course Mth 163 ??????????????Student Name: assignment #015
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22:14:31 `q001. Note that this assignment has 12 questions If you are given $1000 and invest it at 10% annual interest, compounded annually, then how much money will you have after the first year, how much after the second, and how much after the third? Is the the change in the amount of money the same every year, does the change increase year by year, does the change decrease year by year or does it sometimes increase and sometimes decrease?
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RESPONSE --> The initial monetary amount is $1000, with a 10% anual growth, the growth factor is 1.1, gotten from 1+.1. To find the new principle after the first year, multiply the initial amount (1000) by the growth factor (1.1) to get 1100 dollars. To find the new principle after the second year, multiply the initial amount (1100) by the growth factor (1.1) to get 1210 dollars. To find the new principle after the third year, multiply the initial amount (1210) by the growth factor (1.1) to get 1331 dollars. The change in the amount of money increases year by year.
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22:14:46 During the first year the interest will be 10% of $1000, or $100. This makes the total at the end of the first year $1100. During the second year the interest will be 10% of $1100, or $110. At the end of the second year the total will therefore be $1100 + $110 = $1210. During the third year the interest will be 10% of $1210, or $121. At the end of the sphere year the total will therefore be $1210 + $121 = $1331. The yearly changes are $100, $110, and $121. These changes increase year by year.
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22:18:06 `q002. In the preceding problem you obtained amounts $1100, $1210 and $1331. What number would you multiply by $1000 to get $1100? What number we do multiply by $1100 to get $1210? What number would we multiply by $1210 to get $1331? What is the significance of this number and how could we have found it from the original information that the amount increases by 10 percent each year?
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RESPONSE --> you multiply all of the values by 1.1 to arrive at the new amount.This amount is the growth factor, which is found by adding one to the 10% increase each year (1+.1=1.1)
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22:18:27 To get $1100 you have to multiply $1000 by 1100 / 1000 = 1.1. To get $1210 you have to multiply $1210 by 1210 / 1100 = 1.1. To get $1331 you have to multiply $1331 by 1331 / 1210 = 1.1. If the amount increases by 10 percent, then you end up with 110 percent of what you start with. 110% is the same as 1.1.
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22:23:39 `q003. Given the recurrence relation P(n) = 1.10 * P(n-1) with P(0) = 1000, substitute n = 1, 2, and 3 in turn to determine P(1), P(2) and P(3). How is this equation related to the situation of the preceding two problems?
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RESPONSE --> P(n) = 1.10*P(n-1); P(0)=1000 P(1) = 1.10*P(1-1) = 1.10*P(0) = 1.10 * 1000 = 1100 P(2) = 1.10*P(2-1) = 1.10*P(1) = 1.10 * 1100 = 1210 P(3) = 1.10*P(3-1) = 1.10*P(2) = 1.10 * 1210 = 1331 This equation results in the same amounts arrived at through the previous problem, this is because the growth factor from the previous problem (1.1) is multiplied by the principles (1000,1100,1210,and 1331) to get the next step up.
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22:23:45 Substituting 1 into P(n) = 1.10 * P(n-1) we obtain P(1) = 1.10 * P(1-1), or P(1) = 1.10 * P(0). Since P(0) = 1000 we get P(1) = 1.1 * 1000 = 1100. Substituting 2 into P(n) = 1.10 * P(n-1) we obtain P(2) = 1.10 * P(1). Since P(1) = 1100 we get P(1) = 1.1 * 1100 = 1210. Substituting 3 into P(n) = 1.10 * P(n-1) we obtain P(3) = 1.10 * P(2). Since P(2) = 1000 we get P(3) = 1.1 * 1210 = 1331.
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22:26:53 `q004. If you are given $5000 and invest it at 8% annual interest, compounded annually, what number would you multiply by $5000 to get the amount at the end of the first year? Using the same multiplier, find the results that the end of the second and third years.
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RESPONSE --> At the end of a first year with the initial amount of $5000, at an 8% annual interest, you would multiply $5000 by 1.08 (1+.08=1.08) to get the new principle after one year of 5400 dollars. Second year: 5400 * 1.08 = $5832 Third year: 5832 * 1.08 = $6298.56
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22:27:14 If your money increases by 8% in a year, then it the end of the year you will have 108% as much as at the beginning. Since 108% is the same as 1.08, our yearly multiplier will be 1.08. If we multiply $5000 by 1.08, we obtain $5000 * 1.08 = $5400, which is the amount the end of the first your. At the end of the second year the amount will be $5400 * 1.08 = $5832. At the end of the third year the amount will be $5832 * 1.08 = $6298.56.
