course Mth 173 ??????????€?zO??assignment #008
......!!!!!!!!...................................
20:11:58 `q001. Note that there are 5 questions in thie assignment. Sketch a graph of the function y ' = .1 t - 6 for t = 0 to t = 100. Describe your graph.
......!!!!!!!!...................................
RESPONSE --> intercept function y' axis at (0,-6) intercept on the x axis at (60,0). The graph is a straight line with slope of 1. At t = 100 the graph point is (100,4). confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:12:05 The graph of this function has an intecept on the y' axis at (0,-6) and an intercept on the x axis at (60,0). The graph is a straight line with slope .1. At t = 100 the graph point is (100,4).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:14:18 `q002. Now sketch the graph of a function y which starts at the point (0, 100), and whose slope at every t is equal to the value of y ' at t. Thus, for example, at t = 0 we have y ' = .1 * 0 - 6 = -6, so our graph of y will start off a the t = 0 point (0,100) with a slope of -6, and the graph begins by decreasing rather rapidly. But the slope won't remain at -6. By the time we get to t = 10 the slope will be y ' = .1 * 10 - 6 = -5, and the graph is decreasing less rapidly than before. Then by the time we get to t = 20 the slope will be y ' = . 1 * 20 - 6 = -4, etc.. If you sketch a graph of y vs. t with these calculated slopes at the corresponding t values, what does the graph look like? Will it be increasing or decreasing, and will it be doing so at an increasing, decreasing or constant rate? Is the answer to this question different for different parts of the graph? If so over what intervals of the graph do the different answers apply?
......!!!!!!!!...................................
RESPONSE --> slope= -6 at t = 0, -5at t = 10, the slope will remain negative for most of the time so the graph will be decreasing. So the graph be decreasing at a decreasing rate. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:14:24 The graph will have slopes -6 at t = 0, -5 at t = 10, -4 at t = 9, etc.. At least for awhile, the slope will remain negative so the graph will be decreasing. The negative slopes will however become less and less steep. So the graph be decreasing at a decreasing rate. It's obvious that the slopes will eventually reach 0, and since y' is the slope of the y vs. t graph it's clear that this will happen when y' = .1 t - 6 becomes 0. Setting .1 t - 6 = 0 we get t = 60. Note, and think about about the fact, that this coincides with the x-intercept of the graph of y' vs. t. At this point the slope will be 0 and the graph will have leveled off at least for an instant. Past t = 60 the values of y' will be positive, so the slope of the y vs. t graph will be positive and graph of y vs. t will therefore be increasing. The values of y' will also be increasing, so that the slopes of the y vs. t graph will be increasing, and we can say that the graph will be increasing at an increasing rate.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:16:08 `q003. The graph of y vs. t corresponding to the given rate function y ' = .1 t - 6 has slope -6 at the point (0,100). This slope immediately begins changing, and becomes -5 by the time t = 10. However, let us assume that the slope doesn't change until we get to the t = 10 point. This assumption isn't completely accurate, but we're going to see how it works out. If the slope remained -6 for t = 0 to t = 10, then starting at (0, 100) what point would we reach when t = 10?
......!!!!!!!!...................................
RESPONSE --> The slope = rise / run. If the slope remains at -6 from t = 0 to t = 10 The run is 10 and the slope is -6, so rise = slope * run = -6 * 10 = -60. The y coordinate of the graph therefore changes by -60, from y = 100 to y = 100 + (-60) = 40. The corresponding point is (10, 40). confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:16:14 The slope of the graph is the ratio slope = rise / run. If the slope remains at -6 from t = 0 to t = 10, then the difference between 10 is the run. Thus the run is 10 and the slope is -6, so the rise is rise = slope * run = -6 * 10 = -60. The y coordinate of the graph therefore changes by -60, from y = 100 to y = 100 + (-60) = 40. The corresponding point is (10, 40).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:17:05 `q004. We see that we reach the point (10, 40) by assuming a slope of -6 from t = 0 to t = 10. We have seen that at t = 10 the slope will be y ' = .1 * 10 - 6 = -5. If we maintain this slope for the interval t = 10 to t = 20, what will be the coordinates of the t = 20 point?
