course Mth 272 ?????e???j??Q??assignment #001
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20:28:52 `questionNumber 10000 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 did not take math 271 at a community college but rather I took two sequences of calculus at a 4 year university. Hope f
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20:29:00 `questionNumber 10000 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 --> adf
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20:29:29 `questionNumber 10000 4.1.16 (was 4.1.14): Solve for x the equation 4^2=(x+2)^2
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RESPONSE --> 1 confidence assessment: 1
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20:29:36 `questionNumber 10000 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. **
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RESPONSE --> 1m or series of problems. self critique assessment: 1
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20:29:44 `questionNumber 10000 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 --> 1 confidence assessment: 1
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20:29:47 `questionNumber 10000 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 --> 1 self critique assessment: 1
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20:29:54 `questionNumber 10000 How does this graph compare to that of 5^-x, and why does it compare as it does?
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RESPONSE --> 1 confidence assessment: 1
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20:30:11 `questionNumber 10000 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 --> 1 self critique assessment: 1
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20:30:27 `questionNumber 10000 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 --> confidence assessment:
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20:30:46 `questionNumber 10000 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 --> 1 self critique assessment: 1
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20:30:55 `questionNumber 10000 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 --> confidence assessment: 1
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20:31:02 `questionNumber 10000 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 --> 1 self critique assessment: 1
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20:31:08 `questionNumber 10000 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 --> 1 confidence assessment: 1
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20:31:17 `questionNumber 10000 *& 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 --> 1 self critique assessment: 1
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????xg???????assignment #001 001. `query 1 Applied Calculus II 06-12-2007
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20:40:09 `questionNumber 10000
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RESPONSE -->
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20:40:15 `questionNumber 10000 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 --> 1
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20:40:17 `questionNumber 10000 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 --> 1
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20:40:21 `questionNumber 10000 4.1.16 (was 4.1.14): Solve for x the equation 4^2=(x+2)^2
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RESPONSE --> 1 confidence assessment: 1
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h???? 001. `query 1 Applied Calculus II 06-12-2007
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20:59:35 `questionNumber 10000 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 didn't take math 271 but I took two semesters of pre-calculus and calculus respectively. It's been a few months since I've done any calculus but I think I should do moderately well in this course.
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21:00:52 `questionNumber 10000 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'm quite familiar with most of the concepts with the exception of implicit differentiation. I never leared this concept fully. Although I might be a little rusty with the other concepts, I will be able to remember them.
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21:03:32 `questionNumber 10000 4.1.16 (was 4.1.14): Solve for x the equation 4^2=(x+2)^2
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RESPONSE --> x= 2 x= -6 The first step was to take the square root of both sides. This gave me 4= x+2 and -4= x+2. There are two answers because of the square root sign. After that, I simply solved for x to get the answers. confidence assessment: 3
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21:03:53 `questionNumber 10000 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. **
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RESPONSE --> Ok self critique assessment: 3
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21:06:48 `questionNumber 10000 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 line is decreasing. The graph is concave up. There are no asymptotes. confidence assessment: 2
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21:12:50 `questionNumber 10000 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 did not know that the graph was asymptotic to the x axis. I only considered horizontal and vertical asymptotes. I'm not familiar with an axis being an asymptote. self critique assessment: 2
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21:17:02 `questionNumber 10000 How does this graph compare to that of 5^-x, and why does it compare as it does?
