course Mth 272 I have no idea if I did this assignment correctly. The material is difficult to remember, but on top of that, the extemely confusing format of the program coupled with the material makes for not the best first impression of the course. Any words of advice/encouragement are welcome. I completed Assignment 1 on various days so I hope I am submitting all my information correctly without being repetitive. Let me know. }ýg
......!!!!!!!!...................................
15:44:55 INTRODUCTORY NOTE: The typical student starting out a second-semester calculus course it typically a bit rusty. It is also common that students you tend to use the calculator in appropriately, replacing analysis with calculator output. The calculator is in this course to be used to enhance the analysis but not to replace it, as you will learn on the first assignments. Some first-semester courses emphasize calculator over analysis rather than calculator as an adjunct to analysis, and even when that is not the emphasis the calculator tricks are all some students com away with. A student who has completed a first-semester course has the ability to do this work, but will often need a good review. If this is your case you will need to relearn the analytical techniques, which you can do as you go through this chapter. A solid review then will allow you to move along nicely when we get to the chapters on integration, starting with Ch 5. Calculator skills will be useful to illuminate the analytical process throughout. THis course certainly doesn't discourage use of the calculator, but only as an adjunct to the analytical process than a replacement for it. You will see what that means as you work through Chapter 4. If it turns out that you have inordinate difficulties with the basic first-semester techniques used in this chapter, a review might be appropriate. I'll advise you on that as we go through the chapter. For students who find that they are very rusty on their first-semester skills I recommend (but certainly don't require) that they download the programs q_a_cal1_1_13... and q_a_cal1_14_16... , from the Supervised Study Current Semester pages (Course Documents > Downloads > Calculus I or Applied Calculus I) and work through all 16 assignments, with the possible exception of #10 (a great application of exponential functions so do it if you have time), skipping anything they find trivial and using their own judgement on whether or not to self-critique. The review takes some time but will I believe save many students time in the long run. For students who whoose to do so I'll be glad to look at the SEND files and answer any questions you might have. Please take a minute to give me your own assessment of the status of your first-semeseter skills.
......!!!!!!!!...................................
RESPONSE --> I took calculus I over 4 years ago and have not had a math course since. ""Rusty"" is a definite understatement.
.................................................
......!!!!!!!!...................................
15:45:32 You should understand the basic ideas, which include but are not limited to the following: rules of differentiation including product, quotient and chain rules, the use of first-derivative tests to find relative maxima and minima, the use of second-derivative tests to do the same, interpreation of the derivative, implicit differentiation and the complete analysis of graphs by analytically finding zeros, intervals on which the function is positive and negative, intervals on which the function is increasing or decreasing and intervals on which concavity is upward and downward. Comment once more on your level of preparedness for this course.
......!!!!!!!!...................................
RESPONSE --> I will re-learn things along the way I guess, because I have little memory of all of those things mentioned. What else was I to do, though? I did not realize I would need this course for my diploma until about 3 months ago. I will work as hard as I can.
.................................................
......!!!!!!!!...................................
15:46:09 4.1.16 (was 4.1.14): Solve for x the equation 4^2=(x+2)^2
......!!!!!!!!...................................
RESPONSE --> 4^2=16 16=(x+2)^2 take sqrt of both sides to eliminate (^2) 4=(x+2) x=2, or x= -4 (because -4^2 can also equal 16) confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:46:22 The steps in the solution: 4^2 = (x+2)^2. The solution of a^2 = b is a = +- sqrt(b). So we have x+2 = +- sqrt(4^2) or x+2 = +- 4. This gives us two equations, one for the + and one for the -: x+2 = 4 has solution x = 2 x+2 = -4 has solution x = -6. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:46:56 4.1.28 (was 4.1.32) graph 4^(-x). Describe your graph by telling where it is increasing, where it is decreasing, where it is concave up, where it is concave down, and what if any lines it has as asymptotes.
......!!!!!!!!...................................
RESPONSE --> The graph is decreasing towards the x-axis, concave down, and is decreasing at a decreasing rate. It appears as though the y-values for the graph decrease as x approaches infinite. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:47:26 Many students graph this equation by plugging in numbers. That is a start, but you can only plug in so many numbers. In any case plugging in numbers is not a calculus-level skill. It is necessary to to reason out and include detailed reasons for the behavior, based ultimately on knowledge of derivatives and the related behavior of functions. A documented description of this graph will give a description and will explain the reasons for the major characteristics of the graph. The function y = 4^-x = 1 / 4^x has the following important characteristics: For increasing positive x the denominator increases very rapidly, resulting in a y value rapidly approaching zero. For x = 0 we have y = 1 / 4^0 = 1. For decreasing negative values of x the values of the function increase very rapidly. For example for x = -5 we get y = 1 / 4^-5 = 1 / (1/4^5) = 4^5 = 1024. Decreasing x by 1 to x = -6 we get 1 / 4^-6 = 4096. The values of y more and more rapidly approach infinity as x continues to decrease. This results in a graph which for increasing x decreases at a decreasing rate, passing through the y axis at (0, 1) and asymptotic to the positive x axis. The graph is decreasing and concave up. When we develop formulas for the derivatives of exponential functions we will be able to see that the derivative of this function is always negative and increasing toward 0, which will further explain many of the characteristics of the graph. **
......!!!!!!!!...................................
RESPONSE --> I forgot to explain the asymptote but I believe that the picture I described still layed out the ""asymptotic to the positive x-axis"" fairly well. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:49:09 How does this graph compare to that of 5^-x, and why does it compare as it does?
......!!!!!!!!...................................
RESPONSE --> It appears to be minimally different. The ""5"" of the latter function is a higher number than the 4, so one would expect that the graph would be at an increased level for the second equation, though minimally. This is because ""-x"" just means an inversion of denominator and numerator. For example, 4*(1/x) (as in the last graph) is does not appear that different as 5*(1/x). The graphs are very similar. confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:49:21 the graphs meet at the y axis; to the left of the y axis the graph of y = 5^-x is higher than that of y = 4^-x and to the right it is lower. This is because a higher positive power of a larger number will be larger, but applying a negative exponent will give a smaller results for the larger number. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:50:50 4.2.20 (was 4.1 #40) graph e^(2x) Describe your graph by telling where it is increasing, where it is decreasing, where it is concave up, where it is concave down, and what if any lines it has as asymptotes.
......!!!!!!!!...................................
RESPONSE --> The graph is increasing at an increasing rate, concaving up as it crosses the positive y-axis along the positive direction along the x-axis. The line is asymptotic towards the negative x-axis. To read the graph from right to left, the line is decreasing at a decreasing rate, but still asymptotic to the negative x-axis. The line crosses the y-axis around positive 1. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:51:07 For large numbers x you have e raised to a large power, which gets extremely large. At x = 0 we have y = e^0 = 1. For large negative numbers e is raised to a large negative power, and since e^-a = 1 / e^a, the values of the function approach zero. } Thus the graph approaches the negative x axis as an asymptote and grows beyond all bounds as x gets large, passing thru the y axis as (0, 1). Since every time x increases by 1 the value of the function increases by factor e, becoming almost 3 times as great, the function will increase at a rapidly increasing rate. This will make the graph concave up. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:53:27 The entire description given above would apply to both e^x and e^(2x). So what are the differences between the graphs of these functions?
......!!!!!!!!...................................
