#$&* course Mth 173 004. `query 4
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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): I was a little woprried that there may be an easier way to do this that trial and error, such as doing algebra on hte function and solving for t, but with numbers this small, I felt trial and error would be the quickest way to solve the problem. ------------------------------------------------ Self-critique Rating:3 ********************************************* 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: First we must find the t = 20 value, this is easy as we just substitute the number for the variable. 200 * 1.1^20 = 1345.50. Half of that is roughly 672. Using algebra to get the formula to find the time it will take to have $672 we find that t = ln(3.36)/ln(1.1) t = 1.21/.095 t = 12.74 years. confidence rating #$&*: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 $627.69 and at t = 13 you get 690.45).ÿ At 12.75=674.20 so it would probably be about12.72.ÿ This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr. This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40% YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can find the coordinates of each of the four years easily by using the compound interest formula. With the 10% rate we get the coordinates as follows... 200(1.1^1) = 220 200(1.1^2) = 242 200(1.1^3) = 266.20 200(1.1^4) = 337.79 So the coordinates are (1, 220), (2, 242), (3, 266.20), (4, 337.79). We use similar calculations for the next 3 rates to get the following coordinates. For 20%... (1, 240), (2, 288), (3, 345.60), (4, 414.72) 30% (1, 260), (2, 338), (3, 439.40), (4, 571.22) 40% (1, 280), (2, 392), (3, 548.80), (4, 768.32) confidence rating #$&*: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 $384 after 4 years and require 2.06 years to double. The final 4-year amount increases by more and more with each 10% increase in interest rate. The doubling time decreases, but by less and less with each 10% increase in interest rate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q query #11. equation for doubling time YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The opriginal equation is P0 * (1 + r)^t = P To find when exactly the time at which the principle doubled we can make the equation as follows. P0(1 + r)^t = 2P0 At which point the P0 will cancel each other out, simplifying to (1 + r) ^ t = 2. Once you plug in the rate you can solve for t to find out when the principle will have doubled. ln(2)/ln(1+r) = t is the equation you get when you solve for t confidence rating #$&*: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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5000 * (1.08)^t = 10,000 nl(2)/nl(1.08) = Right about 9 years. 5000 * (1.08) ^ 9 = 9995, which is close enough for me to be confident in my formula. confidence rating #$&*: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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q Desribe how on your graph how you obtained an estimate of the doubling time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: On a graph you can find the doubling point, in this case 10,000. Once you find the point, you can move down the to the horizontal axis to see at what point in time $10,000 was hit. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I wasn't sure if I was reading the question right when I answered it, but looking t the given solution, I was pretty well spot on. ------------------------------------------------ Self-critique Rating:3"