course Mth 173 This work may be a little out of order or messed because there were some problems in the questions. I tried to point out anything confusing I saw. ͜Q¥xbassignment #008
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15:55:36 What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?
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RESPONSE --> g(t) = 3t-5 f(z) = 2 ^ z f(g(t)) = 2 ^ (3t - 5)
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15:55:45 ** g(t) = 3t - 5, f(z) = 2^z. The z is a 'dummy' variable; when we find f(g(t)), the g(t) is substituted for z and we get f(g(t)) = 2^(g(t)) = 2^(3t-5). **
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16:04:45 describe in some detail how we can numerically solve a differential equation dy /dt = f(x), given a point (x0, y0) on its solution curve, an interval (x0, xf) and an increment `dx
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RESPONSE --> The interval (x0, xf) can be used as such xf - x0 to find the dt. We would then plug in xf to f(x) to get yf. Once we have both (xf, yf) and (x0, y0) we can find the slope of the line connecting the two points using (yf - y0)/(xf - x0). Then we can find a linear function using y - y0 = (slope)(x - x0)
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16:05:36 ** You start with a point (x0, y0) on the y vs. x graph. You evaluate the function y' for x=x0 and y=x0, which gives you a slope for your y vs. x graph. Using the chosen increment `dx you then multiply `dx by the slope to determine how much change there will be in y, and you use this information to obtain a new approximate point on your y vs. x graph by adding the change in y to y0 and the change in x to x0. You then repeat the process starting with the new point. **
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RESPONSE --> Everything I put seems to have been wrong. Although it might not make a big difference, i like it better with numbers.
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16:06:57 explain why a numerical solution to differential equation is only an approximate solution in most cases
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RESPONSE --> There may be a calculation that has to be done that results in a decimal that is too long to use easily; and that value may be used in another calculation, which makes that calculation's value skewed.
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16:07:44 ** You assume the slope at the initial point, but that slope generally changes at least a bit by the time you get to the second point. So you are assuming a constant slope when the slope actually changes. If your interval is small enough the change in slope will have a small effect. **
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RESPONSE --> That is something that bugs me sometimes. Most of the time we use the average slope between two points.
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16:25:13 query Problem 1.4.10 Solve 2 * 5^x = 11 * 7^x
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RESPONSE --> 2 * 5^x = 11 * 7^x log(2) + xlog(5) = log(11) + xlog(7) xlog(5) - xlog(7) = log(11) - log(2) x(log(5) - log(7)) = log(11) - log(2) x = (log(11) / (log(5) - log(7))) - (log(2) / (log(5) - log(7))) x = approx.-5.0665341
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16:25:32 ** Taking logs of both sides and applying the laws of logarithms we get log 2 + x log 5 = log 11 + x log 7. Rearranging we obtain x log 5 - x log 7 = log 11 - log 2 so that x ( log 5 - log 7) = log 11 - log 2 and x = (log 11 - log 2) / (log 5 - log 7). This can be approximated as -5.07. ** DER
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16:30:14 Problem 1.4.6 simplify 2 ln(e^A) + 3 ln(B^e)
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RESPONSE --> 2 ln(e^A) + 3 ln(B^e) 2 * A + 3 * e * ln(B)
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16:31:13 ** Starting with 2 ln (e^A) + 3 ln (e^B) we use the fact that the natural log and exponential functions are inverses, expressed by the law of logarithms ln(e^x) = x, to get 2 * A + 3 * B or just 2A + 3B. **
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RESPONSE --> The question in the text had 2 ln (e^A) + 3 (B^e)
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16:37:47 ** 174 * .9^t = 174 * e^(kt) if e^(kt) = .9^t, which is the case if e^k = .9. Taking the natural log of both sides we get ln(e^k) = ln(.9) so that k = ln(.9) = -.105 approx. So the function is P = 174 e^(-.105 t). **
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RESPONSE --> I would have answered but the Next Question/Answer button was highlighted and I hit enter. I do understand the natural logs though.
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16:47:10 Query problem 1.4.44 (was 1.6.24) population function for exponential growth if 40 meg in 1980 and 56 meg in 1990; doubling timegive a population function and the doubling time
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RESPONSE --> P = 40meg e^kt 56meg = 40meg e^k10 1.4 = e^k10 ln(1.4) = ln(e^k10) .3364 = 10k k = .03364 P = 40meg e^(.03364t) In the year 2000: 40meg e^(.03364*20) = 78,388,674 80meg = 40meg e^(.03364*t) 2 = e^(.03364*t) ln(2) = ln(e^(.03364*t)) 2 = .03364t Doubling Time = t = 59.453 years.