RESPONSE --> Though they both have the same y-intercept (1), the e^(x) graph appears to have double the concaveness (I may have just made up that terminology) than the e^(2x). What I am trying to say is that the curve on the first graph is far more broad, but they still have the same asymptotic character and the overall descriptions of the graphs are similar (concave up, increasingly increasing rate, etc.). The difference in shape of the graphs is due to the difference in slope. The second equation has a derivative of 2, and the first a derivative of one. This means that the second graph will have twice the slope of the first, and that explains the steepness of the curve. confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:53:35 Note that the graphing calculator can be useful for seeing the difference between the graphs, but you need to explain the properties of the functions. For example, on a test, a graph copied from a graphing calculator is not worth even a point; it is the explanation of the behavior of the function that counts. By the laws of exponents e^(2x) = (e^x)^2, so for every x the y value of e^(2x) is the square of the y value of e^x. For x > 1, this makes e^(2x) greater than e^x; for large x it is very much greater. For x < 1, the opposite is true. You will also be using derivatives and other techniques from first-semester calculus to analyze these functions. As you might already know, the derivative of e^x is e^x; by the Chain Rule the derivative of e^(2x) is 2 e^(2x). Thus at every point of the e^(2x) graph the slope is twice as great at the value of the function. In particular at x = 0, the slope of the e^x graph is 1, while that of the e^(2x) graph is 2. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:54:36 How did you obtain your graph, and what reasoning convinces you that the graph is as you described it? What happens to the value of the function as x increases into very large numbers? What is the limiting value of the function as x approaches infinity?
......!!!!!!!!...................................
RESPONSE --> I obtained my graph using the calculator, because I am unaware of a better method. The reasoning behind it is not just from what I saw (one line having twice the slope of the other, and that the latter was a broad concave line), but that remembering some basics of the calculus I took a long while ago, the e^2x graph is going to be steeper due to a larger slope than the e^x graph. It is a slope ratio of 2:1, which explains the visual graphs. I could also plug in numbers to the equation to get another representation of the previous explanation. A x increases to large numbers, the value of the function does as well, because as x increases, y approaches infinite at an increasing rate. I am kind of unsure what the question is asking. confidence assessment: 1
.................................................
ԒŦeax㗕cY~ assignment #001 001. `query 1 Applied Calculus II 09-05-2008
......!!!!!!!!...................................
11:31:41 *& These questions are answered in the solutions given above. From those solutions you will ideally have been able to answer this question. *&*&
......!!!!!!!!...................................
RESPONSE --> ok. self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:36:40 4.2.32 (formerly 4.2.43) (was 4.1 #48) $2500 at 5% for 40 years, 1, 2, 4, 12, 365 compoundings and continuous compounding
......!!!!!!!!...................................
RESPONSE --> P[1 + (r/n)]^nt P= 2500 r=.05 n= 1,2,4,12,365 t=40 when plugging in the numbers, the answers are (respective with ""n""): 17599.97, 18023.92, 18245.05, 18396.04, 18470.11 confidence assessment: 3
.................................................
......!!!!!!!!...................................
11:36:56 A = P[1 + (r/n)]^nt A = 2500[1 + (0.05/1]^(1)(40) = 17599.97 A = 2500[1 + (0.05/2]^(2)(40) = 18023.92 A = 2500[1 + (0.05/4]^(4)(40) = 18245.05 A = 2500[1 + (0.05/12]^(12)(40) = 18396.04 A = 2500[1 + (0.05/365]^(365)(40) = 18470.11
......!!!!!!!!...................................
RESPONSE --> ok, simple plug and chug confidence assessment: 3
.................................................
......!!!!!!!!...................................
11:39:00 How did you obtain your result for continuous compounding?
......!!!!!!!!...................................
RESPONSE --> A different equation A=Pe^rt P=2500 r= .05 t=40 A=2500*e^(.05*40) A=18.472.64, for continuous compounding confidence assessment: 3
.................................................
......!!!!!!!!...................................
11:39:18 For continuous compounding you have A = Pe^rt. For interest rate r = .05 and t = 40 years we have A = 2500e^(.05)(40). Evaluating we get A = 18472.64 The pattern of the results you obtained previously is to approach this value as a limit. **
......!!!!!!!!...................................
RESPONSE --> ok, again simple plug and chug self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:46:31 4.2.40 (was 4.1 #60) typing rate N = 95 / (1 + 8.5 e^(-.12 t)) What is the limiting value of the typing rate?
......!!!!!!!!...................................