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22:28:23 `q005. How would you write the recurrence relation for a $5000 investment at 8 percent annual interest?
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RESPONSE --> P(n) = 1.08 * P(n-1) with P(0) = $5000
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22:28:31 Just as the recurrence relation for 10 percent annual interest, as seen in the problem before the last, was P(n) = 1.10 * P(n-1), the recurrence relation for 8 percent annual interest is P(n) = 1.08 * P(n-1).
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22:32:17 `q006. If you are given amount $5000 and invest it at annual rate 8% or .08, then after n years how much money do you have? What does a graph of amount of money vs. number of years look like?
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RESPONSE --> After n years, you would have 1.08*P(n-1), the graph of amount of money vs. number of years starts at y value 5000 and then gradually increases at an increasing rate from that point.
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22:34:46 After 1 year the amount it $5000 * 1.08. Multiplying this by 1.08 we obtain for the amount at the end of the second year ($5000 * 1.08) * 1.08 = $5000 * 1.08^2. Multiplying this by 1.08 we obtain for the amount at the end of the third year ($5000 * 1.08^2) * 1.08 = $5000 * 1.08^3. Continuing to multiply by 1.08 we obtain $5000 * 1.08^3 at the end of year 3, $5000 * 1.08^4 at the end of year 4, etc.. It should be clear that we can express the amount at the end of the nth year as $5000 * 1.08^n. If we evaluate $5000 * 1.08^n for n = 0, 1, 2, ..., 10 we get $5000, $5400, 5832, 6298.56, 6802.45, 7346.64, 7934.37, 8569.12, 9254.65, 9995.02, 10,794.62. It is clear that the amount increases by more and more with every successive year. This result in a graph which passes through the vertical axis at (0, 5000) and increases at an increasing rate.
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RESPONSE --> the form I used, 1.08*P(n-1) is not very helpful when trying to find the amount of money after 200 years or so, since you would have to evaluate all of the previous build ups, so $5000 * 1.08^n expresses the value of n much easier without have to go up the steps.
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22:38:20 `q007. With a $5000 investment at 8 percent annual interest, how many years will it take to double the investment?
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RESPONSE --> With a $5000 investment at 8% annual interest, it will take 10 years to completely double the investment. 5000*1.08^10 = $10794.62
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22:39:55 Multiplying $5000 successively by 1.08 we obtain amounts $5400, 5832, 6298.56, 6802.45, 7346.64, 7934.37, 8569.12, 9254.65, 9995.02, 10,794.62 at the end of years 1 thru 10. We see that the doubling to $10,000 occurs very shortly after the end of the ninth year. We can make a closer estimate. If we calculate $5000 * 1.08^x for x = 9 and x = 9.1 we get about $10,072. So at x = 9 and at x = 9.1 the amounts are $9995 and $10072. The first $5 of the $77 increase will occur at about 5/77 of the .1 year time interval. Since 5/77 * .1 = .0065, a good estimate would be that the doubling time is 9.0065 years. If we evaluate $5000 * 1.08^9.0065 we get $10,000.02.
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RESPONSE --> Yes, 9.0065 is a more precise answer, but since it is annually, I was thinking in complete years, not fractions of them.
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22:41:03 `q008. If you are given amount P0 and invest it at annual rate r (e.g., for the preceding example r would be 8%, which in numerical form is .08), then after n years how much money do you have?
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RESPONSE --> $ = P0*r^n r = annual rate n = years
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22:42:14 If the annual interest rate is .08 then each year we would multiply the amount by 1.08, the amount after n years would be P0 * 1.08^n. If the rate is represented by r then each year then each year we multiply by 1 + r, and after n years we have P0 * (1 + r)^n.
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RESPONSE --> In my mine I was automatically adding the 1 to the rate, however, my answer did not show that. But I understand that you have to add one to the rate because then that will take the former value plus the additional percentage.
