course MTH 271 ؽIDΚ}zDxassignment #002
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15:26:59 `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 --> The graph is steeper between points (7,17) and (10,29) because the distance between the X axis points 7 and 10 are shorter than X axis points 3 and 7. On the average, the distance between points (3,5) and (10,29) would be the steepest because it is asking us for the average. This shows us a gradual climb up the graph. What I said earlier would be true for this statement if we were looking for the steepest climb between two points without the average. confidence assessment: 2
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15:30:31 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 --> Ok. I understand. Shouldn't we use more information from the graph to determine the average? self critique assessment: 1
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15:41:56 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 --> X = 2.1 1 / (X - 2) 1 / (2.1 - 2) 1 / .1 = 10 X = 2.01 1 / (X - 2) 1 / (2.01 - 2) 1 / .01 = 100 X = 2.001 1 / (X - 2) 1 / (2.001 - 2) 1 / .001 = 1000 X = 2.0001 1 / (X - 2) 1 / (2.0001 - 2) 1 / .0001 = 10000 As X gets closer to 2, the values of 1/(x-2) become greater and the squares of each other. The value will eventually exceed and billions and trillions. We will never know if it will exceed the number of particles in the universe because we could keep plugging in numbers and keep going. self critique assessment: 2
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D`Ϲz^[z}hɷg assignment #005 005. Calculus qa initial problems 09-08-2007 rłCyݑt assignment #005 005. Calculus qa initial problems 09-08-2007
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15:54:05 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:
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16:12:57 `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 --> Area is equal to length times width. Since this is a trapezoid we must use a different equation. confidence assessment: 0
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16:15:04 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 --> I thought we would need to use a different formula other than Length times width. I was under the impression that different shapes used different formulas ie. circles, retangles, parralelograms, etc. self critique assessment: 3
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16:28:10 `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 --> f(x) = 2^2 f(x) = 4 f(x) = 5^2 f(x) = 25 This gives us the points (4,25) on the graph.
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16:29:42 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 --> I was on the right track, I suppose. I was not aware that 2 and 5 were already points on the x axis. self critique assessment: 2
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16:37:04 `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 --> 1. The line will be rising because with each new week there will be a new total amount of gold in your backyard. Both time and grams will increase causing the graph to rise. 2. The line on the graph will be rising faster and faster. Time will stay constant and the amount of gold increases each week causing the graph to become steeper and steeper. 3. The line on the graph will fall more slowly because the gold is being dwindled down each week. confidence assessment: 2
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16:38:17 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 --> I was thinking of TOTAL gold when I reponded to questions 1 and 2. self critique assessment: 2
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16:42:52 `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 --> 1. The rate of gold accumulating from week to week will show the line raising more and more slowly. This causes a gradual climb in the graph. 2. If you add 1 more gram each week, it will cause the line to rise faster and faster because you are adding more and more gold each week. 3. This will cause a line that falls more and more slowly. The line will fall drastically at first. Since you are removing half the amount of the previous week it will cause the line to fall more and more slowly. confidence assessment: 2
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16:52:49 ``q007. If the depth of water in a container is given, in centimeters, by 100 - 2 t + .01 t^2, where t is clock time in seconds, then what are the depths at clock times t = 30, t = 40 and t = 60? On the average is depth changing more rapidly during the first time interval or the second?
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RESPONSE --> t=30 100 - 2(30) + .01(30^2) 100 - 2(30) + .01(900) 100 - 60 + 9 100 - 69 = 31
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16:53:49 At t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49. At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36. At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16. 49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average. 36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average.
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RESPONSE --> I thought I had followed the order of operations. Where did I mess up? self critique assessment: 2
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17:00:42 `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 --> I do not feel correct in answering this. confidence assessment: 3
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17:03:42 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 --> I did not know how to answer this question because of my lack of confidence on the previous question. self critique assessment: 2
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