RESPONSE --> an increase in time (t) would cause the value to exponentially decrease as time went on, so the value would of rate would continue to approach zero. when substituting zero in for the equation, you get 95, as shown below. N = 95 / (1 + 8.5 * 0) = 95 / 1 confidence assessment: 2
.................................................
......!!!!!!!!...................................
11:46:42 As t increases e^(-.12 t) decreases exponentially, meaning that as an exponential function with a negative growth rate it approaches zero. The rate therefore approaches N = 95 / (1 + 8.5 * 0) = 95 / 1 = 95. *&*&
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:54:11 How long did it take to average 70 words / minute?
......!!!!!!!!...................................
RESPONSE --> 70=95 / (1+8.5e^(-0.12t)) 95=70(1+8.5e^(-0.12t)) e^(-.12 t) = 25/595 t = ln(25/595) / (-.12) t= 26.414 weeks confidence assessment: 3
.................................................
......!!!!!!!!...................................
11:54:38 *& According to the graph of the calculator it takes about 26.4 weeks to get to 70 words per min. This result was requested from a calculator, but you should also understand the analytical techniques for obtaining this result. The calculator isn't the authority, except for basic arithmetic and evaluating functions, though it can be useful to confirm the results of actual analysis. You should also know how to solve the equation. We want N to be 70. So we get the equation 70=95 / (1+8.5e^(-0.12t)). Gotta isolate t. Note the division. You first multiply both sides by the denominator to get 95=70(1+8.5e^(-0.12t)). Distribute the multiplication: 95 = 70 + 595 e^(-.12 t). Subtract 70 and divide by 595: e^(-.12 t) = 25/595. Take the natural log of both sides: -.12 t = ln(25/595). Divide by .12: t = ln(25/595) / (-.12). Approximate using your calculator. t is around 26.4. **
......!!!!!!!!...................................
RESPONSE --> ok understood, it is just manipulation of the previous equation. self critique assessment: 3
.................................................
......!!!!!!!!...................................
11:57:07 How many words per minute were being typed after 10 weeks?
......!!!!!!!!...................................
RESPONSE --> = 95 / (1+8.5e^(-0.12* 10)) = 26.7 words/min after 10 weeks confidence assessment: 3
.................................................
......!!!!!!!!...................................
11:57:15 *& According to the calculator 26.6 words per min was being typed after 10 weeks. Straightforward substitution confirms this result: N(10) = 95 / (1+8.5e^(-0.12* 10)) = 26.68 approx. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
11:59:10 Find the exact rate at which the model predicts words will be typed after 10 weeks (not time limit here).
......!!!!!!!!...................................
RESPONSE --> I am unsure how the rate would be different than the previous calculation. So I am going to once again say ""26.68 words per minute"" and if I am wrong I will need your comments for help. Thanks. confidence assessment: 2
.................................................
......!!!!!!!!...................................
12:02:28 The rate is 26.6 words / minute, as you found before. Expanding a bit we can find the rate at which the number of words being typed will be changing at t = 10 weeks. This would require that you take the derivative of the function, obtaining dN / dt. This question provides a good example of an application of the Chain Rule, which might be useful for review: Recall that the derivative of e^t is d^t. N = 95 / (1 + 8.5 e^(-.12 t)), which is a composite of f(z) = 1/z with g(t) = (1 + 8.5 e^(-.12 t)). The derivative, by the Chain Rule, is N' = g'(t) * f'(g(t)) = (1 + 8.5 e^(-.12 t)) ' * (-1 / (1 + 8.5 e^(-.12 t))^2 ) = -.12 * 8.5 e^(-.12 t)) * (-1 / (1 + 8.5 e^(-.12 t))^2 ) = 1.02 / (1 + 8.5 e^(-.12 t))^2 ). **
......!!!!!!!!...................................
RESPONSE --> I do not undersand this. Is there a better explanation in the beginning of the book? I will look because this is not coming back to me at all. Like I said it's been 4.5 years. self critique assessment: 1
.................................................
......!!!!!!!!...................................