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22:49:26 `q009. If after an injection of 800 mg an antibiotic your body removes 10% every hour, then how much antibiotic remains after each of the first 3 hours? How long does it take your body to remove half of the antibiotic?
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RESPONSE --> An injection of 800mg of antibiotic removes 10% every hour, so the growth factor would be found by subtracting .1 from 1 = .9 After One hour: 800mg*.9 = 720 mg second hour 720 * .9 = 648 mg third hour 648 * .9 = 583.2 mg To find when half of the anitbiotic will be removed, I solved 400 = 800 * .9^n .5=.9^n and through guessing and testing, I concluded that in about 6.577 hours, half of the anitibotic will be removed.
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22:50:00 If 10 percent of the antibiotic is removed each hour, then at the end of the hour the amount left will be 90 percent of what was present at the beginning of the hour. Thus after 1 hour we have .90 * 800 mg, after a second hour we have .90 of this, or .90^2 * 800 mg, and after a third hour we have .90 of this, or .90^3 * 800 mg. The numbers are 800 mg * .90 = 720 mg, then .90 of this or 648 mg, then .90 of this or 583.2 mg.
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22:51:04 `q010. In the preceding problem, what function Q(t) represents the amount of antibiotic present after t hours? What does a graph of Q(t) vs. t look like?
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RESPONSE --> Q(t) = 800 mg * .9^t The graph of Q(t) will have a y intercept of (0,800) and will increase at an increasing rate from this point.
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22:52:43 After t hours we will have 800 * .9^t mg left. So Q(t) = 800 * .9^t. The amounts for the first several years are 800, 720, 648, 583.2, etc.. These amounts decrease by less and less each time. As a result the graph, which passes through the vertical axes at (0,800), decreases at a decreasing rate. We note that no matter how many times we multiply by .9 our result will always be greater than 0, so the graph will keep decreasing at a decreasing rate, approaching the horizontal axis but never touching it.
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RESPONSE --> I was absent mindedly thinking of this problem as if it were like the previous which the rate was increasing so the graph increased at an increasing rate, however, this rate is negative, so the resulting graph would do the opposite as the previous, it would decrease at a decreasing rate
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23:02:35 `q011. Suppose that we know that the population of fish in a pond, during the year after the pond is stocked, should be an exponential function of the form P = P0 * b^t, where t stands for the number of months after stocking. If we know that the population is 300 at the end of 2 months and 500 at the end of six months, then what system of 2 simultaneous linear equations do we get by substituting this information into the form of the function?
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RESPONSE --> P=PO*b^t with t = months after stocking the population is 300 after 2 months, (2,300) and is 500 after six months (6,500) plugging these coordinates into the equation, gives 300 = PO *b^2 and 500 = PO*b^6. [300 = PO *b^2] / [500 = PO*b^6] => 300/500 = b^2/b^6, subtract exponents, .6 = b^ -4, raise both sides to -1/4 so b is equal to about 1.14, plug this into one of the equations to find the value of PO, 300=PO*(1.14)^2 300=PO(1.29) PO=232.38 approximately So, P=232.38(1.14)^t
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23:02:40 We substitute the populations 300 then 500 for P and substitute 2 months and 6 months for t to obtain the equations 300 = P0 * b^2 and 500 = P0 * b^6.
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23:03:57 `q012. We obtain the system 300 = P0 * b^2 500 = P0 * b^6 in the situation of the preceding problem. If we divide the second equation by the first, what equation do we obtain? What do we get when we solve this equation for b? If we substitute this value of b into the first equation, what equation do we get? If we solve this equation for P0 what do we get? What therefore is our specific P = P0 * b^t function for this problem?
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RESPONSE --> [300 = PO *b^2] / [500 = PO*b^6] => 300/500 = b^2/b^6, subtract exponents, .6 = b^ -4, raise both sides to -1/4 so b is equal to about 1.14, plug this into one of the equations to find the value of PO, 300=PO*(1.14)^2 300=PO(1.29) PO=232.38 approximately So, P=232.38(1.14)^t
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23:04:13 Dividing the second equation by the first the left-hand side will be left-hand side: 500/300, which reduces to 5/3, and the right-hand side will be right-hand side: (P0 * b^6) / (P0 * b^2), which we rearrange to get (P0 / P0) * (b^6 / b^2) = 1 * b^(6-2) = b^4. Our equation is therefore b^4 = 5/3. To solve this equation for b we take the 1/4 power of both sides to obtain (b^4)^(1/4) = (5/3)^(1/4), or b = 1.136, to four significant figures. Substituting this value back into the first equation we obtain 300 = P0 * 1.136^2. Solving this equation for P0 we divide both sides by 1.136^2 to obtain P0 = 300 / (1.136^2) = 232.4, again accurate to 4 significant figures. Substituting our values of P0 and b into the original form P = P0 * b^t we obtain our function P = 232.4 * 1.136^t.