......!!!!!!!!...................................
RESPONSE --> The run from t = 10 to t = 20 is 10. slope= -5 rise = slope * run = -5 * 10 = -50. Starting from point (10,40), a rise of -50 and a run of 10 takes the graph to the point (20, -10). confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:18:32 The run from t = 10 to t = 20 is 10. With a slope of -5 this implies a rise of rise = slope * run = -5 * 10 = -50. Starting from point (10,40), a rise of -50 and a run of 10 takes the graph to the point (20, -10).
......!!!!!!!!...................................
RESPONSE --> The slope at t = 40 is y ' = .1 * 40 - 6 = -2. From t = 40 to t = 50 the run is 10, so the rise is rise = slope * run = -2 * 10 = -20. Starting from (40,-80) a rise of -20 and a run of 10 takes us to (50, -100). The slope at t = 50 y ' = .1 * 50 - 6 = -1. From t = 50 to t = 60 the run is 10, so rise = slope * run = -1 * 10 = -10. The slope at t = 60 is y ' = .1 * 70 - 6 = -0. From t = 60 to t = 70 the run is 10, so the rise is rise = slope * run = -0 * 10 = 0. Starting from (60,-110) a rise of and a run of 10 takes us to (70, -110). self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:19:39 `q005. Continue this process up to the t = 70 point, using a 10-unit t interval for each approximation. Describe the graph that results.
......!!!!!!!!...................................
RESPONSE --> The slope at t = 20 is y ' = .1 * 20 - 6 = -4. From t = 20 to t = 30 the run is 10, so the rise is rise = slope * run = -4 * 10 = -40. Starting from (20,-10) a rise of -40 and a run of 10 takes us to (30, -50). The slope at t = 30 is y ' = .1 * 30 - 6 = -3. From t = 30 to t = 40 the run is 10, so the rise is rise = slope * run = -3 * 10 = -30. Starting from (30,-50) a rise of -30 and a run of 10 takes us to (40, -80). The slope at t = 40 is y ' = .1 * 40 - 6 = -2. From t = 40 to t = 50 the run is 10, so the rise is rise = slope * run = -2 * 10 = -20. Starting from (40,-80) a rise of -20 and a run of 10 takes us to (50, -100). confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:19:44 The slope at t = 20 is y ' = .1 * 20 - 6 = -4. From t = 20 to t = 30 the run is 10, so the rise is rise = slope * run = -4 * 10 = -40. Starting from (20,-10) a rise of -40 and a run of 10 takes us to (30, -50). The slope at t = 30 is y ' = .1 * 30 - 6 = -3. From t = 30 to t = 40 the run is 10, so the rise is rise = slope * run = -3 * 10 = -30. Starting from (30,-50) a rise of -30 and a run of 10 takes us to (40, -80). The slope at t = 40 is y ' = .1 * 40 - 6 = -2. From t = 40 to t = 50 the run is 10, so the rise is rise = slope * run = -2 * 10 = -20. Starting from (40,-80) a rise of -20 and a run of 10 takes us to (50, -100). The slope at t = 50 is y ' = .1 * 50 - 6 = -1. From t = 50 to t = 60 the run is 10, so the rise is rise = slope * run = -1 * 10 = -10. Starting from (50,-100) a rise of -10 and a run of 10 takes us to (60, -110). The slope at t = 60 is y ' = .1 * 70 - 6 = -0. From t = 60 to t = 70 the run is 10, so the rise is rise = slope * run = -0 * 10 = 0. Starting from (60,-110) a rise of and a run of 10 takes us to (70, -110).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
?????~?????? assignment #008 ??K???????? Calculus I 09-28-2008
......!!!!!!!!...................................
20:21:11 What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?
......!!!!!!!!...................................
RESPONSE --> g(t) = 3t - 5, f(z) = 2^z
.................................................
......!!!!!!!!...................................
20:21:21 ** 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). **
......!!!!!!!!...................................
RESPONSE --> right ok
.................................................
......!!!!!!!!...................................
20:22:39 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
......!!!!!!!!...................................
RESPONSE --> start with a point (x0, y0) on the y vs. x graph evaluate the function y' for x=x0 and y=x0, which gives you a slope for your y vs. x graph. Using (dx) you 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
.................................................
......!!!!!!!!...................................
20:22:53 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok then i repeat the process
.................................................
......!!!!!!!!...................................
20:23:55 explain why a numerical solution to differential equation is only an approximate solution in most cases
......!!!!!!!!...................................
RESPONSE --> You assume the slope at the initial point and a secondary point. If your interval is small enough the change in slope will have a small effect
.................................................
......!!!!!!!!...................................
20:24:03 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:25:18 query Problem 1.4.10 Solve 2 * 5^x = 11 * 7^x
......!!!!!!!!...................................
RESPONSE --> log 2 + x log 5 = log 11 + x log 7. x log 5 - x log 7 = log 11 - log 2 x ( log 5 - log 7) = log 11 - log 2 x = (log 11 - log 2) / (log 5 - log 7). x= appr. -5.07
.................................................
......!!!!!!!!...................................
20:25:22 ** 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
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:26:07 Problem 1.4.6 simplify 2 ln(e^A) + 3 ln(B^e)
......!!!!!!!!...................................
RESPONSE --> Starting with 2 ln (e^A) + 3 ln (e^B) I used the fact that the natural log and exponential functions are inverses, expressed by the law of logarithms ln(e^x) = x 2 * A + 3 * B
.................................................
......!!!!!!!!...................................
20:26:10 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:27:15 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)?
......!!!!!!!!...................................
RESPONSE --> 174 * .9^t = 174 * e^(kt) e^(kt) = .9^t e^k = .9 ln(e^k) = ln(.9) si, k = ln(.9) = appr. -.105
.................................................
......!!!!!!!!...................................
20:27:19 ** 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). **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:29:36 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
......!!!!!!!!...................................
RESPONSE --> P = P0 * e^(kt). t = time since 1980 and population be in millions. 40 = P0 e^(k * 0) 56 = P0 e^(k * 10). 40 = P0 56 = 40 * e^(10 k) = 1.4. Taking logs 10 k = ln(1.4) k = ln(1.4) / 10 = .0336 so 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. **
.................................................
......!!!!!!!!...................................
20:29:40 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:30:34 query Problem 1.4.45 (was 1.7.42 but changed) time to get to 10% of strontium 90 if half-life 29 yrs
......!!!!!!!!...................................
RESPONSE --> 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 Q = Q0 * e^(-.0239 t). 10% of the original quantity we have Q = Q0 / 10 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
.................................................
......!!!!!!!!...................................
20:30:37 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:30:47 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> no comments
.................................................
......!!!!!!!!...................................
20:31:52 Problem 1.4.26 P=174 * .9^t What is the function when converted to exponential form P = P0 e^(kt)?
......!!!!!!!!...................................
RESPONSE --> If P=174(.9)^t P = P0 e^(kt) P0 = 174 and e^(k t) = .9.t so e^k = .9. k = ln(.9) = .105. The function is P=174 e^-(.105 t)
.................................................
......!!!!!!!!...................................
20:31:59 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).
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:33:32 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
......!!!!!!!!...................................
RESPONSE --> P=Po b^t is the form of the function. 40 * 10^6 Po = 40 * 10^6. P=40*10^6 b^t. t = 10 P = 56 * 10^6 56*10^6=40*10^6 b^10. 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
.................................................
......!!!!!!!!...................................
20:33:35 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
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:33:55 10:34:19
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:34:50 I did not understand this problem, but this is what I have:
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:35:13 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok I dont know what happened there but ok
.................................................
"