13:45:03 4.3.8 (was 4.2 #8) derivative of e^(1/x)
......!!!!!!!!...................................
RESPONSE --> du/dx u= 1/x du/dx=-1/x^2 e^u(du/dx)=[e^(1/x)]*[-1/x^2]= derivative confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:45:15 There are two ways to look at the function: This is a composite of f(z) = e^z with g(x) = 1/x. f'(z) = e^z, g'(x) = -1/x^2 so the derivative is g'(x) * f'(g(x)) = -1/x^2 e^(1/x). Alternatively, and equivalently, using the text's General Exponential Rule: You let u = 1/x du/dx = -1/x^2 f'(x) = e^u (du/dx) = e^(1/x) * -1 / x^2. dy/dx = -1 /x^2 e^(1/x) **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:58:10 Extra Question: What is the derivative of (e^-x + e^x)^3?
......!!!!!!!!...................................
RESPONSE --> du/dx method u= e^-x+e^x n= 3 (u^n)' = n u^(n-1) * du/dx [(e^-x+e^x)^3]'= 3*(e^-x+e^x)^2*(-e^-x + e^x ) confidence assessment: 2
.................................................
......!!!!!!!!...................................
13:58:38 This function is the composite f(z) = z^3 with g(x) = e^-x + e^x. f ' (z) = 3 z^2 and g ' (x) = - e^-x + e^x. The derivative is therefore (f(g(x)) ' = g ' (x) * f ' (g(x)) = (-e^-x + e^x) * 3 ( e^-x + e^x) ^ 2 = 3 (-e^-x + e^x) * ( e^-x + e^x) ^ 2 Alternative the General Power Rule is (u^n) ' = n u^(n-1) * du/dx. Letting u = e^-x + e^x and n = 3 we find that du/dx = -e^-x + e^x so that [ (e^-x + e^x)^3 ] ' = (u^3) ' = 3 u^2 du/dx = 3 (e^-x + e^x)^2 * (-e^-x + e^x), as before. **
......!!!!!!!!...................................
RESPONSE --> ok, turns out I got it right but I was not 100% confident due to their being so many numbers and steps. self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:05:21 4.3.22. What is the tangent line to e^(4x-2)^2 at (0, 1)?
......!!!!!!!!...................................
RESPONSE --> I am confused. The point is 0,1 but yet the equation (when solving for y by plugging in zero for x) does not compute to that. I don't know how to attempt this problem if there is a way. confidence assessment: 0
.................................................
......!!!!!!!!...................................
14:06:49 FIrst note that at x = 0 we have e^(4x-2) = e^(4*0 - 2)^2 - e^(-2)^2, which is not 1. So the graph does not pass through (0, 1). The textbook is apparently in error. We will continue with the process anyway and note where we differ from the text. }The function is the composite f(g(x)) wheren g(x) = e^(4x-2) and f(z) = z^2, with f ' (z) = 2 z. The derivative of e^(4x-2) itself requires the Chain Rule, and gives us 4 e^(4x-2). So our derivative is (f(g(x))' = g ' (x) * f ' (g(x)) = 4 (e^(4x-2) ) * 2 ( e^(4x - 2)) = 8 ( e^(4x - 2)). Now at x = 0 our derivative is 8 ( e^(4 * 0 - 2)) = 8 e^-2 = 1.08 (approx). If (0, 1) was a graph point the tangent line would be the line through (0, 1) with slope 1.08. This line has equation y - 1 = .0297 ( x - 0), or solving for y y = .0297 x + 1. As previously noted, however, (0, 1) is not a point of the original graph.
......!!!!!!!!...................................
RESPONSE --> oh ok, so i should have just done the problem as if i was not confused by the graph point? self critique assessment: 1
.................................................
......!!!!!!!!...................................
14:11:43 4.3.26 (formerly 4.3.24) (was 4.2.22) implicitly find y' for e^(xy) + x^2 - y^2 = 0
......!!!!!!!!...................................