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16:48:05 ** The population function is exponential and has form P = P0 * e^(kt). Let t be the time since 1980 and population be in millions. Then we have 40 = P0 e^(k * 0) and 56 = P0 e^(k * 10). From the first equation we get 40 = P0 so the second equation becomes 56 = 40 * e^(10 k) or e^(10 k) = 56 / 40 = 1.4. Taking logs we get 10 k = ln(1.4) so that k = ln(1.4) / 10 = .0336, approx. Thus our equation is P = 40 e^(.0336 t). This doubles when e^(.0336 t) = 2. Taking the ln of both sides we have .0336 t = ln(2) so that t = ln(2) / .0336 = 20.6, approx. Doubling time is about 20.6 years. **
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RESPONSE --> Mistake: In last step I divided 2 by .03364, not ln(2)
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16:55:32 query Problem 1.4.45 (was 1.7.42 but changed) time to get to 10% of strontium 90 if half-life 29 yrs
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RESPONSE --> If 50% of the Strontium-90 is gone in 29 years, then there is (50/29) percent leaving the bones each year. I will continue to cite that percentage as a fraction because the resulting value is much less accurate. If we are trying to find out how long it would take for 90% to leave the bones, then we multiply 90 * (50/29) to get 52.2 years.
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16:57:35 ** If the model is Q = Q0 e^(kt) then since half-life is 29 years we know that e^(k * 29) = 1/2. Taking ln of both sides 29 k = ln(1/2) so that k = ln(1/2) / 29 = -.0239. So the model is Q = Q0 * e^(-.0239 t). Now if we want to get 10% of the original quantity we have Q = Q0 / 10 so that Q0 / 10 = Q0 e^(-.0239 t) and e^(-.0239 t) = 1/10. Taking logs of both sides and solving for t we get t = ln(1/10) / -.0239 = 96.3 approx. **
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RESPONSE --> I took a logical shot at it. Didn't think I'd be right. But I wrote down the correct answer for later.
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16:58:21 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> No real surprises. It took me a little while to grasp the growth rate and decay rate functions in high school too.
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16:59:06 Problem 1.4.26 P=174 * .9^t What is the function when converted to exponential form P = P0 e^(kt)?
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RESPONSE --> P = 174 e^(-.10536*t)
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16:59:09 If P=174(.9)^t is put in the form P = P0 e^(kt) then P0 = 174 and e^(k t) = .9.t so that e^k = .9. It follows that e^k = .9 so that ln(e^k) = ln(.9) or k = ln(.9) = .105. The function is therefore P=174 e^-(.105 t).
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16:59:56 problem 1.4.32 population function for exponential growth. If 40 meg in 1980 and 56 meg in 1990 give the equation and find the doubling time
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RESPONSE --> problem 1.4.32 population function for exponential growth. If 40 meg in 1980 and 56 meg in 1990 give the equation and find the doubling time I would normally answer the same question twice so I am sure that I do not miss anything, but this one took me a while to type. I hope my first answer was correct.
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17:00:02 P=Po b^t is the form of the function. Initial quantity is 40 * 10^6 so Po = 40 * 10^6. Substituting Po = 40 * 10^6: P=40*10^6 b^t. At t = 10 we have P = 56 * 10^6 so we substitute for P and t: 56*10^6=40*10^6 b^10. We solve for b: 1.4=b^10 b=1.03 P=40*10^6(1.03)^t is our function. doubling time occurs when the 40^10^6 grows to 80*10^6: 80*10^6=40*10^6(1.03)^t 2=1.03^t log2=tlog1.03 t=23.4498
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17:00:06 10:32:42
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17:04:18 Problem 1.4.45 percent of original strontium -- 90 after century; 2.47% annual decay rate. What percent of the original strontium -- 90 would remain after a century?
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RESPONSE --> e^(-.0239016269*100) = .091 or 9%
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17:04:34 10:34:19
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RESPONSE --> 10:34:19 I am not sure what this means.
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17:04:38 I did not understand this problem, but this is what I have:
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17:04:43 Q=Qoe^(-kt)
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17:04:50 That`s all that I can do with that problem at this point
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17:05:30 ** The model is Q(t) = Qo * e^(kt). You know that you lost .0247 of the quantity in a year. Thus Q(1) = Qo e^(k* 1) = (1 - .0247) Qo. So Qo e^(k* 1) = (1 - .0247) Qo. This equation is easily solved for k. Then you substitute t = 100 back into the function, using your newly found k. **
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RESPONSE --> I do not know why, but I used the decay rate for Strontium - 90 found in the last problem.
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