course Mth 173 Owۈassignment #008
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14:38:10 What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?
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RESPONSE --> This looks like: g(t) = 3t - 5 f(z) = 2^z
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14:38:19 ** 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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RESPONSE --> Ok.
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14:40:08 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 first thing to do would be to find the differential of f(x) at the initial point (x0, y0), which would give us the slope of the line. Then we would take the interval 'dx and find the corresponding y value at that point. Then we simply take the differential at that point again and repeat the process.
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14:45:32 ** 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 --> Ok.
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14:46:12 explain why a numerical solution to differential equation is only an approximate solution in most cases
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RESPONSE --> A numerical solution assumes that the slope does not change during the interval, rendering the solution approximate.
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14:46:20 ** 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 --> Ok.
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14:49:00 query Problem 1.4.8 Solve 2 * 5^x = 11 * 7^x
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RESPONSE --> Here we would do a little rearranging to bring the numbers with exponents on one side and the numbers without on the other: 2 / 11 = 7^x / 5^x This reduces to: 2 / 11 = (7/5)^x We can then takes the logarithm of both sides and solve for x: log(2/11) =x * log(7/5) x = log(2/11) / log(7/5) x = -5.06
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14:49:28 ** 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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RESPONSE --> Interesting. Ok.
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14:51:51 Problem 1.4.28 simplify 2 ln(e^A) + 3 ln(B^e)
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RESPONSE --> This is 1.4.6 in the 4th edition. ln(e^A) can be reduced to A, while ln(B^e) can be reduced to e * ln(B): 2A + 3e * ln(B).
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14:59:17 ** 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 --> Hmm. That equation is slightly different from the one in the problem: 2 ln(e^A) + 3 ln(B^e). I see how the answer was achieved to this problem, however.
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15:04:05 query Problem 1.4.31 (was 1.7.26) P=174 * .9^twhat is the function when converted to exponential form P = P0 e^(kt)?
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RESPONSE --> Here the key is that e^(kt) = 0.9^t. All I need to do is solve for k: e^k = 0.9 ln(e^k) = ln(0.9) k = -.10536 Therefore: P = 174 * e^(-.10536 * t)
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15:04:39 ** 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 --> Ok.
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15:15:58 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 --> To find the function, all I need to do is use the generic form (P = Po * k^t) and solve for k by substituting. 56 = 40 * k^10 56 / 40 = k^10 10th root of 1.4 = k k = 1.03422 Function: P = 40 * 1.03422^t Doubling time: 80 = 40 * 1.03422^t 2 = 1.03422^t log(2) = t * log(1.03422) log(2) / log(1.03422) = t t = 20.6.
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15:16:45 ** 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 --> I did the same thing, I just used the other equation rather than using e.
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15:25:49 query Problem 1.4.50 (was 1.7.42 but changed) time to get to 10% of strontium 90 if half-life 29 yrs
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RESPONSE --> Here the first thing that needs to be done is find the value of k in the formula P = Po * e^(kt). We do this by substituting in known values from the half life: 0.5 = 1 * e^(k*29) Then take the natural log of both sides: ln(.5) = ln(e^(k*29)) ln(.5) = k * 29 ln(.5) / 29 = k k = -.0239 Then to find the time to reduce to 10% we simply substitute .1 for P and 1 for Po and solve for t: .1 = 1 * e^(-.0239*t) ln(.1) = ln(e^(-.0239*t)) ln(.1) = -.0239 * t ln(.1) / -.0239 = t t = 96.34
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15:26:03 ** 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 --> Ok.
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15:27:42 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Well, probably the thing to do is use the P = Po * e^(kt) form.
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15:31:11 Problem 1.4.31 P=174 * .9^t What is the function when converted to exponential form P = P0 e^(kt)?
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RESPONSE --> First thing to do is solve for k: e^k = .9 ln(e^k) = ln(.9) k = -.1054 Then substitute: P = 174 * e^(-.1054 * t)
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15:31:20 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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RESPONSE --> Ok.
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15:32:10 problem 1.4.44 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 --> I did this one already. Here are my prior answers: Equation: P = 40 * 1.03422^t Doubling time: t = 20.6
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15:33:12 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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RESPONSE --> Ok.
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15:33:19 10:32:42
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RESPONSE --> ?
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15:36:31 Problem 1.7.42 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 --> First I need to find a function: P = Po * .9753^t Now to find the quantity remaining after a century, we simply take a quantity (in this case 1) and solve for t=100: P = 1 * .9753^100 = .082
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15:36:37 10:34:19
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RESPONSE --> ?
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15:37:11 Q=Qoe^-.0247t
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RESPONSE --> Ok.
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15:37:16 That`s all that I can do with that problem at this point
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RESPONSE --> Hmm.
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15:37:36 ** 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 --> Bingo. That's what I did.
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