"݀?assignment #015 HѴVʖ滴Ň~ Precalculus I 03-13-2006
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23:41:40 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 --> the growth rate is .10 which means the growth factor is 1.10. using 200*1.10^t=400 our answer would be 7.
to obtain the 300 figure we would use the same function as above 200*1.10^t=300 so we would be left with around 4 years.................................................
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23:41:43 ** 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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23:45:06 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 --> 200*1.1^20=1340 half of this value would be 1340/2=670.00
the original doubling time and the only doubling time i know of.................................................
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23:45:09 ** 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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23:47:53 query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%
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RESPONSE --> i don;t understand how to sketch a graph on this.
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23:48:20 ** 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. **......!!!!!!!!...................................
RESPONSE --> okay i did not understand the question, i thought that you meant to give coordinates etc.
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23:49:30 query #11. What is the equation for doubling time?
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RESPONSE --> po*(1+r)^t=2PO
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23:49:32 ** 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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23:51:12 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 --> 1.08^1+doubling time)=2*1.08^2)
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23:51:15 **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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23:53:17 Desribe how on your graph how you obtained an estimate of the doubling time.
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RESPONSE --> the money amount would be on the x axis with the time being on the y axis.
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23:53:41 ** 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. **......!!!!!!!!...................................
RESPONSE --> ? i don;t know if i got this question right or not
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23:56:08 #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 --> growth rate is .11 the growth factor would be 1+r=1+(.11)=.89 so q(t)=QO*(1+r)^t=550mg(.89)^t
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23:56:11 ** 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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00:04:00 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> i don't know what to put in the function for time'
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00:04:03 ** 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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00:04:28 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> ???????????????????????????????????
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00:04:31 ** 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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00:05:50 What is the equation to find the half-life??What is its most simplified form?
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RESPONSE --> Q(doublingtime)=1/2Q(O)
550*89^doubling time=.5*550mg. so we come up with .89doubling time=.5.................................................
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00:05:53 ?20:50:45
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00:05:56 ** 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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00:05:58 ?20:55:41
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00:08:26 #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?......!!!!!!!!...................................
RESPONSE --> I am lost at this point i would guess t=-24.2 and t=31.4 would be Qt between .05 adn .1QO
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00:08:28 ** 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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00:09:36 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> 1.1^t increases for increasing t as t approcahes a point to where it will become constant.
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00:09:39 ?21:15:17
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00:09:42 ** 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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00:09:45 ?21:19:36
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00:10:34 ?#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 --> 12e^-.5x)=13(e^.5)^x=12*.61^x
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00:11:02 ** 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.. **......!!!!!!!!...................................
RESPONSE --> i forgot to solve for b and i also had a clerical error in which i had 13
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00:11:54 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> 12e^.71x=12(e^.71^x=12*2.04^x so b would equal 2.042
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00:11:57 ** 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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00:12:59 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> 12e(3.9x)=12(e^3.9)^x=12*49.4^x so b would equal 49.4
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00:13:01 ** 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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00:14:00 List these functions, each in the form y = A b^x.
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RESPONSE --> y=12(.6065^x)
y=.007(2.03399^x) y=-13(49.40244^x).................................................
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00:14:02 ?21:24:23
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00:14:04 ** The functions are
y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **......!!!!!!!!...................................
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