course Mth 173 ?g?????~????assignment #002
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11:17:29 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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RESPONSE --> y=7 x=-3.5 confidence assessment: 3
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11:18:08 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
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RESPONSE --> self critique assessment: 3
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11:20:18 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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RESPONSE --> The first problems slope was 2/1. this one is 3/1 confidence assessment: 3
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11:20:38 The graph forms a straight line with no change in steepness.
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RESPONSE --> self critique assessment: 3
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11:20:59 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> 3/1 confidence assessment: 3
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11:21:05 Between any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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RESPONSE --> self critique assessment: 3
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11:22:12 `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> increasing; no it is straight. constant rate confidence assessment:
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11:22:37 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate? INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1? In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.
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RESPONSE --> self critique assessment: 3
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11:24:04 `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> decreasing, no change in steepness, decreasing at a constant rate confidence assessment: 3
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11:24:19 From left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.
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RESPONSE --> self critique assessment: 3
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11:25:58 `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> increasing, increase in steepness, increasing rate. confidence assessment: 3
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11:26:10 If you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate. If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties.
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RESPONSE --> self critique assessment: 3
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11:28:10 `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> both, yes, increasing and decreasing confidence assessment: 3
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11:28:19 ** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. **
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RESPONSE --> self critique assessment: 3
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11:28:46 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> increasing, increasing rate confidence assessment: 3
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11:28:58 ** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **
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RESPONSE --> self critique assessment: 3
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?]?????????assignment #005 005. Calculus qa initial problems 06-04-2007
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11:34:29 `q001. The graph of a certain function is a smooth curve passing through the points (3, 5), (7, 17) and (10, 29). Between which two points do you think the graph is steeper, on the average? Why do we say 'on the average'?
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RESPONSE --> Between (7,17) and (10,29); because an average is not exact. confidence assessment: 3
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11:34:40 Slope = rise / run. Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4. The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3. The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints.
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RESPONSE --> self critique assessment: 3
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11:36:49 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)? 1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)? 2. Will the value ever exceed a billion? Will it ever exceed one trillion billions? 3. Will it ever exceed the number of particles in the known universe? 4. Is there any number it will never exceed? 5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2?
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RESPONSE --> 1/.1, 1/.01, 1/.001, 1/.0001 they get smaller and smaller sure, sure, you could do this all day sure no straight line up and down self critique assessment: 3
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11:37:12 For x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001. Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000. It is important to note that x is changing by smaller and smaller increments as it approaches 2, while the value of the function is changing by greater and greater amounts. As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any number, we can exceed it. Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0). As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical. That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all bounds. Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be negative and approach -infinity.
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RESPONSE --> self critique assessment: 3
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11:45:09 `q003. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.
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RESPONSE --> The first one should be larger because y increase a lot more than the second one, and even though x increase a lot more in the second, it is not enough to make up for the y increase in the first one confidence assessment: 3
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11:45:21 Your sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.
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RESPONSE --> self critique assessment: 3
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11:46:18 `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the x = -1 and x = 7 points on the same graph? Explain the basisof your reasoning.
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RESPONSE --> There is a greater distance between 4 and 25 than there is between -1 and 7 confidence assessment: 2
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11:46:34 The line segment connecting x = 2 and the x = 5 points is steeper: Since f(x) = x^2, x = 2 gives y = 4 and x = 5 gives y = 25. The slope between the points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7. The line segment connecting the x = -1 point (-1,1) and the x = 7 point (7,49) has a slope of (49 - 1) / (7 - -1) = 48 / 8 = 6. The slope of the first segment is greater.
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RESPONSE --> self critique assessment: 3
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11:47:47 `q005. Suppose that every week of the current millenium you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time ( this is so), that the the gold remains undisturbed (maybe, maybe not so), that no other source adds gold to your backyard (probably so), and that there was no gold in your yard before.. 1. If you construct a graph of y = the number of grams of gold in your backyard vs. t = the number of weeks since Jan. 1, 2000, with the y axis pointing up and the t axis pointing to the right, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. {}3. Answer the same question assuming that every week you bury half the amount you did the previous week.
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RESPONSE --> rise, rise, decrease confidence assessment: 2
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11:47:59 1. If it's the same amount each week it would be a straight line. 2. Buying gold every week, the amount of gold will always increase. Since you buy more each week the rate of increase will keep increasing. So the graph will increase, and at an increasing rate. 3. Buying gold every week, the amount of gold won't ever decrease. Since you buy less each week the rate of increase will just keep falling. So the graph will increase, but at a decreasing rate. This graph will in fact approach a horizontal asymptote, since we have a geometric progression which implies an exponential function.
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RESPONSE --> self critique assessment: 3
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11:49:02 `q006. Suppose that every week you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time, that the the gold remains undisturbed, and that no other source adds gold to your backyard. 1. If you graph the rate at which gold is accumulating from week to week vs. tne number of weeks since Jan 1, 2000, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. 3. Answer the same question assuming that every week you bury half the amount you did the previous week.
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RESPONSE --> rising line slowly rising quicker rising slowly confidence assessment: 3
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11:49:14 This set of questions is different from the preceding set. This question now asks about a graph of rate vs. time, whereas the last was about the graph of quantity vs. time. Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case, since you buy the same amount eact week, is constant. The graph would be a horizontal straight line. Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1 gram per week. You thus get a risingstraight line because the increase in the rate is the same from one week to the next. Question 3. Since half the previous amount will be half of a declining amount, the rate will decrease while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate approaches but never reaches zero.
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RESPONSE --> self critique assessment: 3
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????????V??D??? assignment #005 005. Calculus qa initial problems 06-04-2007
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12:04:01 `q008. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval?
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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 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. .1 .2 .1 confidence assessment: 2
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12:04:12 At t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning a rate of 9 cm / sec. At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a rate of 8 cm / sec. The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less than 80 cm. The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater than 90 cm. Any answer that isn't between 80 cm and 90 cm doesn't fit the given conditions.. The rate change is a linear function of t. Therefore the average rate is the average of the two rates, or 9.5 cm/s. The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm.
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RESPONSE --> self critique assessment: 3
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??????????? assignment #001 001. Depth vs. Clock Time and Rate of Depth Change 06-04-2007
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12:12:25 `qNote that there are four questions in this assignment. `q001. If your stocks are worth $5000 in mid-March, $5300 in mid-July and $5500 in mid-December, during which period, March-July or July-December was your money growing faster?
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RESPONSE --> March to July because the increase was 3000 dollars then and only 200 dollars between july and december confidence assessment: 3
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12:12:47 The first period (4 months) was shorter than the second (5 months), and the value changed by more during the shorter first period (increase of $300) than during the longer second perios (increase of $200). A greater increase in a shorter period implies a greater rate of change. So the rate was greater during the first period.
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RESPONSE --> self critique assessment: 3
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12:14:32 `q002. What were the precise average rates of change during these two periods?
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RESPONSE --> Average for the first period was $300/4 which is $75 per month average, and the second period was $200/5 which is $40 per month average confidence assessment: 3
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12:14:40 From mid-March thru mid-July is 4 months. A change of $300 in four months gives an average rate of change of $300 / (4 months) = $75/month. From mid-July through mid-December is 5 months, during which the value changes by $200, giving an average rate of $200 / (5 months) = $40 / month. Thus the rate was greater during the first period than during the second.
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RESPONSE --> self critique assessment: 3
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12:16:10 `q003. If the water in a uniform cylinder is leaking from a hole in the bottom, and if the water depths at clock times t = 10, 40 and 90 seconds are respectively 80 cm, 40 cm and 20 cm, is the depth of the water changing more quickly or less quickly, on the average, between t = 10 sec and t = 40 sec, or between t = 40 sec and t = 90 sec?