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??D€?O??W?`??x??? assignment #015 ??????????????Precalculus I 10-18-2006
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23:08:19 query modeling project #2 #5. $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300?
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RESPONSE --> growth rate =.1 growth factor = 1.1 doubles in about 7.25 years first reaches #300 at about 4 years.
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23:08:46 ** The growth rate is .10 and the growth factor is 1.10 so the amount function is $200 * 1.10^t. This graph passes through the y axis at (0, 200), increases at an increasing rate and is asymptotic to the negative x axis. For t=0, 1, 2 you get p(t) values $200, $220 and $242--you can already see that the rate is increasing since the increases are $20 and $22. Note that you should know from precalculus the characteristics of the graphs of exponential, polynomial an power functions (give that a quick review if you don't--it will definitely pay off in this course). $400 is double the initial $200. We need to find how long it takes to achieve this. Using trial and error we find that $200 * 1.10^tDoub = $400 if tDoub is a little more than 7. So doubling takes a little more than 7 years. The actual time, accurate to 5 significant figures, is 7.2725 years. If you are using trial and error it will take several tries to attain this accuracy; 7.3 years is a reasonable approximation for trail and error. To reach $300 from the original $200, using amount function $200 * 1.10^t, takes a little over 4.25 years (4.2541 years to 5 significant figures); again this can be found by trial and error. The amount function reaches $600 in a little over 11.5 years (11.527 years), so to get from $300 to $600 takes about 11.5 years - 4.25 years = 7.25 years (actually 7.2725 years to 5 significant figures). **
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23:11:01 At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?
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RESPONSE --> t=20 when p = 1345.50, half of this is 672.75 and is reached at about 14 years, the doubling time is 28 years
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23:11:58 ** The t = 20 value is $200 * 1.1^20 = $1340, approx. Half the t = 20 value is therefore $1340/2 = $670 approx.. By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx.. For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45).?At 12.75=674.20 so it would probably be about12.72.? This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr ?12.7 yr = 7.3 yr. This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. **
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RESPONSE --> I was unsure of what doubling time meant, but it is the time it takes for the value to double, which for this case, is about 7.3 years.
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23:15:39 query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%
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RESPONSE --> The graph consisted of four lines that all increased at an increasing rate. 10% consisted of points (1,1.1), (2,1.21), (3, 1331), and (4, 1.4641) 20% consisted of points (1,1.2), (2,1.44), (3,1.728), and (4,2.0736) 30% consisted of points (1,1.3), (2,1.69), (3, 2.197), and (4, 2.8561) 40% consisted of points (1,1.4), (2, 1.96), (3,2.744), and (4, 3.8416)
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23:15:47 ** We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double. for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double. Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double. The final 4-year amount increases by more and more with each 10% increase in interest rate. The doubling time decreases, but by less and less with each 10% increase in interest rate. **
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23:17:07 query #11. What is the equation for doubling time?
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RESPONSE --> P(rate^doubling time)=2P
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23:17:55 ** the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The resulting equation is therefore P0 * (1+r)^t = 2 P0. Note that this simplifies to (1 + r)^ t = 2, and that this result depends only on the interest rate, not on the initial amount P0. **
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RESPONSE --> once again, when I say rate I imply that 1 has already been added to the percentage, which i need to make clear.
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23:23:08 Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%.
