course Mth 272 ݠ|T\᳡G{Dxassignment #001
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22:55:30 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.
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RESPONSE --> I completed MTH271 online from Central VA Comm College this past spring. I did fairly well and ended with a B. I may be a little rusty but it will not take me long to get back into the swing of things. I have a good understanding of the basic rules of differentation, exponetial functions, intergration, and finding the areas under a curve or between two curves.
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22:56:21 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.
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RESPONSE --> I am prepared for this course. My worry is more about careless mistakes.
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23:01:59 4.1.16 (was 4.1.14): Solve for x the equation 4^2=(x+2)^2
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RESPONSE --> x = 2 or x = -6 since it is squared we know that (x+2) can be + and -. So we solve for both. confidence assessment: 3
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23:14:04 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.
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RESPONSE --> The graph is decreasing at a decreasing rate. It is concave up The horizontal asymtope is 0. It crosses y-axis at (0,1). confidence assessment: 3
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23:14:36 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. **
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RESPONSE --> I am correct self critique assessment: 3
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23:18:33 How does this graph compare to that of 5^-x, and why does it compare as it does?
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RESPONSE --> The graph is very similar. Values are larger to the left and smaller to the right of ""y"" confidence assessment: 3
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23:18:43 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. **
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RESPONSE --> Got it self critique assessment: 3
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23:21:37 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.
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RESPONSE --> The graph is increasing at an increasing rate. Y-intercept = (0,1). Conacve up. confidence assessment: 3
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23:21:47 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. **
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RESPONSE --> Okay self critique assessment: 3
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23:25:49 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?
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RESPONSE --> It is similar ,only e^x is increasing at half the rate of e^2x confidence assessment: 3
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23:25:54 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. **
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RESPONSE --> self critique assessment: 3
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23:30:52 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?
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RESPONSE --> At first I plotted points for values of -2,-1,0,1,2,3. This gives me the shape ofthe graph. But using the first derivative I see that the slope fo e^(2x) is twice that of e^x. confidence assessment: 3
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23:31:00 *& These questions are answered in the solutions given above. From those solutions you will ideally have been able to answer this question. *&*&
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RESPONSE --> self critique assessment:
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23:40:24 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
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RESPONSE --> yearly = 17599.97 semi annualy = 18023.92 quarterly = 18245.05 monthly = 18396.04 daily = 18470.11 using compound interest formula confidence assessment: 3
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23:40:30 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
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RESPONSE --> confidence assessment:
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23:44:28 How did you obtain your result for continuous compounding?
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RESPONSE --> Continually compounding forula is A=Pe^(rt) confidence assessment: 3
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23:44:36 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. **
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RESPONSE --> self critique assessment:
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23:46:44 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?
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RESPONSE --> The limit rate would be 95.because the denominator gets closer to 1 as ""t"" increases confidence assessment: 3
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23:46:50 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. *&*&
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RESPONSE --> self critique assessment:
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23:55:29 How long did it take to average 70 words / minute?
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RESPONSE --> About 26 weeks confidence assessment: 2
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23:55:49 *& 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. **
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RESPONSE --> I was close self critique assessment: 2
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23:59:49 How many words per minute were being typed after 10 weeks?
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RESPONSE --> 26.7 wpm confidence assessment: 3
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23:59:59 *& 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. **
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RESPONSE --> got it self critique assessment: 3
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00:00:56 Find the exact rate at which the model predicts words will be typed after 10 weeks (not time limit here).
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RESPONSE --> Not sure what you are asking confidence assessment: 0
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00:02:10 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 ). **
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RESPONSE --> I am not sure about this one self critique assessment: 0
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00:05:13 4.3.8 (was 4.2 #8) derivative of e^(1/x)
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RESPONSE --> f'(x)=1/2e^(1/x) confidence assessment: 3
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00:05:53 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) **
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RESPONSE --> missed it but see it self critique assessment: 0
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00:08:54 Extra Question: What is the derivative of (e^-x + e^x)^3?
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RESPONSE --> 3 (e^-x + e^x)^2 * (-e^-x + e^x) confidence assessment: 3
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00:09:07 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. **
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RESPONSE --> okay self critique assessment:
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