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RESPONSE --> Both of the graphs are decreasing at a decreasing rate. Both graphs are also concave up. The graph of 5^-x starts up higher than the graph of (1/4)^x confidence assessment: 2
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21:17:25 `questionNumber 10000 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 --> I mentioned everything but the part about it being lower at the right. self critique assessment: 3
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21:30:29 `questionNumber 10000 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 concave up because if you fill in negative numbers for the x then the point is small. If you fill in large positive numbers then the point will be very high in the y axis. The x-axis is a asymptote. The graph is decreasing overall. confidence assessment: 2
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21:30:44 `questionNumber 10000 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 --> Ok self critique assessment: 3
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21:34:57 `questionNumber 10000 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 --> One main difference is that the e^x graph extends further to the right. Also the graph of e^(2x) is more affected by large positive and negative numbers. For example, when 2 is used for e^x the y value is 7.3 But when 2 is used in e^(2x) the y value is 54.6 confidence assessment: 1
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21:37:39 `questionNumber 10000 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 --> I did not go into as much detail as the answer. I didn't realize that the graph of e^(2x) is twice as great at the value of the function. self critique assessment: 2
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21:40:47 `questionNumber 10000 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 --> I obtained my graph by plotting points. I described my graph by looking at my drawing and noticing whether or not it was increasing or deacreasing and whether or not it had any asymptotes. The value of the function is approaching infinity as very large numbers are used for x. I did not notice a limiting value of the function as x approaches infinity. confidence assessment: 1
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21:41:01 `questionNumber 10000 *& 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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22:20:52 `questionNumber 10000 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 --> Compounded 1- $17,599.97 Compounded 2- $18,023.92 Compounded 4- $18,245.05 Compounded 12- $18,396.04 Compounded 365- $18,470.11 Compounded Continously- $18,470.11 For this problem I used the equation : 2500 (1+ (.05/t))^(t*40) confidence assessment: 2
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22:21:19 `questionNumber 10000 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 --> Ok confidence assessment: 3
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22:22:03 `questionNumber 10000 How did you obtain your result for continuous compounding?
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RESPONSE --> I used the formula 2500e^(.05x40) confidence assessment: 3
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22:22:16 `questionNumber 10000 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 --> ok self critique assessment: 3
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15:43:33 `questionNumber 10000 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 rate would be approaching 95. I got this by doing N= 95/(1 + 8.5 *0) which gave me 95. confidence assessment: 2
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15:43:38 `questionNumber 10000 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 --> ok self critique assessment: 3
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15:49:49 `questionNumber 10000 How long did it take to average 70 words / minute?
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RESPONSE --> It would take approximately 26.3 weeks. I began with 70= 95/(1+8.5e^(-0.12t)). After that, I solved for t. confidence assessment: 1
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15:50:08 `questionNumber 10000 *& 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 off by 0.1. I rounded wrong. self critique assessment: 2
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15:53:38 `questionNumber 10000 How many words per minute were being typed after 10 weeks?
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RESPONSE --> After 10 weeks, about 26.7 words per minute were being typed. I used 10 = 95 / (1+8.5e^(-0.12x 10)) = 26.68 to get this. confidence assessment: 2
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15:53:43 `questionNumber 10000 *& 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 --> ok self critique assessment: 3
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15:57:57 `questionNumber 10000 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 --> N = 95 / (1 + 8.5 e^(-.12 t)). I didn't remember how to get the derivative of this one. confidence assessment: 0
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15:58:36 `questionNumber 10000 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 knew the rate would be the same but I forgot the chain rule. This helped me remember the chain rule self critique assessment: 2
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??O???????? assignment #001 001. `query 1 Applied Calculus II 06-13-2007
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15:59:59 `questionNumber 10000
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RESPONSE -->
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16:00:02 `questionNumber 10000 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 -->
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16:00:03 `questionNumber 10000 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 -->
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16:00:04 `questionNumber 10000 4.1.16 (was 4.1.14): Solve for x the equation 4^2=(x+2)^2
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RESPONSE --> confidence assessment:
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16:00:57 `questionNumber 10000 4.3.8 (was 4.2 #8) derivative of e^(1/x)
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insightlkjs you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. confidence assessment: 1
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16:02:43 `questionNumber 10000 4.3.8 (was 4.2 #8) derivative of e^(1/x)
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RESPONSE --> The derivative is -(1/x^2)e^(1/x). I used the exponential rule for this one. confidence assessment: 2
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16:02:47 `questionNumber 10000 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 --> ok self critique assessment: 3
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16:02:54 `questionNumber 10000 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 --> self critique assessment: 1
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16:03:00 `questionNumber 10000 Extra Question: What is the derivative of (e^-x + e^x)^3?