RESPONSE --> I am not comfortable enough yet to attempt this problem with any accuracy. I hope the penalty is not significant. confidence assessment: 0
.................................................
......!!!!!!!!...................................
14:12:28 The the q_a_ program for assts 14-16 in calculus 1, located on the Supervised Study ... pages under Course Documents, Calculus I, has an introduction to implicit differentiation. I recommend it if you didn't learn implicit differentiation in your first-semester course, or if you're rusty and can't follow the introduction in your text. The derivative of y^2 is 2 y y'. y is itself a function of x, and the derivative is with respect to x so the y' comes from the Chain Rule. the derivative of e^(xy) is (xy)' e^(xy). (xy)' is x' y + x y' = y + x y '. the equation is thus (y + x y' ) * e^(xy) + 2x - 2y y' = 0. Multiply out to get y e^(xy) + x y ' e^(xy) + 2x - 2 y y' = 0, then collect all y ' terms on the left-hand side: x y ' e^(xy) - 2 y y ' = -y e^(xy) - 2x. Factor to get (x e^(xy) - 2y ) y' = - y e^(xy) - 2x, then divide to get y' = [- y e^(xy) - 2x] / (x e^(xy) - 2y ) . **
......!!!!!!!!...................................
RESPONSE --> OK, i am going to review the q_a_ program for those assignments before attempting any further implicit differentiation. self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:28:36 4.3.34 (formerly 4.3.32) (was 4.2 #30) extrema of x e^(-x)
......!!!!!!!!...................................
RESPONSE --> derivative should be zero x' e^(-x) + x (e^-x)'=0 e^-x + x(-e^-x) = 0 e^(-x) (1-x) = 0 (1-x) = 0 x-1, which means that derivative goes up then down, so the extrema would be a maximum at the critical point (1, e^-1) confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:28:46 Again the calculator is useful but it doesn't replace analysis. You have to do the analysis for this problem and document it. Critical points occur when the derivative is 0. Applying the product rule you get x' e^(-x) + x (e^-x)' = 0. This gives you e^-x + x(-e^-x) = 0. Factoring out e^-x: e^(-x) (1-x) = 0 e^(-x) can't equal 0, so (1-x) = 0 and x = 1. Now, for 0 < x < 1 the derivative is positive because e^-x is positive and (1-x) is positive. For 1 < x the derivative is negative because e^-x is negative and (1-x) is negative. So at x = 1 the derivative goes from positive to negative, indicating the the original function goes from increasing to decreasing. Thus the critical point gives you a maximum. The y value is 1 * e^-1. The extremum is therefore a maximum, located at (1, e^-1). **
......!!!!!!!!...................................
RESPONSE --> sounds good self critique assessment: 3
.................................................
......!!!!!!!!...................................
18:03:01 4.3.42 (formerly 4.3.40) (was 4.2 #38) memory model p = (100 - a) e^(-bt) + a, a=20 , b=.5, info retained after 1, 3 weeks.How much memory was maintained after each time interval?
......!!!!!!!!...................................
RESPONSE --> a=20 b=.5 t=1 p = (100 - 20) e^(-.5 * 1) + 20 = 80 * e^-.5 + 20 p= 68.5 % for 1 week p= 37.9% for 3 weeks confidence assessment: 3
.................................................
......!!!!!!!!...................................
18:03:25 Plugging in a = 20, b = .5 and t = 1 we get p = (100 - 20) e^(-.5 * 1) + 20 = 80 * e^-.5 + 20 = 68.52, approx., meaning about 69% retention after 1 week. A similar calculation with t = 3 gives us 37.85, approx., indicating about 38% retention after 3 weeks. **
......!!!!!!!!...................................
RESPONSE --> ok, those were the exact answers i got, but with rounding. self critique assessment: 3
.................................................
......!!!!!!!!...................................
18:07:44 ** At what rate is memory being lost at 3 weeks (no time limit here)?
......!!!!!!!!...................................
RESPONSE --> (100 - a) e^(-bt) + a t=3 (I am assuming the ""a"" and ""b"" values are the same as the previous problem) = -8.93 (losing about 9% memory) self critique assessment: 3
.................................................
......!!!!!!!!...................................
18:08:39 The average rate of change of y with respect to t is ave rate = change in y / change in t. This is taken to the limit, as t -> 0, to get the instantaneous rate dy/dt, which is the derivative of y with respect to t. This is the entire idea of the derivative--it's an instantaneous rate of change. The rate of memory loss is the derivative of the function with respect to t. dp/dt = d/dt [ (100 - a) e^(-bt) + a ] = (100-a) * -b e^-(bt). Evaluate at t = 3 to answer the question. The result is dp/dt = -8.93 approx.. This indicates about a 9% loss per week, at the 3-week point. Of course as we've seen you only have about 38% retention at t = 3, so a loss of almost 9 percentage points is a significant proportion of what you still remember. Note that between t = 1 and t = 3 the change in p is about -21 so the average rate of change is about -21 / 2 = -10.5. The rate is decreasing. This is consistent with the value -8.9 for the instantaneous rate at t = 3. **
......!!!!!!!!...................................
RESPONSE --> I used the equation and got the right answer without thinking in terms of the slip and derivative. Is that ok? Or am I on the wrong track? self critique assessment: 2
.................................................
铡i؆PՕe~ assignment #001 001. `query 1 Applied Calculus II 09-06-2008
......!!!!!!!!...................................
15:23:44 4.2.48 (formerly 4.2.46) (was 4.2 #42) effect of `mu on normal distribution
......!!!!!!!!...................................
RESPONSE --> Mu is the midpoint of a distribution. A change im mu will skew the distribution to the right or left. The amount or significance of change varies depending on the sigma value that corresponds to the distribution. Professor Smith, I am a little lost. I hope this is the answer to the question. However, the questions of this assignment are confusing because they seem to pop up out of nowhere. The numbers prior to each question do not appear to correspond to anything in the book (unless I am looking in the wrong place, I have version 7.) So when things like ""effect of 'mu on normal distribution"" come up, I am a little unsure of what exactly the question is. I will submit this little blurb here in a question form as well. But this assignment was nothing like I have ever experienced. confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:24:42 The calculator should have showed you how the distribution varies with different values of `mu. The analytical explanation is as follows: The derivative of e^[ -(x-`mu)^2 / (2 `sigma) ] is -(x - `mu) e^[ -(x-`mu)^2 / 2 ] / `sigma. Setting this equal to zero we get -(x - `mu) e^[ -(x-`mu)^2 / 2 ] / `sigma = 0. Dividing both sides by e^[ -(x-`mu)^2 / 2 ] / `sigma we get -(x - `mu) = 0, which we easily solve for x to get x = `mu. The sign of the derivative -(x - `mu) e^[ -(x-`mu)^2 / 2 ] / `sigma is the same as the sign of -(x - `mu) = `mu - x. To the left of x = `mu this quantity is positive, to the right it is negative, so the derivative goes from positive to negative at the critical point. By the first-derivative test the maximum therefore occurs at x = `mu. More detail: We look for the extreme values of the function. e^[ -(x-`mu)^2 / (2 `sigma) ] is a composite of f(z) = e^z with g(x) = -(x-`mu)^2 / (2 `sigma). g'(x) = -(x - `mu) / `sigma. Thus the derivative of e^[ -(x-`mu)^2 / (2 `sigma) ] with respect to x is -(x - `mu) e^[ -(x-`mu)^2 / 2 ] / `sigma. Setting this equal to zero we get x = `mu. The maximum occurs at x = `mu. **
......!!!!!!!!...................................
RESPONSE --> That is a lot more detail than I explained, but did you want all of that in the answer? As previously noted, this question was not an easy one to comprehend, especially if you wanted an answer far from what I gave you. self critique assessment: 2
.................................................
......!!!!!!!!...................................
15:25:13 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> Basically just note my comment/concern at the end of my response for the final question. confidence assessment: 3
.................................................