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RESPONSE --> $5000(1.08^(2+dubtime))=$100000(5000(1.08^2)), divide both sides by 5000 1.08^(2+dubtime) = 2(1.08^2) 1.08^2*1.08^dubtime = 2(1.08^2), divide both sides by 1.08^2 1.08^dubtime = 2 through trial and error i got the doubling time to be about 9.01 years
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23:23:14 **dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get 1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2]. This can be written as 1.08^2 * 1.08^doublingtime = 2 * 1.08^2. Dividing both sides by 1.08^2 we obtain 1.08^doublingtime = 2. We can then use trial and error to find the doubling time that works. We get something like 9 years. **
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23:25:01 Desribe how on your graph how you obtained an estimate of the doubling time.
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RESPONSE --> On the graph, you find the y value of 5000 and of that doubled, 10000, I personally drew horizontal lines at these values that intersected the graph itself, where these intersect, I drew vertical lines down to the x axis. Here, the difference between the x axis values is the doubling time.
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23:25:06 ** In this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis. The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. **
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23:26:05 #17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What is the function Q(t)?
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RESPONSE --> Q(t) = 550 mg (1-.11)^`dt = 550mg(.89)^`dt
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23:26:26 ** Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or Q(t)=550(.89)^t? **
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23:27:23 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> At 3:00 pm, there has been a lapse of 5 hours, so the equaton will be 550mg(.89)^5 = 307.123 mg that remain
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23:27:27 ** 3:00 p.m. is 5 hours after the initial time so at that time there will be Q(5) = 550 mg * .89^5 = 307.123 mg in the blood **
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23:29:24 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> The graph crosses the y axis at (0,550) and the x axis is an asymptote. It decreases at a decreasing rate. Half life can be found by finding a value on the y axis, drawing a horixontal line that intersects the graph at this point. Then do the same for double that value. Where these two lines intersect the graph, draw a vertical line down to the x axis and subtract the two corresponding x values which will tell the half life.
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23:29:36 ** Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point. The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down. The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. **
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23:30:16 What is the equation to find the half-life??What is its most simplified form?
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RESPONSE --> P(1+r)^t
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23:30:20 ?????????????????????????20:50:45
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23:31:24 ** Q(doublingTime) = 1/2 Q(0)or 550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have .89^doublingTime = .5. We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). **
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RESPONSE --> This was asking for the half life of this particular problem, not the generalized one. I did, however, obtain the same answer of .89^dubtime = .5
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23:31:28 ?????????????????????????20:55:41
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23:32:53 #19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0. For what values of t did Q(t) lie between .005 Q0 and .01 Q0?
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RESPONSE --> between .05 QO and .1 QO are t values -25 to -31 between .005 Q0 and .01Q0 are t values -49 to -54 These are both fould through trial and error.
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23:33:16 ** Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0. Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0. Solving Q(t) = .05 Q0 we rewrite this as Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get 1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -31.4 approx. Solving Q(t) = .1 Q0 we rewrite this as Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get 1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -24.2 approx. (The solution for .005 Q0 is about -55.6, for .01 is about -48.3 For this solution any value between about t = -48.3 and t = -55.6 will work). **
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23:35:01 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> As the y value of Q(t) grows smaller and smaller, the t value never does equal zero, therefore, is a horizontal asymptote.
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23:35:03 ?????????????????????????21:15:17
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23:35:12 ** The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero. **
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23:35:14 ?????????????????????????21:19:36
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23:38:59 ??#22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )?
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RESPONSE --> y=Ab^x y=12(e^(-.5x)); b = e^(-.5) = approx .607, so y=12(.607)^x y=.007(e^(.71x)); b=e^(.71) = approx 2.034, so y=.007(2.034^x) y=-13(e^(3.9x)); b=e^3.9 = approx 19.402, so y=-13(49.402^x)
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23:39:09 ** 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx. So this function is of the form y = A b^x for b = .61 approx.. **
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23:39:34 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> y=.007(e^(.71x)); b=e^(.71) = approx 2.034, so y=.007(2.034^x)
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23:39:40 ** 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx. So this function is of the form y = A b^x for b = 2.041 approx.. **
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23:39:44 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> y=-13(e^(3.9x)); b=e^3.9 = approx 19.402, so y=-13(49.402^x)
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23:39:48 ** 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx. So this function is of the form y = A b^x for b = 49.4 approx.. **
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23:40:41 List these functions, each in the form y = A b^x.
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RESPONSE --> y=12(.607)^x y=.007(2.034^x) y=-13(49.402^x)
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23:40:43 ?????????????????????????21:24:23
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23:40:48 ** The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **
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