#$&* course MTH 173 5/31/2012 at 1:52PM 008. `query 8
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Given Solution: ** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: describe in some detail how we can numerically solve a differential equation dy /dx = f(x), given a point (x0, y0) on its solution curve, an interval (x0, xf) and an increment `dx Solution: ** 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. ** STUDENT QUESTION: Could you give me, when you critique this assignment, a problem that I could work out to see how this is actually done? It seems a bit confusing to me still. INSTRUCTOR RESPONSE: The Class Notes give additional examples. If dy/dt = t^2 + y, and we know that y(2) = 6, then we can approximate y(2.1) as follows: y(2) = 6 so when t = 2 we have dy/dt = 2^2 + 6 = 10. Therefore if `dt = .1, we obtain the approximation `dy = (dy/dt) * `dt = 10 * .1 = 1. This approximation assumes that the rate of change dy/dt remains constant between y = 2 and y = 2.1. This isn't completely accurate, but since the interval is small the error is also small. We conclude that y(2.1) = y(2) + `dy = 6 + 1 = 7. Now we approximate y(2.2). We know that y(2.1) is about 7, so when t = 2.1 we have dy/dt = 2.1^2 + 7 = 11.41. So if `dt = .1, we have the approximation `dy = 11.41 * .1 = 1.141. Once again this is an approximation which assumes an unchanging value of dy/dt for the entire interval. Again the error is small, but of course it is added to (and in part based on) the error in the preceding step. Our approximation is thus y(2.2) = y(2.1) + `dy = 7 + 1.141 = 8.141. The process could continue. For example, we could do 7 more steps and obtain an approximation to y(3). This approximation would accumulate errors at every step, so the accuracy would decrease with every step. We could improve our accuracy by using a smaller interval. For example we could assume intervals of .01 rather than .1. This would require10 times as many steps, and would accumulate 10 times as many errors; however the errors would tend to be much smaller, and the total error in approximating, say, y(3) using intervals of .01 would be smaller than the total error that would result from intervals of .1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ? ------------------------------------------------ Self-critique Rating: ? ********************************************* Question: `q explain why a numerical solution to differential equation is only an approximate solution in most cases YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When the slope at an initial point changes when it gets to the second point, you can assume that the slope is not constant but actually changes. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q query Problem 1.4.10 Solve 4 * 3^x = 7 * 5^x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log 4 + x log 3 = log 7 + x log 5 x log 5 - x log 3 = log 4 - log 7 x ( log 5 - log 3) = log 4 - log 7 x = (log 4 - log 7) / (log 5 - log 3). x = -1.095 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ OK
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Given Solution: ** Taking logs of both sides and applying the laws of logarithms we get log 4 + x log 3 = log 7 + x log 5. Rearranging we obtain x log 5 - x log 3 = log 4 - log 7 so that x ( log 5 - log 3) = log 4 - log 7 and x = (log 4 - log 7) / (log 5 - log 3). This can be approximated as -1.095. ** DER Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Problem 1.4.6 simplify 2 ln(e^A) + 3 ln(B^e) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 ln (e^A) + 3 ln (B^e) 2 * A + e ln(B^e). 2 A + 3 e ln(B). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ OK
