query_04

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Given Solution: ** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Let P be the value of the investment at the end of t = 20 years. Then we have P = $200*(1.1^t) where t = 20 P = $200*(1.1^20) = $1345.45 We need to find the time t required to reach half of the value of P = $1345.45/2 = $672.725 Thus again using the exponential equation we have $672.725 = $200*(1.1^t) 3.363625 = 1.1^t t = ln(3.363625) / ln(1.1) = 12.727 years Subtracting the time we obtained just above from the time mentioned we calculate the time required for the principle to double itself, which is 20 years – 12.727 = 7.273 years which is consistent with the value obtained before for doubling the principle amount which proves that for an exponential function the time required for doubling a value does not depend upon the initial value. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: ** 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).ÿ At 12.75=674.20 so it would probably be about12.72.ÿ 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40% YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For the first graph we have rate = 10% , the growth rate = 0.1 and the growth factor = 1.1 . Suppose after the 1st year the principle amount is P0*1.1 = 1.1P0. After the second year principle will be P0*1.1*1.1 = 1.21P0, after the 3rd year the amount turns out to be P0*1.1*1.1*1.1 = 1.331P0 and after the 4th year the principle finally turns P0*1.1*1.1*1.1*1.1 = 1.4641P0. thus the co-ordinates of the graph are ( 0 , P0 ) , ( 1 , 1.1P0 ) , ( 2 , 1.21P0) , ( 3 , 1.331P0) and ( 4 , 1.4641P0). The graph when extended and the doubling time calculated with the help of the graph it turns out to be 7.3 years. For the second graph we have rate = 20%, the growth rate = 0.2 and the growth factor = 1.2 .After the 1st year the principle amount is P0*1.2 = 1.2P0. After the second year principle will be P0*1.2*1.2 = 1.44P0, after the 3rd year the amount turns out to be P0*1.2*1.2*1.2 = 1.728P0 and after the 4th year the principle finally turns P0*1.2*1.2*1.2*1.2 = 2.0736P0. thus the co-ordinates of the graph are ( 0 , P0 ) , ( 1 , 1.2P0 ) , ( 2 , 1.44P0) , ( 3 , 1.728P0) and ( 4 , 2.0736P0). The graph when extended and the doubling time calculated with the help of the graph it turns out to be 3.8 years. For the third graph we have rate = 30%, the growth rate = 0.3 and the growth factor = 1.3. Suppose after the 1st year the principle amount is P0*1.3 = 1.3P0. After the second year principle will be P0*1.3*1.3 = 1.69P0, after the 3rd year the amount turns out to be P0*1.3*1.3*1.3 = 2.197P0 and after the 4th year the principle finally turns P0*1.3*1.3*1.3*1.3 = 2.8561P0. thus the co-ordinates of the graph are ( 0 , P0 ) , ( 1 , 1.3P0 ) , ( 2 , 1.69P0) , ( 3 , 2.197P0) and ( 4 , 2.8561P0).The graph when extended and the doubling time calculated with the help of the graph it turns out to be 2.6 years. For the fourth graph we have rate = 40%, the growth rate = 0.4 and the growth factor = 1.4. Suppose after the 1st year the principle amount is P0*1.4 = 1.4P0. After the second year principle will be P0*1.4*1.4 = 1.96P0, after the 3rd year the amount turns out to be P0*1.4*1.4*1.4 = 2.744P0 and after the 4th year the principle finally turns P0*1.4*1.4*1.4*1.4 = 3.8416P0. thus the co-ordinates of the graph are ( 0 , P0 ) , ( 1 , 1.4P0 ) , ( 2 , 1.96P0) , ( 3 , 2.744P0) and ( 4 , 3.8416P0).The graph when extended and the doubling time calculated with the help of the graph it turns out to be 2.1 years Thus we can conclude that the final amount at the end of 4 years increases continuously as rate increases. The doubling period decreases continuously but at an decreasing rate as rate increases. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: ** 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 $384 after 4 years and require 2.06 years to double. The final 4-year amount increases by more and more with each 10% increase in interest rate. The doubling time decreases, but by less and less with each 10% increase in interest rate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q query #11. equation for doubling time YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Let P0 be the initial investment. Let rate be r and final resulting value be P1. The growth rate thus will be 1 + r. since we need to calculate time for doubling time, thus P1 = 2P0. Let doubling time be t. Thus using the exponential equation we know that P1 = P0*[(1 + r)^t] 2P0 = P0*[(1 + r)^t] 2 = ( 1 + r )^t This is the equation for calculating the double time, which shows that the equation is independent of initial principle. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Doubling time for a investment starting at any time is the same that we get when we double from t = 0. Thus if we need to calculate the doubling time starting from t = 2 , it will be the same as the time to be calculated for doubling starting from t = 0. Initial investment = $5000 Double the initial investment = $5000 * 2 = $10,000 Let t be the time required to double the investment $10,000 = $5000 * (1.08^t) t = ln(2) / ln(1.08) = 9.0064 years confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: **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. ** STUDENT COMMENT I have growth factor ^ time = 2 instead of g.f.^doublingtime =2, I don't understand how the calculation in the solution above was done. INSTRUCTOR RESPONSE If P(t) is the principle at clock time t, then the principle at clock time t = 2 is P(2). To double, starting at t = 2, the principle would have to become 2 * P(2). The clock time at which the doubling occurs, starting at t = 2, can be expressed as 2 + doublingTime. Thus the statement that the principle doubles, starting at t = 2, is interpreted as • P(2 + doublingTime) = 2 * P(2). This functional equation will apply for any function, whether exponential or not. In terms of the exponential function of this problem the equation for the specified conditions becomes • $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] For an exponential function the doubling time is constant, so if P(t) is an exponential function we have • P(t + doublingTime) = 2 * P(t) for any starting time t. Your equation corresponds to starting time t = 0. Since the function is exponential your equation gives you the correct answer; had the function not been exponential your approach almost certainly wouldn't have worked. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Desribe how on your graph how you obtained an estimate of the doubling time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: On the graph supposing that P0 is the initial value at t = 0. Thus 2P0 will be the double investment. The graph has the investment values on the y-axis and time on the x-axis. Locate 2P0 on the y axis and draw a horizontal line to the graph drawn. On the intersection point of the graph and the horizontal line draw a vertical line to the x axis and find the intersection point. If initial time was zero then the point of intersection on x axis tells you the doubling time, if initial time was not zero , doubling time is the difference of the intersection point time and the initial time. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: **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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK "

query_04

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&#Your work looks very good. Let me know if you have any questions. &#