course mth 163 I tried to pull up Query for 27 but got a runtime error.
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19:32:50 Have you completed Test 2? If so give comments, insights, etc.. This is the only question on this Query.
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RESPONSE --> Yes, I took it on Monday, April 23. I was not feeling well nad didn't remember how to do a couple of the problems. confidence assessment: 2
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??x?·?????·????S?·? assignment #026 026. `query 26 Precalculus I 04-26-2007
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20:22:57 Give the rabbit populations for the first 12 months. Explain how each new population is obtained, and what your method for obtaining the new population has to do with the assumed nature of rabbit reproduction.
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RESPONSE --> The way I went about finding the first 12 mos. was by making a table. Month 0: 1 pair (2 rabbits); Month 1: 1 pair (2 rabbits) Month 2: 2 pair (4 rabbits); Month 3: 3 pair (6 rabbits) Month 4: 5 pair (10 rabbits) Month 5: 8 pair (16 rabbits) Month 6: 13 pair (26 rabbits) Month 7: 21 pair (42 rabbits) Month 8: 34 pair (68 rabbits) Month 9: 55 pair (110 rabbits) Month 10: 89 pair (178 rabbits) Month 11: 144 pair (288 rabbits) Month 12: 233 pair (466 rabbits) You look to see how many ""old mature"" sets of rabbits you have. Each set will have a new set of babies. You will have as many ""new mature"" rabbits as you had babies the previous month. The chart I drew makes it much easier to explain and understand confidence assessment: 2
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20:27:28 ** 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. . **
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RESPONSE --> This explaination is what I was trying to say but without copying the notes word for word. self critique assessment: 2
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20:41:14 Explain how the rule a(n) = a(n-1) + a(n-2) is related to the enumeration of the rabbit population.
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RESPONSE --> a(n-1) is the number of mature rabbits at the end of the month (n), which is also the total number of rabbits at the end of the previous month which is (n-2) confidence assessment: 1
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20:48:08 STUDENT 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.
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RESPONSE --> I was almost complete with my answer. I forgot to mention that a(n) was the total number of rabbits. I also should have said that (n-2) was the total for 2 months previous. confidence assessment: 2
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21:05:12 What 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?
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RESPONSE --> my ratios are: 1, 2, 1,5, 1.6667, 1.6, 1.625, 1.615, 1.619, 1.618, 1.618 The graph begins at 1 spikes upward, to 2, then dips back down, spikes back up but not as high, then dips back down then levels off into a horizontal line confidence assessment: 2
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21:05:25 ** 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. **
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RESPONSE --> self critique assessment: 3
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21:18:50 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.
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RESPONSE --> y = 1.618^x confidence assessment: 1
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21:27:13 ** 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. **
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RESPONSE --> I wasn't totally sure what I needed to do. After reading the explaination I totally understand. I just didn't associate data points and dividing the equations. self critique assessment: 2
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21:53:56 What are the values predicted by your model for the first 10 members of the Fibonacci sequence? What is the average error of your approximation?
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RESPONSE --> The totals are as follows (beginning with 1) y = .719 * 1.619^x 1 = 1.164 2 = 1.885 3 = 3.051 4 = 4.928 5 = 7.998 6 = 12.900 7 = 20.963 8 = 33.772 9 = 54.948 10 = 88.961 I averaged all the differences and got an error of .043 confidence assessment: 2
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21:56:15 ** 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. **
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RESPONSE --> These were not the same numbers I got. I don't fully understand why..... self critique assessment: 1
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23:32:50 If 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.
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RESPONSE --> By using the table I made earlier for myself, I found there would be 32 pairs of rabbits or 64 total. I'm not sure how to go about figuring the Fibonacci-type sequence. I am thinking it would be a(n) = a(n-1)+2a(n-2) self critique assessment: 2
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23:37:18 The 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.. **
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RESPONSE --> I got the formula correct, the numbers wron with my chart. I missed a set of bunnies. self critique assessment: 2
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23:38:23 How 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?
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RESPONSE --> a(n) =(n - 1) + (n-3) confidence assessment: 1
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23:38:38 ** 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. **
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RESPONSE --> I had it right :):) self critique assessment: 3
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23:43:42 problems 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.
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RESPONSE --> We want a certain level of a drug in the body. We call this L. The proportion of the drug lost between doses is r. To get the amount of hte drug present after the dose you subtract r from 1 or (1-r). To find the dose (D), we see how much will be present right after the dose which is the level we want to maintain plus the dose (L+D). We also have to find out how much o fhte drug remains. We do this by multiplying (1-r) by (L+D). This gives us the formula (L+D)(1-r) = L then we solve for D. confidence assessment: 2
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23:43:52 ** 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. **
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RESPONSE --> self critique assessment: 3
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20:33:34 Starting 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?
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RESPONSE --> I think that I need to find two data points, divide and solve, but I am not sure self critique assessment: 1
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20:39:27 ** 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. **
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RESPONSE --> I was wrong. I wasn't sure exactly how to go about this. I know how to find exponential functions, but just not ones quite like this. After reading, I get it. self critique assessment: 1
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20:43:34 What dosage is required to maintain a minimum level of 800 mg if 70% of the drug is removed between doses?
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RESPONSE --> Using the equation from the beginning: D=rL/(1-r) D = .7(800)/(1-.7) D = 560/.3 D = 1866.67 confidence assessment: 2
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20:45:15 ** The minimum level occurs just before the next dose, and must be 70% of the maximum level which occurs just after the dose. If D is the dose and 800 mg is the level just before the does, then just after the dose the level is 800 mg + D. This level falls to 70%, i.e., to .70 ( 800 mg + D), just before the next dose. This level should be 800 mg. We therefore want ( D + 800 mg) ( 1 - .7) = 800 mg. Subtracting 1 - .7 to get .3, and multiplying this through the grouped expression on the left we get .3 D + 800 mg * .3 = 800 mg so .3 D = 800 mg - .3 * 800 mg = 560 mg and D = 560 mg / .3 = 1867 mg (check my arithmetic). **
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RESPONSE --> I didn't round, but was right! self critique assessment: 3
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20:46:41 What will be the maintenance dose of a drug, given a dose of 500 mg with 35% of the drug removed between doses?
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RESPONSE --> I used the same equation: D= rL/(1-r) 500 = .35L/(1-.35) 500=.35L/.65 325 = .35L 928.57 = L confidence assessment: 2
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20:46:51 ** The way this question is stated, with the word 'dose' used for two different quantities, is incorrect. With this miswording you have the choice of lett 500 mg stand for the dose, or for the maintenance level. If 500 mg is read as the dose, then the maintenance level is L = ( 1-r) / r * D = (1 - .35) / .35 * D = .65 / .35 * D = 930 mg, approx.. If you read this as a 500 mg maintenence level then we L = 500 mg and r = .35 so we have L = ( 1-r) / r * D so that D = r / (1-r) * L = .35 / (1-.35) * 500 = .35 / .65 * 500 = 269.23. This is the dose required to maintain a 500 mg level. If you read this as saying that I believe you might have added the .3 * 800 mg to both sides, which would have given you something a lot like the 3467 mg you got. **
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RESPONSE --> self critique assessment: 3
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20:55:29 04-27-2007 20:55:29 If a patient starts with no drug in her body and takes a 500 mg dose every six hours, losing 40% of the after-dose amount during that time, then how much drug will remain six hours after the initial dose?
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NOTES -------> Since there is no drug to begin with, we start off with the amount just after does which is equal to 500 mg. Since 40% of the after dose is lost, (500*.4) = 200 mg is removed. So 500 - 200 leaves 300 mg before next dose.
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20:55:30 If a patient starts with no drug in her body and takes a 500 mg dose every six hours, losing 40% of the after-dose amount during that time, then how much drug will remain six hours after the initial dose?
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RESPONSE --> confidence assessment: 2
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20:58:34 ** In the time between doses concentration will fall to .60 of the after-dose concentration. Just before the second dose the level will be .60 * 500 mg = 300 mg. Just after the second dose the level will be 300 mg + 500 mg = 800 mg. Just before the third dose the level will be .60 * 800 mg = 480 mg. Just after the third dose the level will be 480 mg + 500 mg = 980 mg. Just before the fourth dose the level will be .60 * 980 mg = 588 mg. Just after the fourth dose the level will be 588 mg + 500 mg = 1088 mg. Just before the fifth dose the level will be .60 * 1088 mg = 653 mg approx.. Just after the fifth dose the level will be 653 mg + 500 mg = 1153 mg. approx. Just before the sixth dose the level will be .60 * 1153 mg = 692 mg approx.. **
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RESPONSE --> Did I miss something? I though it only asked for the first dose? self critique assessment: 2
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21:38:32 What sort of function do you think would model the just-before-dose drug concentration?
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RESPONSE --> It will be an exponential like the previous one: When t = 6, y will be 40% less than A or .6A. So .6A = Ab^6. y = ab^t .6A = Ab^6 and solving for b : b = .6(1/6) = .918 So the model is y = a * .918^t confidence assessment: 1
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21:38:55 ** The increases in mg are 180, 108, 65, 39, 24. 108/180 = .6; 65/108=/6; 39/65=.6; 24/39=.6. The ratio sequence for the increases is constant. This implies an exponential function. **
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RESPONSE --> self critique assessment: 3
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21:55:05 problem 19 Determine the minimum concentration vs. dosage cycle of a drug for six dosage cycles, starting with the first dose, if the dosage is 750 mg and if 55% of the maximum concentration of the drug is removed during each cycle.
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RESPONSE --> I didn't totally understand this series of questions. I went back and read, but still do not fully understand. confidence assessment: 0
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22:15:49 ** After 55% of the first 750 mg dosage is removed there will be about 338 mg left; after the next 750 mg dose there will be 1088 mg, which after removal of 55% will leave about 490 mg. etc. This gives us minimum concentrations of about 338 mg, 490 mg, 558 mg, 588 mg, 602 mg, 609 mg, and 611 mg. **
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RESPONSE --> After reading the explaination I think I understand it better. All I had to do was to multiply like I had been doing before. self critique assessment: 1
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22:42:18 What is the function that models the minimum concentration during a cycle vs. the number of the cycle?
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RESPONSE --> II am not sure what to use for t because it doesn't shouel the time. confidence assessment: 0
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22:51:01 ** As before you can find that the ratios of the sequence of differences are constant, so the function exponentially approaches the limiting concentration. The limiting after-dose concentration L is that for which the amount removed during a cycle is equal to the dosage. At that level the amount removed will be .55 L, which will equal the 750 mg dose. So we have .55 L = 750 mg, which gives us L = 750 mg / .55 =1363.6 mg approx.. The minimum concentration will be 1363.6 mg - 750 mg = 613.6 mg, approx.. Our concentrations of 0, 338, 490, 558, 588, 602, 609, 611 mg lie 276, 124, 56, 25, 11, 5 and 2 mg below the limiting minimum concentration. We find the exponential function that models these differences. The differences are modeled by an exponential function y = A b^n, with y = 276 when n = 1 and y = 124 when n = 2. Solving the equations 276 = A * b^1 124 = A * b^2 for A and b we obtain A = 613.6 and b = .45. The function for the differences is therefore difference = 613.6 * .45^n. The concentrations lie below the limiting minimum concentration of 613.6 by this amount, so our exponential function is concentration = 613.6 - 613.6 * .45^n, or if we factor out the 613.6 concentration = 613.6 ( 1 - .45^n), where the concentration is in mg and n stands for the number of the last dose taken. **
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RESPONSE --> I didn't think about the ratios. This is when get confused. I will read over the explaination again to make sure I fully understand it. self critique assessment: 1
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23:19:13 ** If we have a(n-1) just after dose n - 1 then just before dose n we have .40 * a(n-1). So just after dose n we will have a(n) = .40 * a(n-1) + 500 mg. This recurrence relation together with the initial condition a(0) = 0 defines the situation. }If a(0) = 0 then we have a(1) = .40 * a(0) + 500 mg = 500 mg a(2) = .40 * a(1) + 500 mg = .40 * 500 mg + 500 mg = 700 mg a(3) = .40 * a(2) + 500 mg = .40 * 700 mg + 500 mg = 780 mg etc. **
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RESPONSE --> I am having trouble with these particular questions. After reading this explaination I think I understand how go about doing this kind of problem. It all makes sense once I read the explaination. self critique assessment: 1
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23:25:30 problem 22 What is the dosage and what proportion of a drug is removed if the recurrence relation for the concentration a(n) just before the nth dose is a(n) = .3(a(n-1) + 800), a(1) = 0 ?
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RESPONSE --> From the previous question the proportion is .7 and hte dose is 800? confidence assessment: 0
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23:25:46 The .3 indicates that .3 of the drug remains between doses so that .7 or 70% is removed. The 800 indicates that a does of 800 units is added to the existing concentration with each new does.
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RESPONSE --> YEAH!!!!!!!!!!!!!! I got one!!!!!!!!!!!!!!!!!! self critique assessment: 3
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23:27:05 problem 23 Write the general recurrence relation for initial concentration 0, dosage D and proportion removed = r.
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RESPONSE --> Is that the same as a(n) - r(a(n-1) + D? confidence assessment: 0
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23:27:28 ** If a(n) stands for the amount just after dose number n then (amt just after dose n) = dose + (amt after proportion r is removed from dose n - 1) we have a(n) = D + (1 - r) * a(n-1). **
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RESPONSE --> I forgot 1-r, not just r. self critique assessment: 1
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