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Given Solution: ** Starting with 2 ln (e^A) + 3 ln (B^e) we first 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 + e ln(B^e). We then use the fact that ln(x^y) = y ln(x), applied to the expression ln(B^e), to get 2 A + 3 e ln(B). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q query Problem 1.4.27 5th; 1.4.26 4th; 1.4.31 (was 1.7.26) P=174 * .9^t What is the function when converted to exponential form P = P0 e^(kt)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 174 * .9^t = 174 * e^(kt) e^(kt) = .9^t e^k = .9 ln(e^k) = ln(.9) k = ln(.9) = -.105 P = 174 e^(-.105 t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: ** 174 * .9^t = 174 * e^(kt) if e^(kt) = .9^t, which is the case if e^k = .9. Taking the natural log of both sides we get ln(e^k) = ln(.9) so that k = ln(.9) = -.105 approx. So the function is P = 174 e^(-.105 t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Problem 1.4.27 was 1.4.26 P=174 * .9^t What is the function when converted to exponential form P = P0 e^(kt)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P = P0 e^(kt) P=174(.9)^t where P0 = 174 and e^(k t) = .9.t so that e^k = .9 e^k = .9 ln(e^k) = ln(.9) k = ln(.9) = .105. P=174 e^-(.105 t) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q problem 1.4.40 was 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P = Po b^t 40 * 10^6 so Po = 40 * 10^6. Po = 40 * 10^6 P = 40 * 10^6 b^t At t = 10 P = 56 * 10^6 56 * 10^6 = 40 * 10^6 b^10. 1.4 = b^10 b = 1.0 P=40*10^6(1.03)^t 80 * 10^6 = 40 * 10^6(1.03)^t 2 = 1.03^t log2 = tlog1.03 t = 23.4498 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: P=Po b^t is the form of the function. Initial quantity is 40 * 10^6 so Po = 40 * 10^6. Substituting Po = 40 * 10^6: P=40*10^6 b^t. At t = 10 we have P = 56 * 10^6 so we substitute for P and t: 56*10^6=40*10^6 b^10. We solve for b: 1.4=b^10 b=1.03 Substituting P0 and b into the original form: P=40*10^6(1.03)^t is our function. doubling time occurs when the 40^10^6 grows to 80*10^6: 80*10^6=40*10^6(1.03)^t 2=1.03^t log2=tlog1.03 t=23.4498 10:32:42 STUDENT QUESTION How should I set up the initial problem???? INSTRUCTOR RESPONSE The form of the function is P = P0 * b^t, where P0 and b are regarded as unknowns. To solve for two unknowns you need two equations. You have the values of P and t at two different points, which allows you to write down two equations, which you then solve simultaneously to get P0 and b. The two equations would be 40 = P0 * b^0 and 56 = P0 * b^10. b^0 = 1 so the first equation gives us P0 = 40. This makes the second equation 56 = 40 * b^10, which is solved for b in the manner shown above. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Problem 1.4.50 was 1.4.45 percent of original strontium -- 90 after century; 2.47% annual decay rate. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Q(t) = Qo * e^(kt). Q(100) = Qo e^(k * 100) = (100 - .0247) Qo Qo e^(k * 100) = (100 - .0247) Qo What next? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
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Given Solution: What percent of the original strontium -- 90 would remain after a century? 10:34:19 I did not understand this problem, but this is what I have: Q=Qoe^(-kt) Q=Qoe^-.0247t That`s all that I can do with that problem at this point ** 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. ** STUDENT QUESTION How would Qo e^(k* 1) = (1 - .0247) Qo??? INSTRUCTOR RESPONSE Q(1) = Q0 e^(k * 1) is the amount after one year. The sample loses 2.47% in a year, so the amount after a year is Q0 - .0247 * Q0 = Q0 * ( 1 - .0247). Setting equal the two expressions for the amount after one year we get Q0 e^(k * 1) = Q0 * (1 - .0247), so that e^(k * 1) = (1 - .0247). STUDENT QUESTION Like above I am having difficulty setting the problem up to solve, I get how to solve the actual problems in the book assignment but I am having trouble with the word problems. How should I get the initial equation, I can solve the equation its just getting it to start that is causing me difficulty INSTRUCTOR RESPONSE You need to know that this situation involves a constant decay rate, and that a constant growth or decay rate gives us the form Q(t) = Q0 * e^(kt). Since we are looking for the percent change rather than the actual amount, it doesn't matter what the initial amount Q0 is. All that matters is Q(t) / Q0. So the only important unknown quantity is k. The given information tells us that when t = 1, 2.47% has been lost, so that Q(1) = (1 - .0247) * Q0 = .9753 Q0. Q(1) = Q0 * e^(- k * 1 ), so we have two expressions for Q(1). Setting them equal we have .9753 Q0 = Q0 * e^(-k). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: