course mth 173 time1217pmdate 2/8/10 008. `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 #$&*3 ********************************************* Question: 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 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):ok ------------------------------------------------ Self-critique rating #$&*2.5 ********************************************* Question: `q explain why a numerical solution to differential equation is only an approximate solution in most cases YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because in the cases that a differential equation is used the numerical data obtained is a result of a small period of time and the numerical figure represents that time. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** You assume the slope at the initial point, but that slope generally changes at least a bit by the time you get to the second point. So you are assuming a constant slope when the slope actually changes. If your interval is small enough the change in slope will have a small effect. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q query Problem 1.4.10 Solve 4 * 3^x = 7 * 5^x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The solution is log4+xlogx=log7+xlogx Or simplified is Log7-log4=xlog5-xlog3 Or the result is .336472=x(.5108260 X=1.51818 confidence rating #$&* 2.5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Taking logs of both sides and applying the laws of logarithms we get log 4 + x log 3 = log 7 + x log 5. Rearranging we obtain x log 5 - x log 3 = log 4 - log 7 so that x ( log 5 - log 3) = log 4 - log 7 and x = (log 4 - log 7) / (log 5 - log 3). This can be approximated as -1.095. ** DER &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):I rearranged the variables out of order and switched the sides inproperly. But do see how to calculate the solution. ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q Problem 1.4.6 simplify 2 ln(e^A) + 3 ln(B^e) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the fact that lne^x=1 2*1+3*1= 2a+3b confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Starting with 2 ln (e^A) + 3 ln (e^B) we use the fact that the natural log and exponential functions are inverses, expressed by the law of logarithms ln(e^x) = x, to get 2 * A + 3 * B or just 2A + 3B. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* 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: The function put into this form would be p=174 e^(-.105361t) Ln of .9=-.105361 confidence rating #$&* 2.5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** 174 * .9^t = 174 * e^(kt) if e^(kt) = .9^t, which is the case if e^k = .9. Taking the natural log of both sides we get ln(e^k) = ln(.9) so that k = ln(.9) = -.105 approx. So the function is P = 174 e^(-.105 t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q 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 time give a population function and the doubling time YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P = 40 e^(.0336 t) Ln2/.0336=20.6294 confidence rating #$&* 2.5 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The population function is exponential and has form P = P0 * e^(kt). Let t be the time since 1980 and population be in millions. Then we have 40 = P0 e^(k * 0) and 56 = P0 e^(k * 10). From the first equation we get 40 = P0 so the second equation becomes 56 = 40 * e^(10 k) or e^(10 k) = 56 / 40 = 1.4. Taking logs we get 10 k = ln(1.4) so that k = ln(1.4) / 10 = .0336, approx. Thus our equation is P = 40 e^(.0336 t). This doubles when e^(.0336 t) = 2. Taking the ln of both sides we have .0336 t = ln(2) so that t = ln(2) / .0336 = 20.6, approx. Doubling time is about 20.6 years. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q query Problem 1.4.45 (was 1.7.42 but changed) time to get to 10% of strontium 90 if half-life 29 yrs YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: E^(k*29)=1/2 29k=1/2 Taking ln of both sides we have 3.3673k=-.693147 .693147/3.3673k=-.205847 Q(0) e^-.205847t is the function and to find 1/10 we use Ln1/10=ln(e^-.205847) So then we take ln1/10/-.205847=11.18 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** If the model is Q = Q0 e^(kt) then since half-life is 29 years we know that e^(k * 29) = 1/2. Taking ln of both sides 29 k = ln(1/2) so that k = ln(1/2) / 29 = -.0239. So the model is Q = Q0 * e^(-.0239 t). Now if we want to get 10% of the original quantity we have Q = Q0 / 10 so that Q0 / 10 = Q0 e^(-.0239 t) and e^(-.0239 t) = 1/10. Taking logs of both sides and solving for t we get t = ln(1/10) / -.0239 = 96.3 approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I see that I have found a solution that was wrong in the third or fourth step which made my final solution wrong and do know why I didn’t take the nat log of that number when I reused it to finalize my calcs. ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q Problem 1.4.26 P=174 * .9^t What is the function when converted to exponential form P = P0 e^(kt)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P=174 e^-(.105 t). confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If P=174(.9)^t is put in the form P = P0 e^(kt) then P0 = 174 and e^(k t) = .9.t so that e^k = .9. It follows that e^k = .9 so that ln(e^k) = ln(.9) or k = ln(.9) = .105. The function is therefore P=174 e^-(.105 t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q 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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The doubling time for this amount would be e^(.0336 t) = 2 which equates to about 20 years confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok don’t understand why doubling time changes from previous equation?
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don’t understand the solution given but will have to refer to the notes for explanation on this.