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RESPONSE --> 11 confidence assessment: 2
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16:05:10 `questionNumber 10000 Extra Question: What is the derivative of (e^-x + e^x)^3?
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RESPONSE --> The derivative is 3 (e^-x + e^x)^2 * (-e^-x + e^x). I used the power rule for this. confidence assessment: 1
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16:05:15 `questionNumber 10000 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 --> ok self critique assessment: 3
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16:05:52 `questionNumber 10000 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 --> self critique assessment: 1
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16:05:57 `questionNumber 10000 4.3.22. What is the tangent line to e^(4x-2)^2 at (0, 1)?
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RESPONSE --> 2 confidence assessment: 2
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16:09:11 `questionNumber 10000 4.3.22. What is the tangent line to e^(4x-2)^2 at (0, 1)?
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RESPONSE --> THe tangent line is y = .013x +2. I got the derivative to be (f(g(x))' = g ' (x) * f ' (g(x)) = 4 (e^(4x-2) ) * 2 ( e^(4x - 2)) = 8 ( e^(4x - 2)). confidence assessment: 2
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16:09:43 `questionNumber 10000 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.
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RESPONSE --> I do not know where I messed up on this problem. It might have been after I took the derivative and when solving for y. self critique assessment: 2
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16:09:49 `questionNumber 10000 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.
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RESPONSE --> self critique assessment: 1
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16:09:55 `questionNumber 10000 4.3.26 (formerly 4.3.24) (was 4.2.22) implicitly find y' for e^(xy) + x^2 - y^2 = 0
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RESPONSE --> 1 confidence assessment: 1
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16:10:27 `questionNumber 10000 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 ) . **
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RESPONSE --> ok self critique assessment: 3
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16:10:35 `questionNumber 10000 4.3.26 (formerly 4.3.24) (was 4.2.22) implicitly find y' for e^(xy) + x^2 - y^2 = 0
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RESPONSE --> confidence assessment:
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16:10:38 `questionNumber 10000 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 ) . **
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RESPONSE --> self critique assessment:
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16:10:40 `questionNumber 10000 4.3.34 (formerly 4.3.32) (was 4.2 #30) extrema of x e^(-x)
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RESPONSE --> confidence assessment:
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16:12:54 `questionNumber 10000 4.3.34 (formerly 4.3.32) (was 4.2 #30) extrema of x e^(-x)
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RESPONSE --> Start out with e^(-x) + (e^-x)=0. Then e^(-x)(1-x)=0. There is a maxium at y= 1x e^-1 confidence assessment: 2
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16:13:00 `questionNumber 10000 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). **
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RESPONSE --> ok self critique assessment: 3
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16:13:14 `questionNumber 10000
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insiasdfghts regarding the situation at this point, notes to yourself, or just an OK. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. self critique assessment: 2
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16:13:20 `questionNumber 10000 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. **
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RESPONSE --> asdf self critique assessment: 1
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16:59:31 `questionNumber 10000 4.2.48 (formerly 4.2.46) (was 4.2 #42) effect of `mu on normal distribution
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RESPONSE --> There is a maximum x = `mu confidence assessment: 2
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16:59:41 `questionNumber 10000 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. **
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RESPONSE --> ok self critique assessment: 3
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17:00:52 `questionNumber 10000 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I was more rusty than I remember. I had to ask for a lot of help. confidence assessment: 3
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17:01:06 `questionNumber 10000 Typical Comment so if you feel very rusty you'll know you aren't along: Good grief, lol where to start!!! Just kidding! I guess I really need to be refreshed on how to handle deriving the exponential function with e. 4.2 was the killer for me here with only minimum examples in the section I had to review my old text and notes. It's just been so long.
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RESPONSE --> Ok confidence assessment: 3
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