course PC 163 ÇÙR}ûÏ‚zŸ³Ó‘WâÇ—âíR½¿Úeassignment #015
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13:26:21 `questionNumber 150000 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 --> .10 rate factor is 1.10 and 400 is double the 200 so just apply to get 670 dollars so time is right around 13 . confidence assessment: 2
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13:26:28 `questionNumber 150000 ** 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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RESPONSE --> i see self critique assessment: 2
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13:27:33 `questionNumber 150000 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 --> Half the value would be 1340/2 approx. 670 dollars The doubling time would be close to 12.5 confidence assessment: 2
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13:27:43 `questionNumber 150000 ** 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 understand self critique assessment: 2
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13:35:01 `questionNumber 150000 query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%
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RESPONSE --> First interest rate 10% is 1(1.10)^t confidence assessment: 2
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13:35:31 `questionNumber 150000 ** 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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RESPONSE --> I see, I didn't finish the problem, but I understand how you got it. self critique assessment: 2
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13:39:58 `questionNumber 150000 query #11. What is the equation for doubling time?
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RESPONSE --> Equation is similar to earlierso you solve for double time to get 1.08^2*1.08^double time=2*1.08^2 Divide by 1.08^2 and we get 1.08^double time=2 Then we get close to 8.7 years. confidence assessment: 2
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13:40:29 `questionNumber 150000 ** 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 --> I see self critique assessment: 2
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