725, 12.90030547, 20.87269424, 33.77201928, 54.6431272. These numbers differ from the Fibonnaci numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 by -0.281, 0.163342. -0.117712644, 0.045540942, -0.072314756, -0.027005275, -0.099694535, -.127305757, -0.22798071, and -0.356872798. These are the errors in the function. The average of these errors is easily found by adding the errors and dividing by 10, obtaining an average of about -.016. ** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf mature rabbits were to produce two baby pairs per month, what then would be the definition of the Fibonacci-type sequence that models the situation where we start with 1 pair of baby rabbits, and what would be population be at the end of each of the first 6 months?{}{}** Remember that the total number of pairs in month n-2 will be the number of mature pairs in month n - 1, which will produce the newborns in month n. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we need to calculate 2 baby pairs at the end of each month instead of 1, the new model would be a(n) = a(n-1) + 2 a(n-2) with a(0) = 1 and a(1) = 1. Substituting 1 – 6 months for the value of n you would have: a(1) = 1 a(2) = a(2-1) + 2 a(2-2) ; a(2) = a(1) + 2 * a(0); a(2) = 1 + 2 * 1 = 3 a(3) = a(3-1) + 2 a(3-2) ; a(3) = a(2) + 2 * a(1); a(3) = 3 + 2 * 1 = 5 a(4) = a(4-1) + 2 a(4-2) ; a(4) = a(3) + 2 * a(2); a(4) = 5 + 2 * 3 = 21 a(5) = a(5-1) + 2 a(5-2) ; a(5) = a(4) + 2 * a(3); a(5) = 21 + 2 * 5 = 31 a(6) = a(6-1) + 2 a(6-2) ; a(6) = a(5) + 2 * a(4); a(6) = 31 + 2 * 21 = 73 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `aThe model would give 2 baby pairs for each mature pair; in month n the number of mature pairs is a(n-2), so there would be 2 a(n-2) baby pairs in month n. The a(n-1) pairs from the previous month would also survive, giving a population of a(n) = a(n-1) + 2 a(n-2). The numbers would be 1, 1, 3, 5, 11, 21, 43, ...... . e.g, 1 + 2 * 1 = 3; 3 + 2 * 1 = 5; 5 + 2 * 3 = 11; 11 + 2 * 5 = 21; 21 + 2 * 11 = 43 etc.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qHow would the Fibonacci model change if rabbits required two months to mature, with each mature pair still producing 1 pair of baby rabbits per month?ith each mature pair still producing 1 pair of baby rabbits per month? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would need to determine the total number of pairs from 2 months previous so you would look back 3 months to the population a(n-3). The surviving rabbits would be a(n-1) so you would have a(n)= a(n-1)+a(n-3). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** The number of baby pairs in month n would be the number of mature pairs from the month before, which would be the total number of pairs from 2 months previous to that, so we would be looking back 3 months to the population a(n-3). There would be a(n-3) mature pairs in month n-1 which would give a(n-3) baby pairs in month n. The rabbits from month n-1 would still survive so we would have a(n) = a(n-1) + a(n-3). Note that to get started with this model we would need the populations for the first three months. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qproblems 12-16 How did you describe how to reason out the dosage D required for a desired level L of a drug if proportion r of the drug present immediately after a dose has been removed at the time of the next dose. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the desired level of the drug is L, and if the proportion lost between doses is r, then (1-r) of the drug present immediately after a dose D will remain when it is time for the next dose. The amount after the dose is L + D multiplied times the amount remaining (l-r). So you have (L+D)(1-r) and then solve for L. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a** If proportion r has been removed to give concentration L, then 1-r of the amount immediately after the dose will remain. The amount immediately after the dose is L + D so we have (L + D) * ( 1 - r) = L. You then solve this equation for L: Expanding the left-hand side we have L(1-r) + D(1-r) = L, or L - r L + D - r D = L. Rearranging the equation to isolate L we get - r L = - ( D - r D), or L = ( D - r D) / r = ( 1-r) / r * D. ** YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aStarting with a dose of 429 mg at the instant the concentration of a drug reaches 1000 mg, and knowing that 30% of the drug is removed in 6 hours, what exponential function models the amount of drug vs. time for the 6-hour interval between doses? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We need an exponential function y = A * b^t (note that when t = 0 y = A) such that when t = 6, y is 30% less than A, or y = .7 A. Thus .7 A = A b ^ 6 and b = .7 ^(1/6) = .942. Our model is thus y = A * .942 ^ t. Since when t = 0 we have y = 1429 mg, the model is y = 1429 mg * .942^t. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `qWhat dosage is required to maintain a minimum level of 800 mg if 70% of the drug is removed between doses? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: