#$&* course Mth 279 7/22 9
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Given Solution: `aSlope = rise / run. Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4. The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3. The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints. 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)? 1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)? 2. Will the value ever exceed a billion? Will it ever exceed one trillion billions? 3. Will it ever exceed the number of particles in the known universe? 4. Is there any number it will never exceed? 5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x=2.1, 1/(2.1-2) = 1/.1 = 10 x=2.01, 1/(2.01-2) = 1/.01 = 100 x= 2.001, 1/(2.001-2) = 1000 x = 2.0001, 1/(2.0001-2) = 10000 1. As x approaches 2, x-2 gets closer and closer to zero. 1 divided by a smaller and smaller number gives larger and larger numbers, so as x gets closer and closer to two, the values of 1 / (x-2) get very large. 2. Yes, as x-2 gets very small, 1 / (x-2) gets very large, with no limit. 3. yes, though we don't know what that number is, there is no limit to how high this function can go 4. No, as x approaches 2, the function approaches infinity. 5. There is a vertical asymptote at x=2. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFor 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `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'. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The midpoint of the height of the first trapezoid is (9+5)/2=7, and the midpoint of the height of the second trapeziod is (4+2)/2 = 3. So the first trapezoid is a little more than twice as tall as the second one. However, the first trapeziod has a width of 7-3=4, while the second trapezoid has a width of 50-10=40. So even though the second trapezoid is about half the height of the first, it is ten times as wide. Since area is the space enclosed by the trapezoid, we know that the second trapezoid must have a greater area. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYour 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `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 basis of your reasoning. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(2)=2^2=4, f(5)=5^2=25. Steepness=slope=change in y / change in x Slope = (25-4)/(5-2) = 21/3 = 7, f(-1) = (-1)^2=1, f(7) = 7^2 = 49, Slope is (49-1)/(7--1) = 48 / 8 =6. The slope of the first is 7, while the slope of the second is 6. Since the first slope is larger , the first line segment must be steeper. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. Suppose that every week of the current millennium you go to the jeweler 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1. The graph will lie on a rising straight line, because I am adding the same amount of gold every week, so the number of grams of gold in my backyard is increasing at a constant rate as time (te x-axis) increases. 2. If I bury 1 more gram than I did the previous week, then the rate of how much gold I am adding per week will increase, which means the slope of my graph will also increase. This will give me a line which rises faster and faster. 3. If I bury half the amount I did the previous week, then the amount of gold I am burying per week (which is the rate, or the slope) will decrease over time. However, the amount of gold will be increasing. This would give me a line that is increasing, but more and more slowly. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a1. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not include in my answer that the graph will approach a horizontal asymptote, but that makes since because as we keep cutting the rate in half, the rate (or increase) will become infinitesimaly small and approach zero, so the amount of gold vs time will become almost a straight horizontal line at a certain point. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q006. Suppose that every week you go to the jeweler 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. the 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1. In this case, we are graphing the rate from the previous question versus time. (This is like graphing the derivative of the first function). If we bury the same amount of gold every week, the rate will stay constant and positive, so the graph of rate vs. time will be a horizonal level straight line. 2. If we bury 1 more gram than you did the previous week, the rate is increasing by one gram per week, so the rate vs time graph would be a rising straight line with a slope of 1 gram/week. 3. If we bury half the amount, the rate is decreasing, but not at a constant interval. As we put less and less gold into the backyard, the new amount will be smaller and smaller. The rate vs time graph will be a line which falls but more and more slowly. As we cut the rate in half, it will get very close to zero, forming a horizontal asymptote at zero. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis 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. STUDENT COMMENT: I feel like I am having trouble visualizing these graphs because every time for the first one I picture an increasing straight line INSTRUCTOR RESPONSE: The first graph depicts the amount of gold you have in your back yard. The second depicts the rate at which the gold is accumulating, which is related to, but certainly not the same as, the amount of gold. For example, as long as gold is being added to the back yard, the amount will be increasing (though not necessarily on a straight line). However if less and less gold is being added every year, the rate will be decreasing (perhaps along a straight line, perhaps not). FREQUENT STUDENT RESPONSE This is the same as the problem before it. No self-critique is required. INSTRUCTOR RESPONSE This question is very different that the preceding, and in a very significant and important way. You should have self-critiqued; you should go back and insert a self-critique on this very important question and indicate your insertion by preceding it with ####. The extra effort will be more than worth your trouble. These two problems go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run. The same is true of the last question in this document. STUDENT COMMENT Aha! Well you had me tricked. I apparently misread the question. Please don’t do this on a test! INSTRUCTOR RESPONSE I don't usually try to trick people, and wasn't really trying to do so here, but I was aware when writing these two problems that most students would be tricked. My real goal: The distinction between these two problems is key to understanding what calculus is all about. I want to at least draw your attention to it early in the course. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating: OK ``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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the depth at a certain time, we just plug the time into the function. t(30)=100-2*30+.01*30^2 = 100-60+.01*900 = 100 - 60 + 9 = 49. t(40) = 100 - 2*40 + .01*40^2 = 100 - 80 + .01*1600 = 20 + 16 = 36. t(60) = 100 - 2*60 + .01*60^2 = 100 - 120 + .01*3600 = -20 + 36 = 16. Over the first interval, the depth is changing at a rate of (t(40)-t(30))/(40-30) = (36-49)/(40-30) = -13/10 = -1.3 cm/second. Over the second interval, the depth is changing at a rate of (t(60)-t(40))/(60-40) = (16-36)/(60-40) = -20/20 = -1 cm/second. 1.3 is larger than 1, so the on average the depth is changing more rapidly over the interval from 30
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Given Solution: `aAt 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At t=10 seconds, the rate at which water descends is 10-0.1*10 = 10-1 = 9 cm/second. At t = 20 seconds, the rate at which water descends is 10-0.1*20 = 8 cm/second. To find the total water level change over the ten second interval, we would multiply the rate by 10 seconds. We know the rate has to be in between 8 and 9 cm/second, so the total water level change must be between 8*10 = 80 and 9*10 = 90 cm. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt 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 8.5 cm/s. The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm. STUDENT RESPONSES The following, or some variation on them, are very common in student comments. They are both very good questions. Because of the importance of the required to answer this question correctly, the instructor will typically request for a revision in response to either student response: I don't understand how the answer isn't 1 cm/s. That's the difference between 8 cm/s and 9 cm/s. I don't understand how the answer isn't 8.5 cm/s. That's the average of the 8 cm/s and the 9 cm/s. INSTRUCTOR RESPONSE A self-critique should include a full statement of what you do and do not understand about the given solution. A phrase-by-phrase analysis of the solution is not unreasonable (and would be a good idea on this very important question), though it wouldn't be necessary in most situations. An important part of any self-critique is a good question, and you have asked one. However a self-critique should if possible go further. I'm asking that you go back and insert a self-critique on this very important question and indicate your insertion by preceding it with ####, before submitting it. The extra effort will be more than worth your trouble. This problem, along with questions 5 and 6 of this document, go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run. You should review the instructions for self-critique, provided at the link given at the beginning of this document. STUDENT COMMENT The question is worded very confusingly. I took a stab and answered correctly. When answering, """"How much would you therefore expect the water level to change during this 10-second interval?"""" It is hard to tell whether you are asking for what is the expected change in rate during this interval and what is the changing """"water level."""" But now, after looking at it, with your comments, it is clearer that I should be looking for the later. Thanks! INSTRUCTOR RESPONSE 'Water level' is clearly not a rate. I don't think there's any ambiguity in what's being asked in the stated question. The intent is to draw the very important distinction between the rate at which a quantity changes, and the change in the quantity. It seems clear that as a result of this question you understand this and will be more likely to make such distinctions in your subsequent work. This distinction is at the heart of the calculus and its applications. It is in fact the distinction between a derivative and an integral. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ####The average rate is can be found by adding the rate at the beginning to the rate at the end and dividing by two. So (9+8)/2 is 8.5 cm/second over 10 seconds is 85 cm. So 85 cm could be an answer for the change in water level. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q009. Sketch the line segment connecting the points (2, -4) and (6, 4), and the line segment connecting the points (2, 4) and (6, 1). The first of these lines if the graph of the function f(x), the second is the graph of the function g(x). Both functions are defined on the interval 2 <= x <= 6. Let h(x) be the function whose value at x is the product of the values of these two functions. For example, when x = 2 the value of the first function is -4 and the value of the second is 4, so when x = 2 the value of h(x) is -4 * 4 = -16. Answer the following based just on the characteristics of the graphs you have sketched. (e.g., you could answer the following questions by first finding the formulas for f(x) and g(x), then combining them to get a formula for h(x); that's a good skill but that is not the intent of the present set of questions). What is the value of h(x) when x = 6? Is the value of h(x) ever greater than its value at x = 6? What is your best description of the graph of h(x)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: f(6) = 4 g(6) = 1 h(6) = 4 * 1 = 4 The graph of f(x) has a constant positive slope of 2, while the graph of g(x) has a slope of -3/4. Yes, the value at h(x) will be greater than its value at x = 6 at a point close to x = 6 but before it. While the absolute value of h(x) is larger near x = 2, f(x) is negative until it crosses zero at x = 4, so h(x) will be negative from x = 2 to x = 4. At x = 4, h(x) will be zero, because f(4) = 0. After x=4, but before x=4.5 (before f(x) is greater than 1), h(x) will be less than g(x). At x = 6, h(x) is equal to f(x) because g(x) is equal to 1. The graph of f(x) has a constant positive slope of 2, while the graph of g(x) has a slope of -3/4. The graph of h(x) is increasing, but slower and slower until it reaches a peak near x=5.5, then it starts to decrease faster and faster. I'm not sure exactly what the reasoning is behind this other than just looking at the numbers. Is there a central concept for calculus that I am supposed to use here that is just not popping out for me???? confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q010. A straight line segment connects the points (3,5) and (7,9), while the points (3, 9) and (7, 5) are connected by a curve which decreases at an increasing rate. From each of the four points a line segment is drawn directly down to the x axis, so that the first line segment is the top of a trapezoid and the second a similar to a trapezoid but with a curved 'top'. Which trapezoid has the greater area? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The trapezoid with a curved top has a greater area, because the curve is concave up and reaches higher than the straight line does. Both 'trapezoids' have the same width and height at the edges, so the only difference in area is the top being straight or curved. The trapezoid with the curved top is essentially the first trapezoid with an extra bit from the curve added on top of it. Confidence Assessment: 3 ********************************************* Question: `q011. Describe the graph of the position of a car vs. clock time, given each of the following conditions: The car coasts down a straight incline, gaining the same amount of speed every second The car coasts down a hill which gets steeper and steeper, gaining more speed every second The car coasts down a straight incline, but due to increasing air resistance gaining less speed with every passing second Describe the graph of the rate of change of the position of a car vs. clock time, given each of the above conditions. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When the car coasts down a straight incline, gaining the same amount of speed every second, the change in velocity (or acceleration) is constant, so the velocity graph is a straight line with a constant, positive slope. When the car coasts down a hill which gets steeper and steeper, gaining more speed every second, the change in velocity is not constant, so the velocity graph will be a curve that is increasing at an increasing rate. As time goes on, the car is getting faster and faster, with the change in speed being greater each time. The rate of change of position vs. time graph will be increasing at an increasing rate. When the car coasts down a straight incline gaining less speed with every passing second, the change in velocity is positive because it is still coasting down the hills, but since air resistance increases with increasing speed, the change in velocity will be less as time goes on. So, the rate of change of position vs. time graph will be increasing at a decreasing rate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q012. If at t = 100 seconds water is flowing out of a container at the rate of 1.4 liters / second, and at t = 150 second the rate is 1.0 liters / second, then what is your best estimate of how much water flowed out during the 50-second interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rate was (1.4-1)/2 = 2.4/2 = 1.2. We don't know how the rate changed, but we can use the average rate to estimate the water flow. 1.2 liters/second * 50 seconds = 60 liters confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:"