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RESPONSE --> more quicly between 10 sec and 40 sec confidence assessment: 3
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12:16:24 Between clock times t = 10 sec and t = 40 sec the depth changes by -40 cm, from 80 cm to 40 cm, in 30 seconds. The average rate is therefore -40 cm / (30 sec) = -1.33 cm/s. Between t = 40 sec and t = 90 sec the change is -20 cm and the time interval is 50 sec so the average rate of change is -20 cm/ 50 sec = -.4 cm/s, approx. The depth is changing more quickly during the first time interval.
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RESPONSE --> self critique assessment: 3
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12:17:13 `q004. How are the two preceding questions actually different versions of the same question? How it is the mathematical reasoning the same in both cases?
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RESPONSE --> you have to work both problems in relatively the same way. it is reasoning in both questions because you must think about the time passed to answer both questions. confidence assessment: 3
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12:17:21 In each case we were given a quantity that changed with time. We were given the quantities at three different clock times and we were asked to compare the rates of change over the corresponding intervals. We did this by determining the changes in the quantities and the changes in the clock times, and for each interval dividing change in quantity by change in clock time. We could symbolize this process by representing the change in clock time by `dt and the change in the quantity by `dQ. The rate is thus rate = `dQ /`dt.
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RESPONSE --> self critique assessment: 3
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??€?????L?????????assignment #001 001. Depth vs. Clock Time and Rate of Depth Change 06-12-2007 J??w}??h?????????assignment #003 003. 06-12-2007
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14:18:16 `qNote that there are four questions in this assignment. `q001. Sketch a graph similar to that you constructed for the stock values, this time for the depth of the water vs. clock time (depths 80, 40, 20 at clock times 10, 40, 90). Your first point, for example, will be (10, 80). Connect these points with straight lines and determine the slopes of the lines.
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RESPONSE --> The first line, from (10, 80) to (40, 40) the slope is -40/30. The second line (40, 40) to (90, 20) the slope is -20/50. confidence assessment: 3
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14:20:18 `q003. To what extent do you think your graph with three points and straight line segments between them accurately depicts the detailed behavior of the water over the 80-second period of observation? How do you think the actual behavior of the system differs from that of the graph? How do you think the graph of the actual behavior of the system would differ from that of the graph you made?
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RESPONSE --> I think that there should be more than just three data points in an 80 sec time period. I think that the actual behavior of the system is slower than this appears for it to be. I think that the slope would be less steap. 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 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. confidence assessment: 3
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14:20:30 The straight line segments would indicate a constant rate of change of depth. It is fairly clear that as depth decreases, the rate of change of depth will decrease, so that the rate of change of depth will not be constant. The graph will therefore never be straight, but will be a curve which is decreasing at a decreasing rate.
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RESPONSE --> ok self critique assessment: 3
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14:21:02 `q004. From the given information, do you think you can accurately infer the detailed behavior of the water depth over the 80-second period? Do you think you can infer the detailed behavior better than you could the values of the stocks? Why or why not?
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RESPONSE --> No, because there just aren't enough data points. no, because there are only three points. confidence assessment: 2
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14:21:15 It will turn out that three data points will be sufficient to infer the detailed behavior, provided the data are accurate. However you might or might not be aware of that at this point, so you could draw either conclusion. However it should be clear that the behavior of the water depth is much more predictable than the behavior of the stock market. We don't know on a given day whether the market will go up or down, but we do know that if we shoot a hole in the bottom of a full bucket the water level will decrease, and we expect that identical holes in identical buckets should result in the same depth vs. clock time behavior.
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RESPONSE --> self critique assessment: 3
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????????T??????assignment #001 ??????????????o? Calculus I 06-12-2007 ???F????????????assignment #002 ??????????????o? Calculus I 06-12-2007
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14:12:41 Were you able to complete the DERIVE exercise?
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RESPONSE --> no
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14:12:50 ** If you weren't able to complete the exercise you should download the 30-day trial version of DERIVE and do so. Note that the instructor neither recommends nor discourages you from purchasing DERIVE. Except for the basic worksheet and other assignments that may occur in the first part of the coruse DERIVE is not required and references to DERIVE in the later part of the course will be optional. This ends Query 2.**
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RESPONSE -->
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