#$&* course Mth 173 1/30 12:21 a.m. 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 $600 takes about 11.5 years - 4.25 years = 7.25 years (actually 7.2725 years to 5 significant figures). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didnt check doubling time from more than the initial principle, and didnt try trial and error with a few decimals to get a little closer, I stayed within integer range. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Rewording to try and understand: at what time t is the princple equal to half its value at t=20? P0 = initial principal r= growth rate P0 = P0(1 + r)^0 P(20)=P0 (1+r)^20 I am confused without having values on what to do, so I need to see whats going on. I looked and see that you are usnig the previous problem, this helps me understand, continuing from there P(20)=200 (1+ .1)^20 P(20)= 1345.49999 1345.49999/2= 672.7499949 not sure any other way to find the value for t at 672.7499949 besides trial and error 200(1+.1)^12 = 627.6856753 --lucky got close on first try 200(1+.1)^13 = over 690 200(1+.1)^12.5 = a little over 658 200(1+.1)^12.7 = 670.9916213 200(1+.1)^12.75=674.196 200(1+.1)^12.73= 672.9129 200(1+.1)^12.728=672.7846 200(1+.1)^12.727= 672.720 200(1+.1)^12.7273= 672.7397934 -eh, close enough so t= 12.7273 is about half of t=20 20-12.7273= 7.2727 doubling time is about 7.2727 I am going to see if i remember how to do a logarithm 200* 1.1^t = 672.7499949 log(1.1^t)=3.363749975 There is a a rule about logarithms or something that lets me do this step, but I don't remember it, i just remember to move the exponent to the front. tlog(1.1)= log(3.363749975) t= log(3.363749975)/ log(1.1) t= 12.72745 heyyy, as long as it doesnt get any harder I think that is right. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $627.69 and at t = 13 you get 690.45).ÿ At 12.75=674.20 so it would probably be about12.72.ÿ This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr. This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40% YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: each one starts the same, but the higher percent, the sooner it starts to drasticaly get steeper, in a positive direction. confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $384 after 4 years and require 2.06 years to double. The final 4-year amount increases by more and more with each 10% increase in interest rate. The doubling time decreases, but by less and less with each 10% increase in interest rate. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didnt mark my points, and thought you just wanted a general description of the graph, I did not however check the times required to double at all. So it would seem, the higher the growth rate, the quicker the value will double, which seems obvious enough since it gets steeper quicker ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q query #11. equation for doubling time YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: growth factor^doubling time = 2 confidence rating #$&*: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): I tried saying 2P(1+r)^t, Im not sure what this would give, just noting the fact I kept trying that for a while. ------------------------------------------------ Self-critique Rating:2 ********************************************* Question: `q Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.08^doublingTime=2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Trying to solve down to the growthFactor^doublingTime = 2 gets me confused, I can only understand the end result. 5000(1.08^(2 + doubleTime))= 2 * (5000 *( 1.08^2) this is P(t+doubleTime)= 2P(t) i believe I dont understand past this point. I believe its to divide by 5000, in typewriter notation it looks like I cant do that on the right side to eliminate 5000, it looks like I have to do exponent first because of the pair of parenthesis around it. ------------------------------------------------ Self-critique Rating